![lntroduction to Diccrete-Time Ontrol Chala. 1
(a) Block diagram of a quantizer and its input-output characteristics;
(b) analog input x(t) and discrete output y(t); (c) probability distribution P(e) of
quantization error e(f).
Suppose that the quantization level Q is small and we assume that the quan-
tization error e(t) is distributed uniformly between -4 Se that this error
acts as a white noise. [This is obviously a rather rough assumption. However, since
the quantization error signal e(t) is o£a small amplitude, such an assumption may
be aiceptable as a first-order approximation.] The probability distribution P(e) of
Sec. 1-4 Dala Acquicition, Conversion, and
signar e(t) may be plotted as shown in Figure 1-4(c).The average value of e(t) iszem,
or e(t) = O. Then the variance u h f the quantization noise is
n level Q is small co pared with the average amplitude o£
variance of the quantization noise is seen to be one-twelfth
of the square of the quantization level.
ith the rapid growth in the use of digital computers to perform digital control
actions, both the data-ac uisition system and the distribution system have become
an important part of the entire control system.
The signalconversion that takes place in the digital control system involvest
following operations:
ultiplexing and demultiplexing
.Sample and hold
Analog-to-digital conversion (quantizing and encoding)
.Digital-to-analog conversion (decodlng)
Figure 1-5(a) shows a block diagram o£a acquisition system, and Figure 1-5(b)
shows a block diagram of a data-distrib
n the data-acquisitionsystem the input to the system is a physicalvariable such
asposition, velocity, acceleration, temperature, orpressure. uch a physicalvariable
is first converted into an electrical signal (a voltage or current signal) by a suitable
Physical To digital
variable controller
From digital
controller
To actuator
Figure 1-5 (a) Block diagram of a data-acquisition system; (b) block diagram of a data-
distribution system.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-12-320.jpg)
![The z Transform
The 2 transform applies to the continuous-time signal x(t), sampled signal
x(kT), and the number sequence x(k). dealing with the z transform~if no
confusion occurs in the discussion, we occasionallyuse x(kT) and x(k) interchange-
ably. [That is, to simplify the presentation, we occasionallydrop the explicit appear-
ance of T and write x(kT) as x(k).]
ection 2-1. has presented introductory remarks.
ection 2-2 presents the definition of the z transform and associated subjects.
Section 2-3 gives z transforms of elementary functions. mportant properties
theorems of the z transform are presented in Section 2-4.
computational methods for finding the inverse z transform are discussed in
2-5. Section 2-6 presents the solution of difference equations by the z transform
method. Finally, Section 2-7 gives concluding commeats.
The z transform method is an operational method that is very
working with discrete-time systems. In what followswe shall define
of a time function or a nurnber sequence.
In considering the z transform of a time function x(t), we consider
sampled values o£x(t), that is,x(O),x(T), x(2T), ...,where Tisthe samplin
The z transform of a time function x(t),where t isnonnegative, os o£a
of valuesx(kT), where k takes zero or positive integers and Tis the sampli
is defined by the following equation:
m
For a sequence of numbers x(k), the z transform is defined by
m
The z transforrn defined by Equation (2-1) or (2-2) is referred to as the one-sided
z transform.
The symbol7 denotes "the z transform of." In the one-sided z transform,
we assumer(t) = O fort < O orx(k) = Ofork < O. Note that z is a complexvariable.
Note that, when dealing with a time sequence x(kT) obtained by sampling a
time signal x(t), the z transform X(z) involves T explicitly. Owevery £m a Imnber
sequence x(k), the z transform X(z) does not involve T explicitly.
The z transform of x(t), where -.a < t < .a, or of x(k), where k takes integer
values (k = 0, +-1, t 2 , ), is defined by
m
X(z) = 27 [x(t)] = Z [x(kT)] =
k=-m
Sec. 2-3 z Transformc of Elementary Functions
The z transform defined by Equation (2-3) or (2-4) is referred to as the two-sided
z transform. In the two-sided z transfor e function x(t) is assumed to be
nonzero for t < O and the sequence x(k) is considered
h the one-sided and two-sided z transforms
involvesboth positive and negative powers of z
one-sided z transform is considered in detail.
For most engineering applications the one-sided z transform
venient closed-form solution in its r
an infinite series in z-', converges
radius of absolute convergente, in u ansform method for
time problems it is not necessary each cify the values of z over whichX(z)
ks convergent.
Notice that expansion of the right-hand side of Equation (2-1) gives
any continuous-timefunction x(t) may
zmkin this series indicates the position
inversisn integral method (see Section 2-5 for details.)
n the following we shall present z transfor S of several elementary functions. It is
noted that in one-sided z transform theory, in sampling a discontinuous function x(t),
we assume that the function is continuous from the right; that is, if discontinuity
occurs at t = O, then we assume that x(O) is equal to x(O+) rather than to the average
at the discontinuity, [x(O-) + x(O+)]/2.
Unit-StepFunc~on. ket us find the z transforrn of the unit-step function
S just noted, in sampling a unit-step function we assume that this function is
continuous from the right; that is, l(0) = l . Then, referring to Equation (2-l), we
have](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-19-320.jpg)
![The z Transform Chap. 2
Since the z transform of the exponential function is
7[e-"'] =
1
&
1- e-aTZ-l
we bave
- 1
X ( z ) = ;S[sinwt] = ;S -(elw' - eiw')]
l'j
- z-l sin wT
1- 2z-l cos oT + z - ~
-- z sinoT
z 2 - 2 z coswT+ 1
Obtain the z transform of the cosine function
coswt, O C l t
t < O
If we proceed in a manner similar to the way we treated the z transform o£the
sine function, we have
x ( ~ )= Y wt] = $Y[e1." + e-]" 1
- 1 - z-' COS wT
1 - 2.2-' cos wT + z-'
Obtain the z transform of
Whenever a function in S is given, one approach for finding the corresponding z
transform is to convert X(s) into x(t) and then find the z transform of x(t). Another
approach is to expand X(s) into partial fractions and use a z transform table to find the
z transforms of the expanded terms. Still other approaches will be discussed in
Cec. 2-3 z Transforms of Elementary Functions
The inverse Laplace transform of X e ) is
( t ) = 1- e O 5 t
Hence,
1 1
X(Z)= Z[l - e-'] = --
1- Z - l 1 - e-'Z-l
en&. Just as in working with t lace transformatio*, a table
transforrns of commonly encountered functions is very useful for solving problerns
in the field of discrete-time systems. Table 2-1 is such a table.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-21-320.jpg)
![The z Transform Chap. 2
x(t) = 0, for t < 0.
x(1cT) = x(k) = 0, for lc < 0.
Unless otherwise noted, 1: = 0,1,2,3,. . . .
Cec. 2-4 lmportant Properties and Theorerns of the z Transform
The use of the z transform method in the analysis of discrete-time control systems
may be facilitated if theorems of the z transform are referred to. In this section we
present importantproperties and useful theor
the time function x is z-transformable al
where a is a constant.
To prove this, note thatm by definition
m
. The z transform possesses an Pmportant prop-
erty: linearity. This means that, if f(k) and g(k) are z-transformable and a and P
are scalars, then x(k) formed by a linear combination
x(k) = af(W + 13gW
where F(z) and G(z) are the z tr
The linearity property can be
X(z) = Z[x(k)] = Z[af(k) + pg(k)]
This can be proved as follows:m
m
. The shifting theorern presented here is also ieferred to as
on theorem. If x(t) = O for t < O and x(t) as Wz),](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-22-320.jpg)
![and
The z Transform Chap. 2
where n is zero or a positive integer.
To prove Equation (2-7), note that
m
By defining m = k - n, Equation (2-9) can be written as follows:
m
Sincex(mT) = O for m < O, we may change the lower limit of the summation from
m = -n to m = 0.
Thus, multiplication o£ a z transform by z-" has t e 0f delaying the time
function x(t) by time nT. (That is, move the funct lo the right by time nT.)
To p;ove-~quation(2-8)? we note that
-
For the number sequence x(k), Equation (2-8) can be written as follows:
From this last equation, we obtain
7[ ~ ( k+ l)]= zX(z) - zx(Q) (2-1 1)
Z[r(k + 211 = z l[x(k + 111 - zx(1) = z2X(Z) - z2x(0) - zx(1) (2-12)
lrnportant Properties and Theorems of the z Trancform
Similarly,
Z[x(k + n)]
= znX(z) - znx(0) - zn-'~(1)- zn-24 2 ) - .- - - zx(n-1) (2-13)
where n is a positive integer.
emember that multiplication of the z transform
advancing the signal x(kT) by one step (1sarnpiing per
of the z transform X(z) by z-' has the effect of delaying the signalx(kT) by one step
(1sampling period).
Find the z transforms of unit-step functions that are delayed by 1 sampling period
and 4 sampling periods, respectively, as shown in Figure 2-2(a) and (b).
Using the shifting theorem given by Equation (2-7), we have
1 - z-'
r[i(t - TI] = Z-~Z[I(~)]= Z-l- 1 - *-1 - -1 - z-l
(Note that z-' represents a delay of 1sampling period T, regardless of the value of T.)
Obtain the z transform of
o 2T 3T 4T 5T 6T 7T 8T ' Faginre2-2 (a) UnitSstepfunction
delayed by 1sampling period; (b) unit-
step function delayed by 4 sampling
(b) periods.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-23-320.jpg)
![The z Transform
Weferring to Equation (2-7), we have
Z [ x ( k - l)]= z-%(z)
The z transform of ak is
and so
1 - z-'
z [ f ( a ) ]= Z[ak-']= z-'- - -
1-az-' 1-az-'
where k = 1,2,3, .. . .
Consider the function y(k),which is a sum of functions x(h),where h = 0,1,2,
such that
k
where y ( k ) = O for k < O. Obtain the z transform of y (k).
First note that
Therefore,
Z [ y ( k )- y(k - l ) ]= Z [ x ( k ) ]
which yields
Chap. 2 Cec. 2-4 lmportant Properties and Theorems of the z Transform
where X ( z ) = 2' [x(k)].
If x(t) has the z transform X(z), then the z
transform of e-"'x(t)can be given by X(zeuT).This is known as the complex trans-
lation theorern.
Noting that
Z-' sin o T
Z [sin wt] =
1 - 22-' coswT +
we substitute zeaTfor z to obtain the z transform of e-"' sin wt, as follows:
e-aT -1
z sinwT
Z [e-"' sin wt] =
1 - 2e-aT~-'cos o T + e-2nT~-2
Similarly, for the cosine function, we have
1 - Z-' cos wT
2' [cos wt] =
1 - 22-' coswT + zW2
y substituting zeaTfor z in the z transform of cos wt, we obtain
1 - e-"Tz-' cos wT
Z [e-"' cos ot] =
1 - 2e-aT~-'C O S ~ T+ e-2aTz-2
Obtain the z transform of te-"'.
Notice that
Thus,
. If x(t)has the z transfor X ( z ) and if limX ( z )exists,
then the initial value x(O) of x(t) or x(k) is given by z-t m
x (O) = limX ( z )
2 4 m
is theorem, note that
ketting z --+m in this last equation, we obtain Equation (2-15). The be
signal in the neighborhood of t = O or k = O can thus be determined by the behavior
of X ( z ) at z = m.
The initial value theorem is convenient for checking z transform calculations
for possible errors. ince x(0) is usually known, a check of the initiaI value by
limX(z) can easily spot errors in X(z), if any exist.
z-tm
To prove this theorern, note that
x
"[e-"'x(t)] = 2~ ( k T ) e - " ~ ' z - ~= x(kT)(~e"')-~= X(zeaT) (2-14)
k=O k=O
Thus, we cee that replacing z in X ( z ) by zeaTgives the z transform of e-"'x(t).
Determine the initial value x(0) if the z transform of x(t) is given by
Examplie 2-6 By using the initial value theorem, we find
Given the z transforms of sin wt and cos wt, obtain the z transforms of e-"' sin wt and
e- "' cos wt , respectively, by using the complex translation theorem.
x(0) = iim
(1 - e-')z-'
z-m(l - z-l)(l - CTz-')= O](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-24-320.jpg)
![The z Transforrn Chap. 2
Referring to Example 2-2, notice that this X ( z ) was the z transform of
x(t) = 1 - e-'
and thus x(0) = 0, which agrees with the result obtained earlier.
ence,
Taking the limit as z approaches unity, we have
m 1
ecause of the assumed stability condition and t e condition that x(k) = O for
k < 0, the left-hand side of this iast equation becomes
m
[x(k) - x(k - l)] = [x(O) - x(- l)] + [X (1)- x(O)]
k=O
t-[x(2) - x(l)l + - e * = X ( W ) = limx(k)
k-+ m
ence,
limx(k) = lim [(l- z-1)X(z)]
k-*m 1-2 1
which is Equation (2-16). The finalvalue theorem is very useful in determining the
behavior of x(k) as k - + frorn its z transform X(z).
Determine the final value x(m) of
1
X ( z ) = ----- -
1
1 - Z-1 1 - e - a T Z - 1
a > O
by using the final value theorem.
By applying the final value theorem to the given X(z), we obtain
ec. 2-5 The Inverse z Transforrn
Pt is noted that the given X ( z ) is actually the z transform of
y substituting t = m in this equation, we have
X ( W ) = lim (1 - e-"') = 1
f-m
As a matter of course, the two resdts agree.
. lin this section we have resented important roperties and theo-
rems of the z transforrn that will e to be useful in solving many z transforrn
problerns. For the purpose of conv nt referente, these important properties and
theorems are sumrnarized in Table 2-2. y of the theoserns presented in this
table are discussed in this section. cussed here but included in the table
are derived or proved in Ap
The z transforrnation serves the same role for discrete-time control systemsthat the
Laplace transformation serves for continuous-ti control systems. For the z trans-
form to be useful, we must be familiar with thods for finding the inverse z
transform.
The notation for the inverse z transform is Z-l. The inverse z transform of
X(z) yields the corresponding time sequence x(k).
Irt should be noted that only the time sequence at the sarnpling instants is
obtained frorn the inverse z transform. Thus, the inverse z transform of X(z) yields
a unique x(k), but does not yield a unique x(t). This rneans that the inverse z
transform yields a time sequence that specifies the values of x(t) only at discrete
instants of time, t = O, T,2T,. . . ,and says not ing about the values of x(t) at al1
other times. That is, many different ti e functions x(t) can have the sarne x(kT).
X(z), the z transform of x(kT) or x(k), is given, the operation that
onding x(kT) or x(k) is called the inverse z transformation.
n obvious method for finding the inverse z transform is to refer to a z transform
extensive table of z tra
a sum of simpler z tr
presented in this section.)
Other than referring to z transform fables, four methods for obfaining the
inverse z transform are commonly available:](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-25-320.jpg)
![1f oints (open circles, o)
need to plot(k,y,'ol)to plot(k,y,'o',k,y,'-').
we can convert this equation into t
k) = O for k f O,
(O) and y(l) can be
k = -2 into Eqaiation (2-21), we fin$
Finding the inverse z transfor of Y(z) IXxV b3=2o
following difference equation for y(k):
the initial data y (O) = ) = O for k # O.
ation (2-22) can be solv ,FBRTRAN, or
other.
presented here, which is paralle
Laplace transformation, is widely used in
The method requires that all terms in
recognizable in the table sf z tr
To find the invesse z eran
Sec. 2-5 The Inverse r Transforrn
efore we discuss the partial-fraction-expansion method, we shall review the shifting
theorem. Consider fhe following X(z):
By writing zX(z) as Y ( z ) ,we obfain
1
zX(z) = Y ( z )= -1 - az-'
Referring to Table 2-1, the inverse z transforrn of Y ( z )can be obtained as follows:
T 1 [ Y ( z ) ]= y(k) = ak
Hence, the inverse z transform of X ( z ) = z-' Y ( z )is given by
2-'[X(z)] = x(k) = y(k - 1)
Since y(k) is assumed to be zero for al1 k < O, we have
Consider X(z) as given by
To expand X(z) into partial fractions, we first factor t
of X(z) and find the poles of X(z):
then expand X(z)/z into partial fractions so t at each term is easily recognizable
in a table o£ z transforms. If the shifting theorem is utilized in
transforms, however, X(z), instead of X(z)/z, may be expanded
tions. The inverse z transform of X(z) is obtained as the sum
transforms of the partial fractions.
A comrnonlyused procedure for the cas here allthepoles are o£simpleorder
and there is at least one zero at the origin ( t is, bm = O) is to divide both sides
o£X(z) by z and then expand X(z)/z into partial fractions. OnceX(z)/zis expanded,
it will be of the form
The coefficient aican be determined by multiplying both sides of this last equation
by z - p, and setting z = pi.Thiswill result in zero for allthe terms on the right-hand
side except ehe ai term, in which the multiplicative factor z - pihas been canceled
by the denominator.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-30-320.jpg)
![The r Transform Chap. 2
Note that such determination of a, is valid only for sim
olves a multiple pole
n X(z)/z will have t
The coefficients cl and c2are determined from
Tt is noted that if X(z)/zinvolves a triple
must include a term (z +pJ(z - ~ 7 ~ ) ~ .(
Given the z transform
where a is a constant and T is the sampling period, determine the inverse z transform
x(kT) by use of the partial-fraction-expansion method.
The partial fraction expansion of X(z)/z is found to be
1X o = L - -
z z - 1 z - e-aT
Thus,
From Sable 2-1 we find
r 1
Hence, the inverse z transform of X ( z ) is
x(kT) = 1 - k = 0,1,2,. ..
Let us obtain the inverse z transform of
by use of the partial-fraction-expansion method.
We may expand X ( z ) into partial fractions as follows:
Sec. 2-5 The lnverse z Transform
Noting that the two poles involved in the quadratic term of this last equation are
complex conjugates, we rewrite X ( z ) as follows:
Since
1 - e-aTz-' cos wT
2' [e-"" TOS wkT] = 1 - 2eWaTz-'cos o T + e-2aTz-2
e - a T z - l
sin wTZ [e-"" sin wkT] = 1 - 2e-aT~-'cos o T + e-2aTz-2
by identifying e-"T = 1 and cos oT = $ in this case, we have wT = ~ 1 3and
sin wT = 1/3/2. Hence, we obtain
- -
and
Thus, we have
( k - 1 ) ~ 1 (k - 1 ) ~
x ( k ) = 4(lkP1)- 3(lk-') c o s + (1.-') sin--
3
Wewriting, we have
The first severa1values of x(k) are given by
Note that the inverse z transform of X ( z ) can also be obtained as follows:
Since](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-31-320.jpg)
![and
y-1
I z-l
krr
m 1 - z - 1 + z - 2 ]= ;sí(1.) sin-3
krr 4 ( k - l ) r r
- 2 ~ ?sin- + -sin-
3 v'3 3 '
k = l , 2 ,3,. .
x ( k ) =
k 6 0
Although this solution may look difierent from the one obtained earlier, both solutions
are correct and yield the sarne values for x(k).
Pn is a useful techniq
verse z al for the ztra~sfo
its center at the srigin of t
of complex variables. It c
m
[residue o£x(z)z"-' at pole z = ziaf (2-24)
i=l
(2-25)
Ef X(z)zk-' contains a mult
Sec. 2-5 The Inverse z Transform
partial-fraction-expansion method rnay prove lo be simpler to apply. On the other
hand, in certain problems the partial-fraction-expansion approach may become
laborious. Then, the inversion integral method proves to be very convenient.
Qbtain x(kT) by using the inversion integral method when X ( z ) is given by
Note that
For k = O, 1,2,. . .,X(z)zk-' has two simple poles, z = zl = 1 and z = z2 = e-OT.
Hence, from Equation (2-24), we have
x(k) =
--
Kl =
--
K2 =
--
Hence,
[residue of (1 - e-OT)zk at pole z = z,
1 = 1 ( Z - 1)(z - ePaT) 1
where
[residue at simple pole z = 11
[residue at simple pole z = e-OT]
Qbtain the inverse z transform of
by using the inversion integral rnethod.
Notice that
For k = 0,1,2, .. . ,X(z)zk-' has a simple pole at z = zl = e-"Tand a double pole at
z = z2 = 1. Hence, from Equation (2-24), we obtain
Z k t l
x ( k ) = [residue of at pole z = z,
I = I ( z - I ) ~ ( z- e-OT)
= Ki + KZ
1](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-32-320.jpg)
![where
The z Transform
M, = [residue at simple pole z = e-."?
.k+l
]=
,-a(k+l)T
(Z - ~ ) ~ ( z- e-"') (1 - e-aT)2
K2 = [residue at double pole z = 11
(k + l)zk(z - e-aT)- zk+l
= lim
z-. I ( z - ePaT)'
Difference equations can be solved easily by use o£a
nurnerical values of al1coefficients and parameters are
expressions for x(k) cannot be obtained fr com~uters0lLhX-l7 e*cePt forvepY
special cases. The usefulness of the z "can ethod 6t h t it enables tQ
the clased-form expressisn for x(k).
Consider the linear time-invariant discrete-time system characterized by the
following linear difference equation:
where u(k) and x(k) are the systern's input and output, respedively, at the kth
iteration. In describing such a difference equation in the z plane, we take the z
transform of each term in the equation.
Eet us define
q x(k)]= X(z)
Thenx(k + l),x(k + 2),x(k + 3),.. . andx(k - l),x(k - 2),x(k - 3),...can be
expressed in terms of X(z) and the initial conditions. Their exact z transforms were
derived in Section 2-4 and are surnrnarized in Table 2-3 for convenient referente.
Next we present two exarnple problems for solvingdifference equations by t
z transfcwm method.
-6 P Transform Method for
P TRANSFORWLS OF x(k + m) AND x(k - m)
Discrete function z Pansform
1
Soive the following difference equation by use of the z transform method:
x ( k + 2 ) + 3 x ( k + I ) + & ( k ) = O , x(O)=O, x ( l ) = l
First note that the z transforms of x(k + 2), x(k + l j , and x ( k ) are given,
respectively, by
Z [ x ( k + 2)] = z 2 X ( z )- z2x(0)- zx(1)
Z [ x ( k + 1)]= z X ( z ) - zx(0)
Z[x(k)]= X ( z )
Taking the z transforms of both sides of the given difference equation, we obtain
z2X(zj - z2x(0)- zx(1) + 3zX(z) - 3zx(O) + 2%(z) = O
Substituting the initial data and simplifying gives
1 1=---
1 + z-' 1 + 22-'
Noting that
Obtain the solution of the following difference equation in terms of x(0) and x(1):
x(k + 2) + (a + b)x(k + 1) + abx(k) = O
where a and b are constants and k = 0,1,2,. . . .](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-33-320.jpg)
![The z Transform Chap. 2
?'he z transform of this difference equation can be given by
[z2X(z)- z2r(0)- rx(l)]+ (a + b)[zX(z)- zx(O)] + abX(z) = O
Solving this last equation for X(z) gives
[z2+ (a + b)z]x(O)+ zx(1)
X ( z ) =
z2 + (a + b)z + ab
Notice that constants a and b are the negatives of the two roots of the characteristic
shall now consider separately two cases: (a) a # b and (
(a) For the case where a # b , expanding.X(z)/zPto partial fractions, we obtain
from which we get
The inverse z transform of X ( z ) @ves
bx(0) +x(1) ax(0) + x(1)
x ( k ) = (-a)" + (-b)k, a # b
b - a a - b
where k = 0,1,2, .. . .
) For the case where a = b , the z transform X(z) becomes
The inverse z transforrn of X ( z ) gives
x(k) = ~ ( 0 ) ( - a ) ~+ jax(0) +~ ( l ) ] k ( - a ) ~ - l , a = b
where k = 0,1,2,. . ..
In this chapter the basic theory of the z transform rnet een presenteda T'he
z transform serves the same purpose for linear time-i Saek-tirne sY5&Xls
asthe Laplace transform provides for linear time-invariant continuous-time systems.
method of analyzing data in discrete time results in difference
z transforrn rnethod, linear time-invariant difference equations
into algebraic equations. This facilitates the transient response
analysis of the digital control system. Also, the z transform e t k d all0VJsL E t* We
Chap. 2 Example Problemc and Solutions
conventional analysis and design techniques available to analog (continuous-time)
control systems, such as the root-locus technique. Frequenc
design can be carrie rting the z plane into
z-transforrned char
as the Sury stability
3 and 4.
Obtain the z transform of Gk,where G is an n x n constant matrix.
y definition, the z transform of G ~ S
m
h[ G ~ ]= E~ ~ z - ~
k=O
= 1 + GZ-' + G ~ ~ - " + 3 z - 3 + ...
Note that G~can be obtained by taking the inverse z transform of (
or (21 - G)-' Z . That is,
Obtain the z transform of k2
y definition, the t transform of k2is
m
Z [ k 2 ]= 2 k2z-k = Z-' + 4z-2 + 9z-3 + l6zp4+ e
k=O
= z-'(1 + z-l)(l + 32-' + 6z-2 + IOZ-~+ 1 5 ~ - ~+ )
ere we have used the closed-form ex
involved in the problem. (See Appendix
Obtain the z transforrn of kak-' by two methods.
Method l. By definition, the z transform of kak-' is given by
m](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-34-320.jpg)
![ethod 2. The summation expression for the z transform of kak-' can also be written
as follows:
-Ir
)7[kak-ll = 5kak-lz-k = a-1 k a k z k = 2 (a)k=O k=O ak=O
o~blemA-2-
Show that
and
Also show that
where 15 i 5 k - 1.
Solintaon Define
k
y(k) = 2x(h), k = O, 1,2,. . .
h =O
Then, clearly
Y ( k ) - Y (k - 1) = x ( k )
y writing the z transforms of x ( k ) and y(k) as X ( z ) and Y ( z ) ,respectively, and by
taking the z transform of this last equation, we have
Y ( z ) - z-1 Y ( z ) = X ( z )
Chap. 2 Exarnple Problerns and Solutions
By using the final value theorem, we find
k- m
2 ~ ( h )= Ex ( k ) = limX ( z )
h=O k=O 2-1
Next, to prove Equation (2-29), first define
k
j ( k ) = h=iEx(h) = x(i) + x(i i- 1) + + x ( k )
where 1 r i r k - 1. Define also
X ( z ) = x(i)zWi+ x(i + 1)z-'"" + .- + x(k)z7*+ - -
Then, noting that
we obtain
1-1
Since
the z transform of this last equation becomes
Y ( Z ) - z-' P(z) = Z ( z )
[Note that the z transform of x(k),which begins with k = i, is X(z),not X(z).]Thus,
1
B ( z ) = -1 - 2-1 [,(Z) - h = ~i ( h ) z h1-2-5
Obtain the z transform of the curve x(t)shawn in Figure 2-6. Assume that the sampling
period T is 1 sec.
ution From Figure 2-6 we obtain
x(0) = O
x(1) = 0.25](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-35-320.jpg)
![The z Transform Chap. 2
Then the z transform of x(k) can be given by
Notice that the curve x(t) can be written as
~ ( t )= $ t - g t - 4j1~t- 4)
where I(t - 4) is the unit-step function occurring at f = 4. Cince the sampling period
T = 1sec, the z transform of x(t) can also be obtained as follows:
X(z) = Z [x(t)] = Z [$f ] -7[a(t - 4)l(t - 4)]
Consider X(z), where
Obtain the inverse z transform of X(z).
Sgalntigan We shall expand X(z)/z into partial fractions as follows:
ap. 2 Example
Then
The inverse z transforms- of the individual- terms give
and therefore
~ ( k ) = 9 k ( 2 ~ - ' ) - 2 ~ + 3 , k = 0 , 1 , 2,...
Obtain the inverse z transform of
olution Expanding X(z) into partiaf fractions, we obtain
(Note that in this example X(z) involves a double pole at z = O. Hence the partial
fraction expansion must include the terms ll(z2) and llz.] By referring to Table 2-1,
we find the inverse z transform of each term of this last equation. That is,
ence, the inverse z transform of X(z) can be given by
o - o - o = o , k = O
1 - o - 1 = 0 , k = 1
x(k) =
2 - 1 - 0 = 1 , k = 2
2k-'-O-O=2k-19 k = 3 , 4 , 5 , ...
Rewriting, we have
k = 0 , 1
/O9 k = 2x(k)= 1,
2k-1, k = 3 , 4 , 5,...
Toverify this result, the direct division method rnay be applied to this problem. Noting
that](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-36-320.jpg)
![The z Transforrn Chap. 2
we find
Obtain the inverse z transform of
z-2
>
X(z) = (1 - z-1 >3
utionr The inverse z transform of ~ - ~ / ( 1- z-')~is not available from most z
transform tables. It ispossible, however, to write the givenX(z) as a sum of z transforms
that are commonly available in z transform tables. Since the denominator of X(z) is
(1 - z-')~and the z transform of k2is z-l(l + z-')!(1 - z-')~,let us rewrite X(z) as
from which we obtain the following partial fraction expansion:
The z transforms of the two terms on the rigkt-hand side of this last equation can be
found from Table 2-1. Thus,
It is noted that if the given X(z) is expanded into other partial fractions then the
inverse z transform rnay not be obtained.
As an aiternative approach, the inverse z transform of X(z) may be obtained by
use of the inversion integral method. First, note that
Hence, for k = O, 1,2, .. .,X(z)zk-' has a triple pole at z = 1.Referring to Equation
(2-24), we have
! zk
x(k) = residue of ----at triple pole z = 1
(Z - 113
Chap. 2 Exarnple Probierns and
=- l lirn? (z - 1)3-
(3 - q !z - ~ dzd2 ' (Z -zk113I
Using the inversion integral method, obtain the inverse z transform of
Sola~tion Note that
For k = 0, notice that X(z)zk-' becomes
Hence, for k = O, X(z)zk-' has t h ~ e esimple poles, z = zl = 1, z = z2 = 2, and z =
z3 = O. For k = 1,2,3, . .. ,however, X(z)zbl has only two simple poles, z = zl = 1
and z = z2= 2. Therefore, we must consider x(0) and x(k) (where k = 1,2,3,...)
separately.
For k = O. For this case, referring to Equation (2-24), we have
lo
at pole z = zi
(Z - 1)(z - 2)z 1
where
K1 = [residue at simple pole z = 11
= -10
(2 - 1)(z - 2)z
K2= [residue at simple pole z = 21
z-2 (z - 1)(z - 2)z
K3 = [residue at simple pole z = O]](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-37-320.jpg)
![The r Transform
ence,
x(0) = K1 + K2 + K3 = -10 + 5 + 5 = O
For k = 1,2,3, . . . . Eor this case, Equation (2-24) becomes
10zk--'
at pole z = zi
( z - 1)(2 - 2)
= Ki + K2
where
K1 = [residue at simple pole z = 11
K2 = [residue at simple poIe z = 21
Thus,
x(k)=K1+K2=-10+10(2k-1)=10(2k-1-1), k = 1 , 2 9 3 9
Mence, the inverse z transform of the given X ( z ) can be written
An alternative way to write x(k) for k 2 O is
x(k) = 560(k) + 10(2~-'- l ) , k = 0,1,2,...
where &(k) is the Kronecker delta function and is given by
for k = 0
"(w=jt: t o r k i o
'4-2-1
Obtain the inverce z transform of
Chap. 2
(2-30)
by use of the four methods presented in Section 2-5.
utiora
Method 1: Direcr division method. We first rewrite X ( z ) as a ratio of two polynomiak
in z-':
Dividing the numerator by the denominator, we get
X ( Z )= 1 + 4 ~ 4+ 7 ~ - 2+ ~ o z - ~+- . e .
ence,
x(0) = 1
Chap. 2 Example roblemc and Solutions
Method 2: Compuiational method (MATLAB approach). %(z) can be written as
Hence, the inverse z transform of X ( z ) can be obtained with
Define
num = [l 2 O]
d e n = [ l -2 11
If the values of x(k) for k = O,1,2, . . . ,30 are desired, then enter the Kronecker delta
input as follows:
Then enter the command
Program 2-3. [The screen will show the output x(k) from k = 0 té,
k = 30.1 (MATLAB computations begin from column 1 and end at column 31, rather
Columns 13 through 24
37 40 43 46 49 52 55 58 61 64 67 70
Columns 25 through 31](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-38-320.jpg)
![The z Transform Ckiap. 2
than from column O to column 30.) Thevalues x(k)givethe inverse z transform of X(z).
That is,
x(0) = 1
x(1) = 4
x(2) = 7
Method 3: Partial-fraction-expansion method. expand X ( z ) into the foliowing
partial fractions:
Then, noting that
we obtain
x(0) = 1
x ( k ) = 3 k + l , k = 1 , 2 , 3 ,...
which can be combined into one equation as follows:
x ( k ) = 3 k + l , k = 0 , 1 , 2,...
Note that if we expand X ( z ) into the following partia1 fractions
then the inverse z transform of X ( z ) becomes
x(0) = 1
which is the same as the result obtained by expanding X ( z ) into the other partial
fractions. [Remember that X ( z ) can be expanded into different partial fractions, but
the final result for the inverse z transform is the same.]
Method 4: Inversion integral rnethod. First, note that
Chap. 2 Exarnple Problerns and
For k = O, 1,2, ... ,X(z)zk-' has a double pole at z = P.
(2-24), we have
( 2 + 2)zk
residue of -----
(2 - 1)2
at double pole z = 1
Thus,
- -
x ( k ) = -----i i m - [ ( z - I ) 2 W ]
(2 - l ) !Z-i dz
d
= lim-[(zZ-+I dz + 2)zk]
= 3 k + 1 , k = 0 , 1 , 2,...
Solve the following difference equation:
2x(k) - 2x(k - 1) + x(k - 2) = u(k)
where x(k) = O for k < O and
y taking the z transform of the given difference equation,
1
2X(z) - 22-l X ( z ) + Z - ~ X ( Z )= -1 - z-'
Solving this last equation for X(z), we obtain
Expanding X ( z ) into partial fractions, we get
Notice that the two poles involved in the quadraticterm in this last equation are compiex
conjugates. Hence, we rewrite X ( z ) as follows:
y referring to the formulas for the z transforms of damped cosine and damped sine
functions, we identify e-2aT= 0.5 and cos oT = 11V5 for this problem. Hence, we get
wT = ~ 1 4 ,sin wT = l i d , and e-"= = l i d . Then the inverse z transform of X ( z )can
be written as
x ( k ) = 1 - Se-akTcos wkT + $ewakTsin wkT
from which we obtain](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-39-320.jpg)
![The z TTansform Chap. 2
Consider the difference equation
x(k + 2) - 1.3679x(k + 1) + 0.3679x(k) = 0.3679u(k + 1) + 0.2642u(k)
where x(k) is the output and x ( k ) = O for k 5 O and where u ( k ) is the input and is
given by
u ( k ) = O , k < O
u(0) = 1
u ( l ) = 0.2142
u(2) = -0.2142
u ( k ) = O , k = 3 , 4 , 5 ,...
Determine the output x(k).
olintlon Taking the z transform of the given difference equation, we obtain
[ z 2 X ( z )- z2x(0)- zx(1)l - 1.3679[zX(z)- zx(O)]+ 0.3679X(z)
= 0.3679[zU(z)- zu(0)l + 0.2642U(z) (2-31)
By substituting k = -1 into the given difference equation, we find
x(1) - 1.3679x(O) + 0.3673x(-1) = 0.3679u(O) + 0.2642u(-1)
Since x(0) = x(- 1) = O and since u(- 1) = O and u(0) = 1, we obtain
x(1) = 0.3679u(O) = 0.3679
By substituting the initial data
x(0) = O, x(1) = 0.3679, u(0) = 1
into Equation (2-31), we get
~ " ( 2 )- 0.36792 - 1.3679zX(z) + 0.3679X(z)
= 0.3679zU(z) - 0.36792 + 0.2642U(z)
Solving for X(z), we find
The z transforrn of the input u ( k ) is
U ( Z )= Z [ u ( k ) ]= 1 + 0.21422-' - 0 . 2 1 4 2 ~ - ~
Chap. 2 Example roblems and Solutions
ence,
Thus, the inverse z transform of X ( z ) gives
Consider the difference equation
where x(0) = O and x(1) = 1. Note that x(2) = 1,x(3) = 2,x(4) = 3, .... The series
U, 1,1,2,3,5,8,13, ... is known as the Fibonacci series. Obtain the general solution
x ( k ) in a closed form. Show that the limiting value of x(k + l)/x(k)as k approaches
infinity is (1 + 1/5)/2, or approximately 1.6180.
olution By taking the z transform of this difference equation, we obtain
olving for X ( z ) gives
By substituting the initial data x(0) = O and x(1) = 1into this last equation, we have
The inverse 2 transform of X ( z ) is
Note that although this lact equation involves '~6the square roots in the right-hand side
of this last equation cancel out, and the values of x ( k ) for k = 0,1,2, . .. turn out to
be positive integers.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-40-320.jpg)
![z-Plane Analycis of Discrete-TimeControl Cysterns Chap. 3
Carrier
signal 1
1
Figure 3-2 Impulse sampler as a modulator.
Next, consider the Eaplace transform of Equation (3-1):
X"(s) = 2[x"'(t)] = x(O)%[6(i)] + x(T)Ce[G(t - T)]
+x(2T>ce[8(t- 2T)] +
= x(O) + x ( T ) ~ - ~ '+ ~ ( 2 T ) e - ~ ~ '+ .
m
= ~ ( k T ) e - ~ "
k=O
Notice that if we define
eTs -- z
then Equation (3-2) becornes
I m
The right-hand side of Equation (3-3) ic exactly the same as the right-hand side of
t is the z transform o£the seque e x(O),x(T),x(2T), . .. ,gener-
t = kT, where k = 0,1,2,. . .. ence we may write
and Equation (3-3) becomes
Note that the variable z is a cornplexvariable and Tis the sarnplingperiod. [ht should
be stressed that the notation X(z) does not signify X(s) with s replaced by z, but
ratber X' (S = T.-' Ira z).]
Cec. 3-2 Impulse Sarnpling and Data Hold
. Let us summarizewhat we have just
signal x(t) is impulse sampled in a periodic manner,
signal may be represented by
m
x*(t) = 2x(t)s(t - kT)
k=O
converts a continuous-ti S occurring at the sampling
instants t = 0, T,2T,. .., od. (Note that between any
no information. Two signals
sampled signal.)
Data-hold is a process of generating a continuous-time signal h(t) from a
discrete-time sequence x(kT). A hold circuit converts the sampled cignal into a
continuous-time signal, which approximately reproduces the signaf applied to the
sampler. %hesignalh(t) during the time interval kT 5 t < (k + 1)Tmaybe approx-
imated by a polynomial in r as follows:
where O 5 T < T. Note that signal h(kT) must equal x(kT), or
ence, Equation (3-5) can be written as follows:
Hf the data-hold circuit is an nth-order polynomial extrapolator, it is called an
nth-order hold. Thus, if n = 1, it is called a first-order hold. [The nth-order hold
uses the past n + 1discrete data x((k - n)T),x((k - n + 1)T),. ..,x(kT) to gen-
erate a signal h(kT t r).]
ecause a higher-order hold uses past sarnples to extrapolate a continuous-
nal between the present sampling instant and the next sampling instant, the
accuracy of approximating the ntinuous-time signal improves as the number of
past samples used is increased. wever, this better accuracy is obtained at the cost
of a greater time delay. Hn closed-loop cont ystems, any added time delay in the
loopwill decrease the stability of the system in some casesrnay even cause system
instability.
The simplest data-hold is obtained when n = O in Equation (3-6), that is, when](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-45-320.jpg)
![can be given by
z-Plane Analysis of Discrete-Time Control
where the integration is along the Iine from c - jm to c +j a and this line is paralle1
ne and separates the poles of X(p) from those of
is the convolution integral. Iat is a well-known fact
e evaluated in terms of residues by forrning a closed
contour cansistingof the linefrom c - j a to c +jm a~ida se icircleofinfiniteradius
in the left or right half-plane, provided that the integral along the added semicircle
is a constant (zero or a nonzero constant). That is, we rewrite E
follows:
where T is a semicircle of infinite radius in the left or right half p plane.
ñhere are two ways to evaluate t is integral (me using an infinite semicircle
in the M-half plane and the other an infinite semi
shall describe the results obtained by the two case
analysis here, we assume that the poles
S) can be expressed as a ratio of polly
where g (S)and p (S) are polynomials in s. also assume that p(s) is of a higher
degree in s than q(sj, which means that
shall evalu-
ate the convolution integral given by Equation (3-21) using a closed contour in the
left half of the p plane as shown in Figure 3-8. Using this closed contour, Equation
(3-21) may be evaluated as follows:Noting that the denominator of X(s) is of higher
degree in s than the numerator, the integral along fivanishes.
This integral is equal to the sum of the residues of (p) in the &sed contour.
the z Transform by the Convolution Integral Method
2: p plane
1 ure 3-8 Ciosed contour in the leftOf
1 - e-Tts - p )
half of the p plane.
roblem A-3-6 for the derivation of Equation (3-22).]
for eTsin Equation (3-22), we have
By changing the com lex variable notation from p to s , we obtain
residue of -X(s)z at pole of ~ ( s)
z - eTs iAssume that X(s) has poles SI,s2, . ,S,. f a pole at s = sjis a simple pele,
then the corresponding residue Kj is
If a pole at s = siis a multiple pole of order ni,then the residue Miis
Given
obtain X ( z ) by use 0.6 the convolution integral in ihe left half-plane.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-49-320.jpg)
![z-Plane Analysis of Discrete-Time Control ystems Chap. 3 Sec. 3-3 Obtaining the a Trancform by the Convolution Integral Method
Note that X(s) has a double pole at s = O and a simple pole at S = -1.
Equation (3-23) becomes
X(s>zresidue of at poole of X(s)
1---
1
lim- sZ7 ---
l dL 1
+ Iim (S + 1)7----
- o s s s + z ~ e ~ j--i[ s (S + 1) z -z IeTs
-z[z - eTS+ (S + l)(-T)eTs] 1 z
= lim +--*-+ O (S + 1)2(z- eTSl2 (-1)2 z - e-T
p plane. Let us choose the closed contour shown in Figure 3-9, whi
line from c - jm to c +jm and rR,the portion of the sernicircle of infinite radius
in the right half of the p plane that lies to the right of t is h e . The closed contour
endoses al1poles of 141- e-T("p)],but it does not enclsse any poles of X(p). Now
X"(s) can be written as
Figure 3-9 Closed contour in the right half of the p piane.
evaluating the integrals on the right e of Equation (3-26), we need
deir two cases separately: one case w denominator of X(s)is two or
more degrees higher in s than the numerat other case where the denom-
inator of X(s) is only one degree higher in s than the numerator.
Case 1: X(s) has a denominator two or more degrees hig
numerator. For this case, since X(s) possesses at least two more
we have
Then the integral along & is zero. Thus, in the present case
X"( S ) can be obtained as follows:
erivation of Equation (3-27), see Probiem A-3-7.1 Thus
Note that this expression of the z transforrn is useful in proving the samplingtheorem
(see Section 3-4). owever, it is very tedious to obtain z transforrn expressions of
commonly encountered functions by this method.
Case2: X(s)has a denominator one degree higher in s than the numerator. For
this case, lirirsX(s) = x(O+) =# O < and the integral along TRis not zero. [The
nonzero value is associated with the initial value x(O+)of x(t).]It can be shown that
the contribution of the integral along TRin Equation (3-26) is -fx(O+). That is,
Then the integral term on the right-hand si e of Equation (3-26) becomes
Show that X*(s) is periodic with period 271.l~~.
Referring to Equation (3-29),](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-50-320.jpg)
![ec. 3-3 Obtaining the z Transform by the Convolution Integral Method
Hence,
1 m 1
X*(s +j-k) = - X(s +jw,k +jwh) + -x(O+) rz[gl(t)]= G ( z )
T h = - - 2
the z transform o£xl(t)becomes
Let k + h = m. Then this last equation becomes
1 - 1
X 4 ( s+ jmik) = - X(s +jw,m) + x ( 0 - t ) = X'(s)
T,,-,
Therefore, we have X ( z ) = ZIGl(s) - e-Ts&;*(S)]
X * ( s ) = X * ( s + j w s k ) , k = 0 , 1 , 2 ,... = /Z'[g1(t)l- Z[x1(t)]
Thus, X* ( S ) is periodic, with period 2.lr/wS.This means that, if a function X(s) has a = G1(z)- Z - l G1(z)
pole at s = si in the s plane, then %"(S)has poles at s = si5 jwsk (k = 0,1,2, ...).
= (1 - Z - ~ ) C ~ ( Z )
G(s)follows the zero-order hold. Then the product of the transfer function of the ( z = ( S = (1 - ) Z [ ? ] (3-3 2)
zero-order hold and C(s)becomes
1 - e-Ts
X(s) = y a s )
In what follows we sha'il obtain the z transform of such an X(s).
Note that X(s) can be written as follows:
then the z transform ciif the function
where
fxs)G1(s)= -S
can be obtained as follows. Since
Consider the function
X1(s)= e-TsG(s) x(s)= (1 - e - f i ) 2 T S + '
Ts2 G(s)
x1(t) = Iotgo(t- 4544d~
where = (1 - z-1)2Z[TTs + 1G(s)j (3-33)
go(t)= %-'[e-fi]= 6(t - 7')
Equation (3-33) can be used for obtaining the z transform of the function involving
&(t)= %-'[G(s>] the first-order hold circuit.
Thus,
xl(t)= [6(t - T - r)gl(r)dr
tain the z transform of
1 - e-Ts 1
= g10 - TI
X(s) = --S s + l](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-51-320.jpg)
![r-Plane Analysis of iscrete-Time Control
Referring to Equation (3-32), we have
sampling frequency is sufficiently high compar
onent involved in the continuous-time signal, t
To reconstruct the original
minimum frequency that the sa
frequency is specified in the sam
time signal x(t) has a frequency spectrum as shown in
does not contain any frequency com
, IIf o,, defined as 2 d T , where T is the sampling period,
is greater than 2wl, or
where wl is the highest-frequency component present ontinauous-time sig-
nal x(t), then the signal x(t) can be reconstructed co
sigraal x*(t).
The theorem impIies that if ws > 2w1then, from t dge of the sampled
signal, it is theoretically possible to reconstruct exactl al continuous-time
signal. In what follows, we shall use an intuitive graphical approach to explain the
sampling theorem. For an analytical approach, see
To show the validity of this sampling theorem, we need to find the frequency
-wl O 0, w Figure 3-10 A frequency spectrum.
riginal Signals from
spectrum of the sampled signal x"(t). The Laplace transform of x
obtained in Section 3-3 and is given by uation (3-27) or (3-29),
whether x(O+) = O or quency spectrum,
s in Equation (3-27). uency spectra, we nee
with the value of x(O+).] Thus,
Equation (3-34) gives the frequency
is periodic with period 2n-/O,,
if a function X(s) has a pole at s = sl, then
(k = o, 1,2, . ..).
Figures 3-11(a) and (b) show plots of t e frequency spectra X* (jw) versus
Figure 5-11 Plols of the frequency spectra / X":(jw)l versus w for two values of sampling
frequency w,:(a) ws > 2wl; (b)w, < 2wl.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-52-320.jpg)
![a-Plane Analysis of Discrete-Time Control Sycterns Chap. 3
for twovalues of the samplingfrequency os.Figure 3-11(a) correspondsto o, > 2wl,
ile Figure 3-11(b) corresponds to ws < 2wl. Eac
consists of IX(jw)lTrepeated every ws = 2~12"radlsec
of IX*(jw)]the component IX(jo)l/Tis called theprima
components, Ix(j(o it o,k))llT, are called comple
If os> 2wl, no two components of IX*(jo)/
ill be repeated every w, sadlsec.
original shape of IX(jw)l o longer appears in the plot of
use of the superposition sf spectra.Therefore, we see bhat
ignal x(t) can be reconst ted from the impulse-sampled
signal x"(t) by filtering if and only if os> 2wl.
It is noted that although the
is specified by the sampling theor
component present in the signal,
closed-loop system a
at a frequency much
éo be lOo, to 20w,.)
r, The amplitude frequenc
pass filter GI(jw) is shown in Figure 3-12. The magnitu
over the frequency range - Sos5 o 5 So, and is zero ola
The samplingprocess introduces an infinite number
nents (sideband components) in addition to the primary c
will attenuate al1 such complementary components to
primary component, provided the s g frequency osis greater than twice the
highest-frequency component of t nuous-time ségnal.
reconstructs the continuous-time signal represented by the S
shows the frequency spectra of the signals before and after ideal filtering.
The frequency spectrum at the output of the ideal filter is 112"times the
frequency spectrum of the original continuous-time signalx(t). Since t
has constant-magnitude characteristics for the frequency region -Sw, 5 w 5 Sw,,
there is no distortion at any frequency within this frequency range. That is, there
is no phase shift in the frequency spectrum of the ideal filter. (The phase shift of the
ideal fifter is zero.)
- S 0 ws- - Figure 3-12 Amplitude frequency
2 2 spectrum of the ideal low-pass filter.
econstructing Original Signals from Sampled Signals
Figpaie 3-13 Frequency spectra of the signals before and after ideal filtering.
It is noted that if the sampling frequency is less than twice the highest-
frequency component of the original continuous-time signal, then because of the
frequency spectrum overlap of the pri ary component and complementary com-
ponents, everr the ideal filter cannot recons ct the original continuous-timesignal.
(In practice, the frequency spectrum of continuous-time signal in a control
system may extend beyond k iw,, even though the amplitudes at the higher frequen-
cies are small.)
realizable.
Since the frequency spectrum of the ideal filter is given by
the inverse Folarler transform of the frequency spectrum gives
1 o t- - sin-"-
lrt 2
Equation (3-35) gives the unit-impulse response of the ideal filter. Figure 3-14
shows a plot ofgI(t)versus t. Notice that the response extends from t = -a to t = '30.
This implies that there is a response for t < O to a unit impulse applied at t = O. (That
is, the time response begins before an input is appllied.)This cannot be true in the
physical world. Hense, such an ideal filter Is physlcally unrealizable. [lin many](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-53-320.jpg)
![z-Plane Analysis of Discrete-Time Control
use 3-14 Impulse response gI(t) of ideal filter.
communications systems, however, it is possible to
adding a phase lag, which rneans adding a delay to t
systerns, increasing phase lag is not desirable from the viewpo
fore, we avoid adding a phase lag to approximate the ideal filter.]
Because the Ideal filter is unrealizable and because signalsin practical.control
systems generally have higher-frequency components and are not id
limited, it is not possible, in practice, to exactly reconstruct a coné
from the sarnpled signal, no matter what sarnpling frequency is
words, practically speaking, it is not possible to recon ruct exactly a continuous-
time signali in a practical control system once it is sa
To compare the zero-order hold with the ideal filter, we shall obtain the frequency-
response characteristics of the transfer function of the zero-order hold.
ing jo for s in Equation (3-36), we obtain
The amplitude of the frequency spectrum of Gho(jw)is
The magnitude becomes zero at the frequency equal to the sarnplingfrequency and
at integral rnultiples of the sarnpling frequency.
ec. 3-4 Reconstructing Original Signals from Campled Signals
Figure 3-15(a) shows the frequency-response characteristics of the zero-order
.AScan be seen from Figure 3-15, ere are undesirable gain peaks at
of 3wS/2,5as/2, and so on. Notice
of the zero-order hold are not constant,
and zero-order hold, distortion of the
frequency spectra occurs in the system.
e phase-shift characteristics of the zero-order hold can be obtained as
follows. Note that sin (oU2) alte S positive and negative values as (U increases
from O to os,osto 2ws,2wSto 30.4,
bottom] isdiscontinuous at w = k &/a, where k = 1,2,3, ....Sucha &con-
tinuity or a switch from a positive value to a negative value, or vic
considered to be a phase s ift of i180". lín Figure 3-15(a), phase
to be -180". (It could be assumed to be 4-180" as well.) Thras,
Figure 3-15 (a) Frequency-response curves for the zero-order hold; (b) equivalent Bode
dictgram when T = 1sec.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-54-320.jpg)
![where
z-Plane Analysis of iscrete-Time Control Cystems C
A modification of the presentation o£ the ncy-response diagram of
-15(a) is shown in Figure 3-15(b). The d shown in Figure 3-15(b)
ode diagram of the z o-order hold. The sampiing p
to be 1sec, or T = 2. Notice tude curve approa
ency points that are inte les of the sampling frequency o, =
phase curve [Figure 3-15(b), bottom]
TQsummarize what we have stated, the frequency spectrmm o£t
the zero-order hold includes complementary components, since th
characteristics show that the magnitude of ChO(jo) is not zero for lo1 > 4os,except
at points where o = &os,o = &2os,o = +i3wS,.... Hn the phase curve there are
phase discontinuities of +i180" at frequency points that are multiples of o,. Except
for these phase discontinuities, the phase characteristic is line
Figure 3-16 shows the comparison of the ideal filter and
For the sake of comparison, the magnit
zero-order hold is a low-pass filter, alt quite good. e)£ten,
additional low-pass filtering of the sign essary to effectively
remove frequency components highea than 4o,.
The accuracy of the zero-arder epends on the sam-
pling frequency o,, That is, the output of the hold may be made as close to the
original continuous-time signal as possibk by letting the sarnplingperiod Tbecome
as small as practically possible.
. The phenomenon of the overlap in the frequency spectra is known
asfolding .Figure 3-17 shows the regions where folding error occurs. Tkie frequency
;o,is called the folding frequency or Nyq~tsrfrepency UN.That is,
-3ws -2w, w w, O W, Ws
- - - 2w, 30, w
2 2
Comparison of the ideal filter and the zero-order hold.
econstructing Original
I W s - WS 0 W Figure 3-17 Diagram showing the- " S Ws
-
2 2 regions where folding error occurs.
Bn practice, signals in control systems have igh-frequency components, and some
folding effect will ahost always e e, in an electromechanical syste
some signal may be conta frequency spectrum of the signal,
therefore, may include lo as well as high-frequency noise
comgonents (that is, nois e sampling at frequencies higher
than 400 Hz is not practical, the
a Iow frequency. Rernember that al1signals ith kequencies than U, ap
as signals of frequencies between O and o,. fa&,in certaihicases, a signal of
frequency may appear in the output.
n the frequency spectra of an im sampled signalx*(t),where
own in Figure 3-18, consider an arb frequencypointoz that falls
in the region of the overlap of the frequency spectra. The frequency S
o = o,comprises two components, IX*(jo2)l and Ix* (j(o, - @))l. The latter com-
ponent comes from the frequency spectrum centered at o = w,. Thus, the frequency
Figure 3-18 Frequency spectra of an impulse-sampled signal x*(t).](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-55-320.jpg)
![naiysis of Discrete-TimeControl Systems Chap. 3
In the absence o£the output sampler, if we consi
chronized in phase with the input sampler an
iod) atthe output and observe the sequenceo£values taken byy(t)only at instants
t = kT,then the z transforrn of theoutputy" (t)can alsobe givenby Equation (3-38).
For the continuous-time system, it is a well-known fact that the output y(t) of
e system is related to the input x(t) by the convolution integral, or
g(t - ?-)x(?-)d~=lg>x(t- ~ ) g ( ? - ) d ~
where g(t)is the weighting function of the system or the impulse-response function
of the systern. For discrete-ti e sYstems we have a onvolution summation, whk
is similar to the convofution integral. Since
m
me y(t) of the system to tbe input x*(t)is the sum
O r t < T
g(t)x(O)+ g(t - T)x(T), T i t < 2 T
g(t - T)x(T)+ g(t - 2T)x(2T), 2 T ~ t < 3 T
g(t)x(0)ig(t - T)n(T)+ + g(t - kT)x(kT), kT i t < (k + 1)T
The values o£the output y (t)at the sampling instants t = kT (k = 0,1,2, ...) are
given by
k
where g(kT) is the system's weighting sequence. [The inverse z transform of C ( z )
is called the weighting sequence.] The summation in Equation (3-39) or (3-40) is
called the convolution surnmation. Note that the sirnplified notation
is often used for the convolution summation.
Sincewe ascurned that xft) = O for t < O, we have x(kT - hT) = O for h > k .
Also, since g(kT - hT) = O for h > k, we may assume that the values o£ h in
ec. 3-5 The Pulse Bransfer Function
Equations (3-39) and (3-40) can be taken from
changing the value of the summation. Therefore, Equations (3-39) and (3-40) can
be rewritten as follows:
Ht is noted that if G(s)is a ratio of polynomials in s and if the degree of the
denominator polynomial exceeds the degree of the numerator pofynomial onlly by
1the output y(t)Ps discontinuous, as shown in Figure 3-21(a).
tinuous, Equations (3-41) and (3-42) yield the values immed
pling instants, that is, y(O+),y(T+),... ,y (kT+).Such value
actual response curve.
If the degree of the denominator polynomial exce
by 2 or more, however, the output y(t)is con
hen y(t) is continuous, Equations (3-41) and (3
Figure 3-21 (a) Plot of output y(t)
(impulse response) versus t when the
degree of the denominator polynomial
of G(s) is higher by 1than that of the
numerator polynomial; (b) plot of
2T 3T 4T 5T output y(r) versus t when the degree of
the denominator polynomial of G(s) is
higher by 2 or more than that of the
(W numerator polynomial.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-57-320.jpg)
![r-Plane Analysis of Discrete-Time Controf
Then the z transform of y(t) becomes
where m = n - k. Thus,
Chap.
(3-48)
z transform can be understood as the starre Laplace transform with eTs
y z , the z transform may be considered to be a shorthand notation for the
starred Laplace transform. Thus, Equation (3-48) may be expressed as
which is Equation (3-47). have thus shown that by taking the starred Laplace
transform of both sides of uation (3-45) we obtain
To summarize what we have obtained, note that E
state that in taking the starred Laplace transfo
some are ordinary Laplace transforms and ot
the functions already in starred transforms can be factored out of the starred Laplace
transform operation.
It is noted that systems becorne periodic under starred Laplace transform
operations. Such periodic systems are generally more complicated to analyze than
the original nonperiodic ones, but the former may be analyzed without difficulty if
carried out in the z plane (that is, by use of the pulse-transfer-function approach).
present general procedures for obtaining the pulse transfer function of a system thal
has an impulse sampler at the input to the system, as shown in Figure 3-24(a).
The pulse transfer function G(z) of the system shown in Figure 3-24(a) is
ec. 3-5 The Pulse Transfer Function
v*W
_____)_
S, Y ( z )
Fictitious
sarnpier
(a) Continuous-time system with an impulse sampler at the input; (b)
continuous-time system.
Next, consider the system shown in Figure 3-24(b). The transfer function G(s) is
given by
ortant fact to rernernber is that the pulse transfer function for this system
[G(s)], because of the absence of the input sampler.
The presence or absence of the input sampler is crucial in determining the
pulse transfer function of a system, because, for exarnple, for the systern shown in
igure 3-24(a), the Laplace transform of the output y (t) is
ence, by taking the starred Laplace transform of Y(s), we have
Y*( S ) = C*(s)X*( S )
or, in terms of the z transform,
Y(z) = G(z)X(z)
while, for the system shown in Figure 3-24(b), the Laplace transform of the output
m is
Y(s) = G(s)X(s)
which yields
Y" (S) = [G(s)X(s)f* = [GX(s)]*](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-59-320.jpg)
![? z-Plane Analysis of Discrete-Time Control Systerns Chap. 3
V(z) = Z[Y(s)] = z[c(s)x(s)] = Z[&;X(s)] =
The fact that the z transforrn of G(s)X(s) is not e ual to G(z)X(z) will be discussed
in detaPI Iater in this section.
In discussingthe pulse transfer function, we a e that there is a sarnpler at
ut o£the elernent in consideration. The pres
e output o£the elernent (or the systern) does snot affect the pulse transfer function,
because, if the sarnpler is not physicalfy presesnt a6 th
always possible to assurne that a fictitious sarnpler
eans that, although the output signal is con
e output only at 1 = k?" (k = 0,1,2,. ..) a get sequence y(kT).
Note that only for the case where the input to khe*systemC(s) is an i
sannpled signal is the pulse transfer function given by
ExarnpIes 3-4 and 3-5 demonstrate t e methods h r obtaining the pulse
transfen:function.
Obtain the pulse transfer function G(z) of the system shown in Figure 3-24(a), where
G(s) is given by
Note that there is a sampler at the input of G(s) and therefore the pulse transfer function
is C(z) = h[G(S)].
Method T. By referring to Table 2-1, we have
Hence,
Method 2. The impulse response function for the system is obtained as follows:
Hence,
Therefore,
ec. 3-5 The Pulse Transfer Feinction
Obtain the pulse transfer function of the system shown in Figure 3-24(a), where G(s)
is given by
Note that there is a sampler at the input of G(s).
Method d. G(s) involvesthe term (1 - e-Ts);therefore, referring to Equation (3-32),
we obtain the pulse transfer fttnction as foilows:
G(z) =ZfG(s)] = Z
From Table 2-1, the z transform of each o£the partial-fraction-expansion terms can be
found. Thus,
ethod 2. The given transfer function G(s) can be written as follows:
Therefore, by taking the inverse Laplace transforrn, we have the following impulse
response function:
e-kT+ T - k = 1,2,3,. ..
= {o, k = o
Then the pulse transfer function G(z) can be obtained as follows:
w](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-60-320.jpg)
![z-Plane Analysis of iscrete-Time Control Systems Chap. 3
nce, by taking the starred La of ea& of these two equations, we
Consequently,
fn terms of the z transform notation,
e pulse transfer function between the output y*(t) and input x*(t) is t
given by
Figure 3-25 (a) Sampled system with a sarnpler between cascaded elements G(s) and H(s);
(b) sampied system with no sampler between cascaded elements G(s) and H(s).
ec. 3-5 The Pulse Transfer Function
er the system shown in Figure 3-25(b). From t
where
e starred Lapiace t
In terrns of the z transform notation,
and the pulse transfer function y* (t)and input x* (t)is
Note that
( z ) $. G H ( z ) = Z[C;H(s)]
ransfer functions of t
willl now verify this
Consider the systerns shown in Figures 3-26(a) and (b). Obtain the pulse transfer
function Y(z)IX(z) for each of these two systerns.
Figure3-26 (a) Sampled system with a sampler between elements G(s) = l/(s +a)
and H(s) = l/(s + b); (b) sampled system with no sarnpler between elements G(s)
and H(s).](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-61-320.jpg)
![1 P-PlaneAnaiysis of Discrete-Time Control Systerns Chap. 3
system the computer (digital controller) solves a difference equation whose input-
ut relationship is given by the pulse transfer function CD(z).
In the present system the output signal c(t) is fed
input signal r(t). The error signale(t) = r(t
is converted to a digital signal thr
fed to the digital controller, which
desirable manner to produce the
This desirable relationship between t
by the pulse transfer function GD(z)
selectingthe poles and zeros of GD(z),a number of input-output characteristics can
be generated.]
Weferring to Figure 3-28(b), let us define
Erom Figure 3-28(b), notice that
Bn eerms o£the z transform notation,
Since
and, therefore,
Equation (3-53) gives the closed-loop pulse transfer function of the digital control
system shown in Figure 3-28(b). The performance of such a closed-loop system can
be improved by the proper choice of GD(z),the pulse transfer function of the digital
shall later discuss a variety of forms for GD(z)to be used in obtaining
optimal performance for various given performance indexes.
In the following,we shall consider only a simple case, where the pulse transfer
function GD(z)is of the PID (proportional plus integral plus derivative) type.
ontmller. She analog
ustrial control systems for over half
a century. The basic princ 01 scheme is to act on the variable
to be rnanipulated through a proper combination of three control actions: propor-
tional control action (where the control action is proportional to the actuating error
ec. 3-5 The Pulse Transfer Function
signal, which is the difference between the in t and the feedback signal), integral
control action (where the control action i
ere e(t) is the input to the control1
Define
Figure 5-29 shows the function f(hT). Then
Taking the z transform of this last eequation, we obtain
(Por the derivation of this last equation, refer to Problern A-2-4.) Notice that](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-64-320.jpg)
![iscrete-Time Control Sec. 3-5 The Pulse Trancfer Function 1
instants. In ordinar
e output will not vary very much between any two consecutive
sampling instants. Tn certain cases, however, we may need to find e
between consecutive sampling
Three rnethods for provi sponse between two consecutive sarnpiing
e transform method
ed z transform metkod
State-space method
Here we shall briefly discuss the ethod. The modified z trans-
form method is presented in App ers interested in the modified
z transforrn should read ection B-4.) The state-space method wili,be discussed in
Section 5-5.
for exarnple, the systern shown in
Equation (3-59) will give the continuous-time response c(t). ence, the response
al any time between two consecutive sampling instants can be calculated by the use
of Equation (3-59). [See Problem A-3-18 for sample calculationasof the right-hand
side of Equation (3-59).]](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-67-320.jpg)
~(z)= AiX(Z)
rnay be drawn as shown in Figure 3-41(b). Note that
1
6;$yz) = -
A,
y combining component digital filters as shown in Figure 3-42(a), it is
posslble to draw the block diagrarn of the digital filter G(z) as sho
3-42(b). [Note that Figures 3-42(a) and (b) correspond to the case
en&. Digital filters based on la
with respect to coefficient sensitivity and accuracy. Realizatio
ture is achieved by expanding G(z) into continued fmctions ar
It is noted that the continued-fraction expansion given by
not the only way possible. There are a few different ways to
Figure 3-42 (a) Component block diagrams for ladder programming of G(z) given by
Equation (3-69) when t~ = 2; (b) combination of component block diagrams showing ladder
programming of G(z).
ealimation of Digital Controllers and
structure. For example, a digital filter C(z) may be structured as a continue
fraction expansion form around the origin in terrns sf z-l, as follows:
Also, instead of G(z), its inverse l/G(z) may be anded into continued-Eraction
s in terrns of z or z-1 in order to carry out ladder programming.
Obtain the block diagrams for the following pulse-transfer-function system (a digital
filter) by (1) direct programming, (2) standard programming, and (3) ladder program-
ming:
. Directprogramming. Since the given pulse transfer function can be written as
direct programming yieldsthe block diagram showriin Figure 3-43. Notice that we need
two delay elements.
2. Standard programming. We shall firct rewrite the pulse transfer function as
follows:
where
and
Figure 3-43 Block diagram realization of Y(z)lX(z) = (2 - 0.62-')1(1 + 0.5z-')
(direct programmins).](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-73-320.jpg)
![aria! z-Piane Analvsis of Discrete-Time C o n t r ~ lSystems Chap. 3
Figure 3-47 Block diagram for the digital filter given by Equation (3-77) (non-
recursive form).
y(4T) = 0.2
y(5T) = -0.1
y(6T) = 0.05
y(7T) = -0.025
The impulse response sequence for this digital filter is shown in Figure 3-48.
Input and output curves
for a zero-order hoid.-
Chap. 3 Exarnple Problerns and Solutions
sequence for the digital filter given by
Equation (3-77).
Consider a zero-order hold preceded by a sampler. Figure 3-49 shows the input x(t)
to the sarnpler and the output y(t) of the zero-arder hold. In the zero-arder hold the
valrie ~f the last sample ic retained until the next sample Is talien.
Obtain the expression for y(t). Then find Y ( s )and obtain the transfer function
of the zero-order hold.
olution From Figure 3-49 we obtain
y(t) = x(O)[l(t)- l(t - T)]+ x(T)[l(t- T ) - l(t - 2T)]
+~ ( 2 T ) [ l ( t- 2T) - l ( t - 3T)I +
The Laplace transform of y (t) is
where
m
The transfer function of the zero-order hold is thus
Consider a first-order hold preceded by a sampler. The input to the sampler is x(t) and
the output of the first-order hold is y(t). In the first-order hold the output y(t) for
kT 5 t < (k + l ) T is the straight line that is the extrapolation of the two preceding
sampled values, x((k - l ) T )and x(kT),as shown in Figure 3-50. The equation for the
output y (t) is
t - kT
y (t) = -T
[x(kT)- ~ ( ( k- 1)T)I + x(kT), kT 5 t < (k + l ) T (3-78)
Obtain the transfer function of the first-order hold, assuming a simple fianction
such as an impulse function at t = 0 as the input x(t).](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-76-320.jpg)
![e ,en 7-Plane Analvsis of Discrete-Time Control Systems Chap. 3
ure 3-50 Input and output curves
&on. For an impulse input of magnitude x(0) such that x*(t) = x(0)6(t),the
,--+ .,/A n ; x r ~ nhxr Fnila+inn(3-781 becomes as shown in Figure 3-51. The mathemat-
Hence,
#(O)
Slope = - -T
Figure 3-51 Output curve of the first-
order hold when the input is a unit-
impulse function.
Chap. 3 Example Problems and Solutions
Since
X*(s) = T[x"((t)]= T[x(O)G(t)]= x(0)
the transfer function of the first-order hold is obtained as follows:
Consider the function
Show that s = O is not a pole of X(s). Show also that
has a simple pole at s = 0.
utisn Pf a transfer function involves a transcendental term eCTS,then it may be
replaced by a series valid in the vicinity of the pole in question.
For the function
let us obtain the Laurent series expansion about the pole at the origin. Since, in the
vicinity of the origin, e-Ts may be replaced by
substitution of Equation (3-80) into Equation (3-79) gives
which is the Laurent series expansion of X(s).From this last equation we see that S = O
is not a pole of X(s).
Next, consider Y(s).Since
1 -
Y(s)= -s2
it may be expanded into the Laurent series as
We see that the pole at the origin (S = O) is of order 1, or is a simple pole.
Show that the Laplace transform of the product of two Laplace transformable functions
f(t) and g(t) can be given by](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-77-320.jpg)
![z-Plane Analysis of Discrete-Time Control Cystems Chap. Chap. 3 Exampie Problems and Solutions
1 ciJm we have
.qf(t)g(t)l = ,ij-,- F(P)G(s - P) dP
SoiutB~sn The Lapiace transform of the product of f(t) and g(t) is given by
%[f (0s('11 = jomf(t)g(')e-s' 'ft
Note that the inversion integral is
f(r) = LjcYmF(s)efl ds, t > O we have
277-j c-lm
G(s - p ) =
1
where c is the abscissa of convergence for E(s). Thus, 1 - e - ~ ( ~ - ~ )
1 m C+Jm
.qf(f)g(t)l = T;;;j" I,,.. F(p)eP'+ &)e--" dt
Notice that the poles of 141- e-T(s-P'] may be obtained by solving the equation
Because of the uniform convergence of the integrals considered, we may invert the
order of integration:
-T(s - p ) = kj27~k, k = O,1,2,...
so that the poles are
Noting that 2 ~ r
p = s k j - k = s k j w s k , k = O , 1 , 2,...
T
j~g(i)e-'s-p"dt = G(s - p )
where ws = 2 d T . Thus, there are infinitely many simple poles along a line parallel to
the jw axis.
we obtain
The Laplace transform of x"(t) can now be written as
1 c+im
%If(t)g(t)l = ,ijc-jmF ( P ) W - P) dp
- 1 "+'"- - 1
2Trjj,-,mX(p) 1 - e-ns - PI d~ (3-86)
Show that the Laplace transform of
where the integration is along the line from c - jw to c +jw, and this line is parallelm m
~ " ( t )= x(t)6(t - kT) = ~ ( t ) 6(t - kT) to the imaginary axis in the p plane and separates the poles of X(p) from those of
k=o k=o 1/[1 - e-T(s-P)].Equation (3-86) is the convolution integral. It is a well-known fact that
can be given by such an integral can be evaluated in terms of residues by forming a closed contour
consisting of the line from c - jw to c +jw and a semicircle of infinite radius in the
left or right half-plane, provided that the integral along the added semicircle is a
constant (zero or a nonzero constant). There are two ways to evaluate this integral (one
using an infinite semicircle in the left half-plane and the other an infinite semicircle in
the right half-plane); we shall consider these two cases separately in Problems A-3-6
and A-3-7.
ution Referring to Equation (3-83), rewritten as
1 c+lm
%[f(t)g(t)l = ,ijc-,-F(P)G(s - P) dp
Referring to Equation (3-86), rewritten as
1 '+'" 1
m x w = ,ij,-,m X(pI1 - e-T(s-,>d~
f(t) = x(t) and g(t) = 6(t - kT)
k=o show that, by performing the integration in the left half-plane, X"(S) may be given by
2 [ 8 ( t - kT)] = e-"'
where
and noting that](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-78-320.jpg)
![z-Piane Analysís of Discrete-Time Control Systerns
By substituting z for eTsin Equation (3-87), we have
- -
X(z) = [residue of -X(p)z at poie of ~ ( p )
z - e T p 1By changing the compIex variable notation from p to S, we obtain
X(z) = [residue of
1
Chap. 3
where we assurned that X(z) has h different multiple poles and m - h simple poles
assume that the poles of X(s) lie in the left half-plane and that X(s) can
be expressed as a ratio of polynomials in S, or
where q(S)andp (S)are polynomialsin s. We also assume thatp (S)is of a higher degtee
in S than q(s),which means that
lirnX(s) = O
s-+m
shall evaluate the convolution integral given by Equation (3-86) using
a closed contourin the left half of thep plane as shownin Figure 3-52. Using this closed
contour, Equation (3-86) may be written as
where the closed contour consists of the line from c - jm to c +jm and rL9which án
turn consists of a semicircle of infinite radius and the horizontal lines at jm and -jm,
which connect the line from c - j m to c +j m with the semicirclein the left half of the
choose a value of c such that al1the poles of X(p) lie to the left of the iine
from c - j a to c +jm and al1 the poles of 1/[1- e-T(s-P)] lie to the right of this line.
The closed contour endoses al1 poles of X(p), while the poles of 1/[1- e-T("p)] are
outside the closed contour.
Because we have assumed that the denominator of X(s) is of a higher degree in
s than the riurnerator, the integral along TL(the infinite semicirclein the Ieft half-plane
plus the horizontal lines at j m and -jm, which connect the line frorn c - jm to c + jm
with the semicircle) vanishes. Hence,
This integral is equal to the sum of the residues of X(p) in the closed contour. (Refer
to Appendix B for the residue theorem.) Therefore,
X(P>
1 - e-T('-p) at pele 0f X ( ~ )
1 (3-89)
Chap. 3 Example Problerns and Sol~atíons
I m
X p plane
Figure 3-52 Closed contour in the left half of the p plane.
By substituting z for eTsin Equation (3-89), we have
[ X(piz at p o i of i.(p jX(z) = residue of -----z - eTp
By changing the complex variable notation from p to S, we obtain
Let us assurnethat X(s) has poles si,s2,. .. ,S,. Zf a pole ats = S, is a simplepole,
then the corresponding residue Kiis
If a pole at S = si Is a multiple pole of order ni, then the residue Kiis
Therefore, if X(s) has a multiple pole sl of order nl, a multiple pole s2of order n2,. . . ,
a multiple pole shof order nh,and simple poles s,,+I,~h+2,...,sm,then X(z) given by
Equation (3-90) can be written as
X(s)z at pole of ~ ( s )h(z = 2[residue of -z - eTs 1](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-79-320.jpg)
![nalycis of Discrele-Time Control Chap. 3
(3-93)
where n, is the order of the multiple pole at s = si.
Referring to Equation (3--86), rewritten as
show that by performing this integration in the right haif p plane, X*(S) may be given
provided that the denominator of X(s) is two or more degrees higher in s than the
numerator. Show that if the denominator of X(s) is only one degree higher in s than
the numerator then
Solution Eet us evaluate the convolution integral given by Equation (3-86) in the right
half of the p plane. Let us choose the closed contour shown in Figure 3-53, which
consists of the line fromc - jw to c +jm and r R ,the portion of the semicircle of infinite
radius in the right half of thep plane that lies to the right of this h e . The closed contour
encloses al1 poles of 141 - e-T("p)], but it does not enclose any poles of X(p). Now
X"(s) can be written as
Let us investigate the integral along r R , the portion of the infinite semicircle to
the right of the line from c - jm to c +jm. Since infinitely many poles of 1/[E - e-T(s-P']
lie on a lineparallel to thejw axis, the evaluation of the integral along TRis not as simple
as in the previous case, where the closed contour enclosed a finite number of poles of
X(p) in the left half of the p plane.
In almost al1physical control systems, as s becomes large, X(s) tends to zero at
kast as fast as 11s. Hence, in what follows, we consider two cases, one where the
denominator of X(s) is two or more degrees higher in s than the numerator and another
where the denominator of X(s) is only one degree higher in s than the numerator.
Case 1: X(s) Possesses at Least Two More Poles Than Zeros. Referring to the theory
of complex variables, it can be shown that the integral along TRis zero if the degree
of the denominatorp(s) of X(s) is greater by at least 2 than the degree of the numeralor
q(s); that is, if X(s) possesses at least two more poles than zeros, which implies that
Iim sX(s) = x(O +) = O
S- =
Chap. 3 Example Problemc and Solutions
C---y--JPoles o f
X ( P )
p plane
ure 3-53 Closed contour in the right half of the p plane.
then the integral along TRis zero. Thus, in the present case
Therefore, Equation (3-96) simplifies to
The integral along the closed contour given by Equation (3-97) can be obtained by
evaluating the residues at the infinite number of poles at p = s k jws k. Thus,
The minus sign in front of the right-hand side of this last equation comes from the fact
that the contour integration along the path TRis taken in the clockwise direction. Using
L'Hopital's rule, we obtain](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-80-320.jpg)
![z-Plane Analysis of Discrete-Time Control Systems
Noting that
Thus,
Note that this expression of the z transform is useful in proving the sampling theorem
(see Section 3-4). However, it is very tedious Lo obtain z transform expressions of
commonly encountered functions by this method.
Case 2: X(s) Has a Denominator One Degree Higher in s Than the Numerator. For
this case limw,sX(s) = x(O+) # O < m and the integral along TRis not zero. [The
nonzero value is associated with the initial value x(O+) of x(t).]It can be shown that
the contribution of the integral along PR in Equation (3-96) is -$x(O+). That is,
Then the integral term on the right-hand side of Equation (3-96) becomes
Consider the function
Qbtain A7(z) by using the convolution integral in the right half-plane.
Solution The Laplace transform of x(t) is
Clearly, iim-,sX(s) = x(O+) = 1, or the function has a jump discontinuity at t = 0.
Hence we must use Equation (3-95). Referring to this equation, we have
xample Problems and
Weferring to a formula available in mathematical tables,
and noting that
we can rewrite Equation (3-
x * (S
101) in the form
Thus, we have obtained X ( z ) by using the convolution integral in the right half-plane.
[This process of obtaining the z transform is very tedious because an infinite series of
X(s +jws k ) is involved. The example here is presented for demonstration purposes
only. One should use other methods for obtaining the z transform.]
Obtain the z transform of
X(s) =
S
(S + q2(s + 2)
by using (1)the partial-fraction-expansion method and (2)the residue method.
Solntlopa
1. Partial-fraction-expansion method. Since X(s) can be expanded into the form
2 1 2qs)= -- -- -
s + 1 ( ~ + 1 ) ~S + 2](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-81-320.jpg)
![z-Plane Analysis of Discrete-Time Control
ence,
Y ( z )= X ( z )
Ln terms of the starred Laplace transform notation, this last equation can be written as
Obtain the weighting sequence of the system defined by
for n = 2, 2, and 3, respectively.
oleation For n = 1,we have
Hence, the weighting sequence g,(k) is found to be
g ~ ( k )= (-0)"
For n = 2, we obtain
1 - 1 - az-' + a 2 z - 2 - a3zT3+ ..e
G$(Z)= -
(1 + az-l)' 1 4- az-'
= 1 - 2aZ-i + 3aZZ-' - 4a32-3+ -..
Kence, the weighting sequence &(k) is
g 4 k ) = (k -i- ~ ) ( - a ) ~
For n = 3, we get
ence, the weighting sequence g4k) is
Obtain the discrete-time output C ( z )of the closed-loop control system shown in Figure
3-56. Also, obtain the continuous-time output C(s).
Figure 3-56 Discrete-time control system.
hap. 3 Example Problems and Solutions
utlon From the diagram we have
M($) = Gl(s)E(s)
E(s) = R(s) - H(s)C(s)
ence,
Taking tbe starred Laplace transform of this last equation, we obtain
M* ( S )= [GiR(s)]*- [GiG,H(s)]*
In terms of the z transform notation,
This last equation gives the discrete-time output C(z).
The continuous-time output @ ( S ) can be obtained from the following equation:
Notice that [CiR(s)]*/{I+ [GlG2N(s)]*)is a series of impulses. The continuous-time
output C(s)is the response of G2(s)to the sequence of such impulses. [See Problem
A-3-18 for details of áetermining the continuous-time output ~ ( t ) ,the inverse Laplace
transform of C(s).]
Consider the svstem shown in Figure 3-57. Obtain the closed-loop pulse transfer
function G(z)lR(z).Also, obtain the expression for @(S).
Figure 3-57 Discrete-time control system.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-84-320.jpg)
![z-Plane Analysis of Discrete-Time
Analog P action Digital P action
Analog I action Digital 1 action
Analog P/ action Digital P I action
Analog D action Digital D action
Polar plots of analog and digital controllers with (a) proportional
action, (b) integral action, (c) proportional plus integral action, and (d) derivative
action.
hap. 3 Example Problems and
Analog PID controller Digital PID controller
(a) Polar plot of analog PXD controller; (b) polar pIot of digital PID
controller.
= Kp[e(kT)- e((k - l)?")] + KIe(kT)
+K D [e(kT)- 2e((k - 1)T ) + e((k - 2)T)] (3-105)
where we have used the relationships Kp = K - 4KI,& = KTII;, and KD = KTdlT.
(For these reiationships, refer to the derivations of the positional form of the digital$ID
control equation.) Equation (3-105) takes into consideration the variation of the
positional form in one sampling period.
Suppose the actuating error e(kT)is the difference between the input r(kT)and
the output c(kT),or
e(kT) = r(kT) - c(kT)
By substituting this last equation into Equation (3-105), we obtain
Vm(kT)= Kp[r(kT)- r((k - 1)T)- c(kT) + c((k - 1)T)l
+&[r(kT) - c(kT)]+ KD[r(kT)- 2r((k - 1)T)
-t r((k - 2)T) - c(kT) + 2c((k - 1)T) - ~ ( ( k- 2)IF)I (3-106)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-86-320.jpg)
![z-Plane Analysis of Diccrete-Time Control Systems Chap. 3
The velocity-form PID control scheme given by Equation (3-106) may be modified ints
a somewhat different form to cope with sudden large changes in the set point, Siince
the proportional and derivative control actions produce a large change in the controlker
output when the signal entering the controller makes a sudden large change, to suppress
such a large change in the controller output, the digital proportional and derivative
terms may be modified as discussed next.
If changes in the set point [input r(kT)] are a series of step changes, then
immediately after a step change takes place, the input r(kT) stays constant for a w
until the next step change takes place. Hence, in Equation (3-106) we assume that
(Note that this is true if the input stays constant. But we assume that this holds true
even if a step change takes place.) Then Equation (3-106) rnay be modified to
The z transform of Equation (3-107) gives
Simplifying, we obtain
Equation (3-108) gives the velocity-form PID control scheme. The block diagram
realization of the velocity-form digital PIE) control scheme was shown in Figure 4-30.
he system shown in Figure 3-60(a). Obtain the continuous-time output c(t)
so that the output between any two consecutive sampling instants can be determined.
Find the expression fo-rthe continuous-time output c(t). The sampling period Tis 1sec.
ution For the system shown in Figure 3-60(a), we have
C(s) = G(s)E*(s)
E(s) -R(s) - C(s)
Hence,
or
Thus,
The continuous-tiae output c(t) can therefore be obtained as the inverse Laplace
transform of C(s):
ap. 3 Example Problems and
- -
(a)
1.5
Output points obtained
by sarnple calculations
Figure 3-60 (a) Discrete-time control system; (b) plots of individual impulse
responses; (c) plot of continuous-time output c(t) versus t.
c(t) = (e--l[c(s)] = (e-' G(s)
[ 1:Z;$J
For the present system,](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-87-320.jpg)
![z-Piane Analysis of Diserete-Time Control
ence,
F. 7
Eet us define
Then the z transform expression for this last equation is
Referring to Equation (3-58) for the z transform of G(s), we obtain
1 '
1 - z-'
X(z) = (1 - z-')
0.3679~-'+ 0.2642~-~
'(1 - 0.3679~-')(1 - z-')
-- 1 - 1.3679~-1+ 0.3679~-~
1- z-' + 0.6321~-~
Hence, noting that the sampling period T is 1sec or T = 1,we have
1- 1.3679e-" + 0.3679e-2"
X*(s) =
1 - e-" + 0.6321e-&
Therefore,
1 1- 1.3679e-" + 0.3679e-&
c(t, = T1[-s2(s + 1) 1- e-" + 0.6321e->
(1 - 0.3679e-' - 0.6321eT2 - 0.3996eP3"
+ 0e-4" + 0.2526e-~"+ 0.2526e-& +
Since
1 - - --- 1l 1 + _
s2(s + 1) s2 S S + 1
the inverse Laplace transform of this last equation is
r -I
Hence, we obtain
c(t) = (t - 1 + e-') - 0.3679[(t - 1) - 1+ e-('-"]l(t - 1)
Chap. 3
(5-109
le Problems and Solulions
Figure 3-60(b) shows plots of individual impulse responses given by Equation (3-109).
[Observe that c(t) consists of the sum of impulse responses that occur at t = O,t = 1,
t = 2, ... with weighting factors 1, -0.3679, -0.6321, ....]
From Equation (3-109) we see that for time intervals O 5 t < 1 , l r t < 2,
2 r t < 3,. .. the output c(t) is the sum of impulse responses as follows:
t - 1 + e-', O s t < l
(t - 1+ e-') - 0.3679[(t - 1) - 1 + e-('-')]l(t - l), 15 t < 2
e-') - 0.3679[(t - 1) - 1 t- e-"-"]l(t - 1)
-0.6321[(t - 2) - 1 + e-''-2)]1(t - 2), 2 5 t < 3
Sample calculations for several values of t follow:
The continuous-time output c(t) thus obtained is plotted in Figure 3-60(c).
Consider the digital filter defined by
Draw a series realization diagram and a parallel realization diagram. (Use one first-
order section and one second-order section.)
Solutiolra We shallfirct consider the series realization scheme. To limit the coefficients
to real quantities, we group the second-order term in the numerator (whichhas complex
zeros) and the second-order term in the denominator (which has complex poles).
Wence, we group G(z) as follows:
Figure 3-61(a) shows a series realization diagram.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-88-320.jpg)
![P-Plane Analysis of Discrete-Time Control Systems
P Block diagram reaiizations of the digital filter considered in Problem
A-3-19. (a) Series realization; (b) parallel realization.
Chap. 3 Example Problems and Solutions i
Next, we shall consider the parallel realization scheme. Expansion of G(z)/z into
partial fractions gives
Then G(z) can be written as follows:
Figure 3-61(b) shows a parallel realization diagrarn.
A slowly changing continuous-time signal x(t) is sampled every T sec. Assume that
changes in signalx(t) are very slow cornpared to the sampling frequency. Show that in
the z plane (1 - z-l)lT corresponds to "differentiation," just as s corresponds to
"differentiation" in the s plane.
antloa For a siowlychanging signalx(t), the derivative of x(t) can be approxirnated
by
The z transform of this equation gives
1 1
V(z) = -[X(z) - z-l %(z)] = -(1 - z-l)X(z)
T T
from which we obtain the block diagram shown in Figure 3-62(a). This diagram
corresponds to the s plane differentiation shown in Figure 3-62(b). Note tRat the block
diagrarn shown in Figure 3-62(a) can be rnodified to that shown in Figure 3-63. (For
approxirnation of "integration" in the z plane, see Problems -3-25 through B-3-27.)
se 3-62 (a) Block diagram for "differentiation" in the z plane; (b) block
diagram for "differentiation" in the s plane.
Figure 3-63 Approximate
"differentiator" in the z plane.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-89-320.jpg)
![esign of Discrete-TimeControl Systems by Canventional Methods Cha
The absolute stability and relative stability of the linear time-invariant continuous-
time closed-loop control syste ocations of the closed-loop
poles in the s plane. For example, peles in the left half of the
s pXane near the ju axis will exhib ,and closed-loop poles sn
ative real axis will exhibit
mce the complex variables z = eTs,the poje and zero
locations in the z plane are related to the pole arid zero locations in the s plane.
Therefore, the stability of the linear time-invariant discrete-timeclosed-loop system
can be deterrnined in terms of the locations of the poles of the closed-loop pulse
transfer function. It is noted that the dynamic behavior of the discrete-time control
system depends on the samplingperiod T. In terms of poles and zeros in the z plane,
eir locations depend on the sampling period T. In other w
sampling period Tmodifies the pole and zero Ioeations in the
response behavior to change.
ne. ]en the design of a
are very important in predictirig the dynamic behavior of t
designing discrete-time control systems, the locations of th
zplane are very important. In the followingparagraphs we shall investigate how tbe
locations of the poles and zeros in the splane compare with the locations of the poies
rporated into the process, the co
This means that
trancformation z
part w , we have
a pole in the s plane can be located in the z plane through ihe
= e". Cince the complex variable s has real part o and imaginary
and
From this last equation we see that poles and zeros in the s plane, where frequencies
differin integralmultiples of the sampling frequency 2 d T , are mapped into the same
locations in the z plane. This means that there are infinitelymany values of s for each
value of z .
Since u-is negative in the left half of the s plane, the left half of the s plane
corresponds to
Izl = eTu< 1
23-xjw axis in the s plane corresponds to zl = 1.That is, the imaginary axis in the
S p h e (ihe line o-= O) corresponds to the unit circle in the z plane, and the interior
of the unit circle corresponds to the left half of the s plane.
ec. 4-2 Mapping Between the s
s. Note that since/z =
of z m to m. Consider a represen
on the jw axis in the s plane. As this point moves from -j2 asto j i o,on
ere o, is the sampling fre uency, we have /z/ = 1, a n d b varies from -71 to 71
in the counterc1ocl;wise direction in the z plane. As the re
from j i w, to j$ w, on the jw axis, the corresponding
theunit circIeoncein the counterclockwise direction.
moves from -m to m on the jo axis, we trace the unit
number of times. Frsm this añialysis, it is clear that each
ane maps into the inside of the unit circle i
If of the s plane may be divid nt-5an infinite n ~ b e rof periodic
ian Figure 4-1. The primary s extends fromjw.5 =: -j ose0 j+
ntary strips extend from ji j$%, 2%to I ~ w , ... , and fr
, -j2ws to -j$%?. ...
In the primary strip, if we trace the sequence of points 1-2-3-4-5-1 in the s
plane as shown by the circled numbers in Fig
]tecentered at the origin of th
ng points 1,2, 3, 4, and 5 in
nurnbers in Figure 4-2(b).
The area enclosed by any of the compiementary st
unit circle in the z plane. This means that the correspo
Complementarystrip
s plane
Complementary
strip
1m
z plane
Complernentary
strip $j
Figure 4-1 Periodic strips in the s plane and the corresponding region (unit circle
centereti at the origin) in the z piane.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-94-320.jpg)
![esign of Diccrete-Time Control ystemc by Conventional Methods Chap. 4
/z = 271. X 0.5= n== 1800
Thus, the spiral can be graduate
Figure 4-7(b)]. Once the sam
wda1 any point on the spi
4-7(b), wdcan be determi
specified as w, = 1071.radtsec, then at point 69
Hence, wd at point P is
Note that if a constant-damping-ratio line is in the second or t
in the s plane then the spiral decays within the unit circle in the z pl
if a constant-damping-ratioline is in the first or fourth quadrant in th
corresponds to negative damping), then the spiral grows outside
Figure 4-8 chows constant-damping-ratio loci for i= 0, c = 0.2,5 = 0.4,i= 0.6,
( = 0.8, and { = 1. The [ = 1locus is a horizontal line between points r = O and
z = 1.(Note that Figure 4-8 shows only the loci in the upper half of the z
which correspond to O 5 w 5 4ms. 'F e loci corresponding to --So, r 5 0 are the
rnirror irnages of the loci in the upper half of the z plane about the horizontal axic.)
Wotice that the constant {locj are normal to the constant wnloci in the S plane,
as shown in Figure 4-9(a). In the z plane rnapping, constant w,loca intersect constant
i spirais at right ancles, as shown in Figure 4-9(b). A mapping such as this, wbich
preserves both the size and the cense of angles, is called a conforma1 rnupping.
Im
z plane
-1.0 -0.5 O l= 1 0.5 1.O R e
Figure 4-8 Constant-damping-ratio loci in the z plane.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-97-320.jpg)
![E;<srthe system shown in Figure 4-18, define
and
Then we have
and
steady-state values. FTorn the diagram we have the actuating error
e(t) = r(t) - b(t)
e shall consider the steady-state actuating error at the sampling instants. Note that
from the final valiue theorern we have
lime(kT) = lim [(1- z-')E(z)]
k-- 2-1
Figure 4-18 Discrete-time control system.
Sec. 4-4 Transient and
y substituting Equation (4-12) into Equatim ('-PI), we obtain
(4-13)
As in the case of the continuous-time control system, we consider three types oE
inputs: un&-step, unit-ramp, an ~nit-accelerationinputs.
Static Posktion Error Constant. For a un&-ste input r(t) = 1(t),we have
1
( 2 ) = ----
a - Z-l
y substituting this last equation into Equation (4-13) the steady-state actuating
error in response to a unit-step input can be obtained as álollows:
osition error constant
Then the steady-state actuatingerror in response to a unit-step in
from the equation
The steady-state actuating error in response to a unit-step input becomes zero if
& = which requires that GH(z)have at least one pole at z = 1.
Static Velocity Error Constant. For a unit-ramp input r(t) = tI(t),we have
Tz-'
( z ) = ( ~ -- 1 2
2 1
y substituting this last equation into Equation (4-131, we have
Now we define the static velocity error constant Kvas follows:
K, = lim
z-+1 T
(4-16)
Then the steady-state actuating error in response to a unit-ramp input can be
given by
ifK, = S , then the steady-state actuating error in response to a unit-ramp input is
zem. This requires G ( z )to possess a dorsble psle at z = 1.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-106-320.jpg)
![esign sf Discrete-Time ysterns by Conventional Met ec. 4-5 Design
To combine these two forms into one, let us define t
1 + F(z) = O
where
at F(z) is the open-100 pulse transfer function. The characteristic equation
given by Equation (4-20) can then be written as
Since F(z) is a complex quantity, this last equation can be split into two equations
by equating first the angles and then the agnitudes of the two sides &oobtain
ASuGLE CONDITION:
MAGNITUDE CONDITION :
IF(z)l = 1
The values o£z that fu
of the characteristic e
A plot of the point e angle condition calorre
is the root locus. The r e characteristic equation (the closed-loop
corresponding to a given value of the gain can be located on the root loci by
agnitude condition. %hedetails of applyingthe angle and magnitude conditions
to obtain the closed-loop poles are presented in the following.
cated, but actually it is not difficult if the
are applied.
y locating particuhr points and asymptotes and by computing angles 0
departure from complex poles and angles of arrival at complex zeros, it is possib
to construct root Loci without difficulty.Note that whileroot locirnay be conveniewt
drawn with a digital computer, if manual construction of the root locus plot
attempted, we essentiallyproceed on a trial-and-error basis. ut the number of tri
required can be greatly reduced if the establlshed rules are used.
Because the open-loop conjugate complex poles and conjugate complexzems
if any, are dways located symmetrically ut the real axis, the root loci are agway
symmetric with respect to the real axis. rice, we need only construct the uppe
half of the root loci and draw the mirro age of the upper half in the lower ha1
of the z plane. Remember that the angles of the complex quantities originating from
t k open-loop poles and open-loop s and drawn to a test point z are meacured
in the counterclockwise direction. shall now present the general rulec, and
Obtain the characteristic equation
d then rearrange this equation so arameter of interest, such as gain
pears as the multipiying factor in
r of interest is gain K , wh
the closed-loop zeros consist of the
zeros of G(z) and the poles of H(z).]
ind the starting points and terminating points o
er of separate branches of the root loci. The
corresponding to K = O are open-loop poles and those cor
ence, as K is increased from zero to infinity, a root locaas starts
from an open-loop pole and terminates at a finite open-loop zero
zero at infinity. This rneans that a root-locus plot will have just as
as there are roots of the characteristic equation. [If the zeros at infi
in the count, F(z) has the same number of zeros as poies.]
f the number n of closed-1 e same as the number of o
poles, then the nrrmber of ind 1 root Iocus branches termirrating
open-loop zeros is equal to the number m of the loop zeros. The remaining
n - m branches terminate at infinity (at n - m cit zeros at infinity) along
asymptotes.
. Determine the root loci on the real axis. Root ljoci on the real axis are
determined by open-loop poles and zeros lying on it. The conjugate compliex poles
and zeros of the open-loop pulse transfer function have no effect on the Iocation of
the root loci on the real axis because the angle eontribution of a pair of conjugate
complex poles or zeros is 360" on the real axis. Each portion of the root Iocus on
the real axis extends over a range from a pole or zero to another pole or zero.
Bn constructing the root loci on the real axis, choose a test point on it. If the
tal number of real poles and real zeros to the ri of this test point is odd, then
is point lies on a root locus. The root locus an s complement form alternate
gments along the real axis.
Determine the asymptotes of the root lo the test point z is located far
from the origin, then the anglies of al1the complex quantities may be considered the
same. One open-loop zero and one open-loop pole then each cancel the effects of
the other.
Therefore,the root loci for very large values of z must be asymptotic to straight
h e s whose angles are given as follows:
r180°(2Pd + 1)
Angle of asymptote = , N = 0 , 1 , 2, . . .
n - m
L.,
pii-oceduresfor constrclcting root loci.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-110-320.jpg)
![where (z)/A(z) = F(z), then
n = number of finite poles of F(z) A ( 4K = --
N z )
(4-22)
m = number of b i t e zeros of F(z)
and the breakaway and break-in points (which correspond to multiple roots) can be
determined from the roots of
obtained as fo'ollows. Since
--
z ~ - ~+ [(pl +p2 + +pn) - (z1+ Z2 + . + z , ) ] z ~ - ~ - ~+ .
ne the angle of departure (or angle of arrival) of the root loci from
n - m
--ua = - (Pi + P2 + + pn) - (zi + zi + + zm)
n - m
Angle of departure =
480" - (8, 3- 8, - 4)
If the characteristic equation
1 + F(z) = O
is written as
Figure 4-22 Diagram showing angle of departure.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-111-320.jpg)
![Decign of Discrete-Time Control Syctems by Conventional Methods @ha
For the present case, T = 0.5 and this Iast equation becomes
Since the critica1 gain Mccorresponds to point z = -1, we substitute -1 for z in
Equation (4-28) :
or
K = 8.165
The critical gain M,is thus 8.3 65.
The dosed-loop poles correspon
zl = 0.4098 +j0.6623 and z2 = 0.4098 - j0.6623
These closed-loop poles are indicated by dots in the root locus diagram.
Sampling period 1' = I sec: r this case, Equatiorl (4-2 5) beca
follows:
Kence, G ( z )has poles at z = 1 and z = 0.3679 and a zero at z = 0.
The breakaway point and break-in point are found to be z = 0.6065 and
z = -0.6065, respectively. The corresponding gain values are K = 0.2449 and
K = 4.083, respectively.
Figure 4-24(b) shówsthe root locusdiagram when T = 1sec. T
of gain K is 4.328. The closed-loop poles corresponding to K = 2 are found to be
z, = 0.05185 +j0.6043 and z2 = 0.05185 - j0.6043
and are shown in the root locus diagram by dots.
3. Sampling period T = 2 sec: For this case, Equation (4-25) becomes
e see that G ( z )has poles at z = 1 and z = 0.1353 and a zero at z = 0.
The breakaway point and break-in point are found to be z = 0.3678 and
z :-0.3678, with corresponding gain values K = 0.4622 and K = 2.164, respece
tively. The critical value of gain K for this case is 2.626.
Figure 4-24(c) shows the root locus diagram when T = 2 sec. The closed-loop
poles corresponding to K = 2 are found to be
z, = -0.2971 +j0.2169 and z2 = -0.2971 - j0.2169
These closed-loop poles are shown by dots in rhe root locus diagram.
closed-loop pole is i,then in the s plane the closed-loop pule location (in the upper
half-plane) can be given by
s = -bn+j u n V i - C2
e corresponding point in the z plane is
z = exp [T(-iwn +j w n m ) ]
z plane
Figure 4-25 Closed-loop pole locations in the z plane shown with constant 5loci.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-114-320.jpg)
![Design of Discrete-Time Control Systems by Conventional ec. 4-5 Design Based on the Root-Locus Method
Figure 4-27 (a) Unit-ramp response sequence of the system shown in Figure 4-23
when T = 0.5 sec and K = 2; (b) unit-ramp response sequence when T = 1sec and
K = 2; (c) unit-rarnp response sequence when T = 2 sec and K = 2.
The three cases we have considered demonstrate that increasing the samp
period T adversely affects the system's relative stability. (It may even cause insta
ity in some cases.) It is important to remember that the damping ratio i of t
ched-loop poles of the digital control system is indicative of the relative stddit
only if the sampling frequency is sufficiently high (that is, eight or more sarnpk P
ed sinusoidal oscillation). ncy is low (that is, less
les per cycIe of damped ), then predicting the
relative stability from the damping ratio value is erroneous.
Consider the digital control systemshownin Figure 4-28. In the z plane, design a digital
controller such that the dominant closed-loop poles have a damping ratio 5of 0.5 and
a settling time of 2sec. The samplingperiod is assumed to be 0.2 sec,or T = 0.2. Obtain
the response of the designed digital control system to a unit-step input. Also, obtain
the static velocity error constant 6(,of the system.
For the standard second-order system having a pair of dominant closed-loop
poles, the settling time of 2 sec means that
4 4
settling time = -= -= 2
lw, 0 . 5 ~ ~
which gives the undamped natural frequency w, of the dorninant closed-loop poles as
wn = 4
The damped natural frequency wd is determined to be
wd = con- = 4- = 3.464
Since the sampling period T is 0.2 sec, we have
[Notice that there are approximately nine samples per cycle of damped oscillation
(31.4213.464 = 9.07). Thus, the sampling period of 0.2 sec is satisfactory.]
shall first locate the desired dominant closed-loop poles in the z plane.
Weferring to Equations (4-29) and (4-30), for a constant-damping-ratio locus we have
and
@d
12= Twd = 27T-
@S
From the given specifications(5 = 0.5 and wd = 3.464), the magnitude and angle of the
dominant closed-loop pole in the upper half of the z plane are determined as follows:
Figure 4-28 Digital control systern for Example 4-9.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-117-320.jpg)
![Design of Discrete-Time Control Systems by Conventional Methods Chap.
Stable linear
time-invariant
discrete-time
-31 Stable linear time-invariant discrete-time system
The sampled signal u(kT)is
u(kT) = sin kwT
The z transform of the sampled input is
U(z)= Z' [sinkoT] =
z sin wT
(z - eJ w T ) ( ~- e-lwT)
The response of the system is given by
X(z) = C(z)U(z)= G ( z )
z sin wT
(z - elwT)(z- e-~wT)
- az-- a~
2 - elwT +-----z - e-~wTi[tesrnsdueto poles of G(z)]
Multiplying both cides of Equation (4-34) by ( z - epT)/zywe obtain
[terms due to poles of G
The second term on the right-hand side of this Iast equation a
approaches elwT.Since the system considered here is stable, the third term on th
right-hand side also approaches zero as z approaches e'"'. Hence, by letting. ,-
The coefficient E, the complex conjugate of a, is then obtained as follows:
C(e-jwT>
approach elw',we have
Let us define
~f (el"T)P; Mele
Then
Equation (4-34) can naw be written as
ased on the Frequency-Responce Method
The inverse z transform of this Iast equation is
The Iast term on the right-hand side of E uation (4-35) repsesents t
response. Since the system G ( z ) has been assurned to be stable, al1 transient
response terms will disappear at steady state and we wPlI get the following steady-
state response x,,(kT):
bjected to a sinusoidal in
ase angle, is given by
S = @(o)=/C(eJwT)
In terms of G(eiwT),Equation (4-36) can be written as follows:
x,(kT) = 1G(elwT)lsin (kwT +/G(ejwT))
have shown that G(ejwT)indeed gives the magnitude and phase of the frequency
response of C(z).Thus, to obtain the frequency response of G(z),we need only to
substitute elwTfor z in G(z).The function G(eiwT)is coznrnonlycalled the sinusoidal
pulse transfer function. Noting that
we find that the sinusoidal pulse transfer function G(eiwT)is periodic, with the perio
equal to T.
Consider the system defined by
x(kT) = u(kT) + ax((k - 1)T), O < a < 1
where u(kT)is the input and x(kT) the output. Obtain the steady-state output x(kT)
when the input u(kT)is the sampled sinusoid, or u(kT) = A sinkoT.
The z transform of the system equation is
X ( z ) = U(z)+ az-'X(z)
By defining G(z)= X(z)/U(z),we have](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-120-320.jpg)
![Design of Discrete-Time Control ystems by Conventional Metho ased on the Frequency-
The transfer function G(w) is fer function. Hence Thus,
K[0.01873(ii) 1- O.lw + 0.01752]
consideration the nonmi ( ) = (1 + "1.T
- 1.8in7('iu'")+ o 8 1 g
The frequency axis in the w plane is distorted. 1- 0.lw 1- 0 . 1 ~
Eictitious frequency v and the actual frequency w is -- K(-0.000333w2 - 0.09633~+ 0.9966)
2 wT w2 + 0.9969~
v = - tan-
T 2
f, for exarnple, a bandwidth wbis specified, we need to desi --
K ( l + &)(1-5)
w(w + 1)
2 @bT ~
A simplephase-lead compensator will probably satisfy al1 requirements. There-
vb = T tanl fore, we shall try lead compensation. (If Iead compensation does not satisfyal1require-
ments, we need to use a different type of compensation.)
Consider the digital control system shown in Figure 4-36. Design a digital contr
in the w plane such that the phase margin is 50°,the gain margin is at least 10 dB
the static velocity error constant Kv is 2 sec-l. Assume that the sampling peri
0.2 sec, or T = 0.2.
First, we obtain the pulse transfer function G(z) of the plant that is preceded by
the zero-order hold:
-- K(0.01873~+ 0.01752)
z2 - 1.8187~+ 0.8187
Next, we transform the pulse transfer function G(z)into a transfer function G(w
by means of the biliriear transformation given by Equation (4-37):
z =:
1 + (T12)w - 1+ 0 . 1 ~--
1- (2'12)~ 1- 0 . 1 ~
Eipwe 4-36 Digital control system of Example 4-12.
Now let us assume that the transfer function of the digital controller GD(w)has
unity gain for the low-frequency range and has the following form:
(This is a phase-lead compensator.) It is one of the simplest forms of the digital
controller transfer function. (Other forms may be assumed as well for this problem.)
The open-loop transfer function is
The static velocity error constant K, is specified as 2 sec-l. Hence,
Kv = lim wGD(w)G(w)= 16' = 2
w-io
The gain K is thus determined to be 2.
By setting K = 2, we plot the Bode diagram of G(w):
Figure 4-37 shows the Bode diagram for the system. For the magnitude curves
we have used straight-line asymptotes. The magnitude and phase angle of G(jv) are
shown by dashed curves. (Note that the zero at v = 10, which lies in the right half of
the w plane, gives phase lag.) The phase margin can be read from the Bode diagram
(dashed curves) as 30" and the gain margin as 14.5 dB.
The given specifications require, in addition to K, = 2, the phase margin of 50"
and a gain margin of at least 10 dB. Let us design a digital controller to satisfy these
specifications.
Design of lead compensator. Since the specificationcalls for a phase rnargin of
50", the additional phase-lead angle necessary to satisfy this requirement is 20". To
achieve a phase margin of 50"without decreasing the value of K , the lead compensator
must corrtribute the required phace-lead angle.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-125-320.jpg)
![iscrete-Time Control Systerns by Conventional
G + w ( z )
= q z )
dition to the physical bility conditions, we must pay attention to
--
1+ Go(z)G(z)
the stability aspects of the system ifically, we must avoid canceling an unstable
pole of the plant by a zero o£the d
Sinceit is required that the system exh with zero steady-state any error in the pole-zero cancellation
error, the system must exhibit a finite ce, the desired ciosed- will become unstable. Similarly, the di
loop pulse transfer function must be cel plant zeros that lie outside the unit ciscle.
aozN+alzN-'+ + aN at will happen to the closed-loop puIse transfer
F(z) =
zN function F(z) if G(z an unstable (or critic
z = a outside (or o it circie. [Note that t
or equally,if G(z) inv r more unstable-or
E(z) = a. + alz-l + + a N ~ - N define
hat F(z) must not conta G(z) = -GiW
seriesexpansion of F(z) z - a
e for a physically real where Gl(z) does not include a term that cancels with z - a. Then the closed-loo
p pulse transfer function pulse transfer function becomes
the digital controllerpulse transfer function GD(z).That is, we find t
function G,(Z) that willl satisfyEquation (4-41). SolvingEquation (4-41) foa:GD(z) Gl(zb
G D ( ~ ) G ( ~ )= = F(z) (4-44)
G D ( ~ )z_,
we obtain
Gdz)+ G,(z)G(z)
1 + GD(z)__
G D ( ~ )=
G(z)[I - J;(z)l
Since we require that no zero of GD(z)cancel the unstable pole of G(z) at z = a ,
The designed system must be physicailyrealizable. The conditions for we must have
realizabifityplace certain constraints on tbe closed-loop pulse transfer funct
and the digital controHerpulse transfer function GD(z).The conditions for 1 -- z - a
1 - F(z) =
realizability may be stated as follows: Gdz) z - a + GD(z)Gl(z)
1 + G D ( ~ ) z _ .
The order of the numerator of GD(z>must be equal to or lower t
f the denominator. (Otherwise, the controller requires future
roduce the current output.)
f the plant &;,(S) involves a transportation lag e-L" then the designed ciosed
loop systemmust involveat Ieast the same magnitude of the transportation la
(Otherwise, the closed-loop systemwould have to respond before an input w
glven, which is impossible for a physically realizable system.)
If G(z) is expanded into a series in z-l, the lowest-power term of the serie
expansion of E(z) in z-' merst be at least as large as that of G(z). For
if an expansion of G(z) into a series in z-1 begins with %hez-l term,
first term of F(z) given by Equation (4-42) must be zero, or a. must
that is, fhe expansion has to be of the form
F(z) = alz-l + a2z-2+ + aNzlN
where N 2 n and n is the order of the system. This means that the plant carm
respond instantaneously when a control signal of finite magnitude is appk
the response comes at a delay of at least one sampling period if the crri
expansion of C(z) begins with a term in z-l.
at is, 1- F(z) must have z = a as a zero. Also, notice t
if zeros of C(z) do not cancel poles of GD(z),the zeros of G(z) become zeros of
F(z). [F(z) may involve additional zeros.]
Eet us surnmarize what we have stated concerning stability.
31. Since the digital controller GD(z) should not cancel unstable (or critically
stable) poles of G(z), al1unstable (or critically stable) poles of G(z) must be
included in 1- F(z) as zeros.
(z) that lie inside the unit circle may be canceled with poles of
at lie on or outside the unit circle must not
ence, al1zeros o£G(z) that lie on or outside
the unit circle must be included in F(z) as zeros.
Now we shallproceed with the design. Sincee(kT) = r(kT) - c(kT), referring
to Equation (4-41) we have
Note that for a unit-step input r(t) = l(t)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-129-320.jpg)
![iccrete-Time Control Systems by Conventional Met
and the rernainder must be zero, so that
1 - a l - a , - a 3 = 0
Note that from Equation (4-63) we require W(1) = i.Therefore, by su
z-' = 1 into Equation (4-66), we obtain
2a1 + az = 2.75
Also, Equation (4-64) can be rewritten as
F(z) = al 2-' + a2z-' + a32-3 = (1 + O.7181~-~)~-'&(z)
Hence,
al + a2zy1+ a3zWz= (1 + 0.7181~-')R(z)
Division of the left-hand side of this last equation by 1 $ 0.7181~-'yields the qu
[a, + (az - 0.7181al)z-'1 and the remainder [a3- 0.7181(a2 - 0.7181a1)]z-2
equating the quotient with F,(z) and the remainder with zero, we obtain
Fl(z) = al + (a2- 0.7181al)z--'
and
a3 - 0.7181(az - 0.7181al) = O
Soiving Equations (4-67), (4-68), and (4-69) for al, a2,and a3 gives
al = 1.26184, a2 = 0.22633, a3 = -0.48816
Thus, F(z) is determined as
) = 1.261842-' + 0.22633z-' - 0.48816~-~
fi(z) = 1.26184 - 0.679792-'
N ( z ) = 1 - 0.26184~-~- 0.48817~-~
The digital controller pulse transfer function GD(z) is then determined from Equ
tion (4-50) as follows:
F(z
G"(z) = G ( z ) ( l - z-')N(z)
-- (1 + 0.7181~-~)~-'(1.26184- 0.679802-')
+ 0-7181z-')2-'
(1 - 2-1)(1 - 0.261~4~-1- 0.48817z--2)
(1 - zV')(1- 0.36792-')
(1 - 0.53872-')(1 - 0.36792-')
= 3.4298---
( 1 - 0.84182-')(1 + 0 . 5 7 9 9 ~ ~ ' )
With the digital controller thus designed, the system output in response
unit-step input r(t) = 1is obtained as follows:
C ( z ) = F(z)R(z)
1
= (1.2,61842-' + 0.22633.T2 - 0 . 4 8 8 1 6 ~ ' - ~ ) - - ~
1 - z-
= 1.2618~-'+ 1 . 4 8 8 2 ~ ~ ~+ Y 3+ z - ~+ .
F(z
and
Equation (4-66) gives
Theunit-step response sequence has a malramum overshoot of approximately 50%. The
settling time is 3 sec.
Notice that from Equation (4-65) we have
U ( z )= 2.7181(1 - 0.36792-')(1.26184 - 0.679792-')
Thus, the control signal u(k) becornes zero for k 2 3. Consequently, there are no
intersampling ripples in the response. Figure 4-41 shows plots of c(k)versus k, u ( k )
versus k, and u(t) versus f in the unit-step response. Notice that the assumption of
N = 3, that is, the assumption o£F(z) as given by Equation (4-61), is satisfactory.
Figure 4-41 Plots of c(k) versus k ,
u ( k ) versus k , and u(t) versus t in the
unit-step response of the system de-
signed in Example 4-14.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-134-320.jpg)
![an actual breakaway or break-in point if and only if the value of K at this point is
positive value.
Discuss the procedure for designing lead compensators for digital control syste
the root-locus method.
dominant closed-loop poles.
4.This angle must be contributed by the lead compensator if the new
is to pass through the desired locations for the dominant closed-loop po
3. Assume the lead compensator GD(z)to be
1+ az
GD(z) = KDa----- O < a < l
1 + a7-z '
If static error constants are not specified, determine the location of t
zero of the lead compensator so that the lead compensator will con
necessary angle 4. If no other requirements are imposed on the syst
make the value of a as large as possible. A larger value of a generally res
a larger value of Kv,which is desirable. (If a particular static error const
specified, it is generally simpler to use the frequency-response appr
5. Determine the open-loop gain of the compensated system from the
condition.
Once a compensator has been designed, check to see whether or not al1
formance specifications have been met. If the compensated system does not me
performance specifications, then repeat the design procedure by adjusting the
pensator pole and zero until al1 such specifications are met. If a large statkc
constant is required, cascade a lag network or alter the lead compensator to a lag
compensator.
Draw root locus diagrams in the z plane for the system shown in Figure 4-45 f
following three sarnpling periods: T = 1 sec, T = 2 sec, and T = 4 sec.
hap. 4 Exarnple Problerns and Solutions
Figure 4-45 Digital control system.
first obtain the z transform of G(s). Referring to Example 3-5, we get
F- 7
Next we construct root locus diagrams for the three cases considered.
l. Sampling period T = 1: For T = 1, Equation (4-75) becomes
K[(1 - 1 + e-')z-' + (1 - e-' - e - ' ) ~ - ~ ]
G(z) =
(1 - z-l)(l - e-lz-')
Wotice that G(z) possesses a zero at z = -0.7181 and poles at z = 1 and
z = 0.3679. The breakaway point is at z = 0.6479, and the break-in point is at
z = -2.0841. The root-locus diagram for this case is shown in Figure 4-46(a).
The value of gain K of any point on the root Ioci can be determined from the
magnitude condition
If we choose a point z on the root loci, the value of K at that point can be
calculated by substituting the value of z into this last equation. (This means that
with this value of K that particular point becomes a closed-looppole.) The critical
gain is found to be K = 2.3925.
Sampling period T = 2: For the sampling period T = 2, we have from Equa-
tion (4-75)
The pulse transfer function G(z) in this case possesses a zero at z = -0.5232 and
poles at z = 1and z = 0.1353. The breakaway point is at z = 0.4783, and the
break-in point is at z = -1.5247. The root-iocus diagram for this case is shown
in Figure 4-46(b). The critical gain IC for stability is Ik = 1.4557.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-138-320.jpg)
![1.6 1.8 2.0 2.2
kT ísec)
Figure 4-59 Plot of c(kT)versus kT for the system designed in Problem A-4-12.
Consider the digital control system shown in Figure 4-60. The plant transfer fu
involves a transportation lag e-'". The delay time is 5 sec, or L = 5. Th
c(t)in response to a unit-step input is as shown in Figure 4-61(a). The
zero to the final value in 10 sec (measured frorn t = 5 e0 t = 15) and
course, choose the sampling period to be 2.5 sec, 1 sec, or another value. In
example, however, to simplify our presentation, we set the sampling period at 5 S
Figure 4-60 Digital control system.
Chap. 4 Example Problerns and
1 (a) Desired output c(t)in response to a unit-step input; (b)plot of u ( f )
versus t.
The z transform of the plant that is preceded by the zezo-order hold is
Notice that there is no unstable or critically stabie pole involved in G(z). Therefore,
there is no stability problem involved in this case.
Let us define the closed-loop pulse transfer function as F(z):
In the present case the output c(t) in the unit-step response is specified as shown in
Figure 4-61(a). Since h[l - e-0.1(15-5)]= h(l - e-') = 1, we have h = 1.5820. from
the deadbeat response curve shown in Figure 4-61(a), we obtain
c(0) = o
c(1) = o](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-147-320.jpg)
![lesign oC Discrete-Time Control Systems by Conventional Methods Chap.
Because the plant transfer function C;,(s) involves a double integrator, b must be ze
(Otherwise, the output increases parabolically, instead of linearly.) Consequently,
have
U ( Z )= bo + bl 2-' + ... + b ~ - l
From Figure 4-62 U ( Z )can be given by
For U ( z ) to be a series in 2-' with a finite nurnber of terms, F(z) rnust be
by 1 + z-':
F(2) = (1 + z-')Fi(z)
Then U ( z )can be written as follows:
U ( z ) = 2F(2)
where & ( Z )is a polynomial in z-' with a finite number of terms.
By cornparing Equations (4-81) and (4-82) and by making a simple analysis,
see that F(z) must involve a term with at least z - ~ .Hence, we assurne
~ ( z )= al z-' + a2z-2 + a32-'
This assumed form of F(z) involves the minimurn number of terms; the tra
response will settle in three sampling periods.
We shall now determine constants al,a2,and a3.Frorn Equation (4-81), we
1 -. al 2-1 - a22-2 - a32 - 3 = (1 - -1 2
2 1 N ( z )
If we divide the left-hand side of this last equation by (1 - z-')', the quotient is
(2 - a1)z'-'.The remainder is [2(2- al) - (1 + Q ~ ) ] Z - ~- [(2- al) + a 3 ] ~ - 3 .He
N ( z ) is deterinined as
N ( z ) = 1 + (2 - al)z-'
and the rernainder is set equal to zero:
[Y(:!- al) - (1 + a 2 ) ] ~ - ~- (2 - al + a 3 ) ~ - 3= O
To satisfy this last equation, we require that
2(2 - al) - (1+ a2)= 0
2 - a l + a 3 = 0
From Equation (4-K!), we have
a1.Y1+ a22-2 + a3z-3 = (1 + z-')F1(z)
Jf we divide the left-hand side of this last equation by 1 + 2-l, the quotient
al z-' + (uz - ul)z-'. The remainder is (al - a2 + a3)z-3.
F1(z)= alz-' 4 (a2- al)zW2
roblems and Ssllations
and the remainder is set equal to zero:
al - az + a3 = O (4-86)
y solving Equations (4-84), (4-85), and (4-86) for al, a,, and a', we obtain
al = 1.25, a2 = 0.5, a3 = -0.75
Wence,
and
$(z) = 1.252-' - 0.75z-' = 1.252-'(1 - 0.62-')
and F(z) is determined as follows:
F(z) = 1.252-' + O . ~ Z - ~- 0 . 7 5 ~ - ~
The digital controller Gu(z) is then determined from Equation (4-50):
ith the digital controller thus designed, the system output in response to a
unit-ramp input is obtained as follows:
Hence,
c(0) = o
Notice that from Equation (4-83) we have
Thus, the control signal u ( k ) becomes zero for k s 3. Consequently, there are no
intersalñrpling ripples in the response at steady state. Figure 4-63 shows plots of c(k)
versus k, u(k) versus k , and u(t) versus t in the unit-ramp response.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-149-320.jpg)
![ec. 5-1 lntroduction
function such as the impulse function, ste function, or sinusoi
state-space methods enable the engineer to include initial condi
his is a very convenient and useful feature that is not
ble, state vector, and state
Seate, The state of a dynamic syst is the smallest set of variables (called
state variables) such that the knowledge hese variables at t = to,together with
e knowledge of the input for t $: to, c letely determines the behavior of the
system for any time t 2 to.Note that the concept of state is by no
%tis applicable to biological systems, eeonomic syste
maki
system. Ef at least n variables xl,xs, ...,x, ar
behavior of a dynamic system (so that once the
state at t = tois spe
then such n variabl
Note that state variables need not be
quantities. Variables that do not represenh
neither measurable
in choosing state variables is an advantage o
speaking, however, it is convenient to choos
state variables, if this is possible at all, becau
feedback of al1 state variables with suitable
Vee$or, Hf n state variables are needed to com
f a given system, then these n state variables can
components of a vector x. Such a vector is called a state vector. A state vector Bs thus
a vector that determines uniquely the system state x(t) for any time t z to,once the
state at t = to is given and the input u(t) for t 2 tois specified.
aee. The n-dimensional space whose coordinate axes consist of the
XI axis, x2axis, ...,x,, axis is called a staie space. Any state can be represented b
a point in the state space.
ons, %nstate-space analysis we are concerned wit
tYpes are involved in the m ling of dynamic systems: 1
variables, output variables, and state variables.
state-space representation for a given systern is not unique, except that the numb
of state variables is the same for any of the different state-space representations
the same system.
For time-varying (linear or nonlinear) discrete-time systems,the state e q d
may be written as
x(k + 1) = f [x(k),un(k), k]
and the output equation as
For linear time-varying discrete-time systems, t e state equation an
where
x(k) = n-wector (state vectsr)
y(k) = m-vector
u(k) = r-vector
G(k) = n x n matrix (state matrix)
H(k) = n x r matrix (input matrix)
atrix (output matrix)
D(k) = m x r matrix (direct transmission matrix)
arance of the variable k in the arguments of m
implies that these matrices are time vas-ying. If
not appear explicitly in the matrices, they are
constant. That is, if the system is time invariant,
w(k + 1) = Gx(k) + Hu(k) (5-1)
As in the discrete-time case, continuous-time (linear or nonlinear) systems may
be represented by the following state equation and output equation:
For linear time-varying continuous-time syste
equation are given by
If tbe system is time invariant, then the last ~ W Oe uati~nsare simplified to
Figure 5-l(a) shows the block diagram representation of the discrete-ti
system defined by Equations (5-1) and (5-2), and Figure 5-l(b) shows the contin-
UOUS-timecontrol system defined by Equations (5-3) (5-4). Notice that the basic
coarfigurations of the diserete-time and continusus- systems are the same.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-154-320.jpg)
![State-Space Analysic ec. 5-2 State-Space epresentationc of Díscrete-Time Systems
uations (5-8) and (5-9) are the staf
The state-space representation given
-- -
.eequation and output equation,
by Equations (5-8) and (5-9)
called a controllabl ical form. [For the derivatidn of Equations (5-8) a
(5-9), see Problem
Note that if w
1
e the order of the state variables, that is, we define ne
state variables according to the fashion
then the state equation can be written as follows:
-al -a2 -
-an Xl(k) 1
1 O - . e O O i 2 ( k ) O
= O 1 . a . O O X3(k) + O
O O - - S O - - ~ , i k ) ~ b~
The output equation can be given by
kquations (5-10) and (5-11) are also in the controllable canonical form.
The state-space representation of the discrete
time system given by Equation (5-5), (5-6), or (5-7) may be put in the followin
form:
Tbe state-space representation given by Equations (5-12) and (5-13) is called an
observable canonicalform. [For the derivation of Equations (5-12) and (5-13), see
Problem A-5-2.1 Notice that the n X n state rnatrix of the state e
Equation (5-12) is the transpose of that of the state equation defi
(5-8).
Note that if we reverse the order of the state variables, tha
then the state equation and the output equation becorne as follows:
i,-,(k + 1)
Xn(k 4- 1)
Xdk)
X2(k)
y ( k ) = [ 1 O S . - O O] f
i n - l(k)
in(k) .
Equations (5-14) and (5-15) are also
+ bou(k)
in the observable canonical form.
. If the poles of the pulse transfer function given by
Equation (5-5), (5-6), or (5-7) are al1distinct, then the state-space representation
may be put in the diagonal canonical form as follaws:](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-156-320.jpg)
![ec. 5-3 Solving
resent the solution of a discrete-time state equation by t
Consider the discrete-time syste described by Equation (5-28):
x(k + 1) = Gx(k)+ Hu(k) (5-
Taking the z transforrn of both sides of Equation ( 5 - N ) , we get
zX(z) - zx(0) = GX(z)+ HtJ(z)
where X(Z)= Z[x(k)]and U(z) = Z[u(k)J.Then
- G)X(z)= zx(0) + HU(z)
Premultiplying both sides of this last equation by ( z - G)-', we obtain
(5-
Eking; the inverse z transforrn of both sides of Esuation (5-41) gives
Comparing Equation (5-30) with Equation (5-42), we obtain
Gk = Z-'[(zI - 6)-lz ]
and
k- 1
b_=k-j-1
where k = 1,2,3,. .. .
Notice that the solution
inverting the matrix ( z - G), which rnay be accomplisfied by analiytical meam
by use of a computes routine. The solution also reauires the inverse z transfor
Obtain the state transition matrix of the following discrete-time system:
x(k + 1) = Gx(k)+ Hu(k)
y ( k ) = Cx(k)
where
Then obtain the state x(k) and the output y(k) when the input u ( k ) = 1 for k O
2,. . .. Assume that the initial state is given by
From Equations (5-35) and (5-43) the state transition matrix
Therefoire, we flrst obtain ( z
The state transition matrix ( k )is now obtained as follows:
Equation (5-45) gives the state transition matrix.
Next, compute x(k). The z transform of x(k) is given by
( z ) = (zlj - G)-lzw(0) + ( z
Since
we obtain z2
-z2 + 22
z - 1 z - 1
Hence](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-159-320.jpg)
![Thus, the state vector x(k) is given by
Finally, the output y(k) is obtained as follows:
where
= GWn-2 + a,-i
W, = GH,-, -i- a, I =
Note that a,, a2,. . . a, are the coefr'icients appearing in the determinant &'e
by Equation (5-47). The si's can also be given (see Problern A-5-13) by use of th
trace, as follows:
olvirig Discrete-Time S
(The trace of an n x n matrix is t
For a higher-order determina
into the form given by Equation (5
Equation (5-50) lo compute the si's
ent method isconvenient for computer
Determine the inverse of the matrix (zI - G), where
[
0.1 0.1 o
G = 0.3 -0.1 -0.2
O O -0.3
Also, obtain Gk.
From Equation (5-46), we have
Although the determinant /z - 61 can be expanded easily, here for demonstration
purposes let us use Equation (5-50) to compute al, a2, and a3. First, notice that
a, = -trG = -tr 0.3 -0.1 -0.2 = 0.3
[Y a -1.JThen, from Equation (5-49), we obtain
Hence
O.1 0.4 0.1
a- = -!tr GHl = -+ ([H::--o1 -:.2][0.3-0.3 O ii.2 - ~ 2 ]
[:.o7
"03 -1.021
= -4 tr 0.09 0.01 0.02 = -0.04](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-160-320.jpg)
![me-Space Analycic Ch
j= h
= G(k)x(k) + H(k)u(k)
Thus, we have shown that Equation (5-55) is the solution of Equation (5-52).
Once we get the soiution x(k), the output equation, Equation (5-53), becom
as follows:
(k,J + I)H(j)n(j) + D(k)u(k), k >
If @(k)is nonsingular ecf,so that the inverr;eof q ( k ,
exists, then the inverse of (h,k),is given as follows:
= [G(k - I)G(k - 2) e
G(h)]-l
= ~ - l ( h ) ~ - l ( hi 1) e - ~ - y k- 1)
. A sumrnary on the state transition matrix
G(k - 1 ) q k - 2) . G(h), k > h
f G(k) 1s nonsingular for al1 k values considered, then
i ) foranyi,j,k
f G(k)is singular for any value of k , then
( j , ) , f o r k > j > i
A single-input-single-output discrete-time system may be modeled by a pulse tran
fer function. Extension of the pulse-transfer-function concept to a multiple
multiple-output discrete-time system gives us the pulse-transfer-function ma
this section we shall investigate the relationship between state-space represen
and representation by the pulse-transfer-function matrix.
Pulse-Transfgr-Functa'o~ The state-space representation of an
order linear time-invariant discrete-time system with r inputs and m outputs can
given by
x(k + 1) = Gw(k) + Hu(k)
y(k) = Cw(k) + Dn(k)
ec. 5-4 Pulse-Transfer-Function Matrix
where x(k) is an n-vector, u(k) is an r-vector, (k) is an m-vector, G is an n x n
matrix, His an n x r matrix, is an m x n matrix, and D is an m x r
the z transforrns of Equations (5-58) and (5-59), we obtain
Pdoting that the definition of the pulse transfer function calls for the assumption of
zero initial conditions, here &idstate x(O) is zero. Then
we obtain
where
(z) is called the pulse- nsfer-function matrix.
transfer function rnatrix z) characterizes the in
Since the inverse of matrix (z - G) can be written as
ulse-transfer-function matrix (z) can be given by the equation
Clearly, the poles of (z) are the zeros of lz
cbaracteristic equation of the dáscrete-time syst
or
zn + alzn-l + a2zn-2+ . - + an-lz + a, = O
where the coefficients ai depend on the elements of G.
have shown that for the system defined by
x(k + 1)= Gx(k) + Hu(k)
the pulse-transfer-function matrix is
]in Section 5-2 we sh representations for a
&en system are rela defining a new state
vector %(k)by using a si
x(k) = P%(k)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-162-320.jpg)
![Eea
Obtain the continuous-time state-space representation of the system. Then discre
the state equation and output equation and obtain the discrete-time state-space reDr
sentation of the system. Also, obtain the pulse transfer function for the syst&nby u
Equation (5-60).
The continuous-time state-space representation of the system is simply
Now we discretize the state equation and the output equation. Referring to Eqjuatio
(5-73) and (5-74), we have
G(T) = e--aT
r T I ..-aT
H(T)= jo =
a '
Hence, the discretized version of the system equations is
1- e-aT
x(k + 1) = e-OTx(k) +-a u@)
Y(k) = x(k)
Referring to Equation (5-60), the pulse transfer function for this system is
- - /
a a(l - e-aTz-")
This result, of course, agrees with the z transform of C(s) where it is preced
by a sampler and zero-order hold [that is, where the signal u(t) is sampled and fed
a zero-order hold before being applied to G(s)]:
Obtaim the dascrete-time state and output equations and the pulse transfer hnct
(when the sarnpling period T = 1)of the following continuous-time svstern:
which may be represented in state space by the equations
The desired discrete-time state equation will have the form
x((k + 1)T) = G(T)x(kT) + H(T)u(kT)
where matrices G(T) and B(T) are obtained from Equations (5-73) and (5-74)
6- 1 2 .
iscretization of Continuaus-Time State-Space E
Thus, L
r 1
I h e output equation becomes - ...
hen the sampiing period is 1sec, or T = 1, the discrete-time state equation and the
output equation become, respectively,
and - ...
The pulse-transfer-function reprecentation of this system can be obtained from Equa-
tion (5-60), as follows:
Note that the same pulse transfer function can be obtained by taking the z
transform of C(s) when it is preceded by a sampler and zero-order hold. Assuming
T = 1, we obtain
- - F -.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-165-320.jpg)
![equation
into
then, assurning the sampling period to be 0.05 sec, we obtain G and H as E
space equation
x(k + 1) = Gx(k) + Wu(k)
iscretization of Continuous-Time
As another example, consi es the following system:
Assuming that the sampling period T is 0.05 sec and wit out specifyingthe forrnat,
we get the following discrete-time state equation:
x(k + 1) = Gx(k)+
where matrices G and H can be found in the following computer out
A = [O 1 0 0
20.601 O O O
o 0 0 1
-0.4905 O O O];
B = [O;-1;0;0.5];
[G,H] = c2d(AfB,0.05)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-166-320.jpg)
![ec. 5-6 Liapunov Stability Analycis li
Consider the system discussedin Example 5-5. Obtain the discrete-time state equation
and the output equation at t = kT + AT. Also, obtain the specificexpressions for the
state equation and output equation when 2" = 1 sec and
lin Example 5-5, the matrices G(T)and H(T) were obtained as follows:
4, the z transform solution of the discrete-time sy
shall showthat th
Consider the time-invariant continuous-time system defined by
Eet us assume that the input u is sampled and fed to a zero-order hold
u ( ~ )= u(kT)for kT T < kT + T. Referring to Equation (5-651, the solu
the state equation starting with the irnitial state x(to)is
G(AT)= eAAT
Then we obtain
n important role in &hestability analysisof control
equations. There are two methods o£ stability
efirst method and the second method; both apply
mination of the stability of dynamic systems described by ordinary
.The first method consists entirely of procedures
tions of the differential equations or difference
. The second rnethod, on the other han4 does
ential or difference equations. Thisis the reason
second method is so useful in practice.
Although there are many powerful stability criteria available for control sys-
s, such as the Jury stability criterion and the Routh-Hurwitz stability criteria,
puter calcu'tation.) Y are limited to linear time-invariant systerns. The second method of Liaprtnzov,
To obtain the state equation and the output equation at t = kT + da", where
O < AT < T , we first convert G(T)to G(Air) and H(T) to H(AT) and then substitute
G(ibT)and M(dT) into Equations (5-80) and (5-811, as follows:
1 '(1 - e-'''' ) 1
EOr T = 1 and d T = 0.5 we obtain the state equation and output equation as follows:
[ E ]= [ :::(7:][:rl:l] + [ u : ~ P ~ Y ] u ( ~ )
y(k + O S ) = 11 0.31hlj[:~:i] + (00920)u(l)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-167-320.jpg)
![pace Analysis Chap.
where x is a state vector (an n-vecto
functions of xl,x2,. .,x,, and t. (Note that w
a model to present basic materials on stabilit
Ilx - xell 5 r
where llx - x,ll is called the Euclidean norrn and is defined as follows:
I/x - = [(x~- le)' + ( ~ 2- ~ 2 , ) ~+ .e + (xn - x,,)~]~"
Let S(6) consist of al1 points such that
Ilx, - xell 5 6
and let S ( € )consist of al1 points such that
An equilibrium state x, of the system of Equation (5-82) is said to be s
sense of Eiapunov if, corresponding to each S(€),there is an S(6) such t
tories starting in S(6) do not leave S(€) as t increases indefinitely. The re
6 depends on E and, in general, also depends on h. If 6 does not depen
equilibriurn state is said to be uniformly stable.
Wlat we have stated here is that
$ ( E ) , there must be a region S(8) such
S ( E )as t iilcreases indefinitely.
ec. 5-6 Liapunov Stability Analysis
An equilibriumstate xeof the syste
is said to be asyrnptotically stable if it is stable in t
solution starting within S(6) converges, without leaving S(€),to w, as t increases
indefinitely.
In practice, asymptotic stability is more im ortant than mere stability. Also,
since asymptotic stability is a local conce simply to establish as
oes not necessarily mean that the syste
largest region o£ asy
e dornain of attractio
ble trajectories originate
nating in the domain of attraction is asym
totic stability holdsfor allstates (al1
from which trajectories originate, t e equilibriurn state is
said to be asymptotically stable in the large. That e equilibrium state x,of the
system given by Equation (5-82) is said to be as tically stable In the large if
it is stable and if every solution converges to x, as ses indefinitely. Obviously,
a necessary condition Eor asymptotic stability in e is that there be only one
uilibrium state in the whole state space.
In control engineeringproblems, asymptotic stability in t
feature. -hf the equilibrium state is not asymptotically stable in the large, then the
problem becomes one of determináng the largest region of asymptotic stabiiity. This
is usually very difficult. For al1practica1purposes, however, it is sufficient to deter-
mine a region of asyrnptotic stability large enoraghthat no disturbance will exceed it.
n equilibrium state x, is said to be unstable if for sorne real
number E > O and any real number 6 > O, no atter how small, there is always a
state xoin S(6) such that the trajectory starting at this state leaves S(€).
graphical representation of the foregoing definitions will clarify their meanings.
Let us consider the two-dimensional case. Figures 5-2(a), (b), and (c) show
equilibrium states and typical trajectories corresponding to stability, asymptotic
Figure 5-2 (a) Stable equilibrium state and a representative trajectory; (b) asymp-
totically stable equilibrium state and a representative trajectory; (c) unstable equi-
librium ctate and a representative trajectory.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-169-320.jpg)
![ec. 5-6 Liapunov Stability Analysis
V(x) = C
where
Tf there exists a sealar fimction V(x, t )having continuous first partial derivatives
satisfying the conditions
. V(x, t) is positive definite.
Q(x, t) is negative definite.
The conditions of this theorem may be modified as follows:
entically in t 2 tofor any toand any wo =#
lution starting from xoat t = to.
ly stable in the large.
move toward the origin.
To prove stability (but not asymptotic sta-
biliay) of the origin of the systern defined by Equation (5-&2),the following theorem
may be applied.
. Suppose a system is described by
for al1t. If there exists a scalar function VCx,t) having continuoras
atives and satisfying the conditions
V(x, t) is positive definite.
. v(x, t) is negative semidefinite.
then the equilibrium state at the origin is uniformly stable.
It should be noted that the negative semidefiniteness of V(x, t) [ ~ ( x ,t) 5 O
along the trajectories] means that the origin is uniformly stable but not necessarily
~niformlyasymptotically stable. Hence, in this case t e system may exhibit a limit
cycle operation.
. If an equilibriurn state x = of a system is unstable, then there
function W(x, t) tkat determin s the instability of the eqluilibrium
state. We shall present a theorem on instability in t](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-170-320.jpg)
![.%te-%pace Analysis Chap.
) = posltfve definite
e asymptotic stability of the system of E
For a test of positive definiteness of an n x n rnatrlx, we apply Sylvester
criterion, which states th essary and sirfficientcon
rminants of all tke successivepri
r example, the following n x n
are al1real, then the Hermitianmmatrix beco
r- -7
p11 Pl2
'
Fin p 2 n a Pnn
where pi,denotes the complex conjugate of pij.
a13 the successive principal rninors are positive,
Pil Pi2 -
e . .
F12 p22 " '
9 . .. .. .
tead of first specifying a positive definite matri
is positive definite, it is c
then examine whether
e definite. Note that po
shall strrnrnarize what
Consider the system described by
x = AA
where x is a statevector (an n-vector) and A is an n x re co
necessary and sufficient condition for the equilibrium state
totically stable in the Iarge is that, giv any positive definite
positive definite real symmetric) matrix
(or a positive definite real symmetric) rnatrix
The scalar function x*Px is a Liapunov function for this system. [Note that in
r systern considered, if the equilibrium state (theorigin) is asymptoticaliy stabl
it is asyrnptotically stable in the large.]
wks. In applying Theorem 5-4 to the siability analysis of linear tirn
~tinerous-timesysterns, several important rernarks rnay be rnade.
5-6 Liapunov Stability Analysis
ong any trajectory, then
be chosen to be positive semidefinite.
If an arbitrary positive definite matrix is chosen for
idefinitematrix if ~ ( x )doesnot vanishidentical
the matrix equation
as a root of the chasacteristic equarion, and if for every sum of two roots
then the elements of ined. (Note that for a stable matrix
the sum Ai + hk is always nonzero.)
In determining whether or not t ere exists a positive definite Hermitian or
positive definite real sym , it is csnvenient to choose
is the identity matrix. Then tke ele are determined from
is tested for positive definiteness.
Determine the stabiiity of the equilibriurn state of the following system:
The system has only one equilibrium state at the orágin.
into Equation (5-851, we have
Noting that A is a real matrix, must be a real symmetric matrix. This last equation
may then be written as follows:
where we have noted thatpzl = p12and made the appropriate substitution. If the matrix
B turns out to be positive definite, then x*Bw is a Eiapunov function and the origin is
asymptotically stable.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-172-320.jpg)
![Consequectly,
pace Analysis Chap.
= [; $1
By applying Sylvester's criterion for the positive definiteness of matrix
positive definite. Hence, the equilibrium state, the origin x =
. . -
in the large.
Note that instead of choosing to be I we could choose
semidefinite rnatrix, such as
AV(x) = -x:(k)
O -0.5
[l -1 1;: : ] [ -:]- [;:: Y;]= -[::]By solving this last equation, we obtaln
(Direct programming method) Gonsider the discrete-time system defined by
Y ( z ) - bozn+ b,zn-' +-- + bn
U ( Z ) Z" + alzn-1+ . - + a,
Show that a state-space representation of this system may be given by
hap. 5 Example Problems and
ntáasn The given system can be modified to
This last equation can be written as follows:
(61 - al b ~ ) z - 1+ (b2 - a2b o ) ~ - ~+.+
+ (b, - a, bo)z-"
1+ alz-l + a2z-2+ + a,z-" U(z>
Let us define
(bl - al bo)zll + (b2- a2b o ) ~ - 2+ + (b, - a, bo)z-"
Y ( Z ) =
1 + alz-' + a2zW2+ e + u,,z-~ U(z>
Then Equation (5-95) becomes
ket us rewrite Equation (5-96) in the following form:
~ ( 4
(bi - al bo)z-' + (b2 - a2bo)z-' + + (b, - a, bo)z-,
From this last equation the following two equations may be obtained:
Q ( ~ )= -al;-' Q ( z ) - a2z-2Q ( z ) - - - unz-" Q ( z ) + U(z)
and
F(z) = (bl - al b0)z-' Q ( z )+ (b2 - a2bo)z-' Q ( z ) + ..
+ (b, - a, bo)z-" Q ( z )
Now we define the state variables as follows:
Xn-'(2) = z - ~Q ( z )
Xn(z)= 2-l Q ( z )
Then clearly we have
zX*(z)= X2(z)
zXz(z) = &(z)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-175-320.jpg)
![State-Space Analysis C
In terms of difference equations, the preceding n - 1 equations become
xl(k t 1) = x,(k)
xz(k + 1) = x 4 k )
~ , - ~ ( k+ 1) = x,(k)
By substituting Equation (5-100) into Equation (5-98), we obtain
zX,(z) = -alXn(z) - a2Xn--l(z)- . - anX1(z)+ U ( z )
which may be transformed into a difference equation:
x n ( k + 1 ) = - a n x l ( k ) - a n - l x 2 ( k ) - ~ ~ ~ - a l x n ( k ) + ~ ( k )(5
Also, Equation (5-99) can be rewritten as follows:
'
Y ( Z ) = (b1 - al bo)X,(z) + (bi - a,bo)Xn--l(z)+ . . + (b, - c. bo)X1(r)
By use of this last equation, Equation (5-97) can be written in the form
y(k) = (b, - a, bo)xi(k) + (bn-i - a,-, bo)xz(k)
+ ... +(bi-aibo)x,(k)+bou(k) (5
Combining Equations (5-101) and (5-102) results in the state equation giv
Equation (5-93). The output equation, Equation (5-103), can be rewritten in the
given by Equation (5-94).
(Nested programming method) Consider thepulse transfer function system defin
Y ( z ) bo + blz-' + - .+ b,z-"qZ)= -- =
U(z) l + a l z - ' + . - ~ + a n z - "
Show that a state-cpace representation of this system may be given as ~ L ~ o w s :
- 9 -z
x l ( k t 1) O O a - . O O -a, xl(k)
x z ( k + l ) 1 O V . - O O xz(k)
,-
. .- . . . .. .. . . .
~ , - ~ ( k+ 1) O O + - - 1 O -a2 ~ , - ~ ( k )
x n ( k + l ) O O O 1 -al x,(k)- m
- -,
b, - a, bo
bn-i - 6,-1 bo
+
b2 - a, bo
bi - al bo
P
x1( k )
x2W
y(k) = [O O O 11 !
x,-1 ( k )
- xn(k)
Chap. 5 Exarnpie Problerns and Sslutions
aakiora Rewrite the pulse transfer Eianction as follows:
Y ( z ) - boU ( z )+ zw1[alY ( z ) - bl U(z)]
+ z-'[a, Y ( z ) - bZ U(z)]+ - + ~ - ~ [ a ,Y ( z ) - b, U(z)]= O
or
Y ( z ) = boU(z)+ z-'(b1 U(z) - al Y ( z )+ z-'{b2 U ( z )- a, Y ( z )
+ Z-'[b3 ~ ( z )- a 3 ~ ( z )+ . -11) (5-106)
Now define the state variables as follows:
X,(Z) = z-'Lbl U ( z )- al Y ( z ) + Xn-l(z)]
Xn-,(z) = z-l[bZU(Z)- a, Y ( z ) + XnV2(z)]
(5-107)
&(z) = ~ - l [ b , - ~U(z) - Y ( z )+ Xi(z)]
Xl(z) = z-'[b, U(z) - a, Y ( z ) ]
Then Equation (5-106) can be written in the form
Y ( z ) = boU(z) + &(z) (5-108)
By substituting Equation (5-108) into Equation (5-107) and multiplying both sides of
the equations by z , we obtain
~ X n f z )= Xn-~(z)- a l x n ( z )+ (bi - al bo)U(Z)
zXn-i(z) = X,-~(Z)- a2Xn(z)+ (b2 - a2bo)U(z)
zXz(z) = X ~ ( Z )- a,-, Xn(z) + (b,-1 - anTlbo)U(z)
zX1(z)= -a, X,(z) + (b, - a, bo)U(z)
Taking the inverse z transforms of the preceding n equations and writing the resulting
equations in the reverse order, we obtain
xi(k + 1) = -a, x,(k) + (b, - a, bo)u( k )
xz(k + 1) = xl(k) - an-lxn(k)-t- (b,-l - a,-I bo)u(k)
~ , - ~ ( k+ 1) = x,-,(k) - a2xn(k)+ (6, - a2bo)u(k)
x,(k + 1) = ~ , - ~ ( k )- alxn(k)+ (bl - al bo)u(k)
Also, the inverse z transform of Equation (5-108) yields
y ( k ) = x, ( k ) + bo~ ( k )
Rewritingthe state equation and the output equation in the standard vector-matrixform
gives Equations (5-104) and (5-105), irespectively.
(Partial-fraction-expansion programming method) Consider the pulse transfer func-
tion system given by](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-176-320.jpg)
![and
expanded into the following form:
where
Equation (5-112) can be written in the form
Y ( z )= boU(z)+ AU(z) + L?L U(z)+ m - +--LL.
- p i z -p2 z -pn
w1
Let us define the state variables as follows:
1
X,(z) = -
z -p1
u(4
1
X,(z) = -
z -pn
U(z>
Then Equation (5-114) can be rewritten as
zX,(z) = plxl(z) + U ( z )
zX2(z) = p2&(z) + U(z)
State-Spaw Analycis Ch
Y ( z ) bozn+ bi zn-' + - - - + bn- =
U ( Z ) zn + al zn-' + ...+ a,
Show that the state equation and output equation can be givenin the following
canonical form if al1 poles are distinct.
xi(k + 1) 0 . " .
k 2 ( k : 1 ) ] = Fx,(k + 1) O 0 ~ ~ : ~ ' l ] + [ ~ ] u ( k )
Pn xn( k )
x l w
y(k) = [cl c2 c n l / ( k ) ] + bou(k)
xn(k)
Solation The system puise transfer function can be modified as follows:
Y ( z ) buzn+ bl 2"' + e e + bn--
U(2) ~ " + a ~ z " - l + . . - + a ,
, (bl - al bo)zn-l + (b2- a2bo)zn-' + . - + (bn - anbo)
= bo r
(2 - pí)(z - pz) ( z - pn)
Since al1 poles of the puise transfer function Y(z)/U(z)are distinct, Y(z)/U(z)
Chap. 5 Exarnple Problerns and
Also, Equation (5-113) can be written as
Y ( z )=: bo U ( Z )+ C ~ X , ( Z )+ Q X ~ ( Z )+ + cnXn(z) (5-116)
The inverse z transforms of Equations (5-115) and (5-116) become
xl(k + 1) = p,xl(k) + ~ ( k )
x2(k + 1) = p2 ~ 2 ( k )+ ~ ( k )
(5-117)
xn(k + 1) = pn xn( k ) + ~ ( k )
and
y ( k ) = c,xi(k) + c Z X ~ ( ~ )+ + cnxn(k)+ bou(k) (5-113)
Rewriting the state equation and the output equation in the form of vector-matrix
equations, we obtain Equations (5-109) and (5-110).
(Partial-fraction-expansion prograrnming method) Consider the pulse transfer func-
tion system defined by
Y ( z ) - bozn + bl zn-l + + b,--
U ( Z ) zn + alzn-1 + - m + a,
Assume that the system involves a multiple pole of order m at z = p, and that all other- -
poles are distinct.
Show that this system may be represented by the followii
output equation:
g state equation and
Solutisn Since the system pulse transfer function can be written in the form
(bl - al bo)zn-l + (b2- a2b o ) ~ n - 2+ . -
= bo + + (bn - a, bo)
( 2 - P I ) ~ ( Z- pm+l) * (2 - pn)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-177-320.jpg)
![we obtain
Let us define the first m state variables Xl(z),&(z), ... ,&(z) by the equations
and the remaining n - m state variables ;%i,+l(z),Xm+2(z),
1
Xm+1( z) = --M--
Z - pm+i W )
1
Xm+2( Z ) = ------ U(Z)
2 - Pm+2
Notice that the m state variables defined by Equation (5-123) are related each to th
next by the following equations:
Xm-i(z>- 1
Xm(z) - p i
By tahing the inverse z transforms of al1 of Equation (5-125), the last equation
Equation (5-123), and al1 of Equation (5-124), we obtain
ap. 5 Exarnple Problems and Sslutions
xrn+i(k + 1) = pm+ixm+i(k)+ ~ ( k )
The output equation given by Equation (5-122) can be rewritten as follows:
Y ( z ) = C I X ( Z )+ ~ ~ X Z ( Z )+ .. + cmXm(z) + cm+lXrn+l(z)
+ cm+zXm+,(z)+ + cnXn(Z) + bo U ( Z )
By taking the inverse z transform of this last equation, we get
y(k) = clxi(k) + czxz(k) + . + crnxrn(k) + crn+ixrn+,(k)
+ cm+2xm+2(k)+ ' ' + ~n xn(k)+ bo~ ( k ) (5-127)
Rewriting Equations (5-126) and (5-127) in the standard vector matrix form, we obtain
Equations (5-119) and (5-120), respectively.
Using the nested programming method (refer to Problem A-5-2), obtain the state
equation and output equation for the system defined by
Then draw a block diagram for the system showing al1 ctate variables.
sHu&ion The given pulse transfer function can be written as
Y ( z )= z-'{U(z) - 4Y(z) + z-'[SU(z) - 3Y(z)]}
Define
The state equation can therefore be given by
and the output equation becomes
- -.
Figure 5-3 showsthe block diagram forthe systern defined by the state-space equations.
The output of each delay elernent constitutes a state variable.
Obtain a state-space representation of the system shown in Figure 5-4. The sampling
period Tis 1 sec.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-178-320.jpg)
![Figure 5-3 Block diagram for the system considered in Problem A-5-5.
Y
olution Vde shall first obtain the z transform of the feedfonvard transfer h n
0.3679(z + 0.7181)--
(Z - 1)(z - 0.3679)
diagram modification.
Let us expand G(z) into partial fractions:
1
G(z) = --
0.6321 - z -'--- 0.6321~-'
z - 1 z - 0.3679 1 - Z-' 1 - 0.3679~-'
Figure 5-5 shows the block diagram for the system. Let us choose the output ~f e
unit delay element as a state variable, as shown in Figure 5-5. Then we obtain
ap. 5 Example Problems and Solutions
Figure 5-5 Modified block diagram for the system shown in Figure 5-4.
from which we get
x (k + 1) O 0 6 3 2 x (k)
[xi(k + l)] [-1 1' ][xi(k)] +
Obtain a state-space representation of the following pulse-transfer-function system:
Use the partial-fraction-expansion programming method. Also, obtain the initial values
of the state variables in terms of y(O), y(l), and y(2). Then draw a block diagram for
the system.
rihtion Because we need the initialvalues of the state variables in terms of y(O), y (l),
and y (2), we slightly modify the partial-fraction-expansion programming method pre-
sented in Section 5-2. Let us expand Y(z)/U(z), zY(z)/U(z), and z2Y(z)/U(z) into
partial fractions as follows:](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-179-320.jpg)
![State-Space Analysis . 5 Example Broblerns an
Rewriting, we have
Let uc define the state variables X1(z),&(z), and %(z) as follows:
Then we have
Notice that Equation (5-130) can be written as
zX&) = -X,(z) + U(z)
zX2(z)= -2X4z) + lJ(z)
zx3(z)= -3X3(Z) + u ( Z )
from which we obtain
xi(k + 1) = -xi(k) + u(k)
xz(k + 1) = -2x2(k) + ~ ( k )
x3(k + 1) = -3x3(k) -iu(k)
The output Y ( z )Is given by
In vector-matrix notation, the state space equations become
O O x1(k)
[ =~ 3 ( k+ 1) -2O -30 ] [ x 2 ( k ) ] + [ ~ ] u ( k )x3(k)
The initial data are obtained from Equation (5-131) as follows:
Y @ ) -
Y (1)- ~ ( 1 )- 2u(O)
y(2) - u(2) - 2u(l) + 6u(O)
Figure 5-7 shows the block diagram for the present system.
Figure 5-9 Block diagram for the system considered in Problem A-5-8](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-181-320.jpg)
![State-Space Analysis
Determine d(A) as the greatest common divisor of al1 the elemen
). Choose the coefficient of the highest-degree term in A of d
be 1. If there is no common divisor, d(A) = 1.
3. The minimal polynomial +(A) is then given as A
has n distinct eigenvalues, then the minimal polynomia
racteristic polynomial. Also, if the multiple eigenvalues
linked in a Jordan chain, the minimal polynomial and the characteristic polyno
identical. If, however, the multiple eigenvalues of A are not linked in a Jorda
the minimal polynomial is of lower than the characteristic p
'CJsingthefollowingmatrices A S examples, verify theforegoing state
about the minimal polynomial when multipie eigenvalues are involved.
iatioan First, consider the matrix A. The characteristic polynomial is give
A - 2 -1
- A - O A - 2 y ¡=(A-?)'(A-
O -3 A - 1
Thus, the eigenvalues of A are 2, 2, and 1. Ht can be shown that the Jordan c
form of A is
[H d 81and the multiple eigenvalues arelinked in the Jordan chain as shown. (For the
for deriving the Jordan canonical form of A, refer to Appendix A.)
To determine the minirnal polynomial, let us first obtain adj (
given by
(A-2)fA-1) (A+11) 4(A-2)
[ :
(A - 2)(A - 1)
3(A - 2) (A - 2)"
Notice that there is no common divisor of al1 the elements of adj (A
d(h) =: 1.Thuc, the minimal polynomial +(A) is identical with the chara
nornial, s r
- Al (A - 2)2(h - 1)
= h3 - 5h2+ 8A - 4
A simple calculation proves that
A3 - 5a2+ 8A - 41=
but
Fi2 - 3A + 21 f
Thus, we have shown that the minimal polynornial and the characte
of this matrix A are the same.
hap. 5 Exarnple robfems and Solurions
Next, consider the matrix . The characteristic polynomial is given by
simple computation reveals that matrix has three eigenvectors, and the Jordan
[H % i]Thus, the multiple eige ues are not linked. To obtain the minimaí polynomial, we
first compute adj (A1 -
(A - 2)(X - 1) o
(A - 2)(A - 1)
3(A - 2) (A - 212
from which it is evident that
ence,
As a check, let us compute +(
For the given matrix ,the degree of the minimal polynomiai is lower by 1than that
of the characteristic polynomial. As shown here, if the multiple eigenvalues of an n x n
matrix are not linked in a Jordan chain, the minimal polynomial is of lower degree than
the characteristic polynomial.
Show that by use of the minimal polynomial the inverse of a nonsingular matrix A can
be expressed as a polynomial in with scalar coefficients as follows:
where al, a2,... ,amare coefficients of the minimal polynomial
Then, obtain the inverse of the following matrix
A = [:3 -1:-2:]
Solution Fsr a nonsingular matrix A, its minirnal polynomial $(A) can be written as
follows:](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-183-320.jpg)
![Clearly, there is no common divisor d(h) of al1 elements of adj (A1 -
d ( h ) = 1. Consequently, the minimal polynomial +(A) is given by the eq
Thus, the minimal polynomial +(A) is the same as the characteristic polyno
Since the characteristic equation is
- Al = h3 + 3h2- 7A - 17 = O
we obtain
$(A) = h3 + 3h2 - 7A - 17
y identifying the coefficients ai of the minimal polynomial (which is the sam
characteristic polynomial in this case), we have
al = 3, a, = -7, a3 = -17
The inverse of W can then be obtained from Equation (5-134) as follows:
1
+ alA + a,I) = -(A2 + 3
17
7 O -4
-:I+ 3[n -.
6 -4
=A[! -;;]
- 6. -4.
[$ -8 ;]- - -L.17 17 17
Show that the inverse of z - G can be given by the equation
adj (z1 - G)
(z1 - G)-' =
IzP - G/
-- 1zn-' + H1 + I ~ ? J ~ z ~ - ~+ - + Hn-1
lzli - G/
) = A m + alAm-' + . e - +a,-lA+am
where a, iO. Hence,
1
= --(Am+ alikm-' + + a,-,A2 + am-IA)
am
Premultiplying by A-', we obtain
which is Equation (5-134).
Eor the given matrix A, adj (h - A) can be given as follows:
hap. 5 Example Problems and Sdlatisns
where
and al, a2,... ,a, are the coefficients appearing in the characteristic polynomial given
by
- 61 = zn + al zn-' + a2znW2+ + a,
Show also that
al = -trG
a2 = -2 trGHl
7% simplify the derivation, assume that n = 3. (The derivation can be easily extended
to the case of any positive integer n.)
- z 2 G + Z ~ W ~- z G H ~+ zH2- GHz
The Cayley-Hamilton theorem (see Problem A-5-9) states that an n x n matrix G
satisfies its own characteristic equation. Since n = 3 in the present case, the character-
istic equation is
- 61 = z3 + a1z2+ a2z + a3 = O
and G satisfies the following equation:
G3 + al gy2 + a2G 4- as
Hence Equation (5-136) simplifies to
z2 + H1z + H2) = (z3 + alz2+ a2z C a3)I = jz
Consequently,
which is Equation (5-135) when n = 3.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-184-320.jpg)
![Noting that
we have
and
State-Space Analysis
-sin oh cos wh
-.
State-Space Analysis
Qbtain the continuous-time state-space representation of the system. Then
the system and obtain the discrete-time state-space representation. Also o
pulse transfer function of the discretized system.
ution From the given transfer hnction, we have
y + wZy= w2u
Define
X l = y
1
X 2 = - j
w
Then we obtain the followingcontinuous-time state-space representation of the syst
The discrete-time state-space representation of the systern is obtained as fo
Noting that
we have
-1
cos wT sin wT
- - -sin wT cos wT
s2 + w2 s2 + w2
and
1
-sin oh cos wh
1- cos wT
Hence, the discrete-time state-space representation of the oscillator systern be
as follows:
cos COT sin W T ] [ ~ ~ ( ~ T ) ]+ [1 - C O T ] ~ (
-sin oT cos wT x2(k7') sin wT
xl(kT)
Y = 1 o~[x2(ki)]
The pulse transfer function of the discretized system can be obta
Equation (5-60):
F(z) = C(z1 - G)-'H + D
Chap. 5 Example Problems and
Noting that D is zero, we have
s i n w~ 1 - l [1
z - cos wT
-- (1 - cos wT)(z + 1)
z2 - 22 cos wT + 1
Hence,
m= F(r) = (1 - cos O J T ) ( ~+ 2-')z-'
U(z1 1 - 22-1 cos WT+ z-'
Note that the pulse transfer function obtained in this way is the same as that obtained
by taking the z transform of the system that is preceded by a zero-order hold. That is,
1 1- zP1cos wT
-- (1 - cos w"(l + z-')z-l
1 - 22-' cosoT + z-'
Thus, we get the same expression for the pulse transfer function. The reason for this
is that discretization in the state space yields the zero-arder hold equivalent of the
continuous-time system.
Consider the system shown in Figure 5-8(a). This system involves complex poles. Tt is
stable but not asymptotically stable in the sense of Liapunov. Figure 5-8(b) shows a
Figure5-8 (a) Continuous-timesystem of Problem A-5-15; (b) discretized version
of the system.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-186-320.jpg)
![discretized version of the continuous-time system. The discretized systern is also sta
but not asymptotically stable.
Assuming a unit-step input, show that the discretized system may exhibit hid
oscillations when the sampling period T assumes a certain value.
aa The unit-step response of the continuous-time systemshownin Figure5-8
1s
Hence,
y(t) = cosrt
[Notice that the average value of the output y(t) is zero, not unity.] The respsnse y
versus r is shown in Figure 5-9(a).
I
(c)
Figure 5-9 (a) Unit-step response y(t) of the continuous-time system shown in
Figure 5-8(a); (b) plot of y(kT) versus kT of the discretized svstem shown in Figure-
M ( b ) when T = 4rr sec; (c) piot of y(1cT) versus kTof the discretized system when
T = T sec (Kidden oscillations are shown in the diagram.)
Chap. 5 Exarnple Problerns and Soluaions
The pulse transfer function of the discretized system shown in Figure 5-8(b) is
7 - - -
y& =#- e-" " = (1 - z-')+&]
Hence, the unit-step response is obtained as follows:
The responsey(kT) becomes oscillatoryif T f nn-sec (n = 1,2,3,...).For example,
the response of the discretized systern when T = $71. sec becomes as folIows:
Hence,
A plot of y(kT) versus kT when T = n- sec is shown in Figure 5-9(b). Clearly, the
response is oscillatory. If, however, the sampling period T is rr sec, or T = n-, then
The response y(kT) for k = 0,1,2, . ..is constant at unity. A plot of y(kT) versus kT
when T = n- is shown in Figure 5-9(c).
Notice that if T = n- sec (in fact, if T = nrr sec, where n = 1,2,3,...) the
unit-step response sequence stays at unity. Such a response may give us an irnpression
that y(t) is constant. The actual response is not unity but oscillates between 1and -1.
Thus, the output of the discretized systemwhen T = 7 ~ .sec (or when T = nrr sec, where
n = 1,2,3,.. .) exhibits hidden oscillations.
Note that such hidden oscillations (hidden instability) occur only at certain
particular values of the sampling period T . If the value of T is varied, such hidden
oscillations (hidden instability) show up in the output as explicit oscillations.
Even though the double-integrator system is dynamically simple, it represents an
important class of systems. An example of double-integrator systems is a sateilite
attitude control system, which can be described by](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-187-320.jpg)
![lhi,(i)l}max {cjlxj]< max {cj]ql]
= 1/41
which verifies that H(x)x is a contraction.
In case 2, define the norm by
n
IIxll = L-d cilxil
Then
State-Space Analysis
{ i j } ic j / i j< jj~ ~ 1 ~ ~ 1i j=l Cj j= 1 j- 1
= IIxII
which shows that H(x)x is a contraction.
Wow consider a scalar function V ( x )= I/xII.Clearly, V(x)=
nite and
BV(x(k))= V(x(k + 1)) - V(x(k))
= IlH(x(k))x(k)ll- Il4k)ll < 0
and
Thus, dV(x) is negative definite. Hence, V ( x )= ljxlj is a Liapunov function
syctem considered, and by Theorem 5-5 the origin of the cystemis asymptoticall
in the large.
Prove that if al1 colutions of
x(k + 1) = Gx(k)
where x is an n vector and G is an n x n constant matrix, tend to zero as k app
infinlty, then al1 solutionc of the system
x(k + 1) = Gx(k) + Hu(k)
where 1E% is an n x r constant matrix, are bounded, provided that the input vect
r vector, is bounded.
siÉPori Since u(k)is bounded, there exists a positive constant c such that
llu(k)ll<c, k = 0 , 1 , 2,...
The solution of Equation (5-144) is given by
k
x(k) = e k x ( 0 )+ 2G k ' ~ u ( j- 1)
j= 1
hap. 5 Example Problems and
Hence,
k
Since the origin of the hornogeneous system givenby Equation (5-143) is asymptotically
stable, there exist positive constants a and b (O < b < 1) such that
O < 1 1 ~ 1 1 ~< abk
Then
Therefore,
e have thus proved that Ilx(k)ll ie bounded.
Consider the system defined by the equations
Determine the stability of the equilibrium state.
iatio~a Define the equilibrium state as
Then such an equilibrium state can be determined from the following two simultaneous
equations:
The equilibrium state is thus (0,10).
Now let us consider a new coordinate system with the origin at the equilibrium
state. Define
Then the system equations become](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-191-320.jpg)
![State-Space Anaiysis Chap.
To determine the stability of the origin of the system in the new coordinate system, le
us apply the Liapunov stability equation given by Equation (5-90):
2 0.5
.8,,p,, ;::]lo 0 . 8 - 1; ;::J= -1; a351
be a positive definite matrix having e
Solving this fast equation for matrix
hap. 5 Problerns
Obtain a state-space representation of the system described by the equation
y(k + 2) +y(k + 1) + O.l$y(k) = u(k + 1) + 2u(k)
Obtain the state equation and output equation for the system shown in Figure 5-11.
y applying the Sylvester criterion for positive definiteness, we find that matrix P
positive definite. Therefore, the origin (equilibrium state) is not stable.
The instability of the equilibrium state can, of course, be determined by
transform approach. Let us first eliminate X2 from the state equation. Then we
Xr(k + 2) - 2.8X1(k + 1) + 1.6Xi(k) = O
The characteristic equation for the system in the z plane is
z2 - 2.82 + 1.6 = O
or
( z - 2)(z - 0.8) = O
Hence,
z = 2, z = 0.8
Since pole z = 2 is located outside the unit circle in the z plane, the origin (equilib
state) is unstable.
O
Obtain a state-space representation of the following puise-transfer-function syste
the controllable canonical form.
Obtain a state-space representation of ehe following pulse-transfer-function syste
the observable canonical form.
Y ( z ) z - ~+ 4 ~ - ~
Obtain a state-space representation of the following pulse-transfer-functm syste
the diagonal canonical form.
Y ( z ) 1 + 62-' + S Z - ~
Figure 5-11 Block diagram of a control system.
Obtain the state equation and output equation for the system shown in Figure 5-12.
Obtain the state-space representation of the system shown in Figure 5-13.
Figure 5-14 shows a block diagram of a discrete-time multiple-input-multiple-output
system. Obtain state-space equations for the system by considering xl(k),r2(k),and
x3(k)as shown in the diagram to be state variables. Then define new state variables such
that the state matrix becomec a diagonal matrix.
Obtain the state equation and output equation for the system shown in Figure 5-15.
Obtain a state-space representation of the discrete-time control system shown in
Figure 5-1 6.](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-192-320.jpg)
![tate-Space Analysis Chap.
state-space representation of the scalar difference equation system
y (k + n ) + al(k)y(k + n - 1) + . + a,(k)y ( k )
= bo(k)u(k+ n ) + bl(k)u(k + n - 1) + + b,(k)u(k)
wbere k = 0,1,2,. .. ,may be given by
y(k) = xl(k) + bo(k - n)u(k)
Determine hl(k),hz(k),...,h,(k) in terms of ai(k) and b,(k), where i = 1,2,
and j = O,1,...,n , Determine also the initial values of the state variables
xz(O),. ..,x,(O) in terms of the input sequence u(O),u(l),...,u(n - 1)and the
sequence y (O), y (1), ... ,y (n - 1).
" x(zI - 6)-' = 2k-1 Z - Zk
where m is the degree of the minimal polynomial of G and the Xk9$are n X n m
determined from
gj(c) = ~ , ( Z I ) ~ I+ gj(zz)x2+ e + gj(zm)Xm
where
Obtain the pulse transfer function of the system defined by the equations
Rnd the pulse transfer function of the system defined by
where
= [ 1 O O], D = bo
Obtain a state-space representation for the systemdefined by the followingpulse-trans-
fer-function matrix:
Consider the discrete-time state equation
Obtain the state transition matrix
Consider the system defined by
where matrix G is a stable matrix.
Obtain the steady-state values of x(k) and y(k) when u(k)is a constant vector.
Consider the system defined by](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-194-320.jpg)
![Pole Placernent and Observer Design Chap
state controllability.
j=o
we obtain
the rank of the controllability matrix is n, then for an arbitrary state x(nT) =
there exists a sequence of unbounded control signals u(O), u(T), ...,u((n - 1
controllability, let us assume that
rank[H: GHi .GnW1
quently, we have for any i
rank[HiGHi - - .iG'-lH] < n
complete state controllability.
ec. 6-2 Controllability
m defined by Equation (6-1) is completely state controllable, then
ansfer any initial state to any arbitrary state in at most n sampling
however, that this is true if and only if the magnitude of u
magnitude of u(kT) isboun it may take more than n sa
a
system is defined by
where x(kT) is an n-vector, u(kT) is an r-vector, G is an n x n matrix, and H is an
n x r matrix, then it can be proved that the con lete state controllabil-
ity is that the n x nr matrix
be of rank n, or that
is of rank n and u(kT) is a scalar, then it is possible to
scalar equations from which a sequence
(k = 0,1,2, ...,n - 1) can be uniquely det at any initial state x(0)
is transferred to the desired state in n sam
Note also that, if the control signal is not a scalar
sequence of u(kT) is not unique. Then there exists
r u(kT) to bring the initial state w(0) to a state in not more than
. Consider
the s
where
x(kT) = state vector (n-vector) at kth sampling instant
u(kT) = control vector (r-vector) at kth sampling instant
6 = n X n matrix
H = n x r matrix
T = sampling period
If the eigenvectors of Gare distinct,then it is possible to find a transformation matrix
such that](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-197-320.jpg)
![Note that the control signal ~k(kT)does not affect the complete observability of the
system. To examine complete observability, we may sirnply set u(kT) = O. For t
system, we have
14 -6 61
Notice that
Hence, the rank of the matrix [C* i 6;"
system is not completely observable.
In fact,inthis systemapole-zero cancellation occursin the pulsetransfer
of the system. The pulse transfer function between Xl(r) and U(z)is
and the ~ u l s etransfer function between Y ( z )and Xl(z)is
Clearly, the (z $. 1)factors in the numerator and denominator cancel each other. l n
means that there are nonzero initial states x(0) that cannot be determi&d from t
measurement of y (kT).
ents. The pulse transfer function has no cancellation if and only if th
mpletely state controllable and completely observable. (See Probl
A-6-4.) This means that a canceled transfer function does not carry along al1
information characterizing the dynamic system.-
n what follows, we S al1 examine the relations
nd observability. Consider the system SI defined by
equations
x((k + 1)T) = @x(kT) + Ha(kT)
y(kT) = Cx(k'l")
where
x(kT) = state vector (n-vector) at kt
er(kT) = control vector (r-vector) at kth sampling inskant
(kT) = output vector (m-vector) at kth sampling instant
G = n X n matrix
E% = n X r matrix
= m X n matrix
ual counterpart, which we cal1 syste
k((k + 1)T) = G*X(kT) + (6-22)
(kT) = H*%(kT) (6-23)
where
X(kT) = state vector (n-vector) at kt
G(kT) = control vector (m-vector) at kth sampling instant
"(kT) = output vector (r-vector) at kth sampling instant
G* = conjugate transpose of G
H* = conjugate transpose of H
* = conjugate transpose of
shall now examine an analogy between controllability and observability. This
analogy is referred to as the principle of duality, due to Kalman.
The principle of duality states that system SIdefined by Equations (6-20) and
(6-21) is completely state controllable (observable) if and only if system S, defined
) and (6-23) is completely observable (state controllable). To
let us write down the necessary and sufficient conditions for
complete state controllability and complete observability for systems SI and S2,
respectively.
FOR SYSTEM S,:
necessary and sufficient condition for complete state controlIability is that
rank[HiGHi....G"-lH] = n
A necessary and suffi nt condition for complete observability is that
FOR SYSTEM S,:](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-203-320.jpg)
![(6-25)
the continuous-time to the discrete-time case.
Consider the following continuous-time control system:
(6-27)
This system is completely state controllable and completely observable, since the rank
of the controllability matrix
defined by the equations:
X = Ax
is 2 and the rank of the observability matrix
rn = state vector (n-vector)
is also 2. Notice that the eigenvalues of the state matrix are
y = output vector (m-vector)
AI = j, h2 = -j
A = n X n matrix The discrete-time control system obtained by discretizing the continuous-time control
= m X n matrix system defined by Equations (6-26) and (6-27) may be given as follows:
for complete observability is that the rank of the n x nm matrix
(6-29)
be n. (This n X nrn matrix is commonly called the observability matrix fo where T is the sampling period.
continuous-time system.) Let us show that the discretized system given by Equations (6-28) and (6-29) is
completely state controllable and completely observable if and only if
2nrr
Im (Al - A2) = 1 + 1=k -
T
T f nn, n = 1,2,3,. ..
For the discrete-time control systemobtained by discretizingthe continuous-time
pletely observable in the absence of sampling remains completely state co
control system, we have the following controllability matrix:
and completely observable after the introduction of sampling if and only if, 1- cos T cos T + 1 - 2 cos2T
eigenvalue of the characteristic equation for the continuous-time control §y§ sinT -sinT+2cosTsinT
relationship
the rank of the observability matrix
Notice ehat the rank of [H i GH]is 2 if and only if T # n?l-(where n = 1,2,3,. . .). Also,
Re hi= Re A,](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-204-320.jpg)
![Pole Placement and
In many cases we choose al= a2 = - .= a, = 1. [Note that the necessary a
sufficient condition for the system to be completely state controllable is that ii#
(E = 1,2,. .. ,n) and th t the condition for it to be completely observa
piJ;O(i = 1,2,..., n).
esp, of matrix G, then we choose the trans
'S are eigenvectors correspond to eigenvalues) or ge
nvectors (which co d to multiple lues). (For details
generalized eigenvectors, see
= rnatrix in Jordan canonical form
Now if we define
then Equations (6-30) and (6-31) can be given as follows:
es pm+l,pm+2,...,p, t
are al1distinct and different frornpl and, in addition, if th
(which implies that the rninirnal polynornial is identical to the characteristic po
nomial), then the state-space equations in the Jordan canonical form are given
follows:
seful Transformations i n nalysis and Design
n many cases we choose a, = = - - = a, = 1. [Note that the necessary and
sufficient condition for the
(i = m , m + 1,. ..,n) and
m + l , m + 2,...,n).]
Note that if the rank
minimal polynomial is s -
will have a different Jordan c
(an arbitrary n x n nonsingular matrix) be a si
and write
Then
-l H]
Similarly, for the observability matrix define
*1
be an arbitrary n x n onsingular matrix and write](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-207-320.jpg)
![lacement and Observer
sthat there are q linearly independentcolumn vectors in the controlla
t us define such q linearly independent colu
Also, let us choose n - q a
.
Vq+l i Vq+2 : '
:vn]
the transformation matrix, let us
- 1 ~=
Then we have
or
here q linearly independent colum ca
milton theorem to express matrices of
uppose the system of Equation (6-50) is not completely state control1
e rank of the controllability matrix is less than n, or
q vectors. That is,
ence, Equation (6-53) may be written as follows:
i Gv,+~.a i Gv,]
1; ; ..e o
To simplify the notation, let us define
esígn via P d e Placement
[Q %]= G 2= (n - q ) X q -m matrix
o o
1
Then Equation (6-53) can be written as follows:
e .
f GV,+~:- :Gv,] = [
Thus,
Next, referring to Equation (6-54), we have
(6-56)
e written in terms of q
uation (6-56) may be written as follows:
hus,
where](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-209-320.jpg)
![Pole Placement and bserver Design Chap.
The characteristic equation with the eigenvalues is given by
( Z - P ~ ) ( z ~ 2 ) ' ' ( z - ~ n )
Equating the coefficients of equal powers of z of Equations (6-63) and (6-64)
obtain
al = al + 6,
a2 = a2+ 62
a, = a, + S,
ence, frorn Equation (6-62) we
= [a, S,-1 m
SI]
. .
= [a, - a, i anl1- a,-l : :al -
where the ai's and the ai 's are known coefficients and
ave determined the required feedback gain matri
cients and a known matrix of the system. This proves the sufficient c
is, if the system defined by Equation (6-50) is cornpletely state con
it is always possible to de mine the required state feedback gain
arbitrary pole placement . nce, we have proved that a necessary
condition for arbitrary pole placernent isthat the systembe completel
labre.
. The expression given by Equation (6-65) Bs not
only one used for the deterrnination of the state feedback gain rnatrix K. T
other expressions available. In the following, we shall present m e such ex
commonly called Ackermann's formula.
Gonsider the system defined by Equation (6-50).
is cornpletely state controllable. y using the state f
wish to place closed-loop poles at z = pl,z = ,u2,...,z = p,. That i
characteristic equation to be
1 = ( Z - pl)(z - / L 2 ) " ' ( Z - p,)
= zn + alzn-l + a 2 ~ n - 24 .- + anmlz + a, = O
et us define
G = G - H K
Since the Cayley-Harnilton theorem states that 6 satisfies its own charad
equation, we have
6"+ al6"-1+ a2Gn-2 + ..
Ytie shall rmtilize this lase equaticdn to derive Acltermanm's formula.
ec. 6-5 Design via Pole Placement
Consider now the following identities:
-
G = G - H K
G2= (G - H K ) ~= G~- G
ding equations in order by a,, a,-l,. ..,a0 (where a. = 1),
ng the results, we sbtain
which can be written as follows:
Notice that
#G)=
ence, Equation (6-66) may be modified to read
ince the system is completely state controllable, t e controllability matrix
is of rank n and its inverse exists. hen Equation (6-67) can be modified to the form
Premultiplying both sides of this last equation by [O O O 11,we obtain](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-211-320.jpg)
![Pole Placement and Observer
where
Note that
-' 1 = z2 + z + 0.16
0.16 z + l
Hence,
al = 1, az = 0.16
Determine a suitable state feedback gain matrix R such that the system will
closed-loop poles at
z = 0.5 +j0.5, z = 0.5 'j0.5
Let us first examine the rank of the controllabílitv matrix. The rank of
is 2. Thus, the system is completely state controllable, and therefore arbitrary
placement is possible. The characteristic equation for the desired system is
Hence,
al = -1, az = 0.5
We shall demonstrate four different ways to determine matrix K.
Method l. From Equation (6-69), the state feedback gainmatrix K is given asfollo
R = [a2- a2I al - a,]
Notice that the original systemis alread n a controllable canonicalform,and theref
the transformation matrix T becomes
Hence,
ec. 6-5 Design via Pole Placement
Thus,
ethod 3. From Equation (6-72), the desired state feedback gain matrix
mined as follows:
= [l 1][& t 2 1 - l
where
= (G - piP)-lEI7 i = 1,2
= [G - (0.5 +j0.5)
r - 1 1
Similarly, for p2 = 0.5 - j05, we have
ijz = [G - (0.5 - j0.5)
Consequently, we have
Hence, the desired state feedback gain matrix M is determined to be](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-213-320.jpg)
![Pole Placement and esign via Pole Placement 217
If we use the state feedback u(k) = -> , . ,
becomes
Referring to Equation (6-761, we have
= 6 - &[--a, -a,-, - -a,]
Consider the system given by
Determine the state feedback gain matrix K such that when the control signal is
given by
the closed-loop system (regulator system) exhibits the deadbeat response to an initial
state x(0). Assurne that the control signal u(k) is unbounded.
Weferring to Equation (6-76), for the deadbeat response we have
The system given by Equation (6-77) is already in the controllable canonical form.
Therefore, in Equation (6-78), T = .The characteristic equation for the system given
by Equation (6-77) is
Thus,
Consequently, Equation (6-78) becomes
This gives the desired state feedback gain matrix.
Let us verify that the response of this system to an arbitrary initial state x(0) is
indeed the deadbeat response. Since the closed-loop state equation becornes
to discrete-time control systems. here is no such thing as deadbeat r
continuous-time control systems. deadbeat control, any nonzero error
be driven to zero in at most n sam ng periods if the magnitude of the sca if the initial state is given by- - -
u ( k ) is unbounded. The settling time depends on theWsamplingperiod, sin
response settles down in at most n sampling periods. If the sampling perh
chosen very small, the settling time will also be very small, which implies th
control signal must have an extremely large magnitude. Othenvise, it will
possible to bring the error response to zero in a short time period.
In deadbeat control, the sampling period is the only design parameter. T
if the deadbeat response is desired, the designer rnust choose the sampling pe
carefully so that m extremely iarge control magnitude is not required k1-m
operatmn of tke system. Note that Et Bs not physicallgr possible to iescrezse
where a and b are arbitrary constants, then we have
[ = [ ul[::::;] = [u u][:] = [a]](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-215-320.jpg)
![-6 Regulator system with a state observer.
c. 6-6 State Observers
rnator, is a subsystem
the observed s
e (estimate) state variables, we ~nustbe able to obiain
k - n + 1)and ~ ( k ) ,~ ( k- l), ...,~ ( k-
G-l x(k + 1) = ~ ( k )+ 6-lHu(k)
w(k) = G-' x(k + 1) - 6-'Hu(k)
y shifting k by 1, we get
variables plus some (but not all) of the measlrrable state variables is refe
reduced-order state observation.
x(k - 1) = G-' x(k) - G-' Hu(k - 1)
uting Equation (6-81) into Equation (6-82), we obtain
x(k - 1) = G-'[G-'x(k + 1) - G-'Nu(k)] - G-'Hu(k - 1)
we obtain the state and output equations as follows: -C 2 x ( k + 1) - G-2Ha(k) - 6-'Hu(k - 1)
x(k + 1)= Gx(k) + Hn(k)
~ ( k- 2) = G - 2 ~ ( k )- C2Hu(k - 1) -
= 6-3x(k + 1) - g ~ - ~Hn(k) - GM2Hn(k - 1) - 6-'H
x(k) = state vector (n-vector)
u(k) = control vector (r-vector)
- 6-" Mu(k) - 6-""
y(k) = output vector (m-vector)
- . . . - 6-"Wu(k - n + 1)
G = n X n nonsingular matrix
H = n X r matrix g Equation (6-81) into Equation (6-&O), we obtain
e = m X n rnatrix ~ ( k )= ~ x ( k )= CG-'X(~+ 1) - CG-'Wu(k)](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-218-320.jpg)
![e(k) = x(k) - X(k)
then by subtracting Equation (6-87) from Equation (6-79),
x(k + 1) - %(k+ 1)= G[x(k) - X(k)]
or
e(k -t 1) = Ge(k) '
If matrix G is a stable matri
where matrix K, serves as a weighting matrix. (This means that the dyn
presence of discrepancies between the G and H matri
of the actualsystem, the addition of the difference be
the estimated output will help reduce the differences
consists of the differencebetween the measured output and the estim
is called a full-order state observer.
Figure 6-7 State feedback control system.
ared with the actual state ~ ( k ) .Since the
owever, it is possibIe to compare f(k) =
Consider the state feedback control s
equations are
x(k + 1) = Gx(k) + Hn(k) (6-89)
= cx(k) (6-90)
where
x(k) = state vector (n-vector)
u(k) = control vector (r-vector)
(k) = output vector (m-vector)
G = n x n nonsingular matrix
H = n X r matrix
= m X n matrix
K = state feedback gain matrix (n x r
ssume that the system is completely state controllable and completely observ-
but x(k) is not available for direct measurement. Figure
observer incorporated into the system of Figure 6-7. The observe
to form the control vector u(k),or
From Figure 6-23,we have
her re Reis the observer feedback gain rnatrix (an n X m matrix). This last equation
can be rnodified to read](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-220-320.jpg)
![Sec. 6-6 State
z 0.16 + k,
= O
which reduces to
Then the characteristic equation becomes
z2 -t (1+ k2)z + kl + 0.16 = O
Since the desired characteristic equation is
z2 - z + 0.5 = O
by comparing Equation (6-97) with this last equation, we obtain
kl = 0.34, k2 = -2
Note that the dual relationship exists between the systern state equation cons
Example 6-6 and that of the present system. The state feedbackgain matrix K
in Example 6-6 was K = L0.34 -21. The observer feedback gain matrix K, o
here is related to rnatrix K by the relationship K, = K*.
have thus far discussedfull-order pr
tion observers. They are prediction observers because the estimate
sampling period ahead of the measurement y(k),
rning the matrix K, to exist and
+ KeCI = O to have prescribed eigenvaliues. In wh
a more general approach to determine the observe
matrix Ke.
Consider the system defined by
x(k + 1) = Gx(k) + Hafl(k)
y ( k ) = Cx(k)
where
x(k) = state vector (n-vector)
u(k) = control vector (r-vector)
y(k) = output signal (scalar)
G = n Y, n nonsingular matrix
= 1 X n matrix
The system is assumed &obe completely state controllable an
able. Thus, the inverse of
"1
aliso assume that the control law to be used is
where Z(k) is the observed state and K is an r
system configuration is the same as that shown
The state observer dynamics are given by
Z(k + 1)= GZ(k) + Hu(k) + K,ty(k) - j(k)]
(6-100)
First, define
where
and
where a,, a2,...,anwlare coefficients in the characteristic equation of the original
state equation given by Equation (6-(as),
Next, define
(6-104)
where $(k) is an n-vector. y use of Equation (6-104), Equations (6-98) and (6-99)
can be modified to read](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-222-320.jpg)
![from which we get
iacement and Observer Design
8, = a, - a,
Then, frorn Equation (6-113), we
Hence,
system.
Figure 6-9 Alternative representation of the observed-state feedback control sgi
State Observers
Oncewe selectthe desiredeigenvalues (or desired characteristicequation), the
observer can be designed in a way simil
placement problem. The desired chara
onds at least four or five
in some applications, deadbeat response rnay be desired.
we wish to have deadbeat response, the desired characteristic equation
uation (6-114) with Equation (6-118), we require t
he expressiongiven by Equation (6-11
r the determination of the observer fe
matrix K,. In what follows, we shall deriv
letely observable s ned by Equations (6-98) and
systern the output y( scalar. Referring to Equation
(6-112), tbe characteristic equation for th
ere ]E$, is an n x 1 matrix. Define
Then Equation (6-120) becomes
In the observer design we etermine matrix k, so that this last charac-
teristic equation is identical to the desired characteristic equation for the error
vector, which is
zn + alzn-l -t- + anm1z+ a, =O (6-121)
That is,](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-224-320.jpg)
![x(k + 1) = Gx(k) + Hu(k)
wlth state feedback
the desired matrix K was obtained as given by Equation (6-68), re
K = [o o .. o I][H '.GWi a i cn-lH]-l +(G)
ere, iri the observer design proble
with state feedback
u ( k ) = -&: w(k)
the desired matrix &: can therefore be obtained in a fo
(6-122), as follows:
g; = [O O ... o i][@*; :.. :.( & y 1 e*
e conjugate transpose of both sides of this Zast equation, we have
we obtain
Notice that, since & =
Consequently,
use of Equation (6-125), tke desired observer fee
uation (6-124), can be rewritten as follows:
where +(G) is the desired characteristic polynomial of tke error dynamics. The
expression for Ke given by Equation (6-126) is commonly called Ackermann's
formula for the determination of the observer fee back gain matrix Re.
rediction observer is given by Equation (6-92):
The obsewed-state feedback is given by
Ef this last equation is substituted into the observer equation, we obtain](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-225-320.jpg)
![i = 1,2,3,... ,n
zn + alzn-l + + a,-l z + a, = O
(z - pl)(z - ,u2)'" ' ( Z - p,) = zn + alzn-l + * .. + an-lz + a,
Consider the double integrator system given by the equations
given by Equation (6-126):
where
+(@S)= Gn+ alGn-l + + a,-] G + a, for this system. It is desired that the error vector exhibit deadbeat response. Use the
Hf the desired eigenvalues ,u1,,u2,...,pnof matrix G - Re&' are First we check the observability condition. Notice that the rank of
the observer feedback gain matrix Ke may be given by the equ
is 2. Hence, the system is completely observable. Next we examine the characteristic
equation for the system:
Comparing this characteristic equation with
Note that the -7 are the ergenvecto
In the speeial case wkaere we d z2 + a l z + a2 = O
onse, sothat p1 = J U ~= = JL~,= O, Equation (G-
following equation will give the matrix Kefor the
given by
i = 1,2,3,... ,n
r details of the derivations of Equations (6-129) an
the system is low, assume an observer feedback gain
ents. Then the elements of
respectively. The a,'s are the coefficients of the desire e coefficients of like powers of z s f
of the desired characteristic polynomial, w
(z - p1)(z - p2)- " ' ( Z - p,) = zn + alzn-l + * .. + an-lz + a,
where the pi 'S are the desired eigenvalues of G -
Consider the double integrator system given by the equations
x(k + 1) = Gx(k) + Hu(k)
where
and Tis the sampling period. (See Problem 8-5-16 fo
where time state-space equations for the double integrator S
server configuration is the sarne as that shown in Figure 6-8, design a state observer
for this system. It is desired that the error vector exhibit deadbeat response. Use the
four different methods listed in the foregoing
First we check the observability condition. Notice that the rank of
C is 2. Hence, the system is completely observable. Next we examine the characteristic
equation for the system:
Comparing this characteristic equation with](https://image.slidesharecdn.com/discrete-timecontrolsystems-150605085119-lva1-app6891/85/Discrete-time-control-systems-226-320.jpg)
Discrete time control systems
- 2. Library of Congress Cataloging-in-PublicationData Ogata, Katsuhiko. Discrete-time control systems / Katsuhiko Ogata. -2nd ed. p. cm. Hncludes bibliographical references and inciex. HSBN 0-13-034281-5 1. Discrete-time systems. 2. Control theory. H. IEltle. A402.04 1994 94-19896 629,8'Sdc20 CHP Editoriallproduction supervision: Lynda GriEitithsiT Cover design: Karen Salzbach roduction coordinator: David Dickeyl O 1995, 1987 by Prentics-Hall,Inc. Upper SaddleRiver, New Jersey 07458 All rights reserved. No part of this book may be reproduced,in any form or by any means, without permission in writing from the pubiisher. Frinted in the Wnited States of America 1 0 9 8 7 I S B N D-23-034281-5 Prentice-HallInternationaI (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall CanadaInc., Toronto Prentice-HallHispanoamericana,S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, linc., Tokyo Prentbce-HallAsia Re. Ltd., Siizgapore Editora Prentice-Ealldo Brasil, Etda., Rio de Sanerio 1-1 INTRODUCYION,1 9-2 DIGITAL CONTROL SY 1-3 QUANTIZING AND QUANTIZATIOII ERROR, 8 1-4 DATA ACQUISITION, CONVERSION, AND DISTRIBUTIONSYSTEMS, 1 1 1-5 CONCLUDING COMMENTS, 20 2-1 INTRODUCTION, 23 2-2 THE z TRANSFORM, 24 2-3 z TRANSFORMS OF ELEMENTARY FUNCTIONS, 25 2-4 IMPORTANT PROPERTIESAND THEOREMS OF THE z TRANSFORM, 31 2-5 THE INVERSEz TRANSFORM, 37 2-6 z TRANSFORM METHOD FOR SOLVING DIFFERENCEEQUATIONS, 52 2-7 CONCLUDING COMMENTS, 54 EXAMPLE PROBLEMS AND SOLUTIONS, 55 PROBLEMS, 70
- 3. ontents 3-1 INTRODUCTION, 74 3-2 IMPULSESAMPLINGAND DATA HOLD, 75 3-3 OBTAININGTHEzTRANSFORMBYTHECONVOLUTIONINTEGRALMETHOD,83 3-4 RECONSTRUCTINGORIGINAL SIGNALS FROM SAMPLED SIGNALS, 90 3-5 THE PULSE TRANSFER FUNCTION, 98 3-6 REALlZATlONOF DlGlTAL CONTROLLERS AND DlGlTAL FILTERS, 122 EXAMPLE PROBLEMS AND SOLUTIONS, 138 PROBLEMS, 166 4-1 INTRODUCTION, 173 4-2 MAPPING BETWEENTHE s PLANE AND THE z PLANE, 174 4-3 STABlLlTY ANALYSIS OF CLOSED-LOOPSYSTEMS IN THE z PLANE, 182 4-4 TRANSIENT AND STEADY-STATERESPONSE ANALYSIS, 193 4-5 DESIGN BASED ON THE ROOT-LOCUSMETHOD, 204 4-6 DESIGN BASED ON THE FREQUENCY-RESPONSEMETHOD, 225 4-7 ANALYTICAL DESIGN METHOD, 242 EXAMPLE PROBLEMSAND SOLUTIONS, 257 PROBLEMS, 288 INTRODUCTION, 293 STATE-SPACE REPRESENIATIONSOF DISCRETE-TIMESYSTEMS, 297 SOLVING DISCRETE-TIMESTATE-SPACE EQUATIONS, 302 PULSE-TRANSFER-FUNCTIONMATRIX, 310 DlSCRETlZATlON OF CONTINUOUS-TIMESTATE-SPACEEQUATIONS, 312 LIAPUNOV STABlLlTY ANALYSIS, 321 EXAMPLE PROBLEMSAND SOLUTIONS, 336 PROBLEMS, 370 6-1 INTRODUCTION, 377 6-2 CONTROLLABILITY,378 6-3 OBSERVABILITY, 388 6-4 USEFULTRANSFORMATIONS IN STATE-SPACEANALYSIS AND DESIGN, 396 6-5 DESIGNVIA POLE PLACEMENT, 402 6-6 STATE OBSERVERS,421 6-7 SERVO SYSTEMS, 460 EXAMPLE PROBLEMS AND SOLUTIONS, 474 PROBLEMS,510 7-1 INTRODUCTION, 517 7-2 DIOPHANTINE EQUATION, 518 7-3 ILLUSTRATIVE EXAMPLE, 522 7-4 POLYNOMIAL EQUATIONSAPPROACH TO CONTROL SYSTEMS DESIGN, 525 7-5 DESIGN OF MODEL MATCHING CONTROL SYSTEMS, 532 EXAMPLE PROBLEMS AND SOLUTIONS, 540 PROBLEMS,562 8-1 INTRODUCTION, 566 8-2 QUADRATIC OPTIMAL CONTROL, 569 8-3 STEADY-STATE QUADRATIC OPTIMAL CONTROL, 587 8-4 QUADRATIC OPTIMAL CONTROL OF A SERVO SYSTEM, 596 EXAMPLE PROBLEMS AND SOLUTIONS, 609 PROBLEMS, 629 A-1 A-2 A-3 A-4 A-5 A-6 A-7 A-U DEFINITIONS,633 DETERMINANTS, 633 INVERSIONOF MATRICES, 635 RULES OF MATRIX OPERATIONS, 637 VECTORS AND VECTOR ANALYSIS, 643 EIGENVALUES, EIGENVECTORS,AND SlMlLARlTYTRANSFORMATION, 649 QUADRATIC FORMS, 659 PSEUDO1NVERSES, 663 EXAMPLE PROBLEMS AND SOLUTIONS, 666
- 4. 8-1 INTRODUCTION,681 8-2 USEFULTHEOREMS OF THE z TRANSFORMTHEORY, 681 8-3 INVERSEz TRANSFORMATION AND INVERSIONINTEGRALMETHOD, 686 8-4 MODIFIEDzTRANSFORM METHOD, 691 EXAMPLE PROBLEMS AND SOLUTIONS, 697 C-9 INTRODUCTION, 304 C-2 PRELIMINARY DISCUSSIONS, 704 C-3 POLE PLACEMENT DESIGN, 707 EXAMPLE PROBLEMS AND SOLUTIONS, 718
- 5. X Preface Preface publisher. This book can also serve as a self-study book for practicing engineers w discrete-time control theory by themselves. tive cornrnents absut the material ín this book.
- 7. recent years there has been a rapid increase in the gital controllers in ontrol systems. Digital controls are used for achieving performance-f~r example, in the form of maximum productivity, maximu t9minimum ~ 0 %or minimum energy use. recently, the application of cornputer control has made possible "intelli- on in industrial robots, the optimization of fue1economy ' and refinements in the operation of household appliances and m rnicrowave ovens and sewing machines, among others. Decision-m and flexibilityin the controlprogram aremajor advantagesof digitalcontrolsysterns. The current trend toward digital rather than analog ntrd of dynamicsystems is mainly due to the availability of low-cost digital com ters and the advantages found in working with digital signals rather than continuous-time signals. continuous-time signal is a signal defined over a contin- uous plitude may assume a continuous range of values or may assume only a finite number of distinctvalues. The process o£representing a variable by a set of distinct values is called quantization, and the resulting distinct values are called quantized values. The quantized variable changes only by a set of distinct steps. n analog signal is a signal defined over a continuous range of time whose am- plitude can assume a continuous range of values. Figure 1-l(a) shows a continuous- time analog signal, and Figure 1-l(b) shows a continuous-time quantized signal (quantized in amplitude only).
- 8. lntroduction to Bíscrete-Time Control Systems Chap. 1 Figwe 1-1 (a) Continuous-time analog signal; (b) continuous-time quantized signal; (c) sampled-data signal; (d) digital signal. Notice that the anaiog signal is a special case of the continuous-time signal. In practice, however, we frequently use the terminology "continuous-time" in lieu of "analog." Thus in the literature, including this book, the terms "continuous-time signal" and "analog signal" are frequently interchanged, although strictly speaking they are not quite synonymous. A discrete-time signal is a signal defined only at discrete instants of time (that is, one in which the independent variable t is quantized). n a diXrete-h-le signal9 if the amplitude can assume a continuous range of values, then the signal is called a sampled-data sigizal. A sampled-data signal can be generated by sampling an analog signal at discrete instants of time. It is an amplitude-modulated pulse signal. Figure 1-1(c) shows a sarnpled-data signal. A digital signal i:, a dlscrete-time signalwith quantized arnplitude. Sucha signal can be represented by a sequence of nurnbers, for example, in t ec. 1-1 introdlaction numbers. (In practice, many digital signals are obtained by sampling analog signals and then quantizing them; it is the quantization that allows these analog signals to be read as finite binary words.) Figure 1-l(d) depicts a digital signal. Clearly, it is a signal quantized both in amplitude and in time. The use of the requires quantization of signals both in amplitude and in time. The term "discrete-time signal" is broader than the term "digital signal" or the term "sampled-data signal." 1can refer either to a digital signal or to a sampled-data signal. "discrete time7'and ""dgital" are often interchan used in theoretical study, wh ware or software realizations. In control engineering, ysical plant or process or a nonphysical process such as an economic process. t plants and processes involve continuous-time signals; therefore, if digital controllers are involved in the control systems, signal conversions (analog to dibital and digital to analog) become necess y. Standard techniques are available for such signal conversions; we shall discuss em in Section 1-4. Loosely speaking, terminologies such as discrete-time con pled-data control systems, and digital control systems imply the similar types of control systems. Preciselyspeaking, there are, of in these systems. For example, in a sa -data control system both continuous- time and discrete-time signals exist in the system; the discrete-time signals are amplitude-modulated pulse signals. Digital control systemsmay include both contin- us-time and discrete-time signals; here, the latter are in a numerically coded form. th sampled-data control systems and digital control systems are discrete-time any industrial control systemsinclude continuous-time signals, sampled-data signals, and digital signals. Therefore, in this book we use the term "discrete-time control systems" to describe the control systemsthat include sorneformsof sarnpled- data signals (amplitude-modulated pulse signals) and/or digital signals (signals in numerically coded form). . The discrete-time control systems consid- ered in this book are mostly linear and time invariant, although nonlinear and/or time-varying systems are occasionallyincluded in discussions. A linear systemis one in which the principie of superposition applies. Thus, if y, is the response of the system to input xl and y2 the response to input x2, then the system is linear if and only if, for every scalar a and P, the response to input al+ Px2 is cuyl + ,By2. A linear systern may be described by linear differential or linear difference equations. A time-invariant linear system is one in which the coefficients in the differential equation or difference equation do not vary with time, that is, one in which the properties of the system do not change with time. Discrete-time control systemsare control systemsin which one or more variables can change only at discrete instants of time. These instants, which we shall denote by kTor tk(k = 091,2,. . . ),may specify the times at which some physicalrneasurement
- 9. lntroduction to Discrete-TimeControl Ystems Chap. 1 is performed or the times at which the memory of a digital computer is read out. The time interval between two discrete instants is taken to be sufficiently short that the data for the time between them can be approximated by simple interpolation. Discrete-time control systems differ from continuous-time control systems in that signals for a discrete-time control system are in sampled-data form or in digital form. If a digital computer is involved in a control system as a digital controller, any sampled data must be converted into digital data. Continuous-time systems, whose signals are continuous in time, may be de- scribed by differential equations. Discrete-time systems, which involve sample data signals or digital signals and possibly continuous-time signals as well, may be described by difference equations after the appropriate discretization of continuous- time signals. esses. The sarnpling of a continuous-time signal replaces the origi -time signal by a sequence of values at discrete time points. A sampling process is used whenever a control system involves a digital controller, since a sampling operation and quantization are necessary to enter data into such a controller. Also, a sampling process occurs whenever measurements necessary for control are obtained in an intermittent fashion. For example, in a radar tracking system, as the radar antenna rotates, information about azimuth and elevation is obtained once for each revolutio of the antenna. Thus, the scanning operation of the radar produces sampled data n another example, a sampling process is needed whenever a large-scale controller or computer is time-shared by several plants in order to save cost. Then a control signal is sent out to each plant only periodically and thus the signal becomes a sampled-data signal. The sampling process is usually followed by a quantization process. In the quantiration process the sampled analog amplitude is replaced by a digital ampli- tude (represented by a binary number). Then the digital signal is processed by the computer. The output of the cornputer is sampled and fed to a hold circuit. The output of the hold circuit is a continuous-time signal and is fed to the actuator. shall present details of such signal-processing methods in the digital controlle Section 1-4. The term "discretization," rather than "sampling," is frequently used in the analysis of multiple-input-multiple-output systems, although both mean basically tant to note that occasionally the sampling operation or discretiza- fictitious and has been introduced only to simplify the analysis of control systems that actually contain only continuous-time signals. In fact, we often use a suitable discrete-time model for a continuous-time system. An example is a digital-computer simulation of a continuous-time system. Such a digital-computer- simulated system can be analyzed to yield parameters that will optimize a given performance index. ost of the material presented in this book deals with control systemsthat can be modeled as linear time-invariant discrete-time systems. It isim that many digital control systems are based on continuous-time Since a wealth of experience has been accumulated in the design of continuous-time Sec. 1-2 Digital Control Cystems controllers, a thorough knowledge of them is highly valuable in designing discrete- time control systems. Egure 1-2 depicts a block diagram of a tion of the basic control scheme. The sy feedforward control. In designing such "goadness" of the control system depe choose an appropriate performance index for a given case and design a controller so as to optimize the chosen performance index. . Figure 1-3 shows a block diagram of the system are shownby the blocks. ñhe controller operation is controlled by the clock. In such a digital control system, somepoints of the system pass signalsof varyingamplitude in either continuous time or discrete time, while other points pass signalsin numerical code, as de of the plant is a continuous-time signal. The error signal is con- 1 form by the sample -hold circuit and the analog-to-digital converter. The conversion is done at sampling time. The digital computer l i 1 Clock L _I Digital controlier Noise Figure 1-2 Block diagram of a digital control system.
- 10. lntroduction to Discrete-Time Control Cysterns Chap. ? ipre 1-3 Block diagram of adigital control system showingsignalsinbinary or graphic form. rocesses the sequences of numbers by means of an algorithm and produces new quences of numbers. At every sampling instant a coded number (usually a binary number consisting of eight or more binary d ) must be converted to a physical control signal, which is usually a continuous e Or anal% signal- The digital-to- analog converter and the hold circuit convert sevence ~f numbers in numerical code into a piecewise continuous-time signal. The real-time clock in the cornputer synchronizes the events. The output of the hold circuit, a continuous-time signal, is fed to the plant, either directly or through the actuator, to control its The operation that transforms continuous-time signals into discrete-time data is called sampling or discretizaíion. The reverse operation, the operation that transforms discrete-time data into a continuous-time signal, is called data-hold; it amounts to a reconstruction of a continuous-time signal from the sequence of discrete-time data. It isusuallydone using one of the many extrapolation techniques. Pn many cases it is ne by keeping the signal constant between the successive sampling instants. ( shall discuss such extrapolation techniques in Section 1-4.) The sample-and-hold (SIH) circuit and analog-to-digital (AID) converter con- vert the continuous-time signal into a sequence of numerically coded binary words. Such an AID conversion process is called coding or encoding. The combination of the SIH circuit and analog-to-digital converter may be visualized as a switch that closesinstantaneously at every time interval Tand generates a sequence of numbers in numerical code. The digitalcornputer operateson suchnumbers in numerical code and generates a desired sequence of numbers in numerical code. The digital-to- analog (DIA) conversion process is called decoding. efore we discuss digital control systems in detail, we need to define some of the terms that appear in the block diagram of Figure 1-3. Sample-and-Hold (SIH). "Sarnple-and-hold" is a general term used for a sarnple-and-hold amplifier. It describes a circuit that receives an analog input signal and holds this signal at a constant value for a specified period of time. Usually the signal is elebrical, but other forrns are possible, such as optical and mechanical. igital Control Systerns 7 Analog-to-Digital Gonverter (AlD). An analog-to-digital converter, also called an encoder, is a device that converts an analog signal into a digital signal, usually a numerically coded signal. Such a converter is needed as an interface between an analog component and a digital com onent. A sample-an is often an integral part of a commercialliyavailable A/D converter. The conversion of an analog signal into the corresponding digital signal (binary number) is an ation, because the analog signal ke on an infinite number of values, he variety of different nurnbers a finite set of digits This approximation process is c ore on quantization is presented in Section 1-3.) Digital-to-Analog Converter (DIA). A digital-to-analog converter, also called a decoder, is a device that converts a digital signal (numerically co into an analog signal. Such a converter is needed as an interface between a digital component and an analog component. Plant or Process. A plant is any ical object to be controlled. Examples are a furnace, a chemical reactor, and of machine parts functioning together to perform a particular operation, such as a servo system or a spacecraft. A process is generally defined as a progressive operation or develop marked by a series of gradual changes that succeed one another in a relatively fixed way and lead toward a particular result or end. In this book we cal1 any operation to be controlled a process. Examples are chemical, economic, and biolog- ical processes. The most difficult part in the design of control systems may lie in the accurate modeling of a physical plant or process. There are many approaches to the plant or process model, but, even so, a difficulty may exist, mainly because of the absence of precise process dynamics and the pr of poorly defined random parameters in many physical plants or processes. ,in designing a digital controller, it is necessary to recognize the fact that the mathematical model of a plant or process in many cases is onIy an approximation of the physical one. Exceptions are found in the modeling of electromechanical systems and hydraulic-mechanical systems, since these may be modeled accurately. For example, the modeling of a robot arm system may be accompIished with great accuracy. Transducer. A transducer is a device that converts an input signal into an output signal of another form, such as a device that converts a pressure signal into a voltage output. The output signal, in general, depends on the past history of the input. Transducers may be classified as analog transducers, sampled-data transduc- ers, or digital transducers. An analog transducer is a transducer in which the input and output signalsare continuous functions of time. The magnitudes of these signals may be any values within the physical limitations of the system. A sampled-data transducer is one in which the input and output signalsoccur only at discrete instants of time (usually periodic), but the magnitudes of the signals, as in the case of the analog transducer, are unquantized. A igital transducer is one in which the input and output signals occur only at discrete instants of time and the signal magnitudes are quantized (that is, they can assume only certain discrete levels).
- 11. lntroductisn to Discrete-TimeControl ns, As stated earlier, a si discrete-time signal. A sa in transforming a continuous-time g*al into a dkxete- There are several different t es of sampling operations of practica1 impar- tance: n this case, the campling instants are equally spaced, or tk = kT(k =O,1,2,... ). eriodic sampling is the II-lost conventional tYPe of sampling operation. Multiple-order sampling. The pattern of the tk's is repeated is, tk+r- tk is constant for al1 k . ltiple-ratesampling. In a control system havi? stant involved in one loop may be quite ence, it may be advisable to sample slow time constant, while in a loop involvingonly small time constants the sampling rate must be fast. Thus, a digital control system may have different sampling riods in different feedback paths or may have multiple sam ndorn samplirzg. In this case, the sampling instaiats are ra random variable. n this book we shall treat only the case where the sampling is periodic. The main functions involvedin analog-to-digital conversion are sampling, amplitude quantizing, and coding. hen the vahe of any sample falls between two adjacent "'perrnitted" output stat it must be read as the permitted output state nearest the actual value of the signal. The rOWss of representing a ~ontinuousor anal% sigrial by a finite number of discrete states is called amplitude quantization. That is, "quantizing" means transforming a continuous or analog signalinto a set of states. (Note that quantizing occurs whenever a physical quantity is represented numerically.) The output state of each quantized sample is then described by a numerical code. The process of representing a sample value by a numerical code (such as a binary code) is called encoding or coding. Thus, encoding is a process of assigning a digital word or code to each discrete state. The sampling period and quantizing levels affect the performance of digital control systems. Sothey must be determined carefully. . The standard number system used for processing digital signals is the binary number system. In this system the code group consists of n pulses each indicating either "on" (1) or "off" (O). In the case of quantizing, n "on-off" pulses can represent 2" amplitude levels or output states. The quantization level Q is defined as the range between two adjacent decision points and is given by ec. 1-3 Quantizing and Quantization where the FSR is the full-scalerange. Note that the st bit of the natural binary as the most weight (one-half of the full scale) he most significant SB). The rightmost bit has the least weight (112" t e full scale) and is called the least significant bit (LS The least significant bit is t e quantization leve1 Q. Error. Sincethe nu igitalword is finite,AID in a finite resolution. output can assume only a finite number of levels, and therefore an analog number must be rounded off to the nearest digital level. Menee, any AID conversion involves quantization error. Such quantization error varies between O and t $ Q S error depends on the fineness of the quantization level and can be made as s quantization level smaller (that is, by increasing the there is a maximum for the number of bits n, and so there is always some error due to quantization. The uncertainty present in the quantization process is called quan- tization noise. To determinethe desired sizeof the quantization level (or thenumber of output states) in a givendigital control system, the engineermust have a good understanding of the relationship between the size of the quantization level and the resulting error. Thevarianceof the quantization noise isan important measure of quantization error, since the variance is proportional to the average power associated with the noise. Figure 1-4(a) shows a block diagram of a quantizer together with its input- output characteristics. For an analog input x(t), the outputy(t)takes on only a finite number of levels, which are integral multiples of the quantization level Q . In numerical analysis the error resulting from neglecting the remaining digits is called the round-off error. Since the quantizing process is an approximating process in that the analog quantity is approximated by a finite digital number, the quantization error is a round-off error. Clearly, the finer the quantization level is, the smaller the round-off error. Figure 1-4(b) shows an analog input x(t) and the discrete output y(t), which is in the form of a staircase function. The quantization error e(t) is the difference between the input signal and the quantized output, or Note that the magnitude of the quantized error is For a small quantization leve1 ,the nature of the quantization error is similar to that of random noise. nd, in effect, the quantization process acts as a source of random noise. In what followswe shallobtain the variance of the qu Scach variance can be obtained in terms of the quantization level
- 12. lntroduction to Diccrete-Time Ontrol Chala. 1 (a) Block diagram of a quantizer and its input-output characteristics; (b) analog input x(t) and discrete output y(t); (c) probability distribution P(e) of quantization error e(f). Suppose that the quantization level Q is small and we assume that the quan- tization error e(t) is distributed uniformly between -4 Se that this error acts as a white noise. [This is obviously a rather rough assumption. However, since the quantization error signal e(t) is o£a small amplitude, such an assumption may be aiceptable as a first-order approximation.] The probability distribution P(e) of Sec. 1-4 Dala Acquicition, Conversion, and signar e(t) may be plotted as shown in Figure 1-4(c).The average value of e(t) iszem, or e(t) = O. Then the variance u h f the quantization noise is n level Q is small co pared with the average amplitude o£ variance of the quantization noise is seen to be one-twelfth of the square of the quantization level. ith the rapid growth in the use of digital computers to perform digital control actions, both the data-ac uisition system and the distribution system have become an important part of the entire control system. The signalconversion that takes place in the digital control system involvest following operations: ultiplexing and demultiplexing .Sample and hold Analog-to-digital conversion (quantizing and encoding) .Digital-to-analog conversion (decodlng) Figure 1-5(a) shows a block diagram o£a acquisition system, and Figure 1-5(b) shows a block diagram of a data-distrib n the data-acquisitionsystem the input to the system is a physicalvariable such asposition, velocity, acceleration, temperature, orpressure. uch a physicalvariable is first converted into an electrical signal (a voltage or current signal) by a suitable Physical To digital variable controller From digital controller To actuator Figure 1-5 (a) Block diagram of a data-acquisition system; (b) block diagram of a data- distribution system.
- 13. Introduction to Discrete-TimeControl Systems Chap. 1 transducer. Once the physical variable is converted into a voltage or current signal, the rest of the data-acquisition process is done by electronic means. In Figure 1-5(a) the amplifier (frequentl Iows the transducer performs one or more of the voltage output of the transducer; it converts a current signalinto a vo or it buffers the signal. The low-pass filter that follows the amplifier attenuates the high-frequency signalwmponents, such asnoise signals. (Note that are random in nature and may be reduced by low-pass filters. common electrical noises as power-line interference are generally p be reduced by means of notch filters.) The output of the low-pass filter is an analog signal. m i s signal is fed to the analog multiplexer. The output of the multiplexer is fed to the sample-and-hold circuit, whose output is, in turn, digital converter. The output of the converter is the signal in to the digital controller. The reverse of the data-acquisition process is the ata-distributionProcess “%+, shown in Figure 1-5(b), a data-distribution system consists of registers, a demulti- plexer, digital-to-analog converters, and hold circuits. It conver the signalin digital form (binary numbers) into analog form. The output of the DI converter is fed tQ the hold circuit. The output of the hold circuit is fed to the analog actuator, which, in turn, directly control; the plant under consideration. In the following, we shall discuss each in b.+hal ComPonent h ~ ~ l v e din the signal-processing system. lexer. An analog-to-digital converter is the most expensive component in a data-acquisition system. The analog multiplexer is a device that performs the function of time-sharing an AID converter among many analog chan- nels. The processing of a number of channels with a digital controller is posible because the width of each pulse representing the input signal is very narrow, so the empty space during each sampling period rnay be used for other signals. If many signalsareto be processed by a single digital controller, then these input signalsmust be fed to the controller through a multiplexer. Figure 1-6 shows a schematic diagram of an analog multiplexer. The analog To sampler Figure 1-6 Schematic diagram of an analog multiplexer. Data Acquisition, Conversion, and multiplexer is a multiple switch (usually an electronic switc ) that sequentially switches amongmany analoginput channelsin bed fashion. Thenumber instant of time, only one given input channel, the for a specified period of time. During the connection time the sampl ircuit samples the signal voltage (analog signal) and holds its alue, while the analog-to- converts the analog value into digital ta (binary numbers). Each channel is read in a sequential order, and the corresponding values are converted into digital data in tbe same sequence. lexer. The demultiplexer, which is synchronized with the in pling signal, separates the composite output digital data flrom the digital controller into the original channels. Each channel is connected to a DIA converter to produce the output analog signal for that channel. er in a digital system convelrts an analog pulses. The hold circuit holds the value of the sampled pulse signal over a specified period of time. The sample-and-hold is necessary in the AID converter to pro input signal at the sampling instant. available in a single unit, known as however, the sampling operation and (see Section 3-2). It is common practice to use a single analog-to-digital converter and multiplex many sampled analog inputs into it. practice, sarnplingduration isvery short compared with the sampling period the sampling duration is negligible, the sampler may be considered an mpler." An ideal sampler enables us to obtain a relatively simplemathemat- ical model for a sample-and-hold. (Such a mathematical model will be discussed in e 1-7 showsa simplifieddiagramfor the sample-and-hold.Th g circuit (simply a voltage memory device) in which an inpu acquired and then stored on a high-quality capacitor with low leakage and low dielectric absorption characteristics. In Figure 1-7 the electronic switch is connected to the hold capacitor. Opera- tional amplifier 1 is an input r amplifier with a high input impedance. Op- erational amplifier 2 is the o amplifier; it buffers the voltage on the hold capacitor. There are two modes of operation for a sample-and-hold circuit: the tracking mode and the hold mode. n the switch is closed (that is, when the input signal is connected), the operatin ing mode. The charge on the capacitor in the circuit tracks the input voltage. the switch is open (the input signal is disconnected), the operating mode is the hold mode and the capacitor voltage holds constant for a specified time period. Figure 1-8 shows the tracking mode and the hold mode. Note that, practically speaking, switching from the tracking mode to the hold mode is not instantaneous. If the hsld command is given whde the circuit is in the
- 14. Introduction to Discrete-TimeControl Cystems Analog input Chap. 1 Arnp. 1 I 1 Arnp. 2I Analog output Sarnple-and-hoid cornrnand Figure 1-7 Sample-and-hold circuit. tracking mode, then the circuitwill stay in the tracking mode for a short whilebefore reacting to the hold command. The time interval during which the switching takes place (that is, the time interval when the measured amplitude is uncertain) is called the aperture time. The output voltage duringthe hold mode may decrease slightly.The hold mode droop may be reduced by using a high-input-impedance output buffer amplifier. Such an output buffer amplifier must have very low bias current. The sample-and-hold operation is controlled by a periodic clock. es oofAnalog-to-Digital earlier, the PrQcessby which a sampled analog signal is quantized and converted to a binary number is called analog-to-digital conversion. Thus, an AID converter transforms an analog Inpuí Sarnple to Hold rnode I signal hold offset droop Tracking mode 4 kp:;ye- t ttioíd command Figure 1-43 Traclcing mode and hold is given here mode. Sec. 1-4 Data Acquisition, Conversion, and istribution Systemc signal (usually in the formof a voltage or current) into a coded word. In practice, the logic is based on binary dig and the representation has only a finite performs the operations of sample-and-ho in the digital system a pulse is supplied every sampling period T by a clock. %heAID converter sends a digital signal (binary number) to the digital contiroller each time a pulse arrives. Among the many ID circuits available, the following types are used most frequently: ve-approximation type Each of these four types has its own advantagesan application, the conversion speed, accuracy, size be considered in choosing the type of A/D convert for example, the number of bits in the output signal must be increased.) As will be seen, analog-to-digital converters art of thek feedback l o o p digital-to-analogconverters. The simplestty verter is the counter type. The basic principle on which it works is that clock pulses are applied to the digital counter in such a way that the o the feedback loop in the AID con the output voltage hen the output volt the clock pulses are stopped. The counter output voltage is then the digital output. The successive-approximation type of AID converter is much faster than the counter type and is the one most frequently use Figure 1-9 shows a schematic diagrarn of the successive-approximariontype of D/A Analog converter reference Digital output Analog input Figure 1-9 Schematicdiagramof a successive-approximation-typeof BID converter.
- 15. lntroduction to Discrete-TimeControl The principle of operation of this type of AID converter is as follows. T successive-approximation register (SAR) first turns on the most significant bit (ha the maximum) and compares it with the analog input. The com whether to leave the bit on or turn it off. If the analog input voltage is larger, tfae most significant bit is set on. The next step is to turn on bit 2 and t analog input voltage with three-fourths of the maximum. After n coqleted, the digital output of the successive-approximationreg those bits that remain on and produces the desired digital code. ID converter sets 1bit each clock cycle, and so it requires ody n clock cycles t s generate n bits, where n is the resolution of the converter in bits. (The number n of bits employed determines the accuracy of conversion.) The time required for the conversion is approxirnately 2 psec or less for a 12-bit conversion. ctual analog-to-digital signal converters differ from the ideal signal converter in that the former always have some errors, such as offset error, linearity error, and gain error, the characteristics o£which are shown in Figure 1-10. Also, it is important to note that the input-output characteristics change with time and temperature. Finally, it is noted that commerciaI converters are specified for three basic temperature ranges: cornmercial (0°C to 70°C), industrial (-25°C to 85"C), and military (-55°C to 125°C). onverters. At the output of the digital controller the digital signal must be converted to an analog signal by the process called digital-to- analog conversion.A DIA converter is a devicethat transforms a digital input (binary numbers) to an analog output. The output, in most cases, is the voltage signal. For the full range of the digital input, there are 2" corresponding different analog values, induding O. Por the digital-to-analog conversion there is a one-to-one ~orrespondencebetween the digital input and the analog output. Two methods are commonly used for digital-to-analog conversion: the method usirig weighted resistors, and the one using the R-2R ladder network. The former is simple in circuit configuration, but its accuracy may not be very good. The latter is a little more complicated in configuration, but is more accurate, Figure 1-13 shows a schematic diagram o£ a DIA converter using weighted resistors. The input resistors of e operational amplifier havetheir resistance values weighted in a binary fashion. hen the Iogic circuit recebes binary 1, itch (actually an electronic gate) co ects the resistor to the referente voltage. the logic circuit receives binary 0, the switch connects the resistor to ground. The digital-to-analog converters used in common practice are of the parallel type: al1bits act sirnultaneously upon application of a digital input (binary numbers). The D/A converter thus generates the analog output voltage corresponding to the given digital voltage. For the D/A converter shown in Figure 1-11, if the binary number is b3b.,bl bo,where each of the b's can be either a O or a 1, then the output is Sec. 'l-4 Data Acquisition,Conversion, and Errors in AID converters: (a) offset error; (b) linearity error; (c) gain error. Notice that as the nurnber of bits is increased the range of resistor values becomes large and consequently the accuracy becomes poor. e 1-12 shows a schematic diagram of an n-bit DlA converter using an -2 er circuit. Note that with the xception of the feedback resistor (which is 3R) al1 resistors involved are either or 2R. This means that a high Ievel of accuracy can be achieved. The output voltage in this case can be given by Ciircuii. Shesarnplingoperation pro- duces an amplitude-modulated pulse signal. The function of the hold operation is
- 16. lntroduction to Biscrete-Time Control Figure 1-11 Schematic diagram of a D/A converter using weighted resistors. to reconstruct the analog signalthat has been transmitted as a train of pulse sa That is, the purpose o£the hold operation is to fill in the spaces between sampling periods and thus roughly reconstruct the original analog input signal. The hold circuitis designed to extrapolate the output signalbetween successive points according to some prescribed manner. The staircase waveform of the output shownin Figure 1-13 is the simplest way to reconstruct the original input signal. The Id circuit that produces such a staircase waveform is called a zero-order hold. cause of its simplicity, the zero-order hold is commonly used in digital control systems. Figure 1-12 n-Bit DIA converter using an R-2R ladder circuit Output Figure 1-13 hold. Outpur from a zero-order ore sophisticated hold circuits are available than the zero-orde r-order hold circuits and include the first-order hold a gher-order hold circuits wiIl generally reconstnict a signal more a zero-order hold, but with some disadvantages, as explained next. The first-order hold retains the val e of the previous sampk as we present one, and predicts, by extrapolation, the next sam generating an output slope equal to the slo and present samples and projecting it from in J?igure 1-14. As can easilv be seen from the figure, if the slope of the original signal not change much, the prediction is go slope, then the prediction is wrong a causing a large error for the sampli An interpolativefirst-order ho kdd, reconstructsthe nal signal much more accuratel generates a straight-line ut whose slopeis equal to that joining the previous samplevalue and the present samplevalue, but this time the projection ismade fromthe cuirent samplepoint with Output
- 17. Output Figure 1-15 Output from an inter- polative first-order hold (polygonal hold). the amplltude of the prevlous sample. ence7the accuracy in reconstructing the original signal is better than for other hold circuits, but there is a one-sarnpling- period delay, as shown in Figure 1-15. effect,the better aCcuracY is achieved at the expense of a delay of one sampling period. From the viewpoint of the stability of closed-loop systems, such a delay is not desirable, and so the interpolative first-order hold (polygonal hold) is not used in control system applications. In concluding this chapterwe shall compare digitalcontrollers and analogcontrollers used in industrial control systems and review digital control of processes. Then we shall present an outline of the book. rs. Digital controllers operate only on numbers. Decision making is one of their important Tunctions. They are often used to solve problems involved in the optimal overall operation o£industrial plants. Digital controllers are extremely versatile, They can handle nonlinear control equations involving complicated computations or logic operations. A very much wider class of control laws can be used in digital controllers than in analog con- trollers. Also, in the digital controller, by merely issuing a new program the oper- ations being performed can be changed completely. Shis feature is particularly important if the control system is to receive operating information or instructions from some computing center where econornic analysis and optirnization studies are made. Digital controllers are capable of performing complex computations with constant accuracy at high speed and can have almost any desired degree of compu- tational accuracy at relatively little increase in cost. Originally, digital controllers were used as components only in large-scale control systems. At present, however, thanks to the availability of inexpensive microcornputers, digital controllers are being used in many large- and small-scale control systemc. hn fact, digital controllers are replacing the analog controllers that Sec. 7-5 Concluding Cornrtents have been used in many srnall-scale control systems. Digital controllers ase often superior in performance and lower in price than their analog counterparts. Analog controllers represent the variables in an equation by continuous ical quantities. The easily be designed to serve satisfactorily as non-decision- e cost of analog computers or analog controllers increases of the computations increases, if constant accuracy is to There are additional advantages of digital controllers over analog controllers. Egital components,such assample-and-hold circuits, AID and DlA converters, and highly reliable, and often compact ts have high sensitivity, are often cheaper than their analog counterparts, and are less sensitive to noise signals. And, as mentioned earlier, digital controllers are flexible in allowing programming changes. Digihl Ccpnkr~b industrial process control systems, it is gen- erally not practica1t ry long time at steady state, because certain changes rnay occur in production requirements, raw materials, economic factoss, and processing equipments and techniques. Thus, the transient behavlor of indus- trial processes must always be taken into consideration. Sincethere are interactions amsng process variables, using only one p ariable for each control agent is not suitable for really complete control. use of a digital controller, it ks possible to take into account al1 process v together with econornic factors, ents, equipment performance, and al1other needs, and thereby al control of industrial processes. Note that a system capable of controlling a process as coqlete%yas will have to solve complex equations. The more complete the control, t important it is that the correct relations between operating variables be known and used. The system must be capable of accepting instructions from such varied sources as computers and human operators and must also be capable of changing its control subsystem completely in a short time. Digital controllers are most suitable in such situations. %nfact, an advantage of the digital controller is flexibility, that is, ease of changing control schemes by reprogramming. In the digital control of a complex process, the designer must have a good knowledge of the process to be controlled and must be able to obtain its mathemat- ical model. (The mathematical model rnay be obtained in terms of differential equations or difference equations, or in some other form.) The designer must be familiar with the measurement technology associated with the output of the process and other variables involved in the process. e or she must have a good working knowledge of digital computers as well as modern control theory. Iíf the process is complicated, the designer must investigate severa1different approaches to the design of the control system. In this respect, a good knowledge of simulation techniques is helpful. The objective of this book is to present a detailed acco~intof the control theory that is relevant to the analysis an design of discrete- time control systems. Our enphasis is on understanding the basicconceptc involved.
- 18. lntroduction to Discrete-TimeControl Ystems ChaP '1 In this book, digital controllers are often designed in the form of pulse transfer functionsor equivalent difference equations,which can be easilyimplemented in the form of computer programs. The outline of the book is as follows. Chapter 1has presented introductory ma- terial. Chapter 2presents the z transform theory. ter includes z transforms of elementary functions, important properties an inverse z transform, and the solution of differe method. Chapter 3 treats background materials systems. This chapter includes discussions of impulse sampling and reconstruction of original signalsfrom sampled signals, pulse transfer functions, and realization of digital controllers and digital filters. Chapter 4 first presents mapping between the S plane and the z plane and then discusses stability analysis of closed-loop systems in the z plane, followed by tran- sient and steady-state response analyses, design by the root-locus and frequency- response methods, and an analytical design method. Chapter 5 gives state- space representation of discrete-time systems, the solution of discrete-time state- space equations, and the pulse transfer function matrix. Then, discretization of continuous-time state-space equations and Liapunov stability analysis are treated. Chapter 6 presents control systems design in the state s chapter with a detailed presentation of controllability and obs present design techniques based on pole placernent, follow full-order state observen and minimum-order state observe chapter with the design of servo systems. Chapter 7 treats the approach to the design of control systems. e b%in the cha~terwith diX~ssionsof Diophantine equations. Then we present the design of regulator systems and control systems using the solution of Diophantine equations. The approach here is an alternative to the pole-placement approach ombined with minimum-order observ- ers. The design of model-matching control systems is included in this chapter. Finally, Chapter 8 treats quadratic optimal control problems in detail. The state-space analysisand design of discrete-time control systems, presented in Chapters 5,6, and 8,make extensive use of vectors and rices-In studying h % e chapters the reader may, as need arises, refer to App which ~ummarizecthe basic materials of vector-matrix analysis. Appendix ts materiak in z tEUls- form theory not included in Chapter 2. Appendix C treats pole-placement design problems when the control is a vector quantity. In each chapter, except Chapter 1,the main text isfollowedby solvedproblems and unsolved problems. The reader should study al1 solved probl Solved problems are an integral part of the text. Appendixes A, followedby solvedproblems. The reader who studiesthese solvedproblems willhave an increased understanding of the material presented. atical tool commonly used for the analysis an esis of discrete-time systems is similar to that of the Lapla form in continuous-time systems. tions to linear difference equations become algebraic in nature. (Just as the Laplace transformation transforrns linear time-invariant differential equations into algebraic equations in S , the z transformation transforms linear time-invariant difference equations into algebraic equations in z .) The main objective of this chapter is to resent definitionsof the z transform, basic theorems associated with the z transform, and methods for finding the inverse z transform. Solving difference equations by the z transform method is also cussed. Signak, Diserete-t signals arise if the system involves a sampling of continuous-time als. The samplled signal is x(O),x(T), x(2T), ...,where T is the sampling period. Such a sequence of values arising from the sampling operation is usually written asx(kT). If the system involvesan iterative process carried out by a digital computer, the signal involved is a nu x(O),x(l), x(2). . .. The sequence of numbers is usualhy written as x(k), where tbe argument k indicates the order in which the number occurs in the sequence, for example, x(O),x(l),42). .. . Although x(k) is a number sequence, it can be con- sidered as a sampled signal of x(t) when t e sampling period T is 1sec.
- 19. The z Transform The 2 transform applies to the continuous-time signal x(t), sampled signal x(kT), and the number sequence x(k). dealing with the z transform~if no confusion occurs in the discussion, we occasionallyuse x(kT) and x(k) interchange- ably. [That is, to simplify the presentation, we occasionallydrop the explicit appear- ance of T and write x(kT) as x(k).] ection 2-1. has presented introductory remarks. ection 2-2 presents the definition of the z transform and associated subjects. Section 2-3 gives z transforms of elementary functions. mportant properties theorems of the z transform are presented in Section 2-4. computational methods for finding the inverse z transform are discussed in 2-5. Section 2-6 presents the solution of difference equations by the z transform method. Finally, Section 2-7 gives concluding commeats. The z transform method is an operational method that is very working with discrete-time systems. In what followswe shall define of a time function or a nurnber sequence. In considering the z transform of a time function x(t), we consider sampled values o£x(t), that is,x(O),x(T), x(2T), ...,where Tisthe samplin The z transform of a time function x(t),where t isnonnegative, os o£a of valuesx(kT), where k takes zero or positive integers and Tis the sampli is defined by the following equation: m For a sequence of numbers x(k), the z transform is defined by m The z transforrn defined by Equation (2-1) or (2-2) is referred to as the one-sided z transform. The symbol7 denotes "the z transform of." In the one-sided z transform, we assumer(t) = O fort < O orx(k) = Ofork < O. Note that z is a complexvariable. Note that, when dealing with a time sequence x(kT) obtained by sampling a time signal x(t), the z transform X(z) involves T explicitly. Owevery £m a Imnber sequence x(k), the z transform X(z) does not involve T explicitly. The z transform of x(t), where -.a < t < .a, or of x(k), where k takes integer values (k = 0, +-1, t 2 , ), is defined by m X(z) = 27 [x(t)] = Z [x(kT)] = k=-m Sec. 2-3 z Transformc of Elementary Functions The z transform defined by Equation (2-3) or (2-4) is referred to as the two-sided z transform. In the two-sided z transfor e function x(t) is assumed to be nonzero for t < O and the sequence x(k) is considered h the one-sided and two-sided z transforms involvesboth positive and negative powers of z one-sided z transform is considered in detail. For most engineering applications the one-sided z transform venient closed-form solution in its r an infinite series in z-', converges radius of absolute convergente, in u ansform method for time problems it is not necessary each cify the values of z over whichX(z) ks convergent. Notice that expansion of the right-hand side of Equation (2-1) gives any continuous-timefunction x(t) may zmkin this series indicates the position inversisn integral method (see Section 2-5 for details.) n the following we shall present z transfor S of several elementary functions. It is noted that in one-sided z transform theory, in sampling a discontinuous function x(t), we assume that the function is continuous from the right; that is, if discontinuity occurs at t = O, then we assume that x(O) is equal to x(O+) rather than to the average at the discontinuity, [x(O-) + x(O+)]/2. Unit-StepFunc~on. ket us find the z transforrn of the unit-step function S just noted, in sampling a unit-step function we assume that this function is continuous from the right; that is, l(0) = l . Then, referring to Equation (2-l), we have
- 20. The z Transform Chap. 2 Notice that the series converges if lzl > 1. fina"ir%the transform, the Variable ator. It is not necessary to specify the region of z over which suffices to know that such a region exists. The z transform n ~ ( t )obtained in this way is valid throughout the z plane except at poles of X(z). t is noted that 1(k)as defined by k = 0,1,2,.-. is commonly called a unit-step seqraence. on. Consider the unit-ramp function Notice that x(kT) = k T , k = 0,1,2,. Figure 2-1 depicts the signal. The magnitudes of the sam values are proportional iod T .The z transforrn of the unit-r function can be written as m m m O T 2T 3T 4T t Figure 2-1 Sampled unit-ramp signai. Sec. 2-3 z Transforms of Elementary Fcdnctions Note that it is a function of the sampling period T on ak. Let us obtain the z transform of x(k) as where a is a constant. Referring to the definition of t e z mnsform given by Equation (2-2), we obtain m w - z-- z - a on. Let us find the z transform of e-"', O S t x(t) = Since x(kT) = e-"kT, k = 0,1,2,. . . on. Consider the sinusoidal function sin ot, O 5 t x(t) = 10, t<O Noting that e;"' = cos ot +j sin ot e-;"' = cos ot - j sin ot
- 21. The z Transform Chap. 2 Since the z transform of the exponential function is 7[e-"'] = 1 & 1- e-aTZ-l we bave - 1 X ( z ) = ;S[sinwt] = ;S -(elw' - eiw')] l'j - z-l sin wT 1- 2z-l cos oT + z - ~ -- z sinoT z 2 - 2 z coswT+ 1 Obtain the z transform of the cosine function coswt, O C l t t < O If we proceed in a manner similar to the way we treated the z transform o£the sine function, we have x ( ~ )= Y wt] = $Y[e1." + e-]" 1 - 1 - z-' COS wT 1 - 2.2-' cos wT + z-' Obtain the z transform of Whenever a function in S is given, one approach for finding the corresponding z transform is to convert X(s) into x(t) and then find the z transform of x(t). Another approach is to expand X(s) into partial fractions and use a z transform table to find the z transforms of the expanded terms. Still other approaches will be discussed in Cec. 2-3 z Transforms of Elementary Functions The inverse Laplace transform of X e ) is ( t ) = 1- e O 5 t Hence, 1 1 X(Z)= Z[l - e-'] = -- 1- Z - l 1 - e-'Z-l en&. Just as in working with t lace transformatio*, a table transforrns of commonly encountered functions is very useful for solving problerns in the field of discrete-time systems. Table 2-1 is such a table.
- 22. The z Transform Chap. 2 x(t) = 0, for t < 0. x(1cT) = x(k) = 0, for lc < 0. Unless otherwise noted, 1: = 0,1,2,3,. . . . Cec. 2-4 lmportant Properties and Theorerns of the z Transform The use of the z transform method in the analysis of discrete-time control systems may be facilitated if theorems of the z transform are referred to. In this section we present importantproperties and useful theor the time function x is z-transformable al where a is a constant. To prove this, note thatm by definition m . The z transform possesses an Pmportant prop- erty: linearity. This means that, if f(k) and g(k) are z-transformable and a and P are scalars, then x(k) formed by a linear combination x(k) = af(W + 13gW where F(z) and G(z) are the z tr The linearity property can be X(z) = Z[x(k)] = Z[af(k) + pg(k)] This can be proved as follows:m m . The shifting theorern presented here is also ieferred to as on theorem. If x(t) = O for t < O and x(t) as Wz),
- 23. and The z Transform Chap. 2 where n is zero or a positive integer. To prove Equation (2-7), note that m By defining m = k - n, Equation (2-9) can be written as follows: m Sincex(mT) = O for m < O, we may change the lower limit of the summation from m = -n to m = 0. Thus, multiplication o£ a z transform by z-" has t e 0f delaying the time function x(t) by time nT. (That is, move the funct lo the right by time nT.) To p;ove-~quation(2-8)? we note that - For the number sequence x(k), Equation (2-8) can be written as follows: From this last equation, we obtain 7[ ~ ( k+ l)]= zX(z) - zx(Q) (2-1 1) Z[r(k + 211 = z l[x(k + 111 - zx(1) = z2X(Z) - z2x(0) - zx(1) (2-12) lrnportant Properties and Theorems of the z Trancform Similarly, Z[x(k + n)] = znX(z) - znx(0) - zn-'~(1)- zn-24 2 ) - .- - - zx(n-1) (2-13) where n is a positive integer. emember that multiplication of the z transform advancing the signal x(kT) by one step (1sarnpiing per of the z transform X(z) by z-' has the effect of delaying the signalx(kT) by one step (1sampling period). Find the z transforms of unit-step functions that are delayed by 1 sampling period and 4 sampling periods, respectively, as shown in Figure 2-2(a) and (b). Using the shifting theorem given by Equation (2-7), we have 1 - z-' r[i(t - TI] = Z-~Z[I(~)]= Z-l- 1 - *-1 - -1 - z-l (Note that z-' represents a delay of 1sampling period T, regardless of the value of T.) Obtain the z transform of o 2T 3T 4T 5T 6T 7T 8T ' Faginre2-2 (a) UnitSstepfunction delayed by 1sampling period; (b) unit- step function delayed by 4 sampling (b) periods.
- 24. The z Transform Weferring to Equation (2-7), we have Z [ x ( k - l)]= z-%(z) The z transform of ak is and so 1 - z-' z [ f ( a ) ]= Z[ak-']= z-'- - - 1-az-' 1-az-' where k = 1,2,3, .. . . Consider the function y(k),which is a sum of functions x(h),where h = 0,1,2, such that k where y ( k ) = O for k < O. Obtain the z transform of y (k). First note that Therefore, Z [ y ( k )- y(k - l ) ]= Z [ x ( k ) ] which yields Chap. 2 Cec. 2-4 lmportant Properties and Theorems of the z Transform where X ( z ) = 2' [x(k)]. If x(t) has the z transform X(z), then the z transform of e-"'x(t)can be given by X(zeuT).This is known as the complex trans- lation theorern. Noting that Z-' sin o T Z [sin wt] = 1 - 22-' coswT + we substitute zeaTfor z to obtain the z transform of e-"' sin wt, as follows: e-aT -1 z sinwT Z [e-"' sin wt] = 1 - 2e-aT~-'cos o T + e-2nT~-2 Similarly, for the cosine function, we have 1 - Z-' cos wT 2' [cos wt] = 1 - 22-' coswT + zW2 y substituting zeaTfor z in the z transform of cos wt, we obtain 1 - e-"Tz-' cos wT Z [e-"' cos ot] = 1 - 2e-aT~-'C O S ~ T+ e-2aTz-2 Obtain the z transform of te-"'. Notice that Thus, . If x(t)has the z transfor X ( z ) and if limX ( z )exists, then the initial value x(O) of x(t) or x(k) is given by z-t m x (O) = limX ( z ) 2 4 m is theorem, note that ketting z --+m in this last equation, we obtain Equation (2-15). The be signal in the neighborhood of t = O or k = O can thus be determined by the behavior of X ( z ) at z = m. The initial value theorem is convenient for checking z transform calculations for possible errors. ince x(0) is usually known, a check of the initiaI value by limX(z) can easily spot errors in X(z), if any exist. z-tm To prove this theorern, note that x "[e-"'x(t)] = 2~ ( k T ) e - " ~ ' z - ~= x(kT)(~e"')-~= X(zeaT) (2-14) k=O k=O Thus, we cee that replacing z in X ( z ) by zeaTgives the z transform of e-"'x(t). Determine the initial value x(0) if the z transform of x(t) is given by Examplie 2-6 By using the initial value theorem, we find Given the z transforms of sin wt and cos wt, obtain the z transforms of e-"' sin wt and e- "' cos wt , respectively, by using the complex translation theorem. x(0) = iim (1 - e-')z-' z-m(l - z-l)(l - CTz-')= O
- 25. The z Transforrn Chap. 2 Referring to Example 2-2, notice that this X ( z ) was the z transform of x(t) = 1 - e-' and thus x(0) = 0, which agrees with the result obtained earlier. ence, Taking the limit as z approaches unity, we have m 1 ecause of the assumed stability condition and t e condition that x(k) = O for k < 0, the left-hand side of this iast equation becomes m [x(k) - x(k - l)] = [x(O) - x(- l)] + [X (1)- x(O)] k=O t-[x(2) - x(l)l + - e * = X ( W ) = limx(k) k-+ m ence, limx(k) = lim [(l- z-1)X(z)] k-*m 1-2 1 which is Equation (2-16). The finalvalue theorem is very useful in determining the behavior of x(k) as k - + frorn its z transform X(z). Determine the final value x(m) of 1 X ( z ) = ----- - 1 1 - Z-1 1 - e - a T Z - 1 a > O by using the final value theorem. By applying the final value theorem to the given X(z), we obtain ec. 2-5 The Inverse z Transforrn Pt is noted that the given X ( z ) is actually the z transform of y substituting t = m in this equation, we have X ( W ) = lim (1 - e-"') = 1 f-m As a matter of course, the two resdts agree. . lin this section we have resented important roperties and theo- rems of the z transforrn that will e to be useful in solving many z transforrn problerns. For the purpose of conv nt referente, these important properties and theorems are sumrnarized in Table 2-2. y of the theoserns presented in this table are discussed in this section. cussed here but included in the table are derived or proved in Ap The z transforrnation serves the same role for discrete-time control systemsthat the Laplace transformation serves for continuous-ti control systems. For the z trans- form to be useful, we must be familiar with thods for finding the inverse z transform. The notation for the inverse z transform is Z-l. The inverse z transform of X(z) yields the corresponding time sequence x(k). Irt should be noted that only the time sequence at the sarnpling instants is obtained frorn the inverse z transform. Thus, the inverse z transform of X(z) yields a unique x(k), but does not yield a unique x(t). This rneans that the inverse z transform yields a time sequence that specifies the values of x(t) only at discrete instants of time, t = O, T,2T,. . . ,and says not ing about the values of x(t) at al1 other times. That is, many different ti e functions x(t) can have the sarne x(kT). X(z), the z transform of x(kT) or x(k), is given, the operation that onding x(kT) or x(k) is called the inverse z transformation. n obvious method for finding the inverse z transform is to refer to a z transform extensive table of z tra a sum of simpler z tr presented in this section.) Other than referring to z transform fables, four methods for obfaining the inverse z transform are commonly available:
- 26. The z Transform ec. 2-5 The Inverse z Transform Figure 2-3 Two different continuous-time functions, xl(t) and xZ(t),that have the same values at t = O, T, 2T,. ... Direct division method Gomputational method . Partial-fraction-expansion method nversion integral rnethod In obtaining the inverse z transform, we assume, as usual, that the time sequence x(kT) or x(k) is zero for k < 0. efore we present the four methods, however, a few comments on poles and zeros of the pulse transfer function are in order. h e . Hn engineering appiications of the z transform method, X(z) may have the form where the p,'s (i = 1,2,.. . ,n) are the poles of X(z) and the z,'s ( j = 1,2, .. . ,m) the zeros of X(z). The locations of the poles and zeros of X(z) determine the characteristics of x(k), the sequence of values or numbers. S in the case of the plane analysis of linear continuous-time control systems, we often use a graphical display in the z plane of the locations of the poles and zeros of X(z). Note that in control engineering and signal processing X(z) is frequently expressed as a ratio of polynomials in z-', as follows:
- 27. The z Transform Chap. 2 where z-' is interpreted as the unit del properties and theorems of the z tra expressed in terms of powers of z, as given ,or in terms of powers of z-', as given by Equation (2-18), e poles and zeros of als in z . For example, as poles at z = -1 and z = -2 and zeros at z = O and z = -0.5. If X(z) is written as a ratio of polynomials in z-', however, the prece be written as Although poles at z = -1and z = -2 and a zero at z = -0.5 are clearly seen from e expression, a zero at z = O is not explicitlyshown, and so the beginner may fail see the existence of a zero at z = O. Sherefore, in dealing with the pol of X(z),it is preferable to express X(z) as a ratio of polynomials in z, polynomials in z-l. In addition, in obtaining the inverse z transform by use o inversion integral method, it is desirable t Tres§ as a ratio of ~ o l ~ n o m i a l ~ in z, rather than polynomials in z-', to av anY possible erKJrs determininb the number of poles at the origin of function X(z)zk-l. . In the direet division rnethod we obtain the inverse z transform by expanding X(z) into an infinite power series in z-l. This method is useful when it is difficult to obtain the closed-form expression for the inverse z transform or it is desired to find only the first several terms of x(k). The direct division rnethod stems from the fact that if X(z) is expanded into a power series in z-', that is, if then x(kT) or x(k) is the coefficient of the zwkterm. ence7 the vahes of 4 k T ) 01 x(k) for k = 0,1,2, . . . can be determined by inspectlon. If X(z) is givenin the form of a rational function,the expansion into an infinite power series in increasing powers of z-' can be accomplished by simply dividing the numerator by the demminator, where both the numerator and denominato are wrihten in increasing pdswers of z-l. lif the resulting series is conver Sec. 2-5 The lnverse z Transform coefficients of the zWkterm in the series are the values x(kT) of the time sequence or the values of x(k) of the number sequence. h the present rnethod ,x(l),x(2), ... in a se an expression for the general term from a set of values of x(kT) or x(k). Find x(k) for k = 0,1,2,3,4when X(z) is given by First, rewrite X(z) as a ratio of polynomials in z-', as follows: Dividing the numerator by the denominator, we have 18.68~-~- 22.416~-~+ 3 . 7 3 6 ~ ~ ~ Thus, By comparing this infinite series expansion of X(z) with X(z) = we obtain x(0) = o As seen from this example,the direct division method may be carried out by hand calculations if only the first several terms of the sequence are desired. In general, the method does not yield a closed-forrn expression for x(k), except in special cases. Find x(k) when X(z) is given by 1 - z-lX(z) = -- - z + l 1 + z - l By dividing the numerator by the denominator, we obtain
- 28. The r Transform Chap. 2 By comparíng this infinite series expansion of X ( r ) with X ( z ) = obtain x(O) = O x(1) = 1 x(2) = -1 4 3 ) = 3 n(4) = -1 This is an alternating signal of 1and -1, which starts from k = 1. Figure 2-4 shows a plot of this signal. Obtain the inverse z transform of X ( z ) = 14- 22-1 + 3ze2+ 4t-3 The transform X ( z ) is aiready in the form of a power series in z has a finite number of terrns, it corresponds to a signal of finite length. we find x(0) = 1 x(1) -2 4 2 ) = 3 x(3) = 4 Al1 other x(k) vaiues are zero. En what follows, we present two corn proaches to obtain the inverse z transform. Difference equation approach Consider a system G(z) defined by n finding the inverse z transform, we utilize the Kronecker delta function So(kT), wbere -1 0 4 Figure 2-4 Alternating signal of 1and -1 starting from k = 1. Sec. 2-5 The Inverse z Transform = O, for k f O Assume that x(k), the input to t e system G(z), is the Kronecker delta input, or x(k) = 1, for k = O = 0, for k f O The z transform of the Kronecker delta input is X(z) = 1 Using the Kronecker delta input, Equation (2-19) can be rewritten as Approach. MATLA can be used for finding t transforrn. Weferring to -20), the input X(z) is the z transform of the Kronecker delta input. the Kronecker delta input is given by where N corresponds to the end of the discrete-time duration of the process consid- ered. Since the z transforrn of the Kronecker elta input X(z) is equali to unity, the response of the system to this input is ce the inverse z transform of G(z) is given by y(O),y (l),y(2), .. ..Let us obtain up to k = 40. To obtain the inverse z transform of G(z) with follows: Enter the numerator and denominator as follows: num = [O den = [l Enter the Kronecker delta input. X = Then enter the command Y = to obtaán the response y(k) from k = O tcs k = 40.
- 29. The z Transform Chap. 2 e inverse z transforrn or the response to the Kroneck rograrn 2-1. If this program is executed, the screen will show the output y(k) from k = O to 40 as follows: Y = Columns 1 through 7 O 0.4673 0.3769 0.2690 0.1632 0.0725 0.0032 Columns 8 through 14 -0.0429 -0.0679 -0.0758 -0.0712 -0.0591 -0.0436 -0.0277 Columns 15 through 21 -0.01 37 -0.0027 0.0050 0.0094 0.01 11 0.01O8 0.0092 Columns 22 through 28 0.0070 0.0046 0.0025 0.0007 -0.0005 -0.001 3 -0.001 6 Columns 29 through 35 -0.001 6 -0.001 4 -0.001 1 -0.0008 -0.0004 -0.0002 0.0000 Columns 36 through 41 0.0002 0.0002 0.0002 0.0002 0.0002 0.0001 computations begin from column 1and end at column 41, rather than from column O to column 40.) These values give the inverse z transforrn of G(z). That is, Y (0) = 0 y(1) = 0.4673 y(2) = 0.3769 -5 The lnverse z Transform 33 plot the values of t e inverse z transfor given in the folliowing. onding plot is skown in Figure 2-5. Response to Kronecker Delta Input re 2-5 Response of the system defined by Equation (2-20) to the Kronecker delta input.
- 30. 1f oints (open circles, o) need to plot(k,y,'ol)to plot(k,y,'o',k,y,'-'). we can convert this equation into t k) = O for k f O, (O) and y(l) can be k = -2 into Eqaiation (2-21), we fin$ Finding the inverse z transfor of Y(z) IXxV b3=2o following difference equation for y(k): the initial data y (O) = ) = O for k # O. ation (2-22) can be solv ,FBRTRAN, or other. presented here, which is paralle Laplace transformation, is widely used in The method requires that all terms in recognizable in the table sf z tr To find the invesse z eran Sec. 2-5 The Inverse r Transforrn efore we discuss the partial-fraction-expansion method, we shall review the shifting theorem. Consider fhe following X(z): By writing zX(z) as Y ( z ) ,we obfain 1 zX(z) = Y ( z )= -1 - az-' Referring to Table 2-1, the inverse z transforrn of Y ( z )can be obtained as follows: T 1 [ Y ( z ) ]= y(k) = ak Hence, the inverse z transform of X ( z ) = z-' Y ( z )is given by 2-'[X(z)] = x(k) = y(k - 1) Since y(k) is assumed to be zero for al1 k < O, we have Consider X(z) as given by To expand X(z) into partial fractions, we first factor t of X(z) and find the poles of X(z): then expand X(z)/z into partial fractions so t at each term is easily recognizable in a table o£ z transforms. If the shifting theorem is utilized in transforms, however, X(z), instead of X(z)/z, may be expanded tions. The inverse z transform of X(z) is obtained as the sum transforms of the partial fractions. A comrnonlyused procedure for the cas here allthepoles are o£simpleorder and there is at least one zero at the origin ( t is, bm = O) is to divide both sides o£X(z) by z and then expand X(z)/z into partial fractions. OnceX(z)/zis expanded, it will be of the form The coefficient aican be determined by multiplying both sides of this last equation by z - p, and setting z = pi.Thiswill result in zero for allthe terms on the right-hand side except ehe ai term, in which the multiplicative factor z - pihas been canceled by the denominator.
- 31. The r Transform Chap. 2 Note that such determination of a, is valid only for sim olves a multiple pole n X(z)/z will have t The coefficients cl and c2are determined from Tt is noted that if X(z)/zinvolves a triple must include a term (z +pJ(z - ~ 7 ~ ) ~ .( Given the z transform where a is a constant and T is the sampling period, determine the inverse z transform x(kT) by use of the partial-fraction-expansion method. The partial fraction expansion of X(z)/z is found to be 1X o = L - - z z - 1 z - e-aT Thus, From Sable 2-1 we find r 1 Hence, the inverse z transform of X ( z ) is x(kT) = 1 - k = 0,1,2,. .. Let us obtain the inverse z transform of by use of the partial-fraction-expansion method. We may expand X ( z ) into partial fractions as follows: Sec. 2-5 The lnverse z Transform Noting that the two poles involved in the quadratic term of this last equation are complex conjugates, we rewrite X ( z ) as follows: Since 1 - e-aTz-' cos wT 2' [e-"" TOS wkT] = 1 - 2eWaTz-'cos o T + e-2aTz-2 e - a T z - l sin wTZ [e-"" sin wkT] = 1 - 2e-aT~-'cos o T + e-2aTz-2 by identifying e-"T = 1 and cos oT = $ in this case, we have wT = ~ 1 3and sin wT = 1/3/2. Hence, we obtain - - and Thus, we have ( k - 1 ) ~ 1 (k - 1 ) ~ x ( k ) = 4(lkP1)- 3(lk-') c o s + (1.-') sin-- 3 Wewriting, we have The first severa1values of x(k) are given by Note that the inverse z transform of X ( z ) can also be obtained as follows: Since
- 32. and y-1 I z-l krr m 1 - z - 1 + z - 2 ]= ;sí(1.) sin-3 krr 4 ( k - l ) r r - 2 ~ ?sin- + -sin- 3 v'3 3 ' k = l , 2 ,3,. . x ( k ) = k 6 0 Although this solution may look difierent from the one obtained earlier, both solutions are correct and yield the sarne values for x(k). Pn is a useful techniq verse z al for the ztra~sfo its center at the srigin of t of complex variables. It c m [residue o£x(z)z"-' at pole z = ziaf (2-24) i=l (2-25) Ef X(z)zk-' contains a mult Sec. 2-5 The Inverse z Transform partial-fraction-expansion method rnay prove lo be simpler to apply. On the other hand, in certain problems the partial-fraction-expansion approach may become laborious. Then, the inversion integral method proves to be very convenient. Qbtain x(kT) by using the inversion integral method when X ( z ) is given by Note that For k = O, 1,2,. . .,X(z)zk-' has two simple poles, z = zl = 1 and z = z2 = e-OT. Hence, from Equation (2-24), we have x(k) = -- Kl = -- K2 = -- Hence, [residue of (1 - e-OT)zk at pole z = z, 1 = 1 ( Z - 1)(z - ePaT) 1 where [residue at simple pole z = 11 [residue at simple pole z = e-OT] Qbtain the inverse z transform of by using the inversion integral rnethod. Notice that For k = 0,1,2, .. . ,X(z)zk-' has a simple pole at z = zl = e-"Tand a double pole at z = z2 = 1. Hence, from Equation (2-24), we obtain Z k t l x ( k ) = [residue of at pole z = z, I = I ( z - I ) ~ ( z- e-OT) = Ki + KZ 1
- 33. where The z Transform M, = [residue at simple pole z = e-."? .k+l ]= ,-a(k+l)T (Z - ~ ) ~ ( z- e-"') (1 - e-aT)2 K2 = [residue at double pole z = 11 (k + l)zk(z - e-aT)- zk+l = lim z-. I ( z - ePaT)' Difference equations can be solved easily by use o£a nurnerical values of al1coefficients and parameters are expressions for x(k) cannot be obtained fr com~uters0lLhX-l7 e*cePt forvepY special cases. The usefulness of the z "can ethod 6t h t it enables tQ the clased-form expressisn for x(k). Consider the linear time-invariant discrete-time system characterized by the following linear difference equation: where u(k) and x(k) are the systern's input and output, respedively, at the kth iteration. In describing such a difference equation in the z plane, we take the z transform of each term in the equation. Eet us define q x(k)]= X(z) Thenx(k + l),x(k + 2),x(k + 3),.. . andx(k - l),x(k - 2),x(k - 3),...can be expressed in terms of X(z) and the initial conditions. Their exact z transforms were derived in Section 2-4 and are surnrnarized in Table 2-3 for convenient referente. Next we present two exarnple problems for solvingdifference equations by t z transfcwm method. -6 P Transform Method for P TRANSFORWLS OF x(k + m) AND x(k - m) Discrete function z Pansform 1 Soive the following difference equation by use of the z transform method: x ( k + 2 ) + 3 x ( k + I ) + & ( k ) = O , x(O)=O, x ( l ) = l First note that the z transforms of x(k + 2), x(k + l j , and x ( k ) are given, respectively, by Z [ x ( k + 2)] = z 2 X ( z )- z2x(0)- zx(1) Z [ x ( k + 1)]= z X ( z ) - zx(0) Z[x(k)]= X ( z ) Taking the z transforms of both sides of the given difference equation, we obtain z2X(zj - z2x(0)- zx(1) + 3zX(z) - 3zx(O) + 2%(z) = O Substituting the initial data and simplifying gives 1 1=--- 1 + z-' 1 + 22-' Noting that Obtain the solution of the following difference equation in terms of x(0) and x(1): x(k + 2) + (a + b)x(k + 1) + abx(k) = O where a and b are constants and k = 0,1,2,. . . .
- 34. The z Transform Chap. 2 ?'he z transform of this difference equation can be given by [z2X(z)- z2r(0)- rx(l)]+ (a + b)[zX(z)- zx(O)] + abX(z) = O Solving this last equation for X(z) gives [z2+ (a + b)z]x(O)+ zx(1) X ( z ) = z2 + (a + b)z + ab Notice that constants a and b are the negatives of the two roots of the characteristic shall now consider separately two cases: (a) a # b and ( (a) For the case where a # b , expanding.X(z)/zPto partial fractions, we obtain from which we get The inverse z transform of X ( z ) @ves bx(0) +x(1) ax(0) + x(1) x ( k ) = (-a)" + (-b)k, a # b b - a a - b where k = 0,1,2, .. . . ) For the case where a = b , the z transform X(z) becomes The inverse z transforrn of X ( z ) gives x(k) = ~ ( 0 ) ( - a ) ~+ jax(0) +~ ( l ) ] k ( - a ) ~ - l , a = b where k = 0,1,2,. . .. In this chapter the basic theory of the z transform rnet een presenteda T'he z transform serves the same purpose for linear time-i Saek-tirne sY5&Xls asthe Laplace transform provides for linear time-invariant continuous-time systems. method of analyzing data in discrete time results in difference z transforrn rnethod, linear time-invariant difference equations into algebraic equations. This facilitates the transient response analysis of the digital control system. Also, the z transform e t k d all0VJsL E t* We Chap. 2 Example Problemc and Solutions conventional analysis and design techniques available to analog (continuous-time) control systems, such as the root-locus technique. Frequenc design can be carrie rting the z plane into z-transforrned char as the Sury stability 3 and 4. Obtain the z transform of Gk,where G is an n x n constant matrix. y definition, the z transform of G ~ S m h[ G ~ ]= E~ ~ z - ~ k=O = 1 + GZ-' + G ~ ~ - " + 3 z - 3 + ... Note that G~can be obtained by taking the inverse z transform of ( or (21 - G)-' Z . That is, Obtain the z transform of k2 y definition, the t transform of k2is m Z [ k 2 ]= 2 k2z-k = Z-' + 4z-2 + 9z-3 + l6zp4+ e k=O = z-'(1 + z-l)(l + 32-' + 6z-2 + IOZ-~+ 1 5 ~ - ~+ ) ere we have used the closed-form ex involved in the problem. (See Appendix Obtain the z transforrn of kak-' by two methods. Method l. By definition, the z transform of kak-' is given by m
- 35. ethod 2. The summation expression for the z transform of kak-' can also be written as follows: -Ir )7[kak-ll = 5kak-lz-k = a-1 k a k z k = 2 (a)k=O k=O ak=O o~blemA-2- Show that and Also show that where 15 i 5 k - 1. Solintaon Define k y(k) = 2x(h), k = O, 1,2,. . . h =O Then, clearly Y ( k ) - Y (k - 1) = x ( k ) y writing the z transforms of x ( k ) and y(k) as X ( z ) and Y ( z ) ,respectively, and by taking the z transform of this last equation, we have Y ( z ) - z-1 Y ( z ) = X ( z ) Chap. 2 Exarnple Problerns and Solutions By using the final value theorem, we find k- m 2 ~ ( h )= Ex ( k ) = limX ( z ) h=O k=O 2-1 Next, to prove Equation (2-29), first define k j ( k ) = h=iEx(h) = x(i) + x(i i- 1) + + x ( k ) where 1 r i r k - 1. Define also X ( z ) = x(i)zWi+ x(i + 1)z-'"" + .- + x(k)z7*+ - - Then, noting that we obtain 1-1 Since the z transform of this last equation becomes Y ( Z ) - z-' P(z) = Z ( z ) [Note that the z transform of x(k),which begins with k = i, is X(z),not X(z).]Thus, 1 B ( z ) = -1 - 2-1 [,(Z) - h = ~i ( h ) z h1-2-5 Obtain the z transform of the curve x(t)shawn in Figure 2-6. Assume that the sampling period T is 1 sec. ution From Figure 2-6 we obtain x(0) = O x(1) = 0.25
- 36. The z Transform Chap. 2 Then the z transform of x(k) can be given by Notice that the curve x(t) can be written as ~ ( t )= $ t - g t - 4j1~t- 4) where I(t - 4) is the unit-step function occurring at f = 4. Cince the sampling period T = 1sec, the z transform of x(t) can also be obtained as follows: X(z) = Z [x(t)] = Z [$f ] -7[a(t - 4)l(t - 4)] Consider X(z), where Obtain the inverse z transform of X(z). Sgalntigan We shall expand X(z)/z into partial fractions as follows: ap. 2 Example Then The inverse z transforms- of the individual- terms give and therefore ~ ( k ) = 9 k ( 2 ~ - ' ) - 2 ~ + 3 , k = 0 , 1 , 2,... Obtain the inverse z transform of olution Expanding X(z) into partiaf fractions, we obtain (Note that in this example X(z) involves a double pole at z = O. Hence the partial fraction expansion must include the terms ll(z2) and llz.] By referring to Table 2-1, we find the inverse z transform of each term of this last equation. That is, ence, the inverse z transform of X(z) can be given by o - o - o = o , k = O 1 - o - 1 = 0 , k = 1 x(k) = 2 - 1 - 0 = 1 , k = 2 2k-'-O-O=2k-19 k = 3 , 4 , 5 , ... Rewriting, we have k = 0 , 1 /O9 k = 2x(k)= 1, 2k-1, k = 3 , 4 , 5,... Toverify this result, the direct division method rnay be applied to this problem. Noting that
- 37. The z Transforrn Chap. 2 we find Obtain the inverse z transform of z-2 > X(z) = (1 - z-1 >3 utionr The inverse z transform of ~ - ~ / ( 1- z-')~is not available from most z transform tables. It ispossible, however, to write the givenX(z) as a sum of z transforms that are commonly available in z transform tables. Since the denominator of X(z) is (1 - z-')~and the z transform of k2is z-l(l + z-')!(1 - z-')~,let us rewrite X(z) as from which we obtain the following partial fraction expansion: The z transforms of the two terms on the rigkt-hand side of this last equation can be found from Table 2-1. Thus, It is noted that if the given X(z) is expanded into other partial fractions then the inverse z transform rnay not be obtained. As an aiternative approach, the inverse z transform of X(z) may be obtained by use of the inversion integral method. First, note that Hence, for k = O, 1,2, .. .,X(z)zk-' has a triple pole at z = 1.Referring to Equation (2-24), we have ! zk x(k) = residue of ----at triple pole z = 1 (Z - 113 Chap. 2 Exarnple Probierns and =- l lirn? (z - 1)3- (3 - q !z - ~ dzd2 ' (Z -zk113I Using the inversion integral method, obtain the inverse z transform of Sola~tion Note that For k = 0, notice that X(z)zk-' becomes Hence, for k = O, X(z)zk-' has t h ~ e esimple poles, z = zl = 1, z = z2 = 2, and z = z3 = O. For k = 1,2,3, . .. ,however, X(z)zbl has only two simple poles, z = zl = 1 and z = z2= 2. Therefore, we must consider x(0) and x(k) (where k = 1,2,3,...) separately. For k = O. For this case, referring to Equation (2-24), we have lo at pole z = zi (Z - 1)(z - 2)z 1 where K1 = [residue at simple pole z = 11 = -10 (2 - 1)(z - 2)z K2= [residue at simple pole z = 21 z-2 (z - 1)(z - 2)z K3 = [residue at simple pole z = O]
- 38. The r Transform ence, x(0) = K1 + K2 + K3 = -10 + 5 + 5 = O For k = 1,2,3, . . . . Eor this case, Equation (2-24) becomes 10zk--' at pole z = zi ( z - 1)(2 - 2) = Ki + K2 where K1 = [residue at simple pole z = 11 K2 = [residue at simple poIe z = 21 Thus, x(k)=K1+K2=-10+10(2k-1)=10(2k-1-1), k = 1 , 2 9 3 9 Mence, the inverse z transform of the given X ( z ) can be written An alternative way to write x(k) for k 2 O is x(k) = 560(k) + 10(2~-'- l ) , k = 0,1,2,... where &(k) is the Kronecker delta function and is given by for k = 0 "(w=jt: t o r k i o '4-2-1 Obtain the inverce z transform of Chap. 2 (2-30) by use of the four methods presented in Section 2-5. utiora Method 1: Direcr division method. We first rewrite X ( z ) as a ratio of two polynomiak in z-': Dividing the numerator by the denominator, we get X ( Z )= 1 + 4 ~ 4+ 7 ~ - 2+ ~ o z - ~+- . e . ence, x(0) = 1 Chap. 2 Example roblemc and Solutions Method 2: Compuiational method (MATLAB approach). %(z) can be written as Hence, the inverse z transform of X ( z ) can be obtained with Define num = [l 2 O] d e n = [ l -2 11 If the values of x(k) for k = O,1,2, . . . ,30 are desired, then enter the Kronecker delta input as follows: Then enter the command Program 2-3. [The screen will show the output x(k) from k = 0 té, k = 30.1 (MATLAB computations begin from column 1 and end at column 31, rather Columns 13 through 24 37 40 43 46 49 52 55 58 61 64 67 70 Columns 25 through 31
- 39. The z Transform Ckiap. 2 than from column O to column 30.) Thevalues x(k)givethe inverse z transform of X(z). That is, x(0) = 1 x(1) = 4 x(2) = 7 Method 3: Partial-fraction-expansion method. expand X ( z ) into the foliowing partial fractions: Then, noting that we obtain x(0) = 1 x ( k ) = 3 k + l , k = 1 , 2 , 3 ,... which can be combined into one equation as follows: x ( k ) = 3 k + l , k = 0 , 1 , 2,... Note that if we expand X ( z ) into the following partia1 fractions then the inverse z transform of X ( z ) becomes x(0) = 1 which is the same as the result obtained by expanding X ( z ) into the other partial fractions. [Remember that X ( z ) can be expanded into different partial fractions, but the final result for the inverse z transform is the same.] Method 4: Inversion integral rnethod. First, note that Chap. 2 Exarnple Problerns and For k = O, 1,2, ... ,X(z)zk-' has a double pole at z = P. (2-24), we have ( 2 + 2)zk residue of ----- (2 - 1)2 at double pole z = 1 Thus, - - x ( k ) = -----i i m - [ ( z - I ) 2 W ] (2 - l ) !Z-i dz d = lim-[(zZ-+I dz + 2)zk] = 3 k + 1 , k = 0 , 1 , 2,... Solve the following difference equation: 2x(k) - 2x(k - 1) + x(k - 2) = u(k) where x(k) = O for k < O and y taking the z transform of the given difference equation, 1 2X(z) - 22-l X ( z ) + Z - ~ X ( Z )= -1 - z-' Solving this last equation for X(z), we obtain Expanding X ( z ) into partial fractions, we get Notice that the two poles involved in the quadraticterm in this last equation are compiex conjugates. Hence, we rewrite X ( z ) as follows: y referring to the formulas for the z transforms of damped cosine and damped sine functions, we identify e-2aT= 0.5 and cos oT = 11V5 for this problem. Hence, we get wT = ~ 1 4 ,sin wT = l i d , and e-"= = l i d . Then the inverse z transform of X ( z )can be written as x ( k ) = 1 - Se-akTcos wkT + $ewakTsin wkT from which we obtain
- 40. The z TTansform Chap. 2 Consider the difference equation x(k + 2) - 1.3679x(k + 1) + 0.3679x(k) = 0.3679u(k + 1) + 0.2642u(k) where x(k) is the output and x ( k ) = O for k 5 O and where u ( k ) is the input and is given by u ( k ) = O , k < O u(0) = 1 u ( l ) = 0.2142 u(2) = -0.2142 u ( k ) = O , k = 3 , 4 , 5 ,... Determine the output x(k). olintlon Taking the z transform of the given difference equation, we obtain [ z 2 X ( z )- z2x(0)- zx(1)l - 1.3679[zX(z)- zx(O)]+ 0.3679X(z) = 0.3679[zU(z)- zu(0)l + 0.2642U(z) (2-31) By substituting k = -1 into the given difference equation, we find x(1) - 1.3679x(O) + 0.3673x(-1) = 0.3679u(O) + 0.2642u(-1) Since x(0) = x(- 1) = O and since u(- 1) = O and u(0) = 1, we obtain x(1) = 0.3679u(O) = 0.3679 By substituting the initial data x(0) = O, x(1) = 0.3679, u(0) = 1 into Equation (2-31), we get ~ " ( 2 )- 0.36792 - 1.3679zX(z) + 0.3679X(z) = 0.3679zU(z) - 0.36792 + 0.2642U(z) Solving for X(z), we find The z transforrn of the input u ( k ) is U ( Z )= Z [ u ( k ) ]= 1 + 0.21422-' - 0 . 2 1 4 2 ~ - ~ Chap. 2 Example roblems and Solutions ence, Thus, the inverse z transform of X ( z ) gives Consider the difference equation where x(0) = O and x(1) = 1. Note that x(2) = 1,x(3) = 2,x(4) = 3, .... The series U, 1,1,2,3,5,8,13, ... is known as the Fibonacci series. Obtain the general solution x ( k ) in a closed form. Show that the limiting value of x(k + l)/x(k)as k approaches infinity is (1 + 1/5)/2, or approximately 1.6180. olution By taking the z transform of this difference equation, we obtain olving for X ( z ) gives By substituting the initial data x(0) = O and x(1) = 1into this last equation, we have The inverse 2 transform of X ( z ) is Note that although this lact equation involves '~6the square roots in the right-hand side of this last equation cancel out, and the values of x ( k ) for k = 0,1,2, . .. turn out to be positive integers.
- 41. The z Trancform Chap. 2 The limitingvalue of x(k + l)lx(k)as k approachesinfinity isobtained asfollows: Since 1(1 - x(k i1) lim - k-+m ~ ( k ) Hence, ( 1 + v5"" x(k + ') 1 - - 1.6180lim -= lim k - w x(k) 2 Referring to Problem A-2-13, write a ATLAB program to generate the Fibonacci series. Carry out the Fibonacci series to k = 30. Solistiora The z transform of the difference equation is given by Solving this equation for X ( z ) and substituting the initial data x(0) = O and x(1) = 1, we get The inverse ztransform of X ( z ) will give the Fibonacci series. To get the inverse z transform of X(z),obtain the response of this system to the Kronecker delta input. MATLAB Program 2-4 wilf.yield the Fibonacci series. % ***** The Fibona % response of X(z) t O/O x(Z) = 2/(2"2 - z - 1) ***** num = [O 1 01; den = [l -1 -41; u = [l zeros(1,3O)l; x = fiIter(num,den,d Chap. 2 Example Problemc and Solutionc The filtered output y shown next gives the Fibonacci series. X = Columns 1 through 6 o 1 1 2 3 5 Columns 7 through 12 8 13 21 34 55 89 Columns 13 through 18 144 233 377 61O 987 1597 Columns 19 through 24 2584 4181 6765 10946 17711 28657 4301umns 25 through 30 46368 75025 121393 196418 317811 514229 , Column 31 1 832040 Note that column 1 corresponds to k = O and column 31 corresponds to k = 30. The Fibonacci series is given by x(0) = 0 x(29) = 514,229 x(30) = 832,040 Consider the difference equation x(k + 2) + m ( k + 1) + ,4?x(k)= O (2-32)
- 42. The z Transform Chap. 2 Figure 2-7 Region in the cup plane in which the solution series of Equation (2-32), subjected to initial conditions, is finite, Find the conditions on a and P for which the solution series x ( k ) for k = 0,1,2, ..., subjected to initial conditions, is finite. Then, referring to Example 2-19, the solution x ( k )for k = 0,1,2, ... can be given by The solution series x(k) for k = 0,1,2,. . . , subjected to initial conditions x(O) and x(l),is finite if the absolute values o£a and b are less than unity. Thus, on the ap plane, three critica1points can be located: The interior of the region bounded by lines connecting these points satisfies the con- dition /a/< 1,lb1 < 1. The boundary lines can be given by ,6 = 1, a - P = 1, and a! -t- p = -1. See Figure 2-7. If point ( a ,p) lies inside the shaded triangular region, then the solution seriesx ( k )for k = 0,1,2, ...,subjected to initial conditions x(0) and x(l), is finite. Obtain the z transform of where a is a constant. -2 Obtain the z transform of k3. Obtain the z transform of t2e-"'. Obtain the z transform of the following x(k): x ( k ) = 9k(2k-1)- 2k + 3, k = 0,1,2,. . Assume that x ( k ) = O for k < 0. -5 Find the z transform of where a is a constant. -2- Show that -2-7 Qbtain the z transform of the curve x(t) shown in Figure 2-8. -2- Obtaln the inverse z transform of
- 43. The z Transform Chap. 2 Find the inverse z transfornl of Use (1)the partial-fraction-expansion method and (2)the a MATLAB program for finding x(k), the inverse z transform of X(z). Given the z transform z-' X ( z ) = (1 - z-')(l + 1.32-' + O.4zw2) determine the initial and final values of x(k). Also find x(k),the inverse z transform of X(z), in a closed form. 1 Obtain the inverse z transform of Use (1)the invewion integral method and (2)the -12 Obtain the inverse z transform of z - ~ X ( z ) = (1 - 2-l)(1 - 0.22-l) in a closed form. By using the inversion integral method, obtain the inverse z transform of -2-1 Find the inverse z transform of Use (1)the direct division method and (2) the MATLAB method. Obtain the inverse z transform of by use of the inversion integral method. Chap. roblerns -1 Find the solution of the following difference equation: x(k + 2) - 1.3x(k + 1) + 0.4x(k) = u(k) where x(0) = x ( l ) = O and x ( k ) = O for k < O. For the input function u(k),consider the following two cases: and u(0) = 1 u(k) = O, k # O Solve this problem both analytically and computationally with MATLA -2-17 Solve the following difference equation: x(k + 2) - x ( k + 1) + 0.25x(k) = u(k + 2) where x(0) = 1and x ( l ) = 2. The input function u(k) ic given by u ( k ) = 1 , k = 0 , 1 , 2,... Solve thic problem both anaiytically and computationally with Consider the difference equation: x(k + 2) - 1.3679x(k + 1) + 0.3679x(k) = 0.3679u(k + 1) + 0.2642u(k) where x ( k ) = O for k 5 O. The input u(k) is given by u ( k ) = O , k < 0 u(0) = 1.5820 u(1) = -0.5820 u(k)=O, k = 2 , 3 , 4,... Determine the output x(k). Solve this problem both analytically and computationally with MATLAB.
- 44. The z transform method is particularly useful for an input-single-output lineartime-invariantdiscrete-time presents background material necessary for the analysis and desig control systemsin the zplane. The main advantage of the z transf it enables the engineer to apply conventional continuous-time design methods to discrete-time systems that may be partly discrete time and partl Throughout this book we assume that the sampling operation is uniform; that is, only one sampling rate exists in the system and the sampling period is constant. If a discrete-time control system involves two or more samplers in the system, we assume that al1 samplers are synchronized and have the same sarnpling rate or sarnpling frequency. er. The outline of this chapter is as follows. Section 3-1 gives int ks. Section 3-2 presents a rnethod to treat the sampling operation as a mathernatical representation of the operation of taking samplesx(kT) from a continuous-time signal x(t) by impulse modulation. This section includes derivations of the transfer functions of the zero-order hold and first-order hold. Section 3-3 deals with the convolution integral method for obtaining the z transform. Reconstructing the original continu time signal from the sampled signal is the main subject matter of Sectlon 3-4. ed on the fact that the Laplace transform of the sampled slgnal is periodic, resent the sampling the~rem. ection 3-5 discusses the pulse transfer functio thematical modeling of digital controllers in terms of pulse transfer functions is discussed. Section 3-6 treats the realization of digital controllers and digital filters. ec. 3-2 impulse ampling and Data Discrete-time control systems may opera in discrete time and partly in continuous time. Thus, in such control syst signals appear as discr functions (often in the form of a sequence o or a nurnerical code) a signals as continuous-time functions. In analyzing discrete-ti plays an important role. To see why the z transform method is shall consider a fictitious S er cornmonlycalled an impu f this sarnpler is consider that begins with t = O, with the sampling period equal to Tan pulse equal to the sampled value of the continuous-ti pictorial diagrarn of the i sented by an arrow with an Tke impulse-sampled output is a sequence of impulses, with the st~engthof ulse equal to the magnitude of x(t) at the ding instant of time. kT, the impulse is x(kT)6(t - e that 6(t - kT) = O hall use the notation x"(t) to r signalx*(t), a train of impulses, infinite summation m x*(t) = x(kT)S(t - kT) k=O or shall define a train of unit impulses as ST(t),or k=O The sampler output is equal to the product of t e continuous-time input x(t) and the train of unit impulses OT(t).Consequently, the sampler may be considered a rnodu- lator with the input x(t) as the modulating signal and the train of unit impulses &(t) as the carrier, as shown in Figure 3-2. x * ( t ) X(S) 6r x"(s) Figure 3-1 Impulse sampler.
- 45. z-Plane Analycis of Discrete-TimeControl Cysterns Chap. 3 Carrier signal 1 1 Figure 3-2 Impulse sampler as a modulator. Next, consider the Eaplace transform of Equation (3-1): X"(s) = 2[x"'(t)] = x(O)%[6(i)] + x(T)Ce[G(t - T)] +x(2T>ce[8(t- 2T)] + = x(O) + x ( T ) ~ - ~ '+ ~ ( 2 T ) e - ~ ~ '+ . m = ~ ( k T ) e - ~ " k=O Notice that if we define eTs -- z then Equation (3-2) becornes I m The right-hand side of Equation (3-3) ic exactly the same as the right-hand side of t is the z transform o£the seque e x(O),x(T),x(2T), . .. ,gener- t = kT, where k = 0,1,2,. . .. ence we may write and Equation (3-3) becomes Note that the variable z is a cornplexvariable and Tis the sarnplingperiod. [ht should be stressed that the notation X(z) does not signify X(s) with s replaced by z, but ratber X' (S = T.-' Ira z).] Cec. 3-2 Impulse Sarnpling and Data Hold . Let us summarizewhat we have just signal x(t) is impulse sampled in a periodic manner, signal may be represented by m x*(t) = 2x(t)s(t - kT) k=O converts a continuous-ti S occurring at the sampling instants t = 0, T,2T,. .., od. (Note that between any no information. Two signals sampled signal.) Data-hold is a process of generating a continuous-time signal h(t) from a discrete-time sequence x(kT). A hold circuit converts the sampled cignal into a continuous-time signal, which approximately reproduces the signaf applied to the sampler. %hesignalh(t) during the time interval kT 5 t < (k + 1)Tmaybe approx- imated by a polynomial in r as follows: where O 5 T < T. Note that signal h(kT) must equal x(kT), or ence, Equation (3-5) can be written as follows: Hf the data-hold circuit is an nth-order polynomial extrapolator, it is called an nth-order hold. Thus, if n = 1, it is called a first-order hold. [The nth-order hold uses the past n + 1discrete data x((k - n)T),x((k - n + 1)T),. ..,x(kT) to gen- erate a signal h(kT t r).] ecause a higher-order hold uses past sarnples to extrapolate a continuous- nal between the present sampling instant and the next sampling instant, the accuracy of approximating the ntinuous-time signal improves as the number of past samples used is increased. wever, this better accuracy is obtained at the cost of a greater time delay. Hn closed-loop cont ystems, any added time delay in the loopwill decrease the stability of the system in some casesrnay even cause system instability. The simplest data-hold is obtained when n = O in Equation (3-6), that is, when
- 46. z-Plane Analysis af Discrete-Tirne Control ystems Chap. 3 Zero-order hold Figure 3-3 Sampler and zero-order hold where O I( r < T and k = 0,1,2, .. ..Equation (3-7) implies that the circuit holds the arnplitude of the samplefrom one sampling instantto the next. Such a data-hold is cailled a zero-arder hold, or clamper, or staircase generator. The output of the zero-order hold is a staircase function. In this book, unless otherwise stated, we assume that the hold circuit is of zero order. It will be seen later that the transfer function Ghof the zero-order hold may be given by 1 - e-Ts G,, = - S Figure 3-3 shows a sampler and a zero-order hold. The input signal x(t) is sampled at discrete hnstants and the sampled signal is passed through the zero-order hold. The zero-order hold circuit smoothes the sampled signal to produce the signal h(t), which is constant from the last sampled value until the next sample is available. That is, h ( k ~+ t ) = x(kT), for O 5 t < T (3-8) e shall obtain a mathematical model of the combination of a real sampler and zero-order circuit, as shown in Figure 3-4(a). Utilizing the fact that the integral of an impulse function is a constant, we may assume that the zero-order hold is an integrator, and the input to the zero-order hold circuit is a train of impulses. Then a rnathematical model for the real sampler and zero-arder hold may be constructed as shown in Figure 3-4(b), where GhO(s)is the transfer function of the zero-orcler hold and x" (t) is the impulse sampled signal of x(t). Zero-order hold (a) A real sampler and 61 hold; (b) mathematical model that consists of an impulse (b) sampler and transfer function Gh0(s). ec. 3-2 impulse Sampling and Consider the sampler and zero-order hold shown in Figure 3-4(a). Assu that the signalx(t) is zero for t < O. Then the out ut hl(t) is related to x(t) as follows: Since lace transform of Equation (3-9) becsmes Next, consider ehe mathematical mo shown in Figure 3-4(b). The output of this model rnust be the same as tfiat of real zero-order hold, or From Figure 3-4(b), we have Since Equation (3-11) may be written as y comparing Equations (3-12) and (3-E?), we see that the transfer function of the zero-order hold may be given by Note that, mathematically, the system shown in Figure 3-4(a) is equivalent to the system shown in Figure 3-4(b) from the viewpoint of the input-output relationshi That is, a real sampler and zero-order hold can be aeplaced by a rnathernatically equivalent continuous-timesystemthat consists of an impulse sampler and a transfer
- 47. z-Plane Analysis of Discrete-TimeControl function (1 - e-Ts)lse The two sam fingprocesses will be distinguished (as "cey are in Figure 3-4) by the rnanner in which the sampling switches are drawn. o not use first-order holds in control systems, it is worthwhile to see what the transfer function of first-order holds may look like show that the transfer function of the first-order hold may be given by Mext we shafl derive Equation (3-14). ave stated that Equation (3-6) describes the output of an nth-or circuit, For the first-order hold, n = 1.Let us substitute n = 1into Equation (3-6). Then we have whereO c: T < Tand k = 0,1,2,.... y applying the con h((k - 1)2") = x((k - 1)T) the constant al can be determined as Eollows: ence, Equation (3-15) becomes where 0 5 T < T. The extrapolation process of the first-order Equation (3-16). The continuobis-time output signal h(t) obtained by use of the first-order hold is a piecewise-linear signal, as shown in Figure 3-5. First-order hold ec. 3-2 impulse Sampling and ño derive the transfer function of the first-order hold circuit, it is convenient to assume a simple function for x(t). For function, and a unit-ramp function uppose that we choose a unit-step fun -order hold shown in Figure 3-6(a) consists of stsaight lines that are The output h(t) is shown in the follows: aplace transform of this last equation becomes Figure 3-6 (a) Real sampler cascaded with first-order hold; (b) mathematical model Figure 3-5 Input and output of a first-order hold. u , , consisting of impulse sampler and GM(s).
- 48. r-Plane Analysis of iscrete-TimeControl Figure 3-6(b) shows a mathematical model of the real sarnpler cascaded wit the first-order hold shown in Figure 3-6(a). The mathematical model consists of the impulse sampler and GR1(s),the transfer function of the first-order hol pul signaX of this model is the same as the output of the real system. is also given by Eq place transform of ) to the first-order hold Ghl(s)is ence, the transfer function Chl(s) of the first-order hold is given by Note that a real sarnpler combined with a first-order hold is equivalent to an impulse sampler combined with a transfer function (1- e-TS)2(1FJ+ l)/(aS2). Sirnilarly, transfer functions of higher-order hold circuits may be de-rived by followingthe procedure presented. However, sincehigher-order are not practica1 from the viewpoint of delay (which may cause system instability) and noise effects, we shall not derive their transfer functions here. (The zero-order hold is the simpiest and is used most frequently in practice.) a g . Let us summarize what we have presented so far about impulse sarnpling. A real sampler samples input signal periodically and produces a sequence of pulses as the output. ile the sampling duration (pulse sampier is very small ( will never become zero), the as width, which implies that a sequence of pulses becomes a sequence of impulses whose strengths are equal to the continuous-time signal at the sampling in- stants, simplifies the analysis of discrete-time systems. Such an assurnption is valid if the sampling duration is very small compared with the significant time constant of the system and if a hold circuit is connected to the output of the sampler. . When we transform eTsto z, the concept of impulse sampling (which is purely a mathematical process) enables us to analyze by the z transform method discrete-time control systems that involve samplers and hold circuits. This means that by use of the compiex variable z the techniques developed for the Laplace transform methods can be readily applied to analyze discrete-time systems involving sampling operations. 3, As pointed out earlier, once the real sampler and zero-order hold are mathe- rnatically replaced by the impulse sampler and transfer function (1 - e-Ts)ls, the system becomes a continuous-time system. niis simplifies the analysis of the discrete-time control system, since we may apply the techniques available to continuous-time control systems. btaining the r Transform by the Convolution Integral Method It is repeated that the impulse samplieris a fictitious sampler introduc the purpose of mathematical analiysi lement such a sampler that generates Hnthis section we shallobtain the z transform of x(t) by using the convolutionintegral method. Consider the impulse sarnpler shown in Figure 3-7. The output of the i sampler is x(t)¿i(t - kT) = x(t) (3-1 8) k=O k=O Noting that at X*( S ) is the Eaplace transform of product of two time functions, x(t) 6(t - kT). Note that it is not equal t e product of the two corresponding Laplace transforms. The kaplace transform of the product of two Laplace-transformable functions f(t) and g(t) can be given by f [For the derivation of Equatio l9), see Problern A-3-4.1 Let us substitute x(t - kT) for f(t) and g(t), respectively. Then the Laplace transform of -- ..- 6, Figure 3-7 Impulse sampler
- 49. can be given by z-Plane Analysis of Discrete-Time Control where the integration is along the Iine from c - jm to c +j a and this line is paralle1 ne and separates the poles of X(p) from those of is the convolution integral. Iat is a well-known fact e evaluated in terms of residues by forrning a closed contour cansistingof the linefrom c - j a to c +jm a~ida se icircleofinfiniteradius in the left or right half-plane, provided that the integral along the added semicircle is a constant (zero or a nonzero constant). That is, we rewrite E follows: where T is a semicircle of infinite radius in the left or right half p plane. ñhere are two ways to evaluate t is integral (me using an infinite semicircle in the M-half plane and the other an infinite semi shall describe the results obtained by the two case analysis here, we assume that the poles S) can be expressed as a ratio of polly where g (S)and p (S) are polynomials in s. also assume that p(s) is of a higher degree in s than q(sj, which means that shall evalu- ate the convolution integral given by Equation (3-21) using a closed contour in the left half of the p plane as shown in Figure 3-8. Using this closed contour, Equation (3-21) may be evaluated as follows:Noting that the denominator of X(s) is of higher degree in s than the numerator, the integral along fivanishes. This integral is equal to the sum of the residues of (p) in the &sed contour. the z Transform by the Convolution Integral Method 2: p plane 1 ure 3-8 Ciosed contour in the leftOf 1 - e-Tts - p ) half of the p plane. roblem A-3-6 for the derivation of Equation (3-22).] for eTsin Equation (3-22), we have By changing the com lex variable notation from p to s , we obtain residue of -X(s)z at pole of ~ ( s) z - eTs iAssume that X(s) has poles SI,s2, . ,S,. f a pole at s = sjis a simple pele, then the corresponding residue Kj is If a pole at s = siis a multiple pole of order ni,then the residue Miis Given obtain X ( z ) by use 0.6 the convolution integral in ihe left half-plane.
- 50. z-Plane Analysis of Discrete-Time Control ystems Chap. 3 Sec. 3-3 Obtaining the a Trancform by the Convolution Integral Method Note that X(s) has a double pole at s = O and a simple pole at S = -1. Equation (3-23) becomes X(s>zresidue of at poole of X(s) 1--- 1 lim- sZ7 --- l dL 1 + Iim (S + 1)7---- - o s s s + z ~ e ~ j--i[ s (S + 1) z -z IeTs -z[z - eTS+ (S + l)(-T)eTs] 1 z = lim +--*-+ O (S + 1)2(z- eTSl2 (-1)2 z - e-T p plane. Let us choose the closed contour shown in Figure 3-9, whi line from c - jm to c +jm and rR,the portion of the sernicircle of infinite radius in the right half of the p plane that lies to the right of t is h e . The closed contour endoses al1poles of 141- e-T("p)],but it does not enclsse any poles of X(p). Now X"(s) can be written as Figure 3-9 Closed contour in the right half of the p piane. evaluating the integrals on the right e of Equation (3-26), we need deir two cases separately: one case w denominator of X(s)is two or more degrees higher in s than the numerat other case where the denom- inator of X(s) is only one degree higher in s than the numerator. Case 1: X(s) has a denominator two or more degrees hig numerator. For this case, since X(s) possesses at least two more we have Then the integral along & is zero. Thus, in the present case X"( S ) can be obtained as follows: erivation of Equation (3-27), see Probiem A-3-7.1 Thus Note that this expression of the z transforrn is useful in proving the samplingtheorem (see Section 3-4). owever, it is very tedious to obtain z transforrn expressions of commonly encountered functions by this method. Case2: X(s)has a denominator one degree higher in s than the numerator. For this case, lirirsX(s) = x(O+) =# O < and the integral along TRis not zero. [The nonzero value is associated with the initial value x(O+)of x(t).]It can be shown that the contribution of the integral along TRin Equation (3-26) is -fx(O+). That is, Then the integral term on the right-hand si e of Equation (3-26) becomes Show that X*(s) is periodic with period 271.l~~. Referring to Equation (3-29),
- 51. ec. 3-3 Obtaining the z Transform by the Convolution Integral Method Hence, 1 m 1 X*(s +j-k) = - X(s +jw,k +jwh) + -x(O+) rz[gl(t)]= G ( z ) T h = - - 2 the z transform o£xl(t)becomes Let k + h = m. Then this last equation becomes 1 - 1 X 4 ( s+ jmik) = - X(s +jw,m) + x ( 0 - t ) = X'(s) T,,-, Therefore, we have X ( z ) = ZIGl(s) - e-Ts&;*(S)] X * ( s ) = X * ( s + j w s k ) , k = 0 , 1 , 2 ,... = /Z'[g1(t)l- Z[x1(t)] Thus, X* ( S ) is periodic, with period 2.lr/wS.This means that, if a function X(s) has a = G1(z)- Z - l G1(z) pole at s = si in the s plane, then %"(S)has poles at s = si5 jwsk (k = 0,1,2, ...). = (1 - Z - ~ ) C ~ ( Z ) G(s)follows the zero-order hold. Then the product of the transfer function of the ( z = ( S = (1 - ) Z [ ? ] (3-3 2) zero-order hold and C(s)becomes 1 - e-Ts X(s) = y a s ) In what follows we sha'il obtain the z transform of such an X(s). Note that X(s) can be written as follows: then the z transform ciif the function where fxs)G1(s)= -S can be obtained as follows. Since Consider the function X1(s)= e-TsG(s) x(s)= (1 - e - f i ) 2 T S + ' Ts2 G(s) x1(t) = Iotgo(t- 4544d~ where = (1 - z-1)2Z[TTs + 1G(s)j (3-33) go(t)= %-'[e-fi]= 6(t - 7') Equation (3-33) can be used for obtaining the z transform of the function involving &(t)= %-'[G(s>] the first-order hold circuit. Thus, xl(t)= [6(t - T - r)gl(r)dr tain the z transform of 1 - e-Ts 1 = g10 - TI X(s) = --S s + l
- 52. r-Plane Analysis of iscrete-Time Control Referring to Equation (3-32), we have sampling frequency is sufficiently high compar onent involved in the continuous-time signal, t To reconstruct the original minimum frequency that the sa frequency is specified in the sam time signal x(t) has a frequency spectrum as shown in does not contain any frequency com , IIf o,, defined as 2 d T , where T is the sampling period, is greater than 2wl, or where wl is the highest-frequency component present ontinauous-time sig- nal x(t), then the signal x(t) can be reconstructed co sigraal x*(t). The theorem impIies that if ws > 2w1then, from t dge of the sampled signal, it is theoretically possible to reconstruct exactl al continuous-time signal. In what follows, we shall use an intuitive graphical approach to explain the sampling theorem. For an analytical approach, see To show the validity of this sampling theorem, we need to find the frequency -wl O 0, w Figure 3-10 A frequency spectrum. riginal Signals from spectrum of the sampled signal x"(t). The Laplace transform of x obtained in Section 3-3 and is given by uation (3-27) or (3-29), whether x(O+) = O or quency spectrum, s in Equation (3-27). uency spectra, we nee with the value of x(O+).] Thus, Equation (3-34) gives the frequency is periodic with period 2n-/O,, if a function X(s) has a pole at s = sl, then (k = o, 1,2, . ..). Figures 3-11(a) and (b) show plots of t e frequency spectra X* (jw) versus Figure 5-11 Plols of the frequency spectra / X":(jw)l versus w for two values of sampling frequency w,:(a) ws > 2wl; (b)w, < 2wl.
- 53. a-Plane Analysis of Discrete-Time Control Sycterns Chap. 3 for twovalues of the samplingfrequency os.Figure 3-11(a) correspondsto o, > 2wl, ile Figure 3-11(b) corresponds to ws < 2wl. Eac consists of IX(jw)lTrepeated every ws = 2~12"radlsec of IX*(jw)]the component IX(jo)l/Tis called theprima components, Ix(j(o it o,k))llT, are called comple If os> 2wl, no two components of IX*(jo)/ ill be repeated every w, sadlsec. original shape of IX(jw)l o longer appears in the plot of use of the superposition sf spectra.Therefore, we see bhat ignal x(t) can be reconst ted from the impulse-sampled signal x"(t) by filtering if and only if os> 2wl. It is noted that although the is specified by the sampling theor component present in the signal, closed-loop system a at a frequency much éo be lOo, to 20w,.) r, The amplitude frequenc pass filter GI(jw) is shown in Figure 3-12. The magnitu over the frequency range - Sos5 o 5 So, and is zero ola The samplingprocess introduces an infinite number nents (sideband components) in addition to the primary c will attenuate al1 such complementary components to primary component, provided the s g frequency osis greater than twice the highest-frequency component of t nuous-time ségnal. reconstructs the continuous-time signal represented by the S shows the frequency spectra of the signals before and after ideal filtering. The frequency spectrum at the output of the ideal filter is 112"times the frequency spectrum of the original continuous-time signalx(t). Since t has constant-magnitude characteristics for the frequency region -Sw, 5 w 5 Sw,, there is no distortion at any frequency within this frequency range. That is, there is no phase shift in the frequency spectrum of the ideal filter. (The phase shift of the ideal fifter is zero.) - S 0 ws- - Figure 3-12 Amplitude frequency 2 2 spectrum of the ideal low-pass filter. econstructing Original Signals from Sampled Signals Figpaie 3-13 Frequency spectra of the signals before and after ideal filtering. It is noted that if the sampling frequency is less than twice the highest- frequency component of the original continuous-time signal, then because of the frequency spectrum overlap of the pri ary component and complementary com- ponents, everr the ideal filter cannot recons ct the original continuous-timesignal. (In practice, the frequency spectrum of continuous-time signal in a control system may extend beyond k iw,, even though the amplitudes at the higher frequen- cies are small.) realizable. Since the frequency spectrum of the ideal filter is given by the inverse Folarler transform of the frequency spectrum gives 1 o t- - sin-"- lrt 2 Equation (3-35) gives the unit-impulse response of the ideal filter. Figure 3-14 shows a plot ofgI(t)versus t. Notice that the response extends from t = -a to t = '30. This implies that there is a response for t < O to a unit impulse applied at t = O. (That is, the time response begins before an input is appllied.)This cannot be true in the physical world. Hense, such an ideal filter Is physlcally unrealizable. [lin many
- 54. z-Plane Analysis of Discrete-Time Control use 3-14 Impulse response gI(t) of ideal filter. communications systems, however, it is possible to adding a phase lag, which rneans adding a delay to t systerns, increasing phase lag is not desirable from the viewpo fore, we avoid adding a phase lag to approximate the ideal filter.] Because the Ideal filter is unrealizable and because signalsin practical.control systems generally have higher-frequency components and are not id limited, it is not possible, in practice, to exactly reconstruct a coné from the sarnpled signal, no matter what sarnpling frequency is words, practically speaking, it is not possible to recon ruct exactly a continuous- time signali in a practical control system once it is sa To compare the zero-order hold with the ideal filter, we shall obtain the frequency- response characteristics of the transfer function of the zero-order hold. ing jo for s in Equation (3-36), we obtain The amplitude of the frequency spectrum of Gho(jw)is The magnitude becomes zero at the frequency equal to the sarnplingfrequency and at integral rnultiples of the sarnpling frequency. ec. 3-4 Reconstructing Original Signals from Campled Signals Figure 3-15(a) shows the frequency-response characteristics of the zero-order .AScan be seen from Figure 3-15, ere are undesirable gain peaks at of 3wS/2,5as/2, and so on. Notice of the zero-order hold are not constant, and zero-order hold, distortion of the frequency spectra occurs in the system. e phase-shift characteristics of the zero-order hold can be obtained as follows. Note that sin (oU2) alte S positive and negative values as (U increases from O to os,osto 2ws,2wSto 30.4, bottom] isdiscontinuous at w = k &/a, where k = 1,2,3, ....Sucha &con- tinuity or a switch from a positive value to a negative value, or vic considered to be a phase s ift of i180". lín Figure 3-15(a), phase to be -180". (It could be assumed to be 4-180" as well.) Thras, Figure 3-15 (a) Frequency-response curves for the zero-order hold; (b) equivalent Bode dictgram when T = 1sec.
- 55. where z-Plane Analysis of iscrete-Time Control Cystems C A modification of the presentation o£ the ncy-response diagram of -15(a) is shown in Figure 3-15(b). The d shown in Figure 3-15(b) ode diagram of the z o-order hold. The sampiing p to be 1sec, or T = 2. Notice tude curve approa ency points that are inte les of the sampling frequency o, = phase curve [Figure 3-15(b), bottom] TQsummarize what we have stated, the frequency spectrmm o£t the zero-order hold includes complementary components, since th characteristics show that the magnitude of ChO(jo) is not zero for lo1 > 4os,except at points where o = &os,o = &2os,o = +i3wS,.... Hn the phase curve there are phase discontinuities of +i180" at frequency points that are multiples of o,. Except for these phase discontinuities, the phase characteristic is line Figure 3-16 shows the comparison of the ideal filter and For the sake of comparison, the magnit zero-order hold is a low-pass filter, alt quite good. e)£ten, additional low-pass filtering of the sign essary to effectively remove frequency components highea than 4o,. The accuracy of the zero-arder epends on the sam- pling frequency o,, That is, the output of the hold may be made as close to the original continuous-time signal as possibk by letting the sarnplingperiod Tbecome as small as practically possible. . The phenomenon of the overlap in the frequency spectra is known asfolding .Figure 3-17 shows the regions where folding error occurs. Tkie frequency ;o,is called the folding frequency or Nyq~tsrfrepency UN.That is, -3ws -2w, w w, O W, Ws - - - 2w, 30, w 2 2 Comparison of the ideal filter and the zero-order hold. econstructing Original I W s - WS 0 W Figure 3-17 Diagram showing the- " S Ws - 2 2 regions where folding error occurs. Bn practice, signals in control systems have igh-frequency components, and some folding effect will ahost always e e, in an electromechanical syste some signal may be conta frequency spectrum of the signal, therefore, may include lo as well as high-frequency noise comgonents (that is, nois e sampling at frequencies higher than 400 Hz is not practical, the a Iow frequency. Rernember that al1signals ith kequencies than U, ap as signals of frequencies between O and o,. fa&,in certaihicases, a signal of frequency may appear in the output. n the frequency spectra of an im sampled signalx*(t),where own in Figure 3-18, consider an arb frequencypointoz that falls in the region of the overlap of the frequency spectra. The frequency S o = o,comprises two components, IX*(jo2)l and Ix* (j(o, - @))l. The latter com- ponent comes from the frequency spectrum centered at o = w,. Thus, the frequency Figure 3-18 Frequency spectra of an impulse-sampled signal x*(t).
- 56. r-Plane Analysis of iscrete-Time Control led signal at o = 02inch components not only al frequency ncy o,- 02 (in general, no, where n is an integ spectrum is filtered by a low-pass filter, such as a zero-o component at o = n it were a frequency frequency component t frequency &~)2when y o,- 02 (in general, is not satisfied, then frequency high enough osa. ñt is noted that, if the continuous-time signalx(t) involves a frequency component equal to n times the sampling frequency o,(where n is an integer), then that component may not appear in the sampled signal. For example, if the slgnal x(t) = xl(t) + x2(t) = sin t + s i n 3 where x,(t) = sint and x4t) = sin 3t9is sampled at t = 0 , 2 ~ 1 3 , 4 ~ / 3 ,...(the sa pling frequency o,is 3 radlsec), then the sampled signal will not show the frequency component with w = 3 radisec, the frequency equal to o,.(See Figure 3-19.) Even though the signal x(t) involives an oscillation witb o = is, the component x2(t) = sin 311, the sampled signal does not show Such an oscillation existing in x(t) between the sampling periods is called a hidden oscillation. The transfer function for the continuous-time system relates the Laplace transform oE the continuous-time output to that of the continuous-time input, while the pulse transfer function relates the z transform of the output at the sampling instants to that of the sarnpXed input. Before we discuss the pulse transfer function, it is appropriate to discuss convolution summatian. Convol n. Consider the response of a continuous-time system driven by an ed signal (a train of impulses) as shown in Figure 3-20. Suppose that x ( t ) = 0 for t < O. The impulse-sampled signalx*(t) is t continuous-time system whose transfer function is G(s). The output of the system Sec. 3-5 The Pulse Transfer Function sin 3t / x ( t ) sin t + sin 3t Plots of x,(t) = sin t,xZ(t)= sin 3t, and x(l) = sin t + sin 3t. Sampled signalx(k),where sampling frequency w, = 3 radlsec, does not show oscillation with frequency w = 3 radlsec. is assumed to be a conitinuous-time signal y(t). f at the output there is another sampler, which is synchronized in phase with the input r and operates at the same samplingperiod, then the output is a train of impul assume that y(t) = O for t < 0. ñhe z transform of y(t) is Figure 3-20 Continuous-time system G(s) driven by an impulse-sampled signal.
- 57. naiysis of Discrete-TimeControl Systems Chap. 3 In the absence o£the output sampler, if we consi chronized in phase with the input sampler an iod) atthe output and observe the sequenceo£values taken byy(t)only at instants t = kT,then the z transforrn of theoutputy" (t)can alsobe givenby Equation (3-38). For the continuous-time system, it is a well-known fact that the output y(t) of e system is related to the input x(t) by the convolution integral, or g(t - ?-)x(?-)d~=lg>x(t- ~ ) g ( ? - ) d ~ where g(t)is the weighting function of the system or the impulse-response function of the systern. For discrete-ti e sYstems we have a onvolution summation, whk is similar to the convofution integral. Since m me y(t) of the system to tbe input x*(t)is the sum O r t < T g(t)x(O)+ g(t - T)x(T), T i t < 2 T g(t - T)x(T)+ g(t - 2T)x(2T), 2 T ~ t < 3 T g(t)x(0)ig(t - T)n(T)+ + g(t - kT)x(kT), kT i t < (k + 1)T The values o£the output y (t)at the sampling instants t = kT (k = 0,1,2, ...) are given by k where g(kT) is the system's weighting sequence. [The inverse z transform of C ( z ) is called the weighting sequence.] The summation in Equation (3-39) or (3-40) is called the convolution surnmation. Note that the sirnplified notation is often used for the convolution summation. Sincewe ascurned that xft) = O for t < O, we have x(kT - hT) = O for h > k . Also, since g(kT - hT) = O for h > k, we may assume that the values o£ h in ec. 3-5 The Pulse Bransfer Function Equations (3-39) and (3-40) can be taken from changing the value of the summation. Therefore, Equations (3-39) and (3-40) can be rewritten as follows: Ht is noted that if G(s)is a ratio of polynomials in s and if the degree of the denominator polynomial exceeds the degree of the numerator pofynomial onlly by 1the output y(t)Ps discontinuous, as shown in Figure 3-21(a). tinuous, Equations (3-41) and (3-42) yield the values immed pling instants, that is, y(O+),y(T+),... ,y (kT+).Such value actual response curve. If the degree of the denominator polynomial exce by 2 or more, however, the output y(t)is con hen y(t) is continuous, Equations (3-41) and (3 Figure 3-21 (a) Plot of output y(t) (impulse response) versus t when the degree of the denominator polynomial of G(s) is higher by 1than that of the numerator polynomial; (b) plot of 2T 3T 4T 5T output y(r) versus t when the degree of the denominator polynomial of G(s) is higher by 2 or more than that of the (W numerator polynomial.
- 58. z-Piane Analysis of Discrete-Time Control ystems Chap. 3 the sarnpling instants. The values y(k) in such a case portray the actual response curve. continuous-time part of the system has a limsG(s) = O. S"=' ulse on, From Equation (3-41) we where g(kT - hT) = O for h > k . Henee, the z transfor where m = k - h and a g(rnT)z-" = z transform of g(t) m=O Equation (3-43) relates the pulsed output Y(z) of the syst X(z). It provides a means for determining the z transform of for any input sequence. Divi ing both sides of Equation (3-43) by X(z) gives G(z) given by Equation (3-441, the ratio of the output Y(z) and the input X(z), is cafled the pulse trunsfer function of the discrete-time system. It is the z transform of the weighting sequence. Figure 3-22 shows a block diagram for a pulse transfer function C(z), together with the input X(z) and the out Equation (3-431, the z transforrn of the output signalcan be obtained as the product of the system9spulse transfer function and the z transform of the input signal. u pulse-transfer-function system. ec. 3-"he Pulse iransfer Feanction Note that G(z) is also the z transfor necker delta input: x(kT) = S,(kT) = foa k = O {u: f o r k + Since the z transform of the Kronecker delta input is then, referring to Equation (3-44), the response Y(z) to the Kronecker delta input is Thus, tfme system's response to the Kronecker delta input is G(z), the z transform of the weightirag sequence. This fact is parallel e fact that C(S) is the Laplace transform of the systemysweighting function, e signals in the system are starred (meaning that signals are impulse sampled) others are not. To obtain pulse transfer functions and to analyze discrete-time controf systems,therefore, we must be able to obtain the transforms of output signals of systems that contain sampIing ope uppose the impulse sampler is whose transfer function is G(s), as sh 3. In the following analysis we assume that all initial conditions are zero m. Then the output Y($) is Notice that Y(s)is a product of X* (S), periodic with period 2?;r/o,,and G(S), which is riodic. The fact that th e-sampled sagnals are periodic can be seen fro fact that In the following we skall show that in taking the starred Laplace transform of Equation (3-45) we may factor out X*(S) so that This fact is very important in deriving the pulse transfer function and also in simplifying the block diagram of the discrete-time control system. So derive Equation (3-47), note that the inverse Eaplace transforrn of Y(s) given by Equation (3-45) can be written as follows: Figure 3-23 Impulse-sampled system.
- 59. r-Plane Analysis of Discrete-Time Controf Then the z transform of y(t) becomes where m = n - k. Thus, Chap. (3-48) z transform can be understood as the starre Laplace transform with eTs y z , the z transform may be considered to be a shorthand notation for the starred Laplace transform. Thus, Equation (3-48) may be expressed as which is Equation (3-47). have thus shown that by taking the starred Laplace transform of both sides of uation (3-45) we obtain To summarize what we have obtained, note that E state that in taking the starred Laplace transfo some are ordinary Laplace transforms and ot the functions already in starred transforms can be factored out of the starred Laplace transform operation. It is noted that systems becorne periodic under starred Laplace transform operations. Such periodic systems are generally more complicated to analyze than the original nonperiodic ones, but the former may be analyzed without difficulty if carried out in the z plane (that is, by use of the pulse-transfer-function approach). present general procedures for obtaining the pulse transfer function of a system thal has an impulse sampler at the input to the system, as shown in Figure 3-24(a). The pulse transfer function G(z) of the system shown in Figure 3-24(a) is ec. 3-5 The Pulse Transfer Function v*W _____)_ S, Y ( z ) Fictitious sarnpier (a) Continuous-time system with an impulse sampler at the input; (b) continuous-time system. Next, consider the system shown in Figure 3-24(b). The transfer function G(s) is given by ortant fact to rernernber is that the pulse transfer function for this system [G(s)], because of the absence of the input sampler. The presence or absence of the input sampler is crucial in determining the pulse transfer function of a system, because, for exarnple, for the systern shown in igure 3-24(a), the Laplace transform of the output y (t) is ence, by taking the starred Laplace transform of Y(s), we have Y*( S ) = C*(s)X*( S ) or, in terms of the z transform, Y(z) = G(z)X(z) while, for the system shown in Figure 3-24(b), the Laplace transform of the output m is Y(s) = G(s)X(s) which yields Y" (S) = [G(s)X(s)f* = [GX(s)]*
- 60. ? z-Plane Analysis of Discrete-Time Control Systerns Chap. 3 V(z) = Z[Y(s)] = z[c(s)x(s)] = Z[&;X(s)] = The fact that the z transforrn of G(s)X(s) is not e ual to G(z)X(z) will be discussed in detaPI Iater in this section. In discussingthe pulse transfer function, we a e that there is a sarnpler at ut o£the elernent in consideration. The pres e output o£the elernent (or the systern) does snot affect the pulse transfer function, because, if the sarnpler is not physicalfy presesnt a6 th always possible to assurne that a fictitious sarnpler eans that, although the output signal is con e output only at 1 = k?" (k = 0,1,2,. ..) a get sequence y(kT). Note that only for the case where the input to khe*systemC(s) is an i sannpled signal is the pulse transfer function given by ExarnpIes 3-4 and 3-5 demonstrate t e methods h r obtaining the pulse transfen:function. Obtain the pulse transfer function G(z) of the system shown in Figure 3-24(a), where G(s) is given by Note that there is a sampler at the input of G(s) and therefore the pulse transfer function is C(z) = h[G(S)]. Method T. By referring to Table 2-1, we have Hence, Method 2. The impulse response function for the system is obtained as follows: Hence, Therefore, ec. 3-5 The Pulse Transfer Feinction Obtain the pulse transfer function of the system shown in Figure 3-24(a), where G(s) is given by Note that there is a sampler at the input of G(s). Method d. G(s) involvesthe term (1 - e-Ts);therefore, referring to Equation (3-32), we obtain the pulse transfer fttnction as foilows: G(z) =ZfG(s)] = Z From Table 2-1, the z transform of each o£the partial-fraction-expansion terms can be found. Thus, ethod 2. The given transfer function G(s) can be written as follows: Therefore, by taking the inverse Laplace transforrn, we have the following impulse response function: e-kT+ T - k = 1,2,3,. .. = {o, k = o Then the pulse transfer function G(z) can be obtained as follows: w
- 61. z-Plane Analysis of iscrete-Time Control Systems Chap. 3 nce, by taking the starred La of ea& of these two equations, we Consequently, fn terms of the z transform notation, e pulse transfer function between the output y*(t) and input x*(t) is t given by Figure 3-25 (a) Sampled system with a sarnpler between cascaded elements G(s) and H(s); (b) sampied system with no sampler between cascaded elements G(s) and H(s). ec. 3-5 The Pulse Transfer Function er the system shown in Figure 3-25(b). From t where e starred Lapiace t In terrns of the z transform notation, and the pulse transfer function y* (t)and input x* (t)is Note that ( z ) $. G H ( z ) = Z[C;H(s)] ransfer functions of t willl now verify this Consider the systerns shown in Figures 3-26(a) and (b). Obtain the pulse transfer function Y(z)IX(z) for each of these two systerns. Figure3-26 (a) Sampled system with a sampler between elements G(s) = l/(s +a) and H(s) = l/(s + b); (b) sampled system with no sarnpler between elements G(s) and H(s).
- 62. h r the system of Figure 3-26(a), the two transfer functions G(s) and M(s) are separated by a sampler. assume that the two sarnplers shown are synchronized and kave the same sampling period. The pulse transfer function for this system is Hence, - - - For the system shown in Figure 3-26(b), the pulse transfer function Y(z)lX(z) is obtained as follows: r- -.. Hence, Clearly, we see that the pulse transfer functions of the two systems are different; that is, G(z)H(z) $: GN(z) Therefore, we must be careful to observe whether or not there is a sampler between cascaded elements. d ~ eTra~esferFune~a'csnof C s. In a closed-joop system the existente or nonexistence of an output sampler within the loop makes a difference in the behavior of the system. (If there is an output sampler outside the loop, it will make no difference in the closed-loop operation.) Consider the closed-loop control system shown in Figure 3-27. In this system, the actuating error is sampled. From the block diagrarn, Figure 3-27 Closed-loop control system. Sec. 3-5 The Pulse Transfer Function ence, E(s) = R(s) - H(s)G(s)E*(s) Then, by taking the starred kaplace transform, we obtain or Since we obtain In terms of the z transform notation, the output can be given by The inverse z transform of this last equation gives the values of t sampling instants. [Notethat the actual output ~ ( t )of the systemis a .The inverse z transform of G(z)will not givethe continuous- ulse transfer function for the present closed-loop systern is Table 3-1 shows five typkal configurations for closed-loop discrete-time con- ere, the sarnplers are synchronized and have the same sa period. For each configuration, the correspon iscrete-time closed-loop control system (that is, they do not have pulse transfer functions) because the input signal R(s) cannot be separated from the system clynamics.Although the pulse transfer function may not exist for certain systern configurations, the sarnetechniques discussedin this chapter can still be applied for analyzing them. ontmller. The pulse transfer function of a digital controller may be obtained from the required input-output characteristics of the digital controller. Suppose the input to the digital controller is e(k) and the output is m(k). In general, the output m(k) may be given by the followingtype of difference equation:
- 63. z-Plane Analysis of Discrete-Time Control Systems Chap. 3 FIVE NPICAL CONFIGURATIONS FOR CLOSED-LOOP DISCRETE-TIME CONTROL SYSTEMS Sec. 3-5 The Pulse Transfer Funetion e z transform of Equation (3-51) gives (z) + . - -+ a,z-" ulse transfer function GD(z)of the digital controller may then be given by The use of the pulse transfer function GD quation (3-52) enables us to analyze digital control systems im 3-28(a) shows a block diagram of a digital c N D converter, digital controller,zero-order continuous-time (piecewise-constaant) control Figure 3-28(b) shows the transfer functions o The transfer function of the dlgital contr
- 64. 1 P-PlaneAnaiysis of Discrete-Time Control Systerns Chap. 3 system the computer (digital controller) solves a difference equation whose input- ut relationship is given by the pulse transfer function CD(z). In the present system the output signal c(t) is fed input signal r(t). The error signale(t) = r(t is converted to a digital signal thr fed to the digital controller, which desirable manner to produce the This desirable relationship between t by the pulse transfer function GD(z) selectingthe poles and zeros of GD(z),a number of input-output characteristics can be generated.] Weferring to Figure 3-28(b), let us define Erom Figure 3-28(b), notice that Bn eerms o£the z transform notation, Since and, therefore, Equation (3-53) gives the closed-loop pulse transfer function of the digital control system shown in Figure 3-28(b). The performance of such a closed-loop system can be improved by the proper choice of GD(z),the pulse transfer function of the digital shall later discuss a variety of forms for GD(z)to be used in obtaining optimal performance for various given performance indexes. In the following,we shall consider only a simple case, where the pulse transfer function GD(z)is of the PID (proportional plus integral plus derivative) type. ontmller. She analog ustrial control systems for over half a century. The basic princ 01 scheme is to act on the variable to be rnanipulated through a proper combination of three control actions: propor- tional control action (where the control action is proportional to the actuating error ec. 3-5 The Pulse Transfer Function signal, which is the difference between the in t and the feedback signal), integral control action (where the control action i ere e(t) is the input to the control1 Define Figure 5-29 shows the function f(hT). Then Taking the z transform of this last eequation, we obtain (Por the derivation of this last equation, refer to Problern A-2-4.) Notice that
- 65. Hence, z-Plane Analysis of Discrete-Time Control Systems -29 Diagram depicting function f(hT). Chap. e((h - 1)T)+ e(hT) h=l 2 E ( 4 Then the z transforrn o£Equation (3-55) gives This last equation may be rewritten as follows: where KT Kp = K - -= K - K! = proportional gain 211 2 K~= --KT =: integral gain li K~ = --KTd = derivative gain T Pdotice that the proportional gain I;Lp for the phg>controller is smaller than the proportional gain K for the analog PID controller by KJ2. The pulse transfer function for the digital PID controller becornes The pulse transfer function of the digital PID controller given by Equation (3-56) is commonly referred to as rhe positional form of the PID control scheme. Sec. 3-5 The Pulse Transfer Function Figure 3-30 Block diagram realization of the velocity-form digital PID control scheme. trol scheme exhibits better response characteristics than ontrol scheme. Another a vantage of the velocity-for it is uceful in suppressing e cessivecorrectisns in proce Linear control laws in the form of 138 control actions, in both positional form Consider the control systemwith a digital PID conerollershown in Figure 3-31(a). (The PID controller here is in the positional form.) The trancfer function of the plant is assumed to be
- 66. z-Plane Anaiysis of Discrete-Time Control Systerns Chap. 3 Digital PID Zero-order controller hold Plant Figure 3-31 (a) Block diagram of a control system; (b) equivalent block diagram. and the sampling period T is assurned to be 1sec. Then the transfer function of the zero-order hold becomes 1 - e-" C,( S ) = ----- Since we may redraw the block diagram of Figure 3-31(a) as shown in Figure 3-31(b). Let us obtain the unit-step response of this system when the digital controller is a PID controller with & = l9& = 0.2, and KD = 0.2. The pulse transfer function of the digital controller is given by Then the cloced-loop pulse transfer function becomes We shall use the MATLAB approach to obtain the unit-step response. k re 3-32 Unit-step response.
- 67. iscrete-Time Control Sec. 3-5 The Pulse Trancfer Function 1 instants. In ordinar e output will not vary very much between any two consecutive sampling instants. Tn certain cases, however, we may need to find e between consecutive sampling Three rnethods for provi sponse between two consecutive sarnpiing e transform method ed z transform metkod State-space method Here we shall briefly discuss the ethod. The modified z trans- form method is presented in App ers interested in the modified z transforrn should read ection B-4.) The state-space method wili,be discussed in Section 5-5. for exarnple, the systern shown in Equation (3-59) will give the continuous-time response c(t). ence, the response al any time between two consecutive sampling instants can be calculated by the use of Equation (3-59). [See Problem A-3-18 for sample calculationasof the right-hand side of Equation (3-59).]
- 68. 1 z-Plane Analysis of Biscrete-lime Control ec. 3-6 Realization of Digital Controllers and In this section we discuss reali ulse transfer functions t represent digital conttollers and digital filters may involve either software or hardware or both. In general, "realiza- tion" of a pulse transfer function means eterrnining the physical layout for the propriate combination of arithrnetic an storage operations. In a softwarerealization we obtain co puter programs forthe involved. Pn a hardware realization we build a special- circuitry as digital adders, s (shift register§ with E& sampling period T as a unit Pn the field of digital processing, a digital filter is a cornputational algorithm that converts an input sequence of numbers into an output sequence in such a way that the characteristics of the signal are changed in so fashion. That is, a digital filter processes a digital sigrmal by passing quency components of the digital input signal and rejecting undesirable ones. fhw general terms, a digital controller is a form of digital filter. where al = -1 a:! = O bo = Kp + KI + dg, bl = -(Kp + 2KD) b2 = KD shall now discuss the irect propramming an 1 filters. hn these programmings, coeffickn quantities) appear as rnultipliers in the block diagra diagram schemes where the coefficients a, and b, appe called direct structures . Note that there are important differencesbetween the digital signal processing Consider the digital filter given by used in communications and that used in control. In digital control the process sfer function has n poles and m zeros. of signalsmust be done in real time. n communications, signalprocessing need n of the filter. The fact be done in real time, and therefore delays can be tolerated in the processin Equation (3-60) can be seen easily, since frorn improve accuracy. is sectioñ deals with the block n/(Z) = -al=-'Y(z) - a2z-2Y(z) - - a, a-, Y(z) + boX(z) ments, adders, and mulitipliers. + b, z-l X(z) + + b, z-"X(z) diagram realizations will be discussed. Such block diagram realizations can b Rearranging this Iast equation yields Equation (3-60). as a basis fur a software or hardware design. Bn fact, once the block realization is cornpleted, the physical realization in hardware or software is forward. Note that in a block diagram realization a pulse tran repiesents a delay of one time unit (see Figure 3-34.) (Note also that in the s plane z-' corresponds to a pure delay e-".) In what follows we shall deal with the digital filters that are used for filtering and control purposes. The general form of the pulse transfer function between the output Y(z) and input X(z) is given by Y(z) bo+ blz-l + b2z-' + a - m + b , ~ - ~ G(z)= --= X(z) 1 4- alz-' + a2z-2+ e e + a, z-" ' where the al's and b,'s are real coefficients (some of them may be zero). The pulse transfer function is in this form for many digital controllers. For example, the pulse transfer function of the PID controller given by Equation (3-56) can be expressed in the form of Equation (3-60), as follows: Figure 3-34 Pulse transfer function showing a delay of one time unit.
- 69. 1 z-Plane Analycis of Diccrete-Time Control Systems Chap. 3 The type of realization here is called directprogramming .Direct means that we realize the numerator and denominator of the pulse tra using separate sets of delay elements. The numerator uses a set of m delay elements and the denominator uses a different set of n delay elements. Thus, the total number of delay elements used in direct prograrnming is m + n. The number of delay elements used in direct programming can be reduced. In fact, the nurnber of delay elements can be reduced from n + m to n (where n 2 m). The programming method that uses a minimum possible number of delay elernents standard programming . practice, we try to use the minimum number of delay elernents in realizing pulse transfer function. Therefore, the direct programming t more rhan the minimum number of delay elements is more or less of ac rather than of practica1 value. ing. ASpreviously stated, the number of delay elements required in direct programming can be reduced. In fact, the number of delay elernents used in realizing the pulse transfer function given by Equation (3-60) can be reduced frorn n + m to n (where n 2 m) by rearranging the block diagram, as will be discussed here. First, rewrite the pulse transfer function Y(z)/X(z) given by Equation (3-60) as follows: where and Then, draw block diagrams for the systems given by Equations (3-61) and (3-62), respectively. To draw the block diagrams, we may rewrite Equation (3-61) as Y(z) = b,N(z) + b , ~ - ~ N ( z )+ e - .+ b,z-"H(z) (3-63) and Equation (3-62) as H ( z ) = X(z) - alz-1H(z) - a, zW2M(z) - - - a, z-" H(z) (3-64) Then from Equation (3-63) we obtain Figure 3-36(a). Similarly, we get Figure 3-36(b) from Equation (3-64). The combination of these two block diagrams gives the block diagram for the digital filter G(z), as shown in Figure 3-36(c). The block diagram realization as presented here is based on the standard programming. Norice that we use only n delay elements. The coefficients al,a2,. . a, appear as feedback elements, and the coefficients bo,bl, . . . ,b, appear as fee (a) Block diagram realization of Equation (3-63); (b) block diagram f Equation (3-64); (c) block diagram realization of the digital filter given by Equation (3-60) by standard propramminp.
- 70. z-Piane Analysis of Discrete-Time Control ystems Chap. 3 The block diagrams in Figures 3-35 and 3-36(c) are equivalent, but the latter uses n delay elements, while the former uses n + m delay elements. Obviously, the latter, which uses a smaller number of delay elements, is preferred. ents. Note first that the use of a minimal number of delay elements saves memory space in digital controllers. Also, the use of a minimal number of . The error due to the quantization of the input signalinto a finite number of dis- crete levels. (Tn Chapter 1 we discussed this type of error, which may be con- sidered an additive sourceof noise, called quantization noise. The quantization noise may be considered white noise; the variance of the noise is (r2 = Q2/12.) The error due to the accumulation of round-off errors in the arithmetic oper- ations in the digital system. small errors in the coefficients a, and b, cause large errors in the locations of the poles and zeros of the filter. These three errors arise because of the practica1limitations of the number of bits that represent various signal samples and coefficients. Note that the third type way, the system may be made less sensitive to coefficient inaccuracies. For decomposing higher-order pulse transfer functions in order to avoid the coefficient sensitivity problem, the following three approaches are commonly used. Series programming Parallel programming Ladder programming e shall discusc these three programmings next. The first approach used to avoid the sensitivityproblem ransfer function G(z) as a series connection of first-ordes transfer functions. If G(z) can be written as a product caf G(z) = G1(z)G2(z) C;,(z) then the digital filter for G(z) may be given as a series connection of the component digital f h r s G1(z),G~(z),.. . ,Gp(z), as shown in Figure 3-37, In most casesthe C,(z)(i = 1,2, .. . ,p) arechosen to be either first- or second- order functions. If the poles and zeros of G(z) are known, Gl(z),G2(z),. .,C p ( ~ ) can be obtained by grouping a pair of conjugate complex poles and a pair of ec. 3-6 Realization of Di ¡tal Controllers and Fignre 3-37 Digital filter G(z) decomposed into a series connection of G,(z), G ( z ) , ..., Gp(z). conjugate complex zeros poles and real zerosto pro possible to grouptwo real The grouping is, in a sens irable to group severa to see which is best with respect to the num the range of coefficients, and so forth. To summarize, G(z) may be decornposed as follows: -- 1+ eiz-l +f i ~ - ~ ,1 + ciz-l + diz-2 The block diagrarn for are shown in Figures 3-3S(a) and (b), respectively. The block diagram for the digital filter G(z) is a series connection of p component digital filters such as shown in Figures 3-38(a) and (b). Eng. The second ap roacIS to avoiding the coefficientsen- sitivity problem is to expand the pulse transfer function C(z) into partial fractions. f C(z) is expanded as a sum of A , G1(z),G2(z),...,Gq(z), or SO that where A is sirnply a constant, then the block diagram for the digital filter G(z) can be obtained as a parallel connection of q + 1digital filters, as shown in Figure 3-39. ecause of the presence of the const term A , the first- and second-arder functions can be chosen in simpler forms. at is, C(z) may be expressed as The block diagram for
- 71. r-Plane Analysis of iscrete-Time Control Systerns Chap. 3 igore 3-38 (a) Block diagram representation of Equation (3-65); (b) block diagram representation of Equation (3-66). and that for are shown in Figures 3-40(a) and (b), respectively. The parallel connection of q + 1 component digital filters as shown in Figure 3-40 will produce the block diagram for the digital filter G(z). iag, The third approach to avoidingehecoefficientsensitiv- ity probiem is to implernent a ladder structure, that is, to expand the pulse transfer Sec. 3-6 Realimation of Digiral Controllers and Digital Filters decomposed as a parallel connection of A,Wz), W ) , ...,C&). ion G(z) into the following continued-fraction forrn and to The prograrnming method based on this scheme is calledi ladderprogramming . Let us define Then G(z) may be written as
- 72. z-Plane Analysis of Discrete-Time Control ystems Chap. 3 Q (a) Block diapram representation of Equation (3-67); (b) block diagram representation of Equation (3-68). shall explain this programming method by using a simple example where 2. That is, ec. 3-6 Realization of Digital Controllerc and y the use of the feinctions6V1(z),61B)(z),and 6P1(z),t may be written as follows: Notice that GjB)(z)may be written as The block diagram for G ! ~ ) ( z )given by Equation (3-70) is shown in Figure 3-41(a). Similarly, the block diagram for G ( ~ ) ( z ) ,which may be given by (b) Figure 3-41 (a) Block diagram for GIB'(z) given by Equation (3-70); (b) block diagram for G(*'(z) given by Equation (3-71).
- 73. z-Plane Analysis of iscrete-Time Control Systems Chap. 3 (3-7 1) Xi(z) - c(+R](z)~(z)= AiX(Z) rnay be drawn as shown in Figure 3-41(b). Note that 1 6;$yz) = - A, y combining component digital filters as shown in Figure 3-42(a), it is posslble to draw the block diagrarn of the digital filter G(z) as sho 3-42(b). [Note that Figures 3-42(a) and (b) correspond to the case en&. Digital filters based on la with respect to coefficient sensitivity and accuracy. Realizatio ture is achieved by expanding G(z) into continued fmctions ar It is noted that the continued-fraction expansion given by not the only way possible. There are a few different ways to Figure 3-42 (a) Component block diagrams for ladder programming of G(z) given by Equation (3-69) when t~ = 2; (b) combination of component block diagrams showing ladder programming of G(z). ealimation of Digital Controllers and structure. For example, a digital filter C(z) may be structured as a continue fraction expansion form around the origin in terrns sf z-l, as follows: Also, instead of G(z), its inverse l/G(z) may be anded into continued-Eraction s in terrns of z or z-1 in order to carry out ladder programming. Obtain the block diagrams for the following pulse-transfer-function system (a digital filter) by (1) direct programming, (2) standard programming, and (3) ladder program- ming: . Directprogramming. Since the given pulse transfer function can be written as direct programming yieldsthe block diagram showriin Figure 3-43. Notice that we need two delay elements. 2. Standard programming. We shall firct rewrite the pulse transfer function as follows: where and Figure 3-43 Block diagram realization of Y(z)lX(z) = (2 - 0.62-')1(1 + 0.5z-') (direct programmins).
- 74. z-Plane Analysis of Discrete-Time Control Systems Ckap. 3 Block diagram realizations of these last two equations are shown in Figure 3-44(a) and (b), respectively. If we combine these two diagrams, we obtain the block diagram for the digital filter Y(z)/X(z), as shown in Figure 3-44(c). Notice that the number of delayelements required has been reduced to E by the standard programming. Ladder prograrnrnirzg . shallfirst rewrite the given Y(z)/X(z) in the ladder form as follows: Thus, A. = 2 and Figure 3-44 (a) Block diagram realization of Y(z)lH(z) = 1 - 0.32-'; (b) block diagram realization of fI(z)/X(z) = 2/(1 + 0.52-'); (c) combination of block dia- grams in parts (a) and (b) (standard programming). ec. 3-6 Realizarion of Digital Controllers and 5 Block diagram realization of Y(z)/X(z) = (2 - O.62-')/(1 +0.52-l) (ladder programming). ence, we obtain Referring to Figure 3-41(a) for the block diagram of G;~'(Z),we obtain the block diagram of the digitai filter Y(z)/X(z) as shown in Figure 3-45. Notice that we need only one delay element. filtersmay be classified according to the duration of the irnpul a digital filter defined by the following pulse transfer function: where n 2 m. In terms o£the difference equation, The impulse response of the by Equation (3-721, where we assume not al1 ai9sare zero, r of nonzero samples, although their magnitudes rnay becorne negligibly small as k increases. This type of digital filter is called an infinite-impulse responsefilter. Such a digital filter is also called a recursivefilter, because the previoies values of the output together with t of the input are used in processing the signal to obtain the current cause o£the recursive nature, errors in previoies outputs rnay sive filter may be recognized by the presence of both aiand b, block diagram realization. Next, consider a digital filter where the coefficients ai are al1 zero, or where In terms of the difference equation, y(k) = b,x(k) + b,x(k - 1) + e - . + b,x(k - m ) The impulse response of the digital filter defined by Equadion (3-73) is limited to a finite number of samples defined over a finite range of time intervals; dhat is, the
- 75. ec. 3-6 Realimation of Digital Controllerc and impulse response sequence is finite. his type of digitalfilter is calle filter. lit is also called a nonrecursivefilter or a rnoving-averagefilter. resent value of the o . The finite-impulse the block diagram re Let us define the finite-impulse respon digitalfilter asg(kT). fthe inputx (kT)is ap can be given by g(hT)x(kT- ha") = g(0)x(k?")+g(T)x((k - 1)T ) + .- + g(kT)x(O) (3-74) The output y (kT)is a convolutisn summation of the input signal and the impulse response sequence. She right-hand side of Equation (3-74) consists of k + 1terrns. Thus, the output y(kT) is given in terms of the past k inputs x(O), x((k - 1)T)and the current input x(kT).Notice that as k increases it is not posible to process al1 past values of input to produce the current output. need to limit the number of the past values of the input to process. Suppose we decide to employ the N immediate past values of the inp x((k - 1)T),x((k - 2)T),. ..,x((k - N )T ) and the current input x(kT). This equivalent to approximating the right-hand side of Equation (3-74) by the most recent input values including the current one, or y ( k T ) = g ( O ) x ( k T ) i g ( T ) ~ ( ( k - 1 ) ? " ) + * ~ . + g ( N T ) x ( ( k - N ) T )(3-95 SinceEquation (3-75) is a differen uation, the correspondingdigital fálterira tk z plane can be obtained as foliows taking the z transform of Equation (3-75) we have Y(.) = g(O)X(z)+ g(T)z-'.X(z) i. - + ~ ( W T ) Z - ~ X ( Z )(3-76) Figure 3-46 shows the block diagram realization of this filter. Figure 3-46 Block diagram realization of Equation (3-76). The characteristics of the finite-im ulse response filter can be summarized as follows: The finite-impulse response filter Ps nonrecursive. Thus, because of the lack of feedback, the accumulation of errors in past outputs can be avoided in the prscessing of the signal. ementation of the finite-impulse r filter does not require feed- , so the direct programrn programming are identical. Also, implementation may be speed convolution using the fast Fourier transform. . The poles of the pulse transfer function of the finite-impulse response filter are at the origin, and therefore it is always stable. If the input signal involves high-frequency components, then the number sf delay elements needed in the finite-impulse response filter increases and t amount of time delay becomes large. (This is a disadvantage of the finite- impulse response filter compared with the infinite-impulse response filter.) The digital filter discussed in Example 3-8 is a recursive filter and realize it as a nonrecursive filter. Shen obtain the response to a cker delta input. ividing the numerator of the recursive filter G ( z )by the denominator, we obtain By arbitrarily truncating this series at zW7,we obtain the desired nonrecursive filter, as follows: Figure 3-47 shows the block diagram for this nonrecursive digital filter. Notice that we need a large number of delay elements to obtain a good leve1 of accuracy. Noting that the digital filter is the z transform of the impulse response sequence, the inverse z transform of the digital filter gives the impulse response sequence. By taking the inverse z transform of the nonrecursive filter given by Equation (3-77), we obtain y(kT) = 2x(kT) - 1.6x((k - 1)T)+ 0.8x((k - 2)T) - 0.4x((k - 3)T) +0.2x((k - 4)T) - O.lx((k - 5)T)+ 0.05~((k- 6)T) - 0.025x((k - 7 ) T ) For the Kronecker delta input, where x(0) = 1 and x(kT) = O for k # O, this last equation gives Y (0) = 2 y(T) = -1.6
- 76. aria! z-Piane Analvsis of Discrete-Time C o n t r ~ lSystems Chap. 3 Figure 3-47 Block diagram for the digital filter given by Equation (3-77) (non- recursive form). y(4T) = 0.2 y(5T) = -0.1 y(6T) = 0.05 y(7T) = -0.025 The impulse response sequence for this digital filter is shown in Figure 3-48. Input and output curves for a zero-order hoid.- Chap. 3 Exarnple Problerns and Solutions sequence for the digital filter given by Equation (3-77). Consider a zero-order hold preceded by a sampler. Figure 3-49 shows the input x(t) to the sarnpler and the output y(t) of the zero-arder hold. In the zero-arder hold the valrie ~f the last sample ic retained until the next sample Is talien. Obtain the expression for y(t). Then find Y ( s )and obtain the transfer function of the zero-order hold. olution From Figure 3-49 we obtain y(t) = x(O)[l(t)- l(t - T)]+ x(T)[l(t- T ) - l(t - 2T)] +~ ( 2 T ) [ l ( t- 2T) - l ( t - 3T)I + The Laplace transform of y (t) is where m The transfer function of the zero-order hold is thus Consider a first-order hold preceded by a sampler. The input to the sampler is x(t) and the output of the first-order hold is y(t). In the first-order hold the output y(t) for kT 5 t < (k + l ) T is the straight line that is the extrapolation of the two preceding sampled values, x((k - l ) T )and x(kT),as shown in Figure 3-50. The equation for the output y (t) is t - kT y (t) = -T [x(kT)- ~ ( ( k- 1)T)I + x(kT), kT 5 t < (k + l ) T (3-78) Obtain the transfer function of the first-order hold, assuming a simple fianction such as an impulse function at t = 0 as the input x(t).
- 77. e ,en 7-Plane Analvsis of Discrete-Time Control Systems Chap. 3 ure 3-50 Input and output curves &on. For an impulse input of magnitude x(0) such that x*(t) = x(0)6(t),the ,--+ .,/A n ; x r ~ nhxr Fnila+inn(3-781 becomes as shown in Figure 3-51. The mathemat- Hence, #(O) Slope = - -T Figure 3-51 Output curve of the first- order hold when the input is a unit- impulse function. Chap. 3 Example Problems and Solutions Since X*(s) = T[x"((t)]= T[x(O)G(t)]= x(0) the transfer function of the first-order hold is obtained as follows: Consider the function Show that s = O is not a pole of X(s). Show also that has a simple pole at s = 0. utisn Pf a transfer function involves a transcendental term eCTS,then it may be replaced by a series valid in the vicinity of the pole in question. For the function let us obtain the Laurent series expansion about the pole at the origin. Since, in the vicinity of the origin, e-Ts may be replaced by substitution of Equation (3-80) into Equation (3-79) gives which is the Laurent series expansion of X(s).From this last equation we see that S = O is not a pole of X(s). Next, consider Y(s).Since 1 - Y(s)= -s2 it may be expanded into the Laurent series as We see that the pole at the origin (S = O) is of order 1, or is a simple pole. Show that the Laplace transform of the product of two Laplace transformable functions f(t) and g(t) can be given by
- 78. z-Plane Analysis of Discrete-Time Control Cystems Chap. Chap. 3 Exampie Problems and Solutions 1 ciJm we have .qf(t)g(t)l = ,ij-,- F(P)G(s - P) dP SoiutB~sn The Lapiace transform of the product of f(t) and g(t) is given by %[f (0s('11 = jomf(t)g(')e-s' 'ft Note that the inversion integral is f(r) = LjcYmF(s)efl ds, t > O we have 277-j c-lm G(s - p ) = 1 where c is the abscissa of convergence for E(s). Thus, 1 - e - ~ ( ~ - ~ ) 1 m C+Jm .qf(f)g(t)l = T;;;j" I,,.. F(p)eP'+ &)e--" dt Notice that the poles of 141- e-T(s-P'] may be obtained by solving the equation Because of the uniform convergence of the integrals considered, we may invert the order of integration: -T(s - p ) = kj27~k, k = O,1,2,... so that the poles are Noting that 2 ~ r p = s k j - k = s k j w s k , k = O , 1 , 2,... T j~g(i)e-'s-p"dt = G(s - p ) where ws = 2 d T . Thus, there are infinitely many simple poles along a line parallel to the jw axis. we obtain The Laplace transform of x"(t) can now be written as 1 c+im %If(t)g(t)l = ,ijc-jmF ( P ) W - P) dp - 1 "+'"- - 1 2Trjj,-,mX(p) 1 - e-ns - PI d~ (3-86) Show that the Laplace transform of where the integration is along the line from c - jw to c +jw, and this line is parallelm m ~ " ( t )= x(t)6(t - kT) = ~ ( t ) 6(t - kT) to the imaginary axis in the p plane and separates the poles of X(p) from those of k=o k=o 1/[1 - e-T(s-P)].Equation (3-86) is the convolution integral. It is a well-known fact that can be given by such an integral can be evaluated in terms of residues by forming a closed contour consisting of the line from c - jw to c +jw and a semicircle of infinite radius in the left or right half-plane, provided that the integral along the added semicircle is a constant (zero or a nonzero constant). There are two ways to evaluate this integral (one using an infinite semicircle in the left half-plane and the other an infinite semicircle in the right half-plane); we shall consider these two cases separately in Problems A-3-6 and A-3-7. ution Referring to Equation (3-83), rewritten as 1 c+lm %[f(t)g(t)l = ,ijc-,-F(P)G(s - P) dp Referring to Equation (3-86), rewritten as 1 '+'" 1 m x w = ,ij,-,m X(pI1 - e-T(s-,>d~ f(t) = x(t) and g(t) = 6(t - kT) k=o show that, by performing the integration in the left half-plane, X"(S) may be given by 2 [ 8 ( t - kT)] = e-"' where and noting that
- 79. z-Piane Analysís of Discrete-Time Control Systerns By substituting z for eTsin Equation (3-87), we have - - X(z) = [residue of -X(p)z at poie of ~ ( p ) z - e T p 1By changing the compIex variable notation from p to S, we obtain X(z) = [residue of 1 Chap. 3 where we assurned that X(z) has h different multiple poles and m - h simple poles assume that the poles of X(s) lie in the left half-plane and that X(s) can be expressed as a ratio of polynomials in S, or where q(S)andp (S)are polynomialsin s. We also assume thatp (S)is of a higher degtee in S than q(s),which means that lirnX(s) = O s-+m shall evaluate the convolution integral given by Equation (3-86) using a closed contourin the left half of thep plane as shownin Figure 3-52. Using this closed contour, Equation (3-86) may be written as where the closed contour consists of the line from c - jm to c +jm and rL9which án turn consists of a semicircle of infinite radius and the horizontal lines at jm and -jm, which connect the line from c - j m to c +j m with the semicirclein the left half of the choose a value of c such that al1the poles of X(p) lie to the left of the iine from c - j a to c +jm and al1 the poles of 1/[1- e-T(s-P)] lie to the right of this line. The closed contour endoses al1 poles of X(p), while the poles of 1/[1- e-T("p)] are outside the closed contour. Because we have assumed that the denominator of X(s) is of a higher degree in s than the riurnerator, the integral along TL(the infinite semicirclein the Ieft half-plane plus the horizontal lines at j m and -jm, which connect the line frorn c - jm to c + jm with the semicircle) vanishes. Hence, This integral is equal to the sum of the residues of X(p) in the closed contour. (Refer to Appendix B for the residue theorem.) Therefore, X(P> 1 - e-T('-p) at pele 0f X ( ~ ) 1 (3-89) Chap. 3 Example Problerns and Sol~atíons I m X p plane Figure 3-52 Closed contour in the left half of the p plane. By substituting z for eTsin Equation (3-89), we have [ X(piz at p o i of i.(p jX(z) = residue of -----z - eTp By changing the complex variable notation from p to S, we obtain Let us assurnethat X(s) has poles si,s2,. .. ,S,. Zf a pole ats = S, is a simplepole, then the corresponding residue Kiis If a pole at S = si Is a multiple pole of order ni, then the residue Kiis Therefore, if X(s) has a multiple pole sl of order nl, a multiple pole s2of order n2,. . . , a multiple pole shof order nh,and simple poles s,,+I,~h+2,...,sm,then X(z) given by Equation (3-90) can be written as X(s)z at pole of ~ ( s )h(z = 2[residue of -z - eTs 1
- 80. nalycis of Discrele-Time Control Chap. 3 (3-93) where n, is the order of the multiple pole at s = si. Referring to Equation (3--86), rewritten as show that by performing this integration in the right haif p plane, X*(S) may be given provided that the denominator of X(s) is two or more degrees higher in s than the numerator. Show that if the denominator of X(s) is only one degree higher in s than the numerator then Solution Eet us evaluate the convolution integral given by Equation (3-86) in the right half of the p plane. Let us choose the closed contour shown in Figure 3-53, which consists of the line fromc - jw to c +jm and r R ,the portion of the semicircle of infinite radius in the right half of thep plane that lies to the right of this h e . The closed contour encloses al1 poles of 141 - e-T("p)], but it does not enclose any poles of X(p). Now X"(s) can be written as Let us investigate the integral along r R , the portion of the infinite semicircle to the right of the line from c - jm to c +jm. Since infinitely many poles of 1/[E - e-T(s-P'] lie on a lineparallel to thejw axis, the evaluation of the integral along TRis not as simple as in the previous case, where the closed contour enclosed a finite number of poles of X(p) in the left half of the p plane. In almost al1physical control systems, as s becomes large, X(s) tends to zero at kast as fast as 11s. Hence, in what follows, we consider two cases, one where the denominator of X(s) is two or more degrees higher in s than the numerator and another where the denominator of X(s) is only one degree higher in s than the numerator. Case 1: X(s) Possesses at Least Two More Poles Than Zeros. Referring to the theory of complex variables, it can be shown that the integral along TRis zero if the degree of the denominatorp(s) of X(s) is greater by at least 2 than the degree of the numeralor q(s); that is, if X(s) possesses at least two more poles than zeros, which implies that Iim sX(s) = x(O +) = O S- = Chap. 3 Example Problemc and Solutions C---y--JPoles o f X ( P ) p plane ure 3-53 Closed contour in the right half of the p plane. then the integral along TRis zero. Thus, in the present case Therefore, Equation (3-96) simplifies to The integral along the closed contour given by Equation (3-97) can be obtained by evaluating the residues at the infinite number of poles at p = s k jws k. Thus, The minus sign in front of the right-hand side of this last equation comes from the fact that the contour integration along the path TRis taken in the clockwise direction. Using L'Hopital's rule, we obtain
- 81. z-Plane Analysis of Discrete-Time Control Systems Noting that Thus, Note that this expression of the z transform is useful in proving the sampling theorem (see Section 3-4). However, it is very tedious Lo obtain z transform expressions of commonly encountered functions by this method. Case 2: X(s) Has a Denominator One Degree Higher in s Than the Numerator. For this case limw,sX(s) = x(O+) # O < m and the integral along TRis not zero. [The nonzero value is associated with the initial value x(O+) of x(t).]It can be shown that the contribution of the integral along PR in Equation (3-96) is -$x(O+). That is, Then the integral term on the right-hand side of Equation (3-96) becomes Consider the function Qbtain A7(z) by using the convolution integral in the right half-plane. Solution The Laplace transform of x(t) is Clearly, iim-,sX(s) = x(O+) = 1, or the function has a jump discontinuity at t = 0. Hence we must use Equation (3-95). Referring to this equation, we have xample Problems and Weferring to a formula available in mathematical tables, and noting that we can rewrite Equation (3- x * (S 101) in the form Thus, we have obtained X ( z ) by using the convolution integral in the right half-plane. [This process of obtaining the z transform is very tedious because an infinite series of X(s +jws k ) is involved. The example here is presented for demonstration purposes only. One should use other methods for obtaining the z transform.] Obtain the z transform of X(s) = S (S + q2(s + 2) by using (1)the partial-fraction-expansion method and (2)the residue method. Solntlopa 1. Partial-fraction-expansion method. Since X(s) can be expanded into the form 2 1 2qs)= -- -- - s + 1 ( ~ + 1 ) ~S + 2
- 83. z-Plane Analysis of Discrete-Time Control By substituting jw for s in Ghl(s),we obtain T - Tjw t 1 -,,4 sin2(Twl2)-- T e o2 = tan-' Tw - T o where we have used the relationship T = 2dwS. At a few selected values of o , we have IGh,(jo)/ = T /Ghl(jO) = O0 Figure 3-54 shows plots of the magnitude and phase characteristics of the first-order hold and those of the zero-order hold. From Figure 3-54 it is seen that both the zero-order hold and the iirst-order hold are not quite satisfactory low-pass filters. They allow significanttransmission above the Nyquist frequency, wh. = d T . It is important, therefore, that the signalbe low-pass-filteredbefore the samplingoperation so that the frequency components above the Nyquist frequency are negligible. 1 - e-T" Y(s)= G(s)X*( S ) = -x*( 4 Show that Y*(s)= X*(s) Solutian By taking the starred Laplace transform of Equation (3-102), we have Chap. 3 Example Problems and Solutions Magnitude and phase characteristics of the first-order hold and those of the zero-order hold. Iri terms of the z transform notation, we have where Gís) Figure 3-55 Zero-order hold.
- 84. z-Plane Analysis of Discrete-Time Control ence, Y ( z )= X ( z ) Ln terms of the starred Laplace transform notation, this last equation can be written as Obtain the weighting sequence of the system defined by for n = 2, 2, and 3, respectively. oleation For n = 1,we have Hence, the weighting sequence g,(k) is found to be g ~ ( k )= (-0)" For n = 2, we obtain 1 - 1 - az-' + a 2 z - 2 - a3zT3+ ..e G$(Z)= - (1 + az-l)' 1 4- az-' = 1 - 2aZ-i + 3aZZ-' - 4a32-3+ -.. Kence, the weighting sequence &(k) is g 4 k ) = (k -i- ~ ) ( - a ) ~ For n = 3, we get ence, the weighting sequence g4k) is Obtain the discrete-time output C ( z )of the closed-loop control system shown in Figure 3-56. Also, obtain the continuous-time output C(s). Figure 3-56 Discrete-time control system. hap. 3 Example Problems and Solutions utlon From the diagram we have M($) = Gl(s)E(s) E(s) = R(s) - H(s)C(s) ence, Taking tbe starred Laplace transform of this last equation, we obtain M* ( S )= [GiR(s)]*- [GiG,H(s)]* In terms of the z transform notation, This last equation gives the discrete-time output C(z). The continuous-time output @ ( S ) can be obtained from the following equation: Notice that [CiR(s)]*/{I+ [GlG2N(s)]*)is a series of impulses. The continuous-time output C(s)is the response of G2(s)to the sequence of such impulses. [See Problem A-3-18 for details of áetermining the continuous-time output ~ ( t ) ,the inverse Laplace transform of C(s).] Consider the svstem shown in Figure 3-57. Obtain the closed-loop pulse transfer function G(z)lR(z).Also, obtain the expression for @(S). Figure 3-57 Discrete-time control system.
- 85. 1 r-Plane Analysis of Discrete-Time Control Systems Chap. 3 olution From the diagram we have C(s) = Gz(s)M*(s) M(s) = Gl(s)E*(s) E(s) = R(s) - N(s)C(s) = R(s) - N(s)G~(s)M*(s) Taking the starred Laplace transforrns of both sides of the last three equations gives rdys) = GT(S)E*(S) E*(s) = R*(s) - HG2(s) Solving for C*(S) gives In terms of the z transform notation, we have C(z) - Gl(~)G2(4-- R(z) 1 + Gl(z)HG,(z) The continuous-time output C(s) can be obtained frorn the following equation: Consider the analog PID controller and the digitalPXD controller. The equation far the analog PID controller is ? ... where e(t) is the input to the controller and m(t) is the output of the controller. The transfer function of the analog PID controller is The pulse transfer function of the digital PID controller in the positional form is as given by Equation (3-56): Chap. 3 Example Problems and Solutions Compare the polar plots (frequency-response characteristics) of the analog PID controller with those of the digital PID controller. ution For the analog PID controller, the frequency-response characteristics can be obtained by substituting jo for s in G(s). Thus, For the digital PID controller, the frequency-response characteristics can be obtained by substituting z = e'"= into Go(z): = Kp + & + &(1 - cos wT +j sin wT) 1- c o s o T + jsinwT sin wT = K p + 5 ( l - j2 + KD(l - cos wT +j sinwT) (3-104) 1- coswT shall first compare separately the P action, the P action, and the D action of controller with their counterparts in the digital controller. Notice that in the proportional action (P action) the digital controller has a gain KI/2 less than the corresponding gain in the analog controller, since Kp = K - SKI. See Figure 3-%(a). For the integral action (1action) the real parts of the polar plots of the analog controller and digital controller differ by KJ2, as shown in Figure 3-5$(b). hen the proportional action and integral action are combined, then the real parts of the polar plots for the analog P I action and the digital PI action become the same, as shown in Figure 3-58(c). The polar plots of the derivative action (D action) for the analog controller and the digital controller differ very much, as shown in Figure 3-58(d). Hence, there are considerable differences in the analog D action and the digital D action. The qualitative polar plot of the analog PHD controller can be obtained from Equation (3-103) by varying o from O to m, as shown in Figure 3-59(a). Similarly, the qualitative polar plot of the digital PID controller can be obtained from Equation (3-104) by varying o from O to n-lT, as shown in Figure 3-59(b). Note that, although the polar plots of the analog PI controller and the digital PI controller are similar, there are significant differences between the polar plots of the analog PID controller and the digital PXD controller. In Section 3-5 we derived the pulse transfer function for the PID controller inpositional form. Referring to Figure 3-28, the pulse transfer function for the digitalPID controller was derived as Using Pm(kT) = m(kT) - m((k - 1)T)derive the velocity-formP%Dcontrol equation.
- 86. z-Plane Analysis of Discrete-Time Analog P action Digital P action Analog I action Digital 1 action Analog P/ action Digital P I action Analog D action Digital D action Polar plots of analog and digital controllers with (a) proportional action, (b) integral action, (c) proportional plus integral action, and (d) derivative action. hap. 3 Example Problems and Analog PID controller Digital PID controller (a) Polar plot of analog PXD controller; (b) polar pIot of digital PID controller. = Kp[e(kT)- e((k - l)?")] + KIe(kT) +K D [e(kT)- 2e((k - 1)T ) + e((k - 2)T)] (3-105) where we have used the relationships Kp = K - 4KI,& = KTII;, and KD = KTdlT. (For these reiationships, refer to the derivations of the positional form of the digital$ID control equation.) Equation (3-105) takes into consideration the variation of the positional form in one sampling period. Suppose the actuating error e(kT)is the difference between the input r(kT)and the output c(kT),or e(kT) = r(kT) - c(kT) By substituting this last equation into Equation (3-105), we obtain Vm(kT)= Kp[r(kT)- r((k - 1)T)- c(kT) + c((k - 1)T)l +&[r(kT) - c(kT)]+ KD[r(kT)- 2r((k - 1)T) -t r((k - 2)T) - c(kT) + 2c((k - 1)T) - ~ ( ( k- 2)IF)I (3-106)
- 87. z-Plane Analysis of Diccrete-Time Control Systems Chap. 3 The velocity-form PID control scheme given by Equation (3-106) may be modified ints a somewhat different form to cope with sudden large changes in the set point, Siince the proportional and derivative control actions produce a large change in the controlker output when the signal entering the controller makes a sudden large change, to suppress such a large change in the controller output, the digital proportional and derivative terms may be modified as discussed next. If changes in the set point [input r(kT)] are a series of step changes, then immediately after a step change takes place, the input r(kT) stays constant for a w until the next step change takes place. Hence, in Equation (3-106) we assume that (Note that this is true if the input stays constant. But we assume that this holds true even if a step change takes place.) Then Equation (3-106) rnay be modified to The z transform of Equation (3-107) gives Simplifying, we obtain Equation (3-108) gives the velocity-form PID control scheme. The block diagram realization of the velocity-form digital PIE) control scheme was shown in Figure 4-30. he system shown in Figure 3-60(a). Obtain the continuous-time output c(t) so that the output between any two consecutive sampling instants can be determined. Find the expression fo-rthe continuous-time output c(t). The sampling period Tis 1sec. ution For the system shown in Figure 3-60(a), we have C(s) = G(s)E*(s) E(s) -R(s) - C(s) Hence, or Thus, The continuous-tiae output c(t) can therefore be obtained as the inverse Laplace transform of C(s): ap. 3 Example Problems and - - (a) 1.5 Output points obtained by sarnple calculations Figure 3-60 (a) Discrete-time control system; (b) plots of individual impulse responses; (c) plot of continuous-time output c(t) versus t. c(t) = (e--l[c(s)] = (e-' G(s) [ 1:Z;$J For the present system,
- 88. z-Piane Analysis of Diserete-Time Control ence, F. 7 Eet us define Then the z transform expression for this last equation is Referring to Equation (3-58) for the z transform of G(s), we obtain 1 ' 1 - z-' X(z) = (1 - z-') 0.3679~-'+ 0.2642~-~ '(1 - 0.3679~-')(1 - z-') -- 1 - 1.3679~-1+ 0.3679~-~ 1- z-' + 0.6321~-~ Hence, noting that the sampling period T is 1sec or T = 1,we have 1- 1.3679e-" + 0.3679e-2" X*(s) = 1 - e-" + 0.6321e-& Therefore, 1 1- 1.3679e-" + 0.3679e-& c(t, = T1[-s2(s + 1) 1- e-" + 0.6321e-> (1 - 0.3679e-' - 0.6321eT2 - 0.3996eP3" + 0e-4" + 0.2526e-~"+ 0.2526e-& + Since 1 - - --- 1l 1 + _ s2(s + 1) s2 S S + 1 the inverse Laplace transform of this last equation is r -I Hence, we obtain c(t) = (t - 1 + e-') - 0.3679[(t - 1) - 1+ e-('-"]l(t - 1) Chap. 3 (5-109 le Problems and Solulions Figure 3-60(b) shows plots of individual impulse responses given by Equation (3-109). [Observe that c(t) consists of the sum of impulse responses that occur at t = O,t = 1, t = 2, ... with weighting factors 1, -0.3679, -0.6321, ....] From Equation (3-109) we see that for time intervals O 5 t < 1 , l r t < 2, 2 r t < 3,. .. the output c(t) is the sum of impulse responses as follows: t - 1 + e-', O s t < l (t - 1+ e-') - 0.3679[(t - 1) - 1 + e-('-')]l(t - l), 15 t < 2 e-') - 0.3679[(t - 1) - 1 t- e-"-"]l(t - 1) -0.6321[(t - 2) - 1 + e-''-2)]1(t - 2), 2 5 t < 3 Sample calculations for several values of t follow: The continuous-time output c(t) thus obtained is plotted in Figure 3-60(c). Consider the digital filter defined by Draw a series realization diagram and a parallel realization diagram. (Use one first- order section and one second-order section.) Solutiolra We shallfirct consider the series realization scheme. To limit the coefficients to real quantities, we group the second-order term in the numerator (whichhas complex zeros) and the second-order term in the denominator (which has complex poles). Wence, we group G(z) as follows: Figure 3-61(a) shows a series realization diagram.
- 89. P-Plane Analysis of Discrete-Time Control Systems P Block diagram reaiizations of the digital filter considered in Problem A-3-19. (a) Series realization; (b) parallel realization. Chap. 3 Example Problems and Solutions i Next, we shall consider the parallel realization scheme. Expansion of G(z)/z into partial fractions gives Then G(z) can be written as follows: Figure 3-61(b) shows a parallel realization diagrarn. A slowly changing continuous-time signal x(t) is sampled every T sec. Assume that changes in signalx(t) are very slow cornpared to the sampling frequency. Show that in the z plane (1 - z-l)lT corresponds to "differentiation," just as s corresponds to "differentiation" in the s plane. antloa For a siowlychanging signalx(t), the derivative of x(t) can be approxirnated by The z transform of this equation gives 1 1 V(z) = -[X(z) - z-l %(z)] = -(1 - z-l)X(z) T T from which we obtain the block diagram shown in Figure 3-62(a). This diagram corresponds to the s plane differentiation shown in Figure 3-62(b). Note tRat the block diagrarn shown in Figure 3-62(a) can be rnodified to that shown in Figure 3-63. (For approxirnation of "integration" in the z plane, see Problems -3-25 through B-3-27.) se 3-62 (a) Block diagram for "differentiation" in the z plane; (b) block diagram for "differentiation" in the s plane. Figure 3-63 Approximate "differentiator" in the z plane.
- 90. r-Plane Analysis of Discrete-Time Control ystems Chap. 3 Show that the circuit shown in Figure 3-64 acts as a zero-order hold. R, <<R, 4 Circuit approximating a zero-order hold. -2 Consider the circuit shown in Figure 3-65. Derive a difference equation describing the system dynamics when the input voltage applied is piecewise constant, or e(t) = e(kT), kT 5 t < (k i-l ) T O I C9 Figure 3-65 RC circuit. (Derive first a differential equation and then discretize it to obtain a difference equa- tion.) Consider the impulse sampler and first-order hold shown in Figure 3-66. Derive the transferfunction of the first-order hold, assuming a unit-ramp function as the input x(t) to the sampler. First-order Figure 3-66 Impulse sampler and first- L order hold. Chap. 3 Probterns -3- Consider a transfer function system , ~ 2 Obtain the pulse transfer function by two different methods. Obtain the z transform of Use the residue method and the method based on the impulse response function. Obtain the z transform of 1 - e-*" 1 X(s) = ----- S (S + a)" Consider the difference equation system where y (O) = O. Obtain the response y ( k )when the input x ( k )is a unit-step sequence. Also, obtain the MATLA Consider the difference equation system where y ( k ) = O for k < O. Obtain the responsey ( k )when the input x(k)is a unit-step sequence. Also, obtain the MATEAB solution. Obtain the weighting sequence g(k)of the system described by the difference equation y ( k ) - a y ( k - 1 ) = x ( k ) , - 1 < a < 1 If two systems described by this last equation are connected in series, what is the weighting sequence of the resulting system? Consider the system described by y(k) - y(k - 1) + 0.24y(k - 2) = x(k) + x(k - 1) where x(k) is the input and y(k) is the outpur: of the system. Determine the weighting sequence of the system. Assuming that y(k) = O for k < O, determine the response y ( k )when the input x ( k )is a unit-step sequence. Also, obtain the MATLAB solution. fioblem B-3-31 Consider the cystern
- 91. r-Plane Analysis of Biccrete-Time Control Obtain the response of this system to a unit-step sequence input. Also, obtain the MATH,AB solution. Obtain the response y(kIT) of the following system: where x(t) is the unit-step function and x"(t)is its impulse-sampled version. Assume that the sampling period T is 0.1 sec. Consider the system defined by Using the convolution equation obtain the response y(k) to a unit-step sequence input u(k). -3-1 Assume that a sampled signal X*(s)is applied to a system G(s).Assume also that the output of C(s)is Y(s)and y(O+) = 0. Using the relationship show that Obtain the closed-loop pulse transfer function of the system shown in Figure 3-67 Figure 3-67 Discrete-time control system. kap. Problems Obtain the closed-loop pulse transfer function of the system shown in Figure 3-68. Zero-order K- hold s + a Discrete-time control system. Consider the discrete-time control system shown in Figure 3-69. Obtain the discrete- time output C(z)and the continuous-time output C(s)in terms of the input and the transfer functions of the blocks. Discrete-time control system Consider the discrete-time control system shown in Figure 3-70. Obtain the output sequence c(kT)of the system when it is subjected to a unit-step input. Assume that the sampling period T is 1sec. Also, obtain the continuous-time output c(t). Figure 3-70 Discrete-time control system
- 92. z-Plane Analysis of Discrete-Time Control Systems Chap. 3 -3-1 Obtain in a closed formthe response sequence c(kT) of the system shown in Figure 3-71 when it is subjected to a Kronecker delta input r(k). Assume that the sampling period T is 1 sec. Figure 3-71 Discrete-time control system. Consider the system shown in Figure 3-72. Assuming that the sampling period Tis 0.2 sec and the gain constant K is unity, determine the response c(kT) for k = 0, 1,2, 3, and 4 when the input r(t) is a unit-step function. Also, determine the final value c(m). use 3-72 Discrete-time control system. Obtain the closed-loop pulse transfer function C(z)IR(z) of the digital control systern shown in Figure 3--30. Assume that the pulse transfer function of the plant is G(z). (Note that the system shown in Figure 3-30 is a velocity-form PPD control of the plant.) Assume that a digital filter is given by the following difference equation: Draw block diagrams for the filter using (1) direct programming, (2) standard prograrn- rning, and (3) ladder programming. Consider the digital filter defined by Wealize this digital filter in the series scheme, the parallel scheme, and the ladder scheme. Referring to the approximate differentiator shown in Figure 3-63, draw a graph of the output v(k)versus k when the input x(k) is a unit-step sequence. Consider the system shown in Figure 3-73. Show that the pulse transfer function Y(z)lX(z) is given by Figure 3-73 Digital integrator without delay. Assuming that y(kT) = O for k < O, show that Thus, the output y(kT) approximates the area made by the input. Hence, the system acts as an integrator. Because y(0) = Tx(O), the output appears as soon as x(0) enters the system. This integrator is commonly called a digital integrator without delay. Draw a graph of the output y(kT) when the input x(kT) is a unit-ctep sequence. Consider the system shown in Figure 3-74. Show that the pulse transfer function Y(z)/X(z) is @ven by Figure 3-74 Digital integrator with deiay. Assurning that y(kT) = O for k < O, show that y(kT) = T[x(O) + x(T) + . . + x((k - 1)T)I
- 93. The output y(kT) approximates the area made by the input. Since y(0) = O and y(T) = 7jc(0), the output starts to appear at t = T. This integrator is called a digital integrator with delay. Draw a graph of the output y(kT) when the input x(kT) is a unit-step sequen Consider the system shown in Figure 3-75. Show that the pulse transfer function Y(z)/X(z)is given by Figure 3-75 Dipital bilinear integrator. transfer function o£the digital.controller. Design techniques for continuous- trol systems based on conventional summarizes transient and steady-state response characteristics of discrete-time con- trol systems. The design technique based on the root-locus rnethod is presented in Section 4-5. Section 4-6 first reviews the frequency-response method and then presents frequency-response techniques using the w transforrnation for designing discrete-time control systems. Section 4-19 treats an analytical design method. 173
- 94. esign of Discrete-TimeControl Systems by Canventional Methods Cha The absolute stability and relative stability of the linear time-invariant continuous- time closed-loop control syste ocations of the closed-loop poles in the s plane. For example, peles in the left half of the s pXane near the ju axis will exhib ,and closed-loop poles sn ative real axis will exhibit mce the complex variables z = eTs,the poje and zero locations in the z plane are related to the pole arid zero locations in the s plane. Therefore, the stability of the linear time-invariant discrete-timeclosed-loop system can be deterrnined in terms of the locations of the poles of the closed-loop pulse transfer function. It is noted that the dynamic behavior of the discrete-time control system depends on the samplingperiod T. In terms of poles and zeros in the z plane, eir locations depend on the sampling period T. In other w sampling period Tmodifies the pole and zero Ioeations in the response behavior to change. ne. ]en the design of a are very important in predictirig the dynamic behavior of t designing discrete-time control systems, the locations of th zplane are very important. In the followingparagraphs we shall investigate how tbe locations of the poles and zeros in the splane compare with the locations of the poies rporated into the process, the co This means that trancformation z part w , we have a pole in the s plane can be located in the z plane through ihe = e". Cince the complex variable s has real part o and imaginary and From this last equation we see that poles and zeros in the s plane, where frequencies differin integralmultiples of the sampling frequency 2 d T , are mapped into the same locations in the z plane. This means that there are infinitelymany values of s for each value of z . Since u-is negative in the left half of the s plane, the left half of the s plane corresponds to Izl = eTu< 1 23-xjw axis in the s plane corresponds to zl = 1.That is, the imaginary axis in the S p h e (ihe line o-= O) corresponds to the unit circle in the z plane, and the interior of the unit circle corresponds to the left half of the s plane. ec. 4-2 Mapping Between the s s. Note that since/z = of z m to m. Consider a represen on the jw axis in the s plane. As this point moves from -j2 asto j i o,on ere o, is the sampling fre uency, we have /z/ = 1, a n d b varies from -71 to 71 in the counterc1ocl;wise direction in the z plane. As the re from j i w, to j$ w, on the jw axis, the corresponding theunit circIeoncein the counterclockwise direction. moves from -m to m on the jo axis, we trace the unit number of times. Frsm this añialysis, it is clear that each ane maps into the inside of the unit circle i If of the s plane may be divid nt-5an infinite n ~ b e rof periodic ian Figure 4-1. The primary s extends fromjw.5 =: -j ose0 j+ ntary strips extend from ji j$%, 2%to I ~ w , ... , and fr , -j2ws to -j$%?. ... In the primary strip, if we trace the sequence of points 1-2-3-4-5-1 in the s plane as shown by the circled numbers in Fig ]tecentered at the origin of th ng points 1,2, 3, 4, and 5 in nurnbers in Figure 4-2(b). The area enclosed by any of the compiementary st unit circle in the z plane. This means that the correspo Complementarystrip s plane Complementary strip 1m z plane Complernentary strip $j Figure 4-1 Periodic strips in the s plane and the corresponding region (unit circle centereti at the origin) in the z piane.
- 97. esign of Diccrete-Time Control ystemc by Conventional Methods Chap. 4 /z = 271. X 0.5= n== 1800 Thus, the spiral can be graduate Figure 4-7(b)]. Once the sam wda1 any point on the spi 4-7(b), wdcan be determi specified as w, = 1071.radtsec, then at point 69 Hence, wd at point P is Note that if a constant-damping-ratio line is in the second or t in the s plane then the spiral decays within the unit circle in the z pl if a constant-damping-ratioline is in the first or fourth quadrant in th corresponds to negative damping), then the spiral grows outside Figure 4-8 chows constant-damping-ratio loci for i= 0, c = 0.2,5 = 0.4,i= 0.6, ( = 0.8, and { = 1. The [ = 1locus is a horizontal line between points r = O and z = 1.(Note that Figure 4-8 shows only the loci in the upper half of the z which correspond to O 5 w 5 4ms. 'F e loci corresponding to --So, r 5 0 are the rnirror irnages of the loci in the upper half of the z plane about the horizontal axic.) Wotice that the constant {locj are normal to the constant wnloci in the S plane, as shown in Figure 4-9(a). In the z plane rnapping, constant w,loca intersect constant i spirais at right ancles, as shown in Figure 4-9(b). A mapping such as this, wbich preserves both the size and the cense of angles, is called a conforma1 rnupping. Im z plane -1.0 -0.5 O l= 1 0.5 1.O R e Figure 4-8 Constant-damping-ratio loci in the z plane.
- 98. ysternc by Conventional Methods C z plane W = W, Figure 4-11 (a) A desirable region in the s plane for closed-loop pole locations; (b) corresponding region in the z plane. On the basis o£the preceding discussions on mapping fsom the s plane to the z plane, the desirable region can be mapped to the z plane as in Figure 4-11(b). Note that if the dominant closed-loop poles of the continuous-time control cys- tem are required to be in the desirabie region specified in the s plane, then the domi- nant closed-loop poles of the equivalent discrete-time control systern must lie inside the region in the z plane that corresponds to the desirable region in the s plane. Once the discrete-time control system is designed, the system response characteristics must be checked by experiments or sirnulation. If the response characteristics are not satisfac- tory, then closed-looppoIe and zero locationsmust be modified until satisfactoryresults are obtained. entsLI, For discrete-time control systems,it is necessary to pay particular attention to the samplingperiod T. This is because, if the samplingperiod is tos 1ong and the sampling theorem is not satisfied, then frequency folding occurs and fhe effective pole and zero locations will be changed. Suppose a continuous-time control system has closed-loop po s = - ijwl in the s plane. If the sampting operation is involvedin this syst if o1 > 4o,, where ws is the sampling frequency, then frequency folding occ the system behaves as if it had poIes at s = -ulij(wl t nw,), where n = 1,2, 3, . . . . This means that the sampling operation folds the poles outside the primar strip back into the primary strip, and the poles will appear at s = --ul ij(wl - @S see Figure 4-12(a). On the z plane those poles are mapped into onepair of conjuga complex poles, as shown in Figure 4-12(b). hen frequency folding occurs, oscdla- eions with frequency o, - ol, rather than frequency al, are observed. tbe stabiiity of linear time-invariant single-input-single-output discrete-time control systems. Cvnsider the following dosed-loop pulse-transfer function sg~stern: Im z plane -12 (a) Diagram showing s plane poles at -u,k jw, and folded poies appearing at -fi +j(w, + w,), - al IT j(wl F 2w,), . . . ; (b) z plane mapping of s plane poles at -a1 +jw,, - 02 I+ j(wl + u,), - cr, +j(wl 2 2w,), ... The stability of the system defined by Equation (4-3), as well as of other types of discrete-time control systems, rnay be determined from the Iocations of the closed- loop poles in the z plane, or the roots of the characteristic equation 1. For the system to be stable, the closed-ioop poles or the roots of the charac- teristic equation must lie within the unit circie in the z plane. Any closed-loop pole outside the unit circle rnakes the system unstable. If a simple pole lies at z = 1, then the system becomes critically stable. Also, the system becomes critically stable if a single pair of conjugate complex poles lies on the unit circle in the z plane. Any multiple closed-loop pole on the unit circle makes the system unstable. 3. Closed-loop zeros do not affect the absolute stability and therefore may be located anywhere in the z plane. Thus, a linear time-invariant single-input-single-output discrete-time closed- loop control systern becomes unstable if any of the ciosed-loop poles lies outside the unit circle andlor any inultiple closed-loop pole lies on the unit circle in ihe z piane.
- 99. Consider the closed-loop control systern shown in Figure 4-13. Determine the stability of the system when K = 1. The open-loop trancfer function G(s) of the system is Referring to Equation (3-58), the z transforrn of G(s)is Since the closed-loop pulse transfer function for the cystem 1s the characteristic equation is 1 + G ( z ) =O which becomes ( z - 0.3679)(z - 1) + 0.36792 + 0.2642 = O The roots of the characteristic equation are found to be Z I = 0.5 +j0.6181, z2 = 0.5 - j0.6181 the system ic stable. It is important to note that in the absence of the sampler a second-order syskem is alwaysstabie. In the presence of the sampler, however, a second-order system such as this can becorne unstable for large values of gain. n fact, it can be dmwn that the second-order system shown in Figure 4-13 will become unstable if K > 2.3925. (See Example 4-7.) ii@. ñhree stability tests can be applied direc (z) = O without solving for the roots. Two of them are the Schur-Cohn stability test and the Jury stability test. These two tests Fignre 4-13 Closed-loop control system of Example 4-2. nalysís of Closed-loop ysterns in the z Plane reveal the existence of any unstable roots (the roots that lie outside the unit circle in the z plane). However, these tests neither give the locations of unstable roots nor e the effects of parameter changes on the sy ability, except for the always real, the Jury test is preferred to the Schur-Cohn test. ty Test. Hn applying the Jury stability test to a given charac- n P(z) = O, we construct a table whose ele coefficients of P(z). Assume that the characteristic equatio in z as follows: where a. > O. Then the Jury table becomes as given in 'Hable 4-1. Notice that the elements in the first row consist of the coefficients in P(z) arranged in the ascending order of powers of z. The elements in the second row consist of the coefficients of (z) arranged in the descending order of powers of z. The elements for rows 3 through 2n - 3 are given by the following determinants: 4-1 GENERAL FORM OF THE JURY STABILITY TABLE Row z0 z1 z z . . . ~ n - 2 Zn-i Zn
- 100. esign of Discrete-Time Control Systerns by Conventianal Methods Ckap. 4 Note that the last row in the table consists of t ee elements. @om: systerns, 2n - 4 = 1and the Jury table concists otice that the elernents in any even-ra.umberedrow are si irnmediately prece ing odd-numbered row. Test. A system with the c P(z) = O given by Equation (4-51, ~ewrittenas where a. > O, is stable if the following conditions are al1 satisfied: Construct the Jury stability table for the EolIowing characteristic equation: where a0 > O. Write the stability conditions. Referring to the general case of the Jury stability table given by Table 4-1, a SurY stability table for the fourth-order system may be constructed as shown in Table 4-2. This table is slightly modified from the standard form and is convenient for the compu- tations of the b's and c's. The determinant given in middle of each row gives the vahe of b or c written on the right-hand side of the same row. The stability conditions are as follows: Cec. 4-3 Stability Analysis of Closed-Loop Systerns in the r JURY STABILIN TABLE FOR THE FOURTH-ORDER SYSTEM Wow zO z l 2 z3 2 It is noted that the value of cl (or, in the case of the nth-order system, the value of q,) is not used in the stability test, and therefore the computation of c, (or ql)may be omitted. Examine the stability of the following characteristic equation: P(z) = z4 - 1 . 2 ~ ~+ 0 . 0 7 ~ ~+ 0.32 - 0.08 = O Notice that for this characteristic equation ao = 1 Clearly, the first condition, la4/< a,,, is satisfied. Let us examine the second condition for stability:
- 101. JURY SIABILITY TABLE FOR THE SYSTEM OF EXAMPLE 4-4 Row zO z1 z2 z3 z4 ec. 4-3 Stability Analysis of Closed-Loop ystems in the r Plane Examine the stability of the characteristic equation given by P(z) = z3 - 1 . 1 ~ ~- 0 . 1 ~+ 0.2 = o First we identify the coefficients: The second condition is satisfied. The third condition for stability becomes P(-1) = 1 + 1.2 + 0.07 - 0.3 - 0.08 = 1.89 > O, n = 4 = even ence the third condition is satisfied. We now cor~structthe Jury stability table. Referring to Example 4-3, we c the values of b3,b2,bi, and boand c2and co.The result is shown in Table 4-3. (Al the value of ci is shown in the table, cl is not needed in the stability test and th need not be computed.) Frorn this table, we get 1631 =: 0.994 > 0.204 = lbo/ lc2l = 0.946 > 0.315 = Thus both parts of the fourth condition given in Example 4-3 are satisfied. Sin conditions for stability are satisfied, the given characteristic equation is stable, roots lie inside the unit circle in the z plane. As a rnatter of fact, the given characteristic equation P(z) can be factored follows: P(z) = (z - O.8)(z + 0.5)(z - 0.5)(z - 0.4) As a rnatter of course, the result obtained above agrees with the fact that al1roo within the unit circle in the z plane. The conditions for stability in the Jury test for the third-order system are as follows: The first condition, la31 < ao,is clearly satisfied. Now we examine the second condition of the Jury stability test: This indicates that at least une root is at z = 1.Therefore, the system is at best critically stable. The remaining tests determine whether the system is criticaliystable or unstable. (Tf the given characteristic equation represents a control system, critica1 stability will not be desired. The stability test may be stopped at this point.) The third condition of the Jury test gives The third condition is satisfied. Wow we examine the fourth condition of the Jury test. Simple computations give b2 = -0.96 and bo = -0.12. Hence, The fourth condition of the Jury test is satisfied. From the above analysis we conciude that the given characteristic equation has one root on the unit circle (z = 1) and its other two roots within the unit circle in the z plane. Hence, the system is critically stable. Exa A control system has the following characteristic equation: P(z) = z3 - 1.32' - 0 . 0 8 ~+ 0.24 = O Determine the stability of the system. We first identify the coefficients:
- 102. esign of Discrete-Time Control ystems by Conwentional Clearly, the first condition for stability, /as/< ao, is satisfied. Next, we examine $he second condition for stability: P(1) = 1 - 1.3 - 0.08 + 0.24 = -0.14 < O The test indicates that the second condition for stability is violated. The system is may stop the test here. Consider the discrete-time unity-feedback control system (with sampling period T = 1 sec) whose open-loop pulse transfer function is given by Determine the range of gain K for stability by use of the Jury stability test. The closed-loop pulse transfer function becomes w-- K(0.36792 + 0.2642) R(z) - z2 + (0.3679K - 1.367912 + 0.3679 + 0.2642K Thus, the characteristic equation for the system is Since this is a second-order systern, the Jury stability conditions may be written as follows: 1. 14< ao 2. P(1) > O 3. P(-1) > O, n = 2 = even e shall now apply the first condition for stability. Since a, = 0.3679 + 0.2642K and a. = 1, the first condition for stability becomes 10.3679 4- 0.2642KI < 1 The second candition for stability becomes P(1) = 1 + (0.3679K - 1.3679) + 0.3679 + 0.2642K = 0.6321K > 0 which gives K > O The third condition for stability gives which yields For stability, gain constant K must satisfy inequalities (4-61, (4-71, and (4-8). Hence, 2.3925 > K > O 'The range of gain constant K Eor ctability is between O and 2.3925. tabilily Analysis of Closed-Loop ystems in the z Plane f gain K is set equal to 2.3925, then the system becomes critically stable (meaning that sustained oscillations exist at the output). The frequency of the sustained oscilla- tions can be determined if 2.3925 is substituted for K in the characteristic equation and the resulting equation is investigated. With K = 2.3925, the characteristic equation becomes The characteristic roots are at 2 = 0.2439 rt j0.9698. Noting that the sampling period T is equal to 1 sec, from Equation (4-2) we have @S 2 7 ~ ud= - = - /¿ = tan-' ---- 2 7 ~ 271. 0.2439 - 1.3244 radisec The frequency of the sustained oscillations is 1.3244 radisec. time control systems is to use the bi ion couplied with the Routh stability criterion. The method requires transformation from the z complex plane, the w plane. who are familiar with the stability criterion will find the amount of computation require criterion. The bilinear transformation defined by which, when solved for w, gives maps the inside of the unit circle in the z plane into t e left ha&of the w plane. ~ h i s can be seen as foliows. Let the real pan of w be called oand the imaginary part w , so that Since the inside of the unit circle in the z pIane is we get which yields
- 103. iscrete-TimeControl ystems by Conventional Met ec. 4-4 Transient and ,tfie inside of the unit circle in t Absolute stability is a basic r nt of al1 control systems. In reliative stability and steady- acy are also required of any whether continuous time or P(z) = aozn+ alzn-' + . + a,-l z + a, = O as follows: + + a,-l----- Then, clearing the fractions by multiplying both sides of this last equation by (w - l.)",we obtain systems with energy storage cannot respond instantaneously and will always exhibit Q(w) = bown + bl wn-l + - + b,-, w + b, = O transient response whenever they are subjected to inputs or disturbances. systems design, because unstable or critically stable control systems are not de As mentioned earlier, the arnsunt of computation required in this approach is more than that required in the Jury stabifity test. Therefore, we shall not go a further on this subject here. We refer the reader to Problem A-4-3, whe present method is used for stability analysis. Frequently, the performance characteristics of control system are specified in terms of the transient response to a unit-step input, since the unit-step input is easy to generate and is sufficiently e useful inforrnation on both the transient response and the steady-state response characteristics of the systern. e transient response of a system to a unié-step input depends on the initial conditions. For conveniente in comparing transient responses of various systems, it is a cornmon practice to use the standard initial condition: the system is at rest initially and the output and al1its time-derivatives are zero. The response character- istics can then be easily compared. The transient response of a practica1control system, w ere the output signaj is continuous time, often exhibits damped oscillations before reaching the steady state. (Thisis true for the majority of discrete-timeor digital control systemsbecause If we are interested in the effect of a system parameter on the stability of loop control system, a root-loc S diagram may prove to be useful AB may be used to compute an plot a root-locus diagram. It is noted that in testing the stability of a characteristic equation it S,to find the roots of the characteristic equation directl nt out that stability has nothing to do with the syslem9 ability to follow a particulax: input. The error signal in a closed-loop cmtr system rnay increase without bound, even if the system is stable. (Wefer t Section 4-4 for a diccussion of error constanes.)
- 104. -15 (a) Unit-step response of o kT the system shown in Figure 4-14; (b) discrete-time output in the unit-step (b) response. output c(t)of such a system to a unit-step input shown in Figure 4-15(a). Figure 4-15(b) shows Just as in the case of continuous-time control systems, of a digital contr01 system may be characterized not damped natural frequency, but also by the rise time time, and so forth, in response to a step input. In fact, in specifying such transient response characteristics, it is common to specify the following quantities: TRANSIENT RESPONSE SPECIFICATIONS l. Delay time t, . Wise time tr Peak time t, . Maximum overshoot M, Settling time ts The aforementioned transient response specifications in the unit-step response are defined in what follows and are shown graphicaily in Figure 4-16. 1. Delay time t,. The delay time is the time required for the response to reacb half tbe firial value the very first time. Transient and Steady-State Unit-step response cufve showing transient response specifications t,: tr, tp, M,,, ands. ise time t,. The rise time is the time required for the respons to 9O%, or 5% to 95%, or O% lo 100% of its situation. For underdamped second-order syste nly used. For overdamped systems a O% to 90% rise time is cornmoniy Peak time t,. The peak time is the time re for the response to reach the first peak of the overshoot. Maximum overshoot .The maximum overshoot is the maximum peak value of the response curve measured from unity. Hf t e final steady-stat, value of the response differs from unity, then it is common to use the maximum percent overshoot. It is defined by the relation aximum percent overshoot = The amount of the maximum (percent) overs stability of the system. . Settling time t,. The settling time is the time required for the response curve to reach and stay within a range about the final value of a size specified as an absolute percentage of the final value, usually 2%. The settling time is related to the largest time constant of the control system. The time-domain specifications just given are quite important since most control systems are time-domain systems; that is, they must exhibit acceptable time responses. (This means that the control system being designed must be modified until the traíident response is satisfactory.) Not al1 the specifications we have just defined necessarily apply to any given
- 105. iserete-Time Control y Conventional Met Consider the discrete-time control system defined by Obtain the unit-step response o£this system. program for obtaining the unit-step sesponse is shown in MATLA Program 4-1. The resulting plot oE c(k) versus k is shown in Figure 4-17. num = [O 0.4673 -0.339311; den = 11 -9 .XG!i' 0.66071; r = ones(l,$I); k = 8: 40; c = filter(num,den,r 1; pior(k,c,W v = 10 40 O 1.6); axic(vf; grid title('UnitStep Response') xlabel('k') Steady-State Emm Analyrd$is, An i ortant feature associated with transient response is steady-state error. The steady-state performance of a stable control systern is generally judged by tke steady-state error ation inputs. In what foltows we shall investigate a t caused by the inability of a system to followparticular types of inputs. (It shou!d be m e d thit, besides this type o£ steady-state error, there are errors that can be mribnted tcs ather causes, such as imperfections iil cystem climpcsnemts, cí&lc ec. 4-4 Trancient and Steady- tate Response Analysis -17 Unit-step response of the system defined by Equation (4-10). friction, backlash, or deterioration or agingof components. Hn this section, however, we shall not discuss steady-state error due to such causes.) ny physical control system inherently S rs steady-state error Bn response in types of inputs. That is, a system ma e no steady-state error with step inputs, but the same systern may exhibit nonzero stea y-state error in response hether or not agiven systemwill exhibit steady-state error in respo to a given type ofinput depends on the type of open-loop transfer function of system. Gonsider the continuous-time control system whose o tion G(s)H(s) is given by The term sNin the denominator represents a pole of multiplicity N at the origin. It is customary to classify the system according to the number of integrators in the open-loop transfer function. A system is said to be of type O, type 1, type 2,. .. , if W = O, N = 1, N = 2,. .. , respectively. Type O systems will exhibit finite steady-state errors in response to step inputs and infinite errors in response to ramp and higher-order inputs. Type 1systems will exhibit no steady-state error in response to step inputs, finite steady-state errors in response to ramp inputs, and infinite steady-state errors in response to acceleration and higher-order inputs. As the type number is increased, accuracy is improved. However, increasing the type number aggravates the stability problem. A comprornise between steady-state accuracy and relative stability (tran- skat response characteristics) is always necessary. The concepts of static error constants can be extended to the discrete-time control system, as discussed ln what follows.
- 106. E;<srthe system shown in Figure 4-18, define and Then we have and steady-state values. FTorn the diagram we have the actuating error e(t) = r(t) - b(t) e shall consider the steady-state actuating error at the sampling instants. Note that from the final valiue theorern we have lime(kT) = lim [(1- z-')E(z)] k-- 2-1 Figure 4-18 Discrete-time control system. Sec. 4-4 Transient and y substituting Equation (4-12) into Equatim ('-PI), we obtain (4-13) As in the case of the continuous-time control system, we consider three types oE inputs: un&-step, unit-ramp, an ~nit-accelerationinputs. Static Posktion Error Constant. For a un&-ste input r(t) = 1(t),we have 1 ( 2 ) = ---- a - Z-l y substituting this last equation into Equation (4-13) the steady-state actuating error in response to a unit-step input can be obtained as álollows: osition error constant Then the steady-state actuatingerror in response to a unit-step in from the equation The steady-state actuating error in response to a unit-step input becomes zero if & = which requires that GH(z)have at least one pole at z = 1. Static Velocity Error Constant. For a unit-ramp input r(t) = tI(t),we have Tz-' ( z ) = ( ~ -- 1 2 2 1 y substituting this last equation into Equation (4-131, we have Now we define the static velocity error constant Kvas follows: K, = lim z-+1 T (4-16) Then the steady-state actuating error in response to a unit-ramp input can be given by ifK, = S , then the steady-state actuating error in response to a unit-ramp input is zem. This requires G ( z )to possess a dorsble psle at z = 1.
- 107. Static Acceleration Error Constant. For a unit acceleration i n ~ u td r ) = Table 4-4 lists system ty nding steady-state errors in response to step, ramp, and accel diserete-time control system of the configuration shown in Figure 4-18. The steady-state error analysis just presented applies to t iscrete-time control system shown in Figure 4-18. For a differ configuration, it is noted that if the closed-loop discrete-time control system has a closed-loop pulse transfer function, t en the staticerror constants can be determined by an analysis similar to the e just presented. Table 4-5 lists the static error configurations of discrete-time control sy ntrol system does not have a closed-lo static error constants cannot be define d from the system at the terms "position error," "velocity er dy-state deviations in the output po transients have died out the input a at the same velocity, but have a finite position difference. By substituting this last equation into Equation (4-13), we obtain ,. r,, -,, 1 r 2 ( 1 + z-l)z-li .. T2 e dehne the static acceleration error constant Knas follows: Ka -- lim 2 4 1 T2 Shen the steadv-state actuatinp.error The steady-state actuating error in response to a unlt-acceleration input beco zero if Ka = 33. This requires GN(z) to possess a triple Equations (4-15), (4-17), and (4-19) give the ex actuating errors of the discrete-time control system shown in Figure 4-18 at samplirig instants for a unit-step, unit-ramp, and unit-acceleration in tively. . It is irnportant to emphasize that the actuating error is the differ- -5 STATIC ERROR CONSTANTS FOR TYPICAL CLOSED-LOOP CONFIGURATIONS OF DISCRETE-TIME CONTROL SYSTEMS Closed-loop configuration VaIues of Kp, K,, and K, ence between the reference input and the feedback signal, noi t tween the referente input and the out t. From the foregoing analysis we see that a type O systern will exhibit a constant ady-skate actuating step input and an infinite actuating error in response to r higher-order inputs. A type 1system will exhibit a zero stead in response to a step input, a constant steady-state error in response to a ramp input, &, = $5G(z)H(z) E_(, = iim (1 -z-"G(z)H(z) 2-1 T K, =lim ( 1 -~ - l ) ~ G ( z ) H ( z ),241 T 2 and an steady-state error response order inputs. SYSTEWI TYPES AND THE CORRESPONDING STEADY-STATE ERRORS IN RESPONSE TO STEP, RAMP, AND ACCELERATION INPUTS FOR THE DISCRETE-TIME CONTROL SYSTEM SHOWN IN FIGURE 4-18 P;P = G~(z)HGz(z) 1 -z I ) Gl(z)HGz(~) K, = iim ( 2-1 T 1 -Z-')~G~(Z)HGZ(Z) K~ = lim 'Z-+l T 2 1 Steady-state errors in response to Acceleration System Step input Ramp input input r(t) = 1 r(t) = t r(t) = St2 Type O system 1 co co 1 + KP Type 1system O -1 m K,' o O 1 Type 2 system - r, = $mGl(z)G&)H(z) K, = lim ( 1 -z-l)G1(z)G2(z)H(z) 2 4 1 T I(, = lim ( 1 -Z-')~G~(Z)GZ(Z)EI(Z) Z-+l T Z
- 108. Design of Discrete-TimeControl Syslerns by Conventional Methods Chap. 4 n examining transient response characteristics osed-loop pulse transfer function If 1GD(z)C(z)I 1, then we find Figure 4-19 (a) Digital closed-loop control system subjected to reference input and disturbance input; (b) modified block diagram where the disturbance input is con- sidered the input to the system. Sec. 4-4 Transient and Steady-Sbte Responce Analysis Since the system error is E(z) = W(z) - C(z) = -C(z) we find the error E(z) due to the 1 E(z) = -- GD(4 Thus, the larger the gain of GD(z)is, the smaller the error E(z). an integrator [which means that GD(z)has a pole at z = 11,then the steady-state error due to a constant disturbance Bs zero. Thks may be seen as follows. Since for a constant dicturbance of magnitude N we have if GD(z)invoIves a pole at z = 1, then it rnay be written as ere eD(z)does not involve any zeros at z = 1. T be given by Hf a linear system is subjected to both the reference input and a disturbance input, then the resulting error is the sum of the errors due to the reference input and e disturbance input. The total error must be kept within acceptable limits. Note that the point where the disturbance enters the system is very i nt in adjusting the gain of GD(z)G(z).For exarnple, concider the system in Figure 4-20(a). The closed-loop pulse transfer function for the disturbance is To minimize the effects of disturbance N(z) on the system error E(z), the gain of GD(z)G(z) must be made as large as possible. owever, for the system shown in Figure 4-20(b), the closed-loop pulse transfer function for the disturbance is to minimize the effects of disturbance N(z) on the system error E(z), the gain of GD(z)G(z)must be made as small as possible. Therefore, it is advantageous to obtain the expression E(z)/jV(z) before concluding whether the g n of GD(z)G(z)should be large or to minimize the error due to disturbances t is important to remember, howev hat the rnagnitude of the gain cannot be determined solelyfromthe disturbance considerations. It must be determined by considering the responses to both reference and disturbance e frequency regions for the reference input and disturbance input are Wfficientlyapart, a suitable fiiter rnay be inserted in the system. If the frequency
- 110. esign sf Discrete-Time ysterns by Conventional Met ec. 4-5 Design To combine these two forms into one, let us define t 1 + F(z) = O where at F(z) is the open-100 pulse transfer function. The characteristic equation given by Equation (4-20) can then be written as Since F(z) is a complex quantity, this last equation can be split into two equations by equating first the angles and then the agnitudes of the two sides &oobtain ASuGLE CONDITION: MAGNITUDE CONDITION : IF(z)l = 1 The values o£z that fu of the characteristic e A plot of the point e angle condition calorre is the root locus. The r e characteristic equation (the closed-loop corresponding to a given value of the gain can be located on the root loci by agnitude condition. %hedetails of applyingthe angle and magnitude conditions to obtain the closed-loop poles are presented in the following. cated, but actually it is not difficult if the are applied. y locating particuhr points and asymptotes and by computing angles 0 departure from complex poles and angles of arrival at complex zeros, it is possib to construct root Loci without difficulty.Note that whileroot locirnay be conveniewt drawn with a digital computer, if manual construction of the root locus plot attempted, we essentiallyproceed on a trial-and-error basis. ut the number of tri required can be greatly reduced if the establlshed rules are used. Because the open-loop conjugate complex poles and conjugate complexzems if any, are dways located symmetrically ut the real axis, the root loci are agway symmetric with respect to the real axis. rice, we need only construct the uppe half of the root loci and draw the mirro age of the upper half in the lower ha1 of the z plane. Remember that the angles of the complex quantities originating from t k open-loop poles and open-loop s and drawn to a test point z are meacured in the counterclockwise direction. shall now present the general rulec, and Obtain the characteristic equation d then rearrange this equation so arameter of interest, such as gain pears as the multipiying factor in r of interest is gain K , wh the closed-loop zeros consist of the zeros of G(z) and the poles of H(z).] ind the starting points and terminating points o er of separate branches of the root loci. The corresponding to K = O are open-loop poles and those cor ence, as K is increased from zero to infinity, a root locaas starts from an open-loop pole and terminates at a finite open-loop zero zero at infinity. This rneans that a root-locus plot will have just as as there are roots of the characteristic equation. [If the zeros at infi in the count, F(z) has the same number of zeros as poies.] f the number n of closed-1 e same as the number of o poles, then the nrrmber of ind 1 root Iocus branches termirrating open-loop zeros is equal to the number m of the loop zeros. The remaining n - m branches terminate at infinity (at n - m cit zeros at infinity) along asymptotes. . Determine the root loci on the real axis. Root ljoci on the real axis are determined by open-loop poles and zeros lying on it. The conjugate compliex poles and zeros of the open-loop pulse transfer function have no effect on the Iocation of the root loci on the real axis because the angle eontribution of a pair of conjugate complex poles or zeros is 360" on the real axis. Each portion of the root Iocus on the real axis extends over a range from a pole or zero to another pole or zero. Bn constructing the root loci on the real axis, choose a test point on it. If the tal number of real poles and real zeros to the ri of this test point is odd, then is point lies on a root locus. The root locus an s complement form alternate gments along the real axis. Determine the asymptotes of the root lo the test point z is located far from the origin, then the anglies of al1the complex quantities may be considered the same. One open-loop zero and one open-loop pole then each cancel the effects of the other. Therefore,the root loci for very large values of z must be asymptotic to straight h e s whose angles are given as follows: r180°(2Pd + 1) Angle of asymptote = , N = 0 , 1 , 2, . . . n - m L., pii-oceduresfor constrclcting root loci.
- 111. where (z)/A(z) = F(z), then n = number of finite poles of F(z) A ( 4K = -- N z ) (4-22) m = number of b i t e zeros of F(z) and the breakaway and break-in points (which correspond to multiple roots) can be determined from the roots of obtained as fo'ollows. Since -- z ~ - ~+ [(pl +p2 + +pn) - (z1+ Z2 + . + z , ) ] z ~ - ~ - ~+ . ne the angle of departure (or angle of arrival) of the root loci from n - m --ua = - (Pi + P2 + + pn) - (zi + zi + + zm) n - m Angle of departure = 480" - (8, 3- 8, - 4) If the characteristic equation 1 + F(z) = O is written as Figure 4-22 Diagram showing angle of departure.
- 112. tbus found give the location at which the root loci cross the imaginary axis an e value of the corresponding gain . Any point on the root loci is a possible clos -looppole when the value of gain K sa e magnitude condition enables us to root location on the locus. The ma f gain K of the open-loop pulse transfer function is given in the problem, t by applying the magnitude condition, Equation (4-24), it is possible to locate closed-loop poles for a given K on each branch of the root loci by a trial-and-e method. It is important to note th z) and the numerator of H(z involve common factors then the corresponding open-loop poles and zeros cancel each other,reducing the degree of the characteristic equation by one or The root locus plot of G(z)N(z) wi1Í not show al1 the roots of tfie charact equation, but only the roots of the reduced equation. Toobtain the complete set of closed-loop poles, we must add tfie cance or polies of G(z)H(z) to those closed-loop poies obtaine of G'(z)Ei(z). The important thing to reme ber i~that a cancel& pele of gg(z is a closed-loop pole of the system. As an example, consider the case where G(z) and H(z) of the svstern show 2 , --- -J - - - --- --- in Figure 4-21 are given by and Then, clearly, the pole z = -a of G(z) and the zero z = -a of other, resulting in z + c z + a-= z + c G(Z)H(z) = (z + a)(z + b ) z + d (z + b)(z + d) owever, the closed-loop pulse transfer function of the svstem is ased on the Woot-Loctds Method Note, however, that if pole-zers canmcellatáon occurs in the feed-forward pu transfer function, then the same e-zero cancellation occurs in the cl pulse transfer function. Consi hown in Figure 4-21, assume G(z) = G~(z)G~(z)9 (z) = 1 pose pole-zero cancellation occurs in GD(z)Gl(z).Por exa Then the closed-loop pulse transfer function beco ecauseof the pole-zero cancellation,the third-order systembecomes one of second t is important to summarize that the effect of pole-ze ( 2 ) is different from that of pok-zem cancellation transfer function (such as pole-zera cancellation in the ant). In the forrner, the canceled of the closed-loop system ereas in the latter the canceled po system (in the Iatter the order of the S poles). In what Eollows we shall T on the relative stability of the closed-loop control system. C own in Figure 4-23. Assurne that the digital controller is of the integral type, or that Let us draw root lscus diagramsfortke systemfor three values of thesam T:0.5 sec, 1sec, and 2 sec. Let us also determine the critica1value of K for eac case. And finally let us locate the closed-loop poles corresponding to K = 2 for eac of the three cases. Integral controller cFigure 4-23 Digital control system.
- 113. e shall first obtain the z transforrn of Gh(s)GP(s): -- - 1- e-T-- z - e-= The feedforward pulse transfer function becomes Kz 1 - e-* G(z) = G,(z)2?[Gh(s)G,(s)I = ----z - l ~ - e - ~ The characteristic equation is 1+ G(z) = O or 1, Notice T( Surnplingperiod T = 0.5 sec: For this case, Equation (4-25) become G(z) = 0.3935Kz (z - 1)(z - 0.6065) that G(z)has poles at z = 1and z = 0.6065 and a zero at z = 0. 3 draw a root locus diagram, we first locate the poles and then find the breakaway point and break-in point. Not pulse transfer function with two poles and one zero results in a circ centered at the zero. The breakaway point and break-in point are writing the characteristic equation in the form of Equation (4-22), ,= - (z - 1)(z - 0.6065) 0.3935~ and differentiating K with respect to z and equating the result to zero: dK = - z2 - 0.6065- dz 0.3935z2 = o Hence, z2= 0.6065 Or z=0.7788 and z=-0.7788 Equation (4-27) yiel nce both K values are positive, z = 0.7788 Figure 4-24(a) shows the root locus diagrarn value of gain K for this case is obtained by use of t can be obtained from Equation (4-26) as follows: lrn z plane ( = 0.5 sec) I m z plane íT = 2 sec) Figure 4-24 (a) Root locus diagrarn for the system shown in Figure 4-23 when T = 0.5 sec; (b) root locus diagram when T = 1 sec; (c) root locus diagram when T = 2 sec.
- 114. Decign of Discrete-Time Control Syctems by Conventional Methods @ha For the present case, T = 0.5 and this Iast equation becomes Since the critica1 gain Mccorresponds to point z = -1, we substitute -1 for z in Equation (4-28) : or K = 8.165 The critical gain M,is thus 8.3 65. The dosed-loop poles correspon zl = 0.4098 +j0.6623 and z2 = 0.4098 - j0.6623 These closed-loop poles are indicated by dots in the root locus diagram. Sampling period 1' = I sec: r this case, Equatiorl (4-2 5) beca follows: Kence, G ( z )has poles at z = 1 and z = 0.3679 and a zero at z = 0. The breakaway point and break-in point are found to be z = 0.6065 and z = -0.6065, respectively. The corresponding gain values are K = 0.2449 and K = 4.083, respectively. Figure 4-24(b) shówsthe root locusdiagram when T = 1sec. T of gain K is 4.328. The closed-loop poles corresponding to K = 2 are found to be z, = 0.05185 +j0.6043 and z2 = 0.05185 - j0.6043 and are shown in the root locus diagram by dots. 3. Sampling period T = 2 sec: For this case, Equation (4-25) becomes e see that G ( z )has poles at z = 1 and z = 0.1353 and a zero at z = 0. The breakaway point and break-in point are found to be z = 0.3678 and z :-0.3678, with corresponding gain values K = 0.4622 and K = 2.164, respece tively. The critical value of gain K for this case is 2.626. Figure 4-24(c) shows the root locus diagram when T = 2 sec. The closed-loop poles corresponding to K = 2 are found to be z, = -0.2971 +j0.2169 and z2 = -0.2971 - j0.2169 These closed-loop poles are shown by dots in rhe root locus diagram. closed-loop pole is i,then in the s plane the closed-loop pule location (in the upper half-plane) can be given by s = -bn+j u n V i - C2 e corresponding point in the z plane is z = exp [T(-iwn +j w n m ) ] z plane Figure 4-25 Closed-loop pole locations in the z plane shown with constant 5loci.
- 115. from which we get Izl e - T h and & = TU.- = To, = 8 (rad) From Equations (4-29) and (4-30), the value of J can be calculate in the case where the sampling period T is 0.5 sec, we have the elos K = 2 at z = 0.4898 +j0.6623 lz/ = 'd0.4098' + 0.6623' = 0.7788 y solving IzI = = 0.7788 ' for the exponent, we find T b , = 0.25 Also, & = = 58.25" = 1.0167 rad 0.4098 ence, fi= T w , W = 1.0167 rad From Equations (4-31) and (4-32), we obtain TSmn 0.25 ~ ~ ~ ' d i - 5 2= 1.0167 or which ylelds J = 0.2388 (From Figure 4-25 we graphically obtained 0.24 for S, which is actual Svalue of 0.2388.) It is irnportant to point out that in the second-order system t l' is indicative of the relative stability (for erarnple, in respect to the m overshoot in the unit-step response) only if the sa high (so that there are eight to ten samplin frequency is not high enough, the maximu be much higher than would be predicted To compare the effects of different sampling sponse, we shall compare the unit-step response se T considered in the preceding analysis. aced on the Root-Lscus closed-loop pulse transfer function for t rd pulse transfer function C(z) is given For T = 0.5 sec and response can be given by from whicb we obtain t response sequence c(kT) versus kT s Figure 4-26(a). From Figure 4-25 we see that the angle of the Bine connecting the origin and pole at z = 0.4098 +j0 ately 58.25".The angle 8 termines the number of samples per cycIe a£s atian. Note that Hence, for 8 = 58.25", we have 360°/8 = 36Q0/58.25"= 6.18 samples per cycle of ed oscilliation, as seen fro Similarly, for T = 1sec and K = 2, the unit-step response is given by e unit-step response sequence c(kT) versus in Figure 4-26(b). Since e angle of the line connecting the origin and oop pole for the present case is 85.10°, as shown in Figure 4-25, we ha ately 360°/85.10" = 4.23 samples per cycle, which is ve recommend eight or more sa Finally, for T = 2 sec and R = 2, the unit-step response is given by The unit-step response sequence c(kT) versus kT is shown in Figure 4-26(c). Frorn Figure 4-25, the angle of the line connecting the origin and the closed-loop pole for the present case is 143.87", and consequently we have 360"/143.87"= 2.50 samples per cycle, as seen from Figure 4-26(c). (Note that a slow sampling frequency such as 2.50 samples per cycle is unacceptable.) Figure 4-26 has shown three different plots of the unit-step response sequence c(kT) versus kT. As can be seen from these plots, if the sampling period is srnall, a plot of c(kT) versus kT will glve a fairly accurate portaait o£the response c(t). However, if the sa eriod is not sufficiently small, then the plot of c(kT) versus kT will not an accurate resullt. It is very important to choose m adequate sampling d based on the satisfaction of the samplingtheorem, systern
- 116. n of Discrete-TimeControl ystems by Conventional Methodc Chap. O 1 2 3 4 5 6 7 8 k T ( s e c ) 1 2 3 4 5 6 7 8 k T ( s e c ) Figure 4-26 (a) Unit-step response sequence of the system shown in Figure 4-23 when T = 0.5 sec and K = 2; (b) unit-step response sequence when T = 1sec and K = 2; (e) unit-step response sequence when T = 2 sec and K = 2. dynarnics, and actual hardware considerations. Note that barely satisfying the S pling theorern is not sufficient. An acceptable rule of thumb is to have eight to sarnples per cycle (six samples per cycle is marginal) if the system is underdamp and exhibits oscillation in the response. Next let us investigate the effect e sarnplingperiod T on the steady- sball consider the unit-I- case where the samplirr open-loop pulse trancfer furrction is ec. 4-5 Design e static velocity error constant Kv is given by = lim ----- Y 0.78702 Z+I 0.52 (z - l)(z - 0.6065)1 Thus, the steady-state error in res '= 0.25es, = -= - KJ 4 irnilarky,for the case where T = 1secand = 2, the open-loop pulse transfes function is and the static velocity error constant Kv is given by (1 - z-')G(z) Kv= lim m and the steady-state error in response to a unit-ram 1= - = 0.5ess = K, 2 Finaly, for the case where T = 2. sec and K = 2,t function is anta the static velocity error constant unit-ramp input are obtained , respectively,as h r t s (a), (b); a*$ (c) of Figure 4-27 show, respectively, the plots of tke unit-ramp response sequence c(kT) versus kT for the three cases considered.
- 117. Design of Discrete-Time Control Systems by Conventional ec. 4-5 Design Based on the Root-Locus Method Figure 4-27 (a) Unit-ramp response sequence of the system shown in Figure 4-23 when T = 0.5 sec and K = 2; (b) unit-ramp response sequence when T = 1sec and K = 2; (c) unit-rarnp response sequence when T = 2 sec and K = 2. The three cases we have considered demonstrate that increasing the samp period T adversely affects the system's relative stability. (It may even cause insta ity in some cases.) It is important to remember that the damping ratio i of t ched-loop poles of the digital control system is indicative of the relative stddit only if the sampling frequency is sufficiently high (that is, eight or more sarnpk P ed sinusoidal oscillation). ncy is low (that is, less les per cycIe of damped ), then predicting the relative stability from the damping ratio value is erroneous. Consider the digital control systemshownin Figure 4-28. In the z plane, design a digital controller such that the dominant closed-loop poles have a damping ratio 5of 0.5 and a settling time of 2sec. The samplingperiod is assumed to be 0.2 sec,or T = 0.2. Obtain the response of the designed digital control system to a unit-step input. Also, obtain the static velocity error constant 6(,of the system. For the standard second-order system having a pair of dominant closed-loop poles, the settling time of 2 sec means that 4 4 settling time = -= -= 2 lw, 0 . 5 ~ ~ which gives the undamped natural frequency w, of the dorninant closed-loop poles as wn = 4 The damped natural frequency wd is determined to be wd = con- = 4- = 3.464 Since the sampling period T is 0.2 sec, we have [Notice that there are approximately nine samples per cycle of damped oscillation (31.4213.464 = 9.07). Thus, the sampling period of 0.2 sec is satisfactory.] shall first locate the desired dominant closed-loop poles in the z plane. Weferring to Equations (4-29) and (4-30), for a constant-damping-ratio locus we have and @d 12= Twd = 27T- @S From the given specifications(5 = 0.5 and wd = 3.464), the magnitude and angle of the dominant closed-loop pole in the upper half of the z plane are determined as follows: Figure 4-28 Digital control systern for Example 4-9.
- 118. and 3 464 & = 27.r- = 0.6927 rad = 39.69" 31.42 can now locate the desired dominant closed-loop pole in the upper half of the z e, shown in Figure 4-29 as point P. Note that at point P z = 0.6703/39.69" = 0.5158 + j0.4281 Noting that the sampling period T is 0.2 sec, the pulse transfer function G(z) the plant preceded by the zero-order hold can be obtained as follows: LIUS I ~ S Leyuarion can oe written as Next, we locate the poles (z = 1and z = 0.6703)and zero (z = -0.8760) on the z plane, as shown in Figure 4-29. If point P is to be the location for the dominant closed-loop pole in !he upper half of the z plane, then the sum of the angl at point Pmust be equal to t180".However, the surn of the angle contributionsat go P is 17.10" - 138.52" - 109.84" = -231.26" Hence, the angle deficiency is -231.26" + 180" = -51.26" The controller pulse transfer function must provide $51.26". The pulse transfer fun tion for the controller may be assumed to be r plane @re4-29 Root locus diagram of the system considered in Example 4-9. ased on the Woot-Locus Methad is the gain constant of the controller. lf we decide to cancel the pole at z = 0.6703 by the zero of the controller at z = -a, then the pole of the controller can be detesmined (from the condition that the controller must provide +51.26") as a point at z = 0.2543 (@ = -0.2543). Thus, the pulse transfer function for the controller may be determined as The open-loop pulse transfer function now becomes The gain constant K can be determined from the followingrnagnitude condition: IGD(z)G(z)/,= 0.5158 + j0.4281 = 1 Hence, O.O1758(z + 0.8760) (z - 0.2543)(z - 1) z=0.5158+~0.4281 = 1 which gives K = 12.67 The designed digital controller is The open-loop pulse transfer function for the present system is Hence, the closed-loop pulse transfer function is The response to the unit-step input R(z) = 1/(1 - z-l) can be obtained from Figure 4-30 shows the unit-step response sequence c(kT) versus kT. The plot shows that the maximum overshoot is approximately 16% (which means that the damping ratio is approximately 0.5) and the settling time is approximately 2 sec. The digital controller just designed catisfies the given specifications and is satisfactory.
- 119. ecign Gased on the Frequency- Familiarity with the relati ane pole and zero Iocations and the time response charact ning discrete-time control between O and 1i Similarly, in the s plane a closed-loop poPe on the negative real axis neai: the origin increases the settling time. In the z plane, su to a closed-loop pole on the positive real axis betwee pole in the z plane bekween O and 1(Pn particular, n settling time. The psesence of a closed-loop pole or between O and -1 in the z plane, however, affect slightly. 0.4 0.2 O 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 kT ísec) N-30 Unit-step response sequence of the system designed in Exampie 4-9. The static velocity error constant M, of the system is given by = 2.801 1 If it is required to have a large value of Kv, then we may include a lag cornpen- sator. For example, adding a zero at z = 0.94 and a pole at z = 0.98 would raise t- K.. value three times, since (1 - 0.94)1(1 - 0.98) = 3. (It is important that the pole and zero of the lag compensator lie on a finite number of allocable discrete points.) A lag compensator, which has a pole and a zero very close to each other, does not significan ti^ change the root locus near the dominant closed-looppoles. The effect of a lag compen- sator on the transient response is to introduce a srnall but slowly decreasing transienl component. Such a srnall but slow transient, however, is not desirable frorn the vkw- point of disturbance or noise attenuation, since the response to disturbances would not attenuate promptly. Finally, it is noted that although the designed system is of the third order, it acts as a second-order system, since one pole of the plant has been canceled by the zero of the controller. Because of this, the present system has only two closed-loop poles. The dominant closed-loop poles are the only closed-loop poles in this case. If a pole and a zero do not cancel each other, then the system will be of the third order. en& lt is important to note that the poles of thp ~ ~ ~ C P A - ~ A ~ T Pn1llsc transferfunction determine the natural modes of the svstem. frequency response behaviors, however, are stro the c~o~ed-looppulse irancfer function. The freqeaency-response concept plays the same powerful role in digital control systems as it does in continuous-time control systems. s stated earlier, it is assumed in this book that the reader is familiar with conventional frequency- se design techniques for continuouc-time control systems. In fact, familiarity de diagrams (logarithmic plots) is necessary in the extension of the comen- tional frequency-response techni ues to the analysis and design of control systems. Frequency-response methods have very frequently been used in the sator design. She basic reason is the simplicity of the rnethods. In pe frequency-response tests on a dáscrete-time system, it i nt that the system have a low-pass filter before the sampler so that sideban response o£the linear time-invariant system to a sinuso quency and modifies only the amplitude and phase of amplitude and phase are the only two quantities that must be dealt with. In the following, we shall analyze the response of the linear time-invariant discrete-time system to a sinusoidal input; this analysis will be followe definition of the sinusoidal pulse transfer function. Then e discuss the design of a discrete-time control systern in the w plane by use of a Earlier in this book we stated that the frequency response of G(z)can be obtained by substituting z = elwTinto G(z).In what follows we shall show that this is indeed true. Gonsider the stable linear time-invariant discrete-time system shown in Fig- 4-31. The input to the system G(z)before sampling is u(í) = sin wt
- 120. Design of Discrete-Time Control Systems by Conventional Methods Chap. Stable linear time-invariant discrete-time -31 Stable linear time-invariant discrete-time system The sampled signal u(kT)is u(kT) = sin kwT The z transform of the sampled input is U(z)= Z' [sinkoT] = z sin wT (z - eJ w T ) ( ~- e-lwT) The response of the system is given by X(z) = C(z)U(z)= G ( z ) z sin wT (z - elwT)(z- e-~wT) - az-- a~ 2 - elwT +-----z - e-~wTi[tesrnsdueto poles of G(z)] Multiplying both cides of Equation (4-34) by ( z - epT)/zywe obtain [terms due to poles of G The second term on the right-hand side of this Iast equation a approaches elwT.Since the system considered here is stable, the third term on th right-hand side also approaches zero as z approaches e'"'. Hence, by letting. ,- The coefficient E, the complex conjugate of a, is then obtained as follows: C(e-jwT> approach elw',we have Let us define ~f (el"T)P; Mele Then Equation (4-34) can naw be written as ased on the Frequency-Responce Method The inverse z transform of this Iast equation is The Iast term on the right-hand side of E uation (4-35) repsesents t response. Since the system G ( z ) has been assurned to be stable, al1 transient response terms will disappear at steady state and we wPlI get the following steady- state response x,,(kT): bjected to a sinusoidal in ase angle, is given by S = @(o)=/C(eJwT) In terms of G(eiwT),Equation (4-36) can be written as follows: x,(kT) = 1G(elwT)lsin (kwT +/G(ejwT)) have shown that G(ejwT)indeed gives the magnitude and phase of the frequency response of C(z).Thus, to obtain the frequency response of G(z),we need only to substitute elwTfor z in G(z).The function G(eiwT)is coznrnonlycalled the sinusoidal pulse transfer function. Noting that we find that the sinusoidal pulse transfer function G(eiwT)is periodic, with the perio equal to T. Consider the system defined by x(kT) = u(kT) + ax((k - 1)T), O < a < 1 where u(kT)is the input and x(kT) the output. Obtain the steady-state output x(kT) when the input u(kT)is the sampled sinusoid, or u(kT) = A sinkoT. The z transform of the system equation is X ( z ) = U(z)+ az-'X(z) By defining G(z)= X(z)/U(z),we have
- 121. iscrete-Time Control Let us substitule eJwTforz in G(z). Then the sinusoidal pulse transfer hnction G(e1"3 can be obtained as G'(eJ"*) = 1 - 11 - ae-jmT - 1- a coswT +ja sinwT The amplitude of G(eJuT)is IG(eJuT)l= M = 1 VI + a2 - 2a c o s w ~ and the phase angle of G(eJWT)is /G(eJ"T) = 6, = -tan-' a sin wT 1- a coswT Then the steady-state output x,,(kT) can be written as,follows: x,(kT) = AM sin (kwT + 8) -- A isin kwT - tan-' 2/1 + a2 - 2a c o s w ~ ne. Before we ca apply our well-developed frequency-response methods to the ana discrete-time control sgstems, certain rnodifications in the z pl necessary. Since in the z plane the frequency appea frequency response in the z plane, the sim completely lost. Thus, the direct application of freque worthy of consideration. In fact, since the z transformat complementary strips of the left half of the s plane into th conventional frequency-response methods, which d plane, do not apply to the z plane. The difficulty, however, can be overcome by transforming the pulse tran function in the z plane into that in the w plane. The transformation, cornrnonly ca the w transformation, a bilinear transforrnation, is defined by Z = 1 + (TI2)w 1 - (2'12)~ where T is the sampling period involved in the discrete-time control system und consideration. By converting a given pulse transfer function in the z plane into rational function of w, e frequency-responsemethods can be extended to discret time control systerns. y solving Equation (4-37) for w9 we obtain %heinver relationship 2 2 - 1w=--- T z + 1 Through the z transforrnation and the w transforrnation, the prim the kit half of the s plane is first mapped into the inside of the unit circl plane and then mapped into the entire left half of the w plane. The two processes are depicted in Figure 4-32. (Note that in the s plane we conside primary stiip.) lvotice that the origin of the z plane is mapped into w = -2lT in the w plane. 1Votice also that, as s varies from O to joJ2 along Sec. 4-6 Design Based on the Frequency-Responce Method -32 Diagrarns showing mappings fromthe s plane to the z plane and from the ,z plane to the w plane. (a) Primary strip in the left half of thes plane; (b)z plane mapping of the primary strip in the s plane; (c) w plane mapping of the unit circle in the z plane. lane, z varies Pom 1to -1along to 03 along the imaginary axis in Although the left half of the w plane corre o the left half of the s plagie and the imaginary axis of the w plane correspond there are differences between the two plane behavior in the s plane over the frequency rang maps to the range where v is the fictltious frequency e. This means that, frequency response characteristi filter will be repro- discrete or digital filter, the fiequency scale on which the response occurs will be compressed from an infinite interval in t e analog filter to a finite interval in the digital filter. Once the pulse transfer function G ( z )is transformed into G ( w )by means of the w transformation, it may be treated as a conventional transfer function in w. Conventional frequency-response techniques can then be used in the w plane, and SO the well-established frequency-response design tec niqUescan be applied to the design of discrete-time control systerns. noted earlier, v represents the fictitious frequency onal frequency-sesponse techniques may be used to for the transfer function in w. (In the brief review of the section, we shall use the fictitious frequency v as the variable.) Although the w plane resembles the s plane geometrically, the frequency axis in the w plane is distorted. The fictitious frequency v and the actual frequency o are related as follows:
- 122. ecign Based on the Frequency- Equation (4-39) gives the relationship between ihe actual frequency o a fictitious frequency v. Note that as the actual fre uency o moves from -4 @S t uency v moves from -co to O, and as w moves from O to Sw,, v Referring to Equation 44-39), the actual freauencv w can he traaacii~t~dk f r ~th correspondlng bandwidth in the w pllane is (212') tan.( - ,r J corresponds to G(jw,), where wl = (%T)tan-"(v, n2). figure 4-33 s tionship between the fictitious frequency v times iT and the actual frequency w f the frequency range between O and Sw,. Notice that in Equation (4-39) if wT is small then v = o This means that for small wT the transfer functions C(s) and G(w) resemble ea other. Wote that this is the direct result of the inclusion of the scale factor 21T Equation (4-38). The presence of this scale factor in the transformation enables to maintain the same error constants before and after the w transformation. (T means that the transfer function in the w plane will approach that in the s hes zero. See Example 4-11, which follows.) Consider the transfer-function system shown in Figure 4-34. The sampling period assumed to be 0.1 sec. Obtain G(w). The z transform nf Gfr) i<: Figure 4-33 Relationship between fictitious frequency v times 4T and actual frequency w for the frequenc range between O and 4w,. < JY Gís) -34 Transfer-function system of Example 4-11. By use of the bilinear transforrnation given by Equation (4-371, or G(z) can be transformed into G(w) as follows: Notice that the Iocation of the pole of the plant is s = -10 and that of the pole in the w plane is w = -9.241. The gain value in the s plane is 10and that in the wplane is 9.241. (Thus, both the poie locations and the gain values are similar in the s piane and the wplane.) However, G(w) has a zero at w = 2lT = 20, although the plant does not have any zero. As the sampling period Tbecomes smaller, the w plane zero at w = 2l-F approaches infinity in the right half of the w plane. Wote that we have 1olim G(w) = iim- w-+o s-OS + 10 This fact is very useful in checking the numerical calculations in transforming G( S ) into G(w)- To sumrnarize, the w transforrnation, a bilinear transformation, maps the inside of the unit circle of the z plane into the left half of the w plane. The overall result due to the transformations from the s plane into the z plane and from the z plane into the w plane is that the wplane and the S plane are similar over the region of interest in the s plane. This is because some of the distortions caused by the transformation from the s plane into the z plane are partly cornpensated for by the transformation from the z plane into the w plane. Note that if
- 123. esign Based on the Frequency-Wesponse Method where the si's and bi7sare constants, is transformed into the w plane by the trans- formation z = 1 + (TI2)w 1 - (TI2)w then G(w)takes the form where the aiysand the Piysare constants (some of m may be zero). Thus, G(w) is a ratio of polynomials in w ,where the degrees of numerator and denorninator may or may not be equali. Since G(jv) is a rational ction of v, the Nyquist stability criterion can be applied to G(jv). En terms of th iagram, the conventional straight-line approximatiori to the magnitude curve as'well as the concept of the phase margin and gain margin apply to C(jv). Design by means of put-single-sutput lar, if the transfer function is in a facto ode diagram can be drawn and reshaped are well known. As stated earlier, the conventional freq transfer funclions in the w plane. Recalh that the separate plots, the logarithmic magnitude IG / G ( jv) versus log v. Sketchingof the logarith of G(jv), so that it works on the principle instead of multiplying individual terms. Fam be applied, and therefore the magnitude stsaight-line asymptotes. Usirng the Bode di controller rnay be designed with conventio It is irnportant to note that there rnay be a difference in the magnitudes for C(jo) and G(jv). The high-frequency asymptote of magnitude curve for G(ju) rnay be a constant-decibel line (that is, a horizon On the other hand, lim,,,G(s) = O, then the magnitude of G(jo) alw proaches zero (-m d as o approaches infinity. For exarnple, referring to Exarnp 4-11, we obtained G ( w )for G(s)as follows: The high-frequeracy magnitude of G(jv) is whlle the high-frequency magnitude of the plant is 10 lirn 1-1jo + PO = O The differencein the Bode diagrams at the high-frequency end can be explained a fo!iows. First. recail that we are interested only in the frequency range O 5 o 5 whichcorresponds to O 5 v 5 m. Then, noting that v = in t to o = 1o,in thesplane, it canbe said that lim,,, /C(jv)lcorr ( j o + lo)/,which is a constant. (Ht is irnportant to note tha enerally not equal to each otker.) From the pole-zero polnt of view, it can be sai at when IG(jv)l is a non constant cnt v = m it is implied that G(w) contains e same number of pole In general, one or m G(w) lie in the right f the w plane. The resence of a zero in the right half of the w ) is a nonminimum phase transfer function. Therefore, we must be careful in drawing the phase angle curve in the Bode diagram. approach to the analysis and design of control systems is particularly useful for the following reasons: ode diagram the low-frequency asymptote of the magnitude curve is he static error constants KP,Kv, or transient response can be translated into those of the n terms of the phase margin, gain margin, bandwidth, and so forth. These specifications can easily be handled in the particular, the phase and gain margins can be rea directly from the Bode diagrarn. The design of a digital compensator (or digital controller) to satisfy the given in terrns of the phase margin and gain margin) can be carried e diagram in a simple and straightforward manner. phase lag-lead compensation techniques. Phase lead compensation is cornmonly used for irnproving stability margins. The phase lead compensation increases the system bandwidth. Thus, tbe system has a faster speed to respond. However, such a systern usin phase Iead compensation may be subjected to high-frequency noise problems ue to its increased high- frequency gains. hase lag compensation reduces the systern gain at higher frequencies without reducing the system gain at lower frequencies. The S em bandwidth is reduced and thus the system has a slower speed of response. cause of the reduced high- frequency gain, the total system gain can be increased, and thereby low-frequency gain can be increased and the steady-state accuracy can be improved. Also, any bigh-frequency noises involved in the system can be attenuated. In some applications, a phase lag compensator is cascaded with a phase lead compensator. The cascaded compensator is known as aphase lag-lead compensator. By use of the lag-lead compensator, the low-frequency gain can be increased (which means an improvement fn steady-state accuracy), while al the same time the system bandwith and stability margins can be increased. Note that the PID controller is a special case of a phase lag-lead controller. The PD control action, which a£fectsthe high-frecpency region, increases the phase
- 124. techniques for PID controllers basically follow those of phase lag-lead compe tors. (lín industrial control systerns, however, each PID control action in the controller may be adjusted experimentally.) numbers of bits, if the number of bits employed is insufficient, the pole and locations of the filter rnay not be realized exactly as desired and the resulting inaccuracies in the locations of poles and zeros, (The important thing to remernber is that the poles and zeros of the filter in the z plane rnust lie on a finite number of allowable discrete points,) 8, First, obtain G(z), the z transforrn of the plant preceded by a hold. Tbe* transforrn G(z) into a transfer function G(w) through the bllinear transfwma- tion given by Equation (4-37): 1 + (TJ2)w z = 1 - (7-/2)w Figure 4-35 Digital control system. That is, Ht is important that the sampling period T be chose is to sample at the frequency 10times that of the b system. (Although digital controlsand signalprscessing use similar approaches ing continuous-time signals, t the field of signal ,while in the field of re generally low. Sra quencies is mainly due to tbe different trade-offs in these twu fields.) Substitiste w = jv into G(w) and plot the Bode diagra ode diagram the static error constants, t the low-frequency gain of the discrete-time controller (or transfer function GD(w)is unity, determine the system gain by satisfying the requirement for a given static error constant. Then, by using conventional design techniques for continuous-time control systerns, deter- mine the pole(s) and zero(s) of the digital controller transfer function. [CD(w) is a ratio of two polynomials in w.fThen the open-loop transfer function of the designed system is given by GD(w)G(w). .Transform the controller transfer function CD(w) into GD(z) through the bilinear transformation given by Equation (4-38): 2 2 - 1 Then GD(z) = GD(w)~w=(~,r)(z-~)i(z+i, is the pulse transfer function of the digital controller. 6,Realiize the pulse transfer function GD(z)by a computational algorithm. esign procedure just given, it Is important to note the following:
- 125. Design of Discrete-Time Control ystems by Conventional Metho ased on the Frequency- The transfer function G(w) is fer function. Hence Thus, K[0.01873(ii) 1- O.lw + 0.01752] consideration the nonmi ( ) = (1 + "1.T - 1.8in7('iu'")+ o 8 1 g The frequency axis in the w plane is distorted. 1- 0.lw 1- 0 . 1 ~ Eictitious frequency v and the actual frequency w is -- K(-0.000333w2 - 0.09633~+ 0.9966) 2 wT w2 + 0.9969~ v = - tan- T 2 f, for exarnple, a bandwidth wbis specified, we need to desi -- K ( l + &)(1-5) w(w + 1) 2 @bT ~ A simplephase-lead compensator will probably satisfy al1 requirements. There- vb = T tanl fore, we shall try lead compensation. (If Iead compensation does not satisfyal1require- ments, we need to use a different type of compensation.) Consider the digital control system shown in Figure 4-36. Design a digital contr in the w plane such that the phase margin is 50°,the gain margin is at least 10 dB the static velocity error constant Kv is 2 sec-l. Assume that the sampling peri 0.2 sec, or T = 0.2. First, we obtain the pulse transfer function G(z) of the plant that is preceded by the zero-order hold: -- K(0.01873~+ 0.01752) z2 - 1.8187~+ 0.8187 Next, we transform the pulse transfer function G(z)into a transfer function G(w by means of the biliriear transformation given by Equation (4-37): z =: 1 + (T12)w - 1+ 0 . 1 ~-- 1- (2'12)~ 1- 0 . 1 ~ Eipwe 4-36 Digital control system of Example 4-12. Now let us assume that the transfer function of the digital controller GD(w)has unity gain for the low-frequency range and has the following form: (This is a phase-lead compensator.) It is one of the simplest forms of the digital controller transfer function. (Other forms may be assumed as well for this problem.) The open-loop transfer function is The static velocity error constant K, is specified as 2 sec-l. Hence, Kv = lim wGD(w)G(w)= 16' = 2 w-io The gain K is thus determined to be 2. By setting K = 2, we plot the Bode diagram of G(w): Figure 4-37 shows the Bode diagram for the system. For the magnitude curves we have used straight-line asymptotes. The magnitude and phase angle of G(jv) are shown by dashed curves. (Note that the zero at v = 10, which lies in the right half of the w plane, gives phase lag.) The phase margin can be read from the Bode diagram (dashed curves) as 30" and the gain margin as 14.5 dB. The given specifications require, in addition to K, = 2, the phase margin of 50" and a gain margin of at least 10 dB. Let us design a digital controller to satisfy these specifications. Design of lead compensator. Since the specificationcalls for a phase rnargin of 50", the additional phase-lead angle necessary to satisfy this requirement is 20". To achieve a phase margin of 50"without decreasing the value of K , the lead compensator must corrtribute the required phace-lead angle.
- 126. ased on the Frequency-Response Method To find the frequency point where the magnitude is -4.425 dB, we substitute w = jv in G(w) and find the magnitude of C(jv): y trial and error, we find that at v = 1.7 the magnitude becomes approximately -4.4 dB. We select this frequency to be the new gain crossover frequency v,. Noting that this frequency corresponds to 11(V%), or and ar = 0.3534 The lead compensator thus deterrnined is The magnitude and phase angle curves for GD(jv)and the magnitude and phase angle curves of the compensated open-loop transfer function GD(jv)G(jv)are shown by solid curves in Figure 4-37. From the Bode diagram we see that the phase rnargin is 50" and the gain margin is 14 dB. The controller transfer function given by Equation (4-40) will now be trans- formed back to the z plane by the bilinear transformation given by Equation (4-38): Thus, The open-loop pulse transfer function of the compensated system is
- 127. ased on the Frequency- Unit-StepResponseof Desígned System 1.6- a 1 0.0891z2+ 0.0108~- 0.0679 z3- 2.28S5z2+ 1.84602 - 0.5255 1.4 - -- 0.0891(z + 0.9357)cz - 0.8145) (Z- O.S126)(z - 0.7379 - j0.3196)(z - 0.7379 +j0.3196) 1.2 - o O o O o o 1 - O O O O O O O O o O O O O O O O O O o O O O O O O O O o o ~ ~ Y 2 0.8 - O -O 0.6 - o poles acts as dominant closed-loop poles. (Tke systembehaves as if it is a secsn 0.4 - unit-step response of this sy o the unit-step response curve 0.2 - exhibits a mLurimum oversh o 06 O 5 10 15 20 25 30 35 40 k (SamplingperiodT = 0.2 sed under normal operation of this systern. Figure 4-38 Plot of unit-step response of the designed system. the o value varies from O to 4o,. titious frequency v varies from O to m, since v = (21T) tan (oT12).Thus system for O S o 5 4o, is similar hit-step response of designed systern---------- ing analog control system for O r v 5 a. Since C(jv) is a rational function of Y, it is basically the same as G(jo). In determining possible unstable zeros of the cha stability criterion can be applied. Therefore, line approximation to the magnitude curve concept of phase margin and gain margin ap Compare transfer functions G(w) and G(s). As w of the presence of the scale factor 21Tin the w ing static error constants for G(w) and C(s) scale factor 2/T, this will not be true.) The w transformation may generate one or more right half-plane zeros in G(w). If one or more t half-plane zeros exist, then G(w) is a nonminimum phase transfer functio ecause the zeros in the right half-plane are generated by the sample-and-hol peration, the locations of these zeros depend on the sampling períod T. The effects of these zeros in the right half-plane on the response become smaller as the sampling period 1" becomes smaller. In what follows, let us consider the effects on the resgonse of the zero in the right half-plane at w = 2lT. The zero at w = SlTcausesdistortion in the frequency response as v approaches 217'. Since 2 oT v = - tan- ' $ 2
- 128. esign of Discrete-Time Control ysteems by Conventional Methods Chap. then, as u approaches 'EIT, tan (wT12) approac wT 57- and thus The main reason why the control actions of analogcontrollersarelimited isthat ther are physical limitations in pneumatic, hydraulic, and eliectronic components. Suc limitations may be completely ignored in designing digital controllers. Thus, man control schernes that have been irnpossible with analog controls are possible digital controls. In fact, optimal control schemes that are not possiblie with an controllers are made possible by digital control schemes. In this section we specifically present an analytical design method for digi coneroflers "eat will force the error sequence, when subjected to a speci time-domain input, to become zero after a finite number of sampling p in fact, to become zero and stayzero after the minimum possible number of sarn periods. Lf the response of a closed-loop control system to a step input e minimumpossible settling time (that is, the output reaches the finalvalue in the mi mum time and stays there), no steady-state error, and no ripples bet samplinginstants, then this type of response is cornmonly The deadbeat response will be discussed in this section. ( response again in Chapter 6, where we discussthe pole placement technique ana t design of state observers.) The discussions that follow are lirnited to the determination of the con algorithms or pulse transfer functions o£digital controllers for single-input-sin output systems, givea desired optirnal response characteristics. For optimal con of multiple-input-multiple-outputsystems, see Chapter 8, where the state-spa approach is used. Design of Digital CesntroUersfeso- Minimu Error. Consider the digitai control system shown in Figure 4-39(a). The erro signal e([). which is rhe difference between the input r(r) and the output c(f).i ec. 4-7 Analytical Design Method -39 (a) A digital control system; (b) diagram showing equivalent digital control system. sarnpled every time interval T. Lhe input to the digital controller is the error signal e(kT). ñhe output of the digital controller is the control signal u(k2"). The control signal u(kT) is fed to the zero-order hold, and output of the hold, u(t),which is a piecewise continuous-time signal, is fed to t plant. [Although the sampler at the input o£the zero-order hold is not shown, the signal u(k2")is first sampled and fed to the zero-order hold. As mentioned earlier, the zero-order hold shown in the iagram is a sarnple-and-hold esired to design a GD(z)such that the closed-loop control system will exhibit the minimum possible settling time with zero steady-state error in response to a step, a ramp, or an acceleration input. It is required that the output not exhibit intersampling ripples after the steady state is reached. The system must satisfy any other specifications, if required, such as a specification for the static velocity error constant. Let us define the z transform of the plant that is preceded by the zero-order frold as G(z),or Then the open-loop pulse transfer function becomes GD(z)G(z),as shown in Figure 4-39(b). Next, define the desired closed-loop pulse transfer function as F(z):
- 129. iscrete-Time Control Systerns by Conventional G + w ( z ) = q z ) dition to the physical bility conditions, we must pay attention to -- 1+ Go(z)G(z) the stability aspects of the system ifically, we must avoid canceling an unstable pole of the plant by a zero o£the d Sinceit is required that the system exh with zero steady-state any error in the pole-zero cancellation error, the system must exhibit a finite ce, the desired ciosed- will become unstable. Similarly, the di loop pulse transfer function must be cel plant zeros that lie outside the unit ciscle. aozN+alzN-'+ + aN at will happen to the closed-loop puIse transfer F(z) = zN function F(z) if G(z an unstable (or critic z = a outside (or o it circie. [Note that t or equally,if G(z) inv r more unstable-or E(z) = a. + alz-l + + a N ~ - N define hat F(z) must not conta G(z) = -GiW seriesexpansion of F(z) z - a e for a physically real where Gl(z) does not include a term that cancels with z - a. Then the closed-loo p pulse transfer function pulse transfer function becomes the digital controllerpulse transfer function GD(z).That is, we find t function G,(Z) that willl satisfyEquation (4-41). SolvingEquation (4-41) foa:GD(z) Gl(zb G D ( ~ ) G ( ~ )= = F(z) (4-44) G D ( ~ )z_, we obtain Gdz)+ G,(z)G(z) 1 + GD(z)__ G D ( ~ )= G(z)[I - J;(z)l Since we require that no zero of GD(z)cancel the unstable pole of G(z) at z = a , The designed system must be physicailyrealizable. The conditions for we must have realizabifityplace certain constraints on tbe closed-loop pulse transfer funct and the digital controHerpulse transfer function GD(z).The conditions for 1 -- z - a 1 - F(z) = realizability may be stated as follows: Gdz) z - a + GD(z)Gl(z) 1 + G D ( ~ ) z _ . The order of the numerator of GD(z>must be equal to or lower t f the denominator. (Otherwise, the controller requires future roduce the current output.) f the plant &;,(S) involves a transportation lag e-L" then the designed ciosed loop systemmust involveat Ieast the same magnitude of the transportation la (Otherwise, the closed-loop systemwould have to respond before an input w glven, which is impossible for a physically realizable system.) If G(z) is expanded into a series in z-l, the lowest-power term of the serie expansion of E(z) in z-' merst be at least as large as that of G(z). For if an expansion of G(z) into a series in z-1 begins with %hez-l term, first term of F(z) given by Equation (4-42) must be zero, or a. must that is, fhe expansion has to be of the form F(z) = alz-l + a2z-2+ + aNzlN where N 2 n and n is the order of the system. This means that the plant carm respond instantaneously when a control signal of finite magnitude is appk the response comes at a delay of at least one sampling period if the crri expansion of C(z) begins with a term in z-l. at is, 1- F(z) must have z = a as a zero. Also, notice t if zeros of C(z) do not cancel poles of GD(z),the zeros of G(z) become zeros of F(z). [F(z) may involve additional zeros.] Eet us surnmarize what we have stated concerning stability. 31. Since the digital controller GD(z) should not cancel unstable (or critically stable) poles of G(z), al1unstable (or critically stable) poles of G(z) must be included in 1- F(z) as zeros. (z) that lie inside the unit circle may be canceled with poles of at lie on or outside the unit circle must not ence, al1zeros o£G(z) that lie on or outside the unit circle must be included in F(z) as zeros. Now we shallproceed with the design. Sincee(kT) = r(kT) - c(kT), referring to Equation (4-41) we have Note that for a unit-step input r(t) = l(t)
- 130. esign of Discrete-Time Control Cysternc by Convenlional For a unit-ramp input v(t) = tl(t), And for a unit-acceleration input r(t) = it21(t), S, in general, z transforms of such time-domain ten as R(z) = P(z) (1 - 2-1)4+1 where P(z) is a polynomial in z-l. Notice that for a unit-step input q = O; for a unit-ra p input, P(z) = Tz-' and q = 1; and for a uni input, P(z) = T2z-l(l + z-l) and q = 2. By substituting Equation (4-46) into Equation (4-451, we obtain E(z) = 20ensure that the system reaches steady state in a finite number o£sam and maintains zero steady-state error, E(z) must be a polynomiali in z- number of terms. Then, by referring to Equation (4-47), we choose the 1 .- F(z) to be of the form 2 - F(z) = (1 - ~-l)q+~AJ(z) where N(z) is a polynomlal in z-1 with a finite number of terms. Then E(z) = P(z)N(z) which is a polynomial in z.-' with a finite number of terms, This means that signal becomes zero in a finite number of sampling periods, From the preceding analysis,the pulse transfer functlon of the digital co can be deterrnined as follows. y first letting F(z) satisfy the physical realiza and ctabilityconditions and then substituting Equation (4-48) into Equation (4-4 we obtain G"(') = G(z)(l - Z - ' ) ~ + ' N ( ~ ) Equation (4-50) gives the pulse transfer function of the digital controller tkat produce zero steady-state error after a finite number of sarnpling For a stable plant G,(s),the csndition that the output not exhibit inters r&@es after the settlirag time is reached may be written as fcsllows: c(t L nT) = constant, for ste C(t nT) = constant, for ra ?(t nT) = constant, for acceleration inputs ion must be satisfied tion on c(t), C(t), or ontinuous time and time function; therefore, to have no ripples in the out y state must be either constant or mono ing) for step, rarnp, or acceleration i Since the closed-loop pulse transfer function F(z) is a polynomial in z-', all the closed-loop poles are at the origin or at z = O. The mult pole at the origin is very sensitive to syst Although a digital control system design with zero steady-state error in response t transiemt response characteristics for the inferior or sometimes unacceptable tran types of input. (This is always true in control system will exhibit the best resp it is designed for, but will not exhibit o hich an analog controller is discretized, an increase in the sampling period changes the system dynamics and may lead to system instabil- ity. On the other hand, the behavior of the digital control systern we are designing in this section does not depend on the ch Since the inputs r(t) considered here are time-d inputs, ramp inputs, and acceleration inputs), th chosen arbitrarily. H;or a smaller sampling period, an integral multiple of the sarnplingperiod T )be avery small camplingperiod T ,the magnitude of Knltr~lsignalwillbecome excessivelylarge, with the result that saturation nomena will t&e place in m, and the design method presented is section will no longer ence, the sampling period T should not be too small. On the other hand, if the sampling period T is chosen too iarge, the system may behave unsatisfactorily or may even become unstable when it is subjected to suffi- cientlytime varying inputs (such asfrequency-domaininputs).Thus, a compro- mise is necessary. A rule of thumb is to choose the smallest sampling period T such that no saturation phenomena occur in the control signal. Consider the digital control systern shown in Figure 4-39(a), where the plant transfer function %(S) is given by
- 131. esign of Discrete-Time Control yctemc by Conventional Methodc Chap. 4 Design a digital controller GD(z)such that the closed-loop system will exhibit a- . deadbeat response to a unit-step input. (in a deadbeat response the system should n exhibit intersampling ripples in the output after the settling time is reached.) T sarnpling period T is assurned to be 1 sec. Then, using the diktal controller GD[z) designed, investigate the response of this system to a unit-ramp input. The first step in the design is to determine the z transform of the plant that preceded by the zero-order hold: Now redraw the block diagram of the system as shown in Figure 4-39(b). Define t closed-loop pulse transfer function as F(z). or Notice that if G ( z )is expanded into a series in z-' then the first term will be 0.3679~- Hence, F(7) must begin with a term in z-l. Referring to ~quation(4-42) and noting that the system is of the second ord (n = 2), we assurne F(z) to be of the following form: -7 Analytical Design Method For U ( z )to be a series in z-' with only two terrns, F(z)must be of the following form: F(z) = (1 + 0 . 7 1 8 1 ~ - ~ ) z - ~4 (4-54) where Fl is a constant. Then U(z)can be written as follows: Equation (4-55) gives U(z)in terrns of El. Once constant fi is determined, U(z)can be @ven as a series in z-1 with only tws terms. Now we shall determine N(z),F(z),and K. y substituting Equation (4-52) into Equation (4-531, we obtain 1 - alz-' - a2z-" (1 - z-l)N(z) The left-hand side of this last equation must be divisible by 1 - z-'. If we divide the left-hand side by 1 - z-', the quotient is 1 + (1 - al)z-' and the remainder is (1 - al - a2)z-2.Hence, N ( z ) is determined as and the remainder must be zero. This requires that 1 - a l - a 2 = 0 Also, from Equations (4-52) and (4-54) we have F(z) = al z-' + a2zW2= (1 + 0.7181~-')z-~F, Hence, al + a2z-l = (1 + 0.7181z-')fi Division of the left-hand side of this last equation by 1 + 0.7181z-' yields the quotient al and the remainder (az- 0.7181ai)~-1.By equating the quotient with fi and the remainder with zero, we obtain and Solving Equations (4-57) and (4-58) for al and az gives al = 0.5820, a2 = 0.4180 Thus, F(z) is determined as and f i = 0.5820 Equation (4-56) @ves
- 132. Design of Discrete-Time Control yctems by Conventional Methods Chap. The digital controller pulse transfer function G D ( z )is tben determined from Equatio (4-50), as follows. By referring to Equations (4-51), (4-54), and (4-60), With the dioital controller thus desioned. the closed-loo~~ u l s etransfer functio" " I L .... becomes as follows: Hence, ec. 4-7 Analflical Design Method 1 Figure4-40 Responses of the system designed in Example 4-13. (a) Plots of c(k) versus k , u(k)versus k , and u(t)versus t in the unit-step response; (b) plots of c(k) versus k , u(k)versus k , and u(t)versus t in the unit-ramp response. The signal u ( k )becomes constant (b = 1)for k 1 2. Hence, the cystem output will not exhibit intersampling ripples. Figure 4-40(b) shows plots of c ( k )versus k , u ( k )versus k , and u(t) versus t in the unit-ramp response. Note that the static velocity error constant Kv for the present system is 0.58202-' + 0.4180z-' == lim Z-1 1 + 0.41802-'
- 133. esign of Discrete-lime Control ystemc by Conventional Methods Ckap. Thus, the steady-state error in the unit-ramp response is ec. 4-7 Analytical Design Meahod which is indicated in Figure 4-40(b). where we used Equation (4-50) with q = O. Notice that from Gquation (4-62) we have F(1) = 1.Hence, Kv can be written as follows: (If a catisfactory result is not obtained, we must assume N > 3.) Since the input is a step function, from Equation (4-48) we require that 1- n(z) = (1 - Z-')N(z) (4-62) Note that the presence of a critically stable pole at z = 1in the plant pulse transfer function G(z) requires 1 - F(z) to have a zero at z = 1. owever, the fmction 1 - F(z) already has a term 1- z-l and therefore satisfies the stability requirement. The requirement that the static velocity error constant be 4 sec-l can be written as follows: Since the system output shouId not exhibit intersampling ripples after the settiing time is reached, we require U(z) to be of the following form: Because the plant transfer function G,(s) involves an integrator, b must be zero. Consequently, we have U(Z) = bo + blz-' + b2z-' Also, frorn Figure 4-39(b), U(z) can be given by For U(z) to be a series in z-' with three terrns, F(z) must be of the foilowing form: where fi(z) is a first-degree polynomial in z-l. Then U(z) can be written as follows: From Equations (4-61) and (4-62), we have Hf we divide 1 - a, z -'- azzV2- a3z - ~by 1 - z-', the quotient is 1 + (1 - al).¿-' + (1 - a, - a2)zeZand the remainder is (1 - al - az - a 3 ) ~ - ~ .Hence, N(z) is deter- mined as W(z) = 1 + (1 - al)z-' + (1 - al - az)zw2 (4-66)
- 134. iccrete-Time Control Systems by Conventional Met and the rernainder must be zero, so that 1 - a l - a , - a 3 = 0 Note that from Equation (4-63) we require W(1) = i.Therefore, by su z-' = 1 into Equation (4-66), we obtain 2a1 + az = 2.75 Also, Equation (4-64) can be rewritten as F(z) = al 2-' + a2z-' + a32-3 = (1 + O.7181~-~)~-'&(z) Hence, al + a2zy1+ a3zWz= (1 + 0.7181~-')R(z) Division of the left-hand side of this last equation by 1 $ 0.7181~-'yields the qu [a, + (az - 0.7181al)z-'1 and the remainder [a3- 0.7181(a2 - 0.7181a1)]z-2 equating the quotient with F,(z) and the remainder with zero, we obtain Fl(z) = al + (a2- 0.7181al)z--' and a3 - 0.7181(az - 0.7181al) = O Soiving Equations (4-67), (4-68), and (4-69) for al, a2,and a3 gives al = 1.26184, a2 = 0.22633, a3 = -0.48816 Thus, F(z) is determined as ) = 1.261842-' + 0.22633z-' - 0.48816~-~ fi(z) = 1.26184 - 0.679792-' N ( z ) = 1 - 0.26184~-~- 0.48817~-~ The digital controller pulse transfer function GD(z) is then determined from Equ tion (4-50) as follows: F(z G"(z) = G ( z ) ( l - z-')N(z) -- (1 + 0.7181~-~)~-'(1.26184- 0.679802-') + 0-7181z-')2-' (1 - 2-1)(1 - 0.261~4~-1- 0.48817z--2) (1 - zV')(1- 0.36792-') (1 - 0.53872-')(1 - 0.36792-') = 3.4298--- ( 1 - 0.84182-')(1 + 0 . 5 7 9 9 ~ ~ ' ) With the digital controller thus designed, the system output in response unit-step input r(t) = 1is obtained as follows: C ( z ) = F(z)R(z) 1 = (1.2,61842-' + 0.22633.T2 - 0 . 4 8 8 1 6 ~ ' - ~ ) - - ~ 1 - z- = 1.2618~-'+ 1 . 4 8 8 2 ~ ~ ~+ Y 3+ z - ~+ . F(z and Equation (4-66) gives Theunit-step response sequence has a malramum overshoot of approximately 50%. The settling time is 3 sec. Notice that from Equation (4-65) we have U ( z )= 2.7181(1 - 0.36792-')(1.26184 - 0.679792-') Thus, the control signal u(k) becornes zero for k 2 3. Consequently, there are no intersampling ripples in the response. Figure 4-41 shows plots of c(k)versus k, u ( k ) versus k, and u(t) versus f in the unit-step response. Notice that the assumption of N = 3, that is, the assumption o£F(z) as given by Equation (4-61), is satisfactory. Figure 4-41 Plots of c(k) versus k , u ( k ) versus k , and u(t) versus t in the unit-step response of the system de- signed in Example 4-14.
- 135. esign of Discrete-Time Control ystems by Conventional [Vethods Chap. Next, let us investigate the response of this system to a unit-ramp input: C(z) = F(z)R(z) -. 1 = (1.26184z-' + O.22633z-' - 0.4$$16zz3)(1 - , ,z 1 on the objectives of the = 1.2618z-' + 2.7500~-~+ 3.7500~-~+ . response characteristics In the unit-ramp response, the control signal U(z)is obtained as foollows: F(z) 1 2-'F(z) qz)= ---U(Z)= -= - G(z) G ( z ) G ( z ) 1 - z-' 1 - z-l z -l = (3.4298 - 3.10962-' 4- 0 . 6 7 9 8 ~ ~ ~ ) - 1 - 2-' = 3.42982-1 + 0.3202~-~+ zw3+kW4-t- z - ~+ The signal u(k)becomes constant (b = 1)for k 2 3. Hence, the system output Show that geometrically the patterns of the poles near z = 1 in the z plane are similar exhibit intersampling ripples. Figure 4-42 shows plots of c(k)versus k , u(k)v to the patterns of poles in the s plane near the origin. and u(t) versus t in the unit-ramp response. Wotice that the steady-state err unit-ramp response is e, = 1/K, = $, as indicated in Egure 4-42. z = p Near the origin of the s plane, z = eTs = 1+ + . L T ~ ~ ~+ - . - z - l = T s Thus, geometrical patterns of the poles near z = 1 in the z plane are similar to the patterns of poles in the s plane near the origin. y(k) - 0.6y(k - 1) - 0.81y(k - 2) + 0.67y(k - 3) - 0.12y(k - 4) = x ( k ) where x ( k ) is the input and y(k) is the output of the system. Determine the stability of the system. 2 Soliaation The pulse transfer function for the system is o X ( Z ) 1 - 0.62-' - 0.81z-' + 0.672-3 - O . l 2 ~ - ~ -2 -e z4 z4 - 0 . 6 ~ ~- 0.812' + 0.672 - 0.12 P(z) = z4 - 0 . 6 ~ ~- 0.81z2-i- 0.672 - 0.12 2 = a o z 4 + a 1 z 3 + a 2 z 2 + a 3 z - t - a 4 , ao>O O Figure 4-42 Plots of c ( k ) versus k > Then we have ) u(k)versus k , and u(t)versus r in the unit-ramp response of the system ao = 1 -2 designed in Example 4-14. = -0.6 esign of Discrete-Time Control ystems by Conventional [Vethod: Next, let us investigate the response of this system to a unit-ram C(z) = F(z)R(z) we note thát thelatter irnproves the ramp response characteristics at the e -. 1 the settling time. (The latter system requires one extra sampling period to = (1.26184z-' + O.22633z-' - 0.4$$16zz3)(1 - , , steady state.) Note also that the former as beteer step response characteri z 1 is, a shorter settling time and no overshoot n the objectives of the = 1.2618z-' + 2.7500~-~+ 3.7500~-~+ . system, we rnay choose one over the other. response characteristics In the unit-ramp response, the control signal U(z)is obtained as foollovc- are required, then the system should be d the ramp input as the z -l = (3.4298 - 3.10962-' 4- 0 . 6 7 9 8 ~ ~ ~ ) - 1 - 2-' = 3.42982-1 + 0.3202~-~+ zw3+kW4-t- z - ~+ The signal u(k)becomes constant (b = 1)for k 2 3. Hence, the system < exhibit intersampling ripples. Figure 4-42 shows plots of c(k)versus k , and u(t) versus t in the unit-ramp response. Wotice that the steady-stai unit-ramp response is e, = 1/K, = $, as indicated in Egure 4-42. o Figure 4-42 Plots of c(k ) u(k)versus k , and u(t)vc -2 unit-ramp response of the designed in Example 4-lr the system described by Define
- 136. Design of Discrete-Time Control ystems by Conventional Methods a, = -0.81 a3 = 0.67 a4 = -0.12 The Jury stability conditions are: l. la4/< ao. This condition is clearly satisfied. . P(1) > O. Since P(1) = 1 - 0.6 - 0.81 + 0.67 - 0.12 = 0.14 > O the condition is satisfied. 79- P(-1) > O. Since P(-1) = 1 i- 0.6 - 0.81 - 0.67 - 0.12 = O the condition is not satisfied. P(- 1) = O implies that there is one root at z = - lbsi > lbo/.Since the condition is satisfied. 5. /c2j > /COI. Since Clearing the fractions by rnultiplying both sides of this last equation by (w - 1 j 3 9 get -0.14w3 + 1.06w2 + 5 . 1 0 ~+ 1.98 = O Chap. 4 Exarnple roblerns and Soieitions By dividing both sides of this last equation by -0.14, we obtain w 3- 7.571w2 - 3 6 . 4 3 ~- 14.14 = 8 The Wouth array for Equation (4-71) becomes as foliows: one sign,, w3 1 -36.43 c h a n g e L w2 -7.571 -14.14 Routh stability criterion states that the number of roots with positive real parts is equal to the number of changes in sign of the coefficients of the first column of the array. Since there is one sign change for the coefficients in the first column, there is one root in the right half of the w plane. This means that the original characteristic equation given by Equation (4-70) has one root outside the unit circle in the z plane. The system is unstable. (Compare the amount of computation needed in the present method and that needed in the Jury stability test. See in particular Example 4-6.) Lonsider the system defined by The sampling period T is 0.5 sec. Using ,plot the unit-ramp sesponse up to k = 20. slaritioarn The unit-ramp input u may be written as u = k T , k = 0 , 1 , 2,... In the MATLAB program, this input can be given as where N is the end of the process considered. A M-ATLABprogram for plotting the unit-ramp response of the system consid- ered is given in MATLAB Program 4-3. The resulting plot is shown in Figure 4-43. Show that if the characteristic equation for a closed-loop system is written as where A ( z ) and B ( z ) do not contain K, then the breakaway and break-in points can be determined from the roots of where the primes indicate differentiation with respect to z.
- 137. iscrete-rime Control Systems by Convenlional Unit-Ramp Response 1O k Figure 4-43 Unit-ramp response of the system considered in Problem -4-4-4. 4 Exarnple Problems and Colutions "2 Sslution Let us write the characteristic equation as f(z) = A ( z ) + KB(z) = O (4-72) Suppose that f(z) = O has a multiple root o£order r. Then f(z) may be wriRen as f(z) = ( 2 - z,)~(z- 22) ' ' ' (2 - zp) If we differentiate this equation with respect to z and set z = zl, we get This means that multiple roots of f(z) will satisfy the following equation: where Solving Equation (4-73) for K , we obtain This particular value of K will yield rnultiple roots of the characteristic equation. liwe substitute this value of K into Equation (4-72), we obtain If this last equation is solved for z, the points where rnultiple roots occur can be obtained. 8 n the other hand, from Equation (4-72) we have and If dK1dz is set equal to zero, we get the same equation as Equation (4-74). Therefore, the breakaway or break-in points can be simply determined from the roots of It should be noted that not al1 the solutions of Equation (4-74) or of dKidz = O cor- respond to actual breakaway or break-in points. Cuch a point for which dKldz = O is
- 138. an actual breakaway or break-in point if and only if the value of K at this point is positive value. Discuss the procedure for designing lead compensators for digital control syste the root-locus method. dominant closed-loop poles. 4.This angle must be contributed by the lead compensator if the new is to pass through the desired locations for the dominant closed-loop po 3. Assume the lead compensator GD(z)to be 1+ az GD(z) = KDa----- O < a < l 1 + a7-z ' If static error constants are not specified, determine the location of t zero of the lead compensator so that the lead compensator will con necessary angle 4. If no other requirements are imposed on the syst make the value of a as large as possible. A larger value of a generally res a larger value of Kv,which is desirable. (If a particular static error const specified, it is generally simpler to use the frequency-response appr 5. Determine the open-loop gain of the compensated system from the condition. Once a compensator has been designed, check to see whether or not al1 formance specifications have been met. If the compensated system does not me performance specifications, then repeat the design procedure by adjusting the pensator pole and zero until al1 such specifications are met. If a large statkc constant is required, cascade a lag network or alter the lead compensator to a lag compensator. Draw root locus diagrams in the z plane for the system shown in Figure 4-45 f following three sarnpling periods: T = 1 sec, T = 2 sec, and T = 4 sec. hap. 4 Exarnple Problerns and Solutions Figure 4-45 Digital control system. first obtain the z transform of G(s). Referring to Example 3-5, we get F- 7 Next we construct root locus diagrams for the three cases considered. l. Sampling period T = 1: For T = 1, Equation (4-75) becomes K[(1 - 1 + e-')z-' + (1 - e-' - e - ' ) ~ - ~ ] G(z) = (1 - z-l)(l - e-lz-') Wotice that G(z) possesses a zero at z = -0.7181 and poles at z = 1 and z = 0.3679. The breakaway point is at z = 0.6479, and the break-in point is at z = -2.0841. The root-locus diagram for this case is shown in Figure 4-46(a). The value of gain K of any point on the root Ioci can be determined from the magnitude condition If we choose a point z on the root loci, the value of K at that point can be calculated by substituting the value of z into this last equation. (This means that with this value of K that particular point becomes a closed-looppole.) The critical gain is found to be K = 2.3925. Sampling period T = 2: For the sampling period T = 2, we have from Equa- tion (4-75) The pulse transfer function G(z) in this case possesses a zero at z = -0.5232 and poles at z = 1and z = 0.1353. The breakaway point is at z = 0.4783, and the break-in point is at z = -1.5247. The root-iocus diagram for this case is shown in Figure 4-46(b). The critical gain IC for stability is Ik = 1.4557.
- 139. iscrete-Time Control z plane K = 2.3925 I z plane Unit circle /'- z plane 6 Root-locus diagrams for the system shown in Figure 4-45 when (a) T = 1 sec, (b) T = 2 sec, and (c) T = 4 sec. 3. Sampling period T = 4: For the case of T = 4, Equation (4-75) gives The breakaway point ic at z and the break-in point is at z hap. = 1 sec) l = -0.945 The root-locus diagram is shown in Figure 4-46(c). The critica1gain for stability is K = 0.9653. Frorn the three cases considered, notice t period is, the larger the critica1 gain K for stabiiity. Consider the digital control system shown in Figure 4-47, where the plant is of the first order andhas a dead time of 2 sec. T e samplingperiod is assumed to be 1sec, or T = 1. Design a digital PI controller such that the dominant closed-loop poles Rave a damping ratio 5 of 0.5 and the number of samples er cycle of damped sinusoidal oscillation is 10. Obtain the response of the systern to a unit-step input. Also, obtain the static velocity error constant Kv and find the steady-state error in the response to a unit-ramp input. I -47 Digital control system. Soltiiition The pulse transfer fianction of the plant that Is preceded by a zero-order hoId is The digital PI controller has the following pulse transfer function: The open-loop pulse transfer function becomes
- 140. z plane Figure 4-48 Pole and zero locations in the z ~ l a n eof the svstem considered in We locate the open-loop poles in the z plane as shown in Figure 4-48. There is on open-loop zero involved in this case, but its location is unknown at this point. Since it is required to have 10sarnpies per cycle of damped sinusoidal oscillatio the dominant closed-loop pole in the upper half of the z plane must lie on a 1 from the origin having an angle of 360°/10 = 36". Frorn Equations (4-1) and (4 rewritten as the desired closed-loop pole location can be determined as follows. Noting tha or w ~ / w ,= 0.1. Since 5is specified as 0.5, we have The closed-loop pole is located at point P in Figure 4-48, where (at point P ) kap. 4 Example (Note that this point is the intersection of the 5 = 0.5 lacus and the line from the origin having an angle of 36".) If point P is to be the closed-loop pole locatisn in the upper half of the z plane, then the angle deficiency at point B is The controller zero must contribute f93.52". This means that the zero of the controller must be located at z = 0.5881. Therefore, Hence, the PI controller is deterrnined as follows: where K = M, -t K,. Gain constant K is determined from tke magnitude condition: Thus, + Ki = 0.5070 From Equations (4-76) and (4-77), we find that K, = 0.2982 and K, = 0.2088 Hence, the PI controller just designed can be given by Finally, the open-loop pulse transfer function becomes The closed-loop pulse transfer function becomes The response c(kT) to the unit-step input can be obtained easily by use of MATLAB. A MATLAB program for plotting the unit-step response is shown in MATLAB Program 4-4. The resulting plot is shown in Figure 4-49.
- 141. % Unit-step response ---------- num = [O O O 0.3205 -0.18851; den = [l -13 7 9 0.3679 0.3205 -0.1 8851; r = oneo;(1,571; title('Unit-Step Response') xlabel('k'f yIabel('c(k)') - - Unit-StepResponse n of Discrete-Time Control Corisider the system shown in Figure 4-50. that the dominant closed-Ioop poles of the system will have a damping ratho 5 0.5. We also want the number of samples per cycle of damped sinusoidal oscillati~n be 8. Assurne that the sampling period T is 0.2 sec. Wsing the root-locus method in the z plane, determine the pulse of the digital contrdler. Obtain the response of the designed system to klso obtain ehe static velocity error csnstant Kv. Chap. 4 Exarnple Problerns and Solutions Digital controller S ( S + 1) Figure 4-33 Digital control system. We shall first locate the desired closed-loop poles in the z plane. Referring to Equations (4-1) and (4-2), for a constant-damping-ratio locus we have U d 1- = - o-), 8 9 Plot of c(kT)versus kT for the system designed in Problem A-4-8. (Sampling period T = 1sec.) and Since we require eight samplesper cycleof damped sinusoidal oscillation, the dominant closed-loop pole in the upper half of the z plane must be located on a line having an angle of 45" and passing through the origin as shown in Figure 4-51. (Note that the number of samples per cycle is 360'18. Hence, eight samples per cycle requires 8 = 360"/8 = 45".) Thus, which gives Since the sampling period T is specified as 0.2 sec, we have Therefore, y letting 5 = 0.5 and substituting Equation (4-79) into Equation (4-78), we obtain IZI = e-0.4535 = 0.6354 Hence, we can locate the desired closed-loop pole in the upper half of the z plane, as shown by point P in Figure 4-51. Note that at point P Izl/Z = 0.6354/450 = 0.4493 + 10.4493 Next, we obtain the pulse transfer function G(z)of the plant that is preceded by a zero-order hold:
- 142. Design of Discrete-Time Control Cystems by Onventional Methods .4 Exarnpie Problems and 71 and choose the zero of the controller to cancel the pole at z = 0.8187. Then the pole of the controllercan be determined easilyfrom the angle condition as z = 0.1595. Thus, we have The open-loop pulse transfer function of the system is therefore obtained as follows: The gain constant K can be determined from the magnitude condition: Hence, we have determined the pulse transfer function of the digital controller to be The open-loop pulse transfer function is The closed-loop pulse transfer fiinction is Because of the cancellation of a pole of the plant and the zero of the controller, the order of the system is reduced from third to second. The system has only a pair of conjugate complex closed-loop poles. Figure 4-52 shows the unit-step response sequence c(kT) versus kT. The plot shows the maximum overshoot to be approximately 16.5%. Finally, the static velocity error constant Kv is determined as follows: 1 - z-' z-1 1 - 2-' 0.2610(1 + 0.93562-')z-' = lim,i [-------0.2 (1 - 0.15952-')(1 - z-')
- 143. Assume the following form for the lead compensator: 1 + TW GD(w)= KD------ O < a < l 1 + arw9 The open-loop transfer function of the compensated system may be writt -- where Gl(w)= K D G(w).Determine gain KD to satisfy the requirement o given static velocity error constant. Chap. 4 Exarnple Problerns and Figure 4-53 Digital control system in the w piane. Using the gain KDthus determined, draw a ode diagram of Gl(w),the gain- adjusted but uncompensated system. Evaluate the phase margin. Determine the necessary phase lead angle 4 to be added to the system. . Add 5" - 12" to & to compensate for the shift of the gain crossover frequency. Define this added &as 4,. Determinethe attenuation factor cx from the foollowing equation: 1 - a sin cb, = -l + a ermine the frequency point where the magnitude of the uncompensared em Gi@) is equal to -20 log (U&). Select this frequency as the new gain crossover frequency. This frequency corresponds to v, = ll(V"&),and the max- imum phase shift +, occurs at this frequency. Determine the cornvr frequencies of the lead compensator as follows: 1 Zero of lead compensator: Y = -T 1 Bole of lead compensator: Y = -a7 7. Check the gain margin to be sureit is satisfactory. kf not, repeat the design process by modifying the pole-zero Iocation of the compensator until a satisfactory result is obtained. The primary function of a lag cornpensator is to provide attenuation in the high-frequency range to give a system sufficient phase margin. The phase lag character- istic is of no consequence in lag compensation. . Assume the following form for the lag compensator: The open-loop transfer function of the compensated system may be written as
- 145. Exarnple Problerns and z plane Root-locus diagram for the system designed in Problem A-4-11. the root-locus diagram, constant 5loci for 6 = 0.5 and 0.6 are superimposed. From the diagram it can be seen that the damping ratio 5of the closed-loop poles is approx- imately 0.55. The line connecting the closed-loop pole in the upper half of the z plane and the origin has an angle of 37". ence, the number o£samples per cycleof damped sinusoidal oscillation is 360°/37"= 9.73. Consider the digital control system shown in Figure 4-57> where the plant transfer function is 11s'. Design a digital controller in the w plane such that the phase margin is 50" and the gain margin is at least 10 d .The sampling period is 0.1 sec, or T = 0.1. After designing the controller, obtain the static velocity error constant Kv.Also, obtain the response of the designed system to a unit-step input. shall first obtain the z transform of the plant that is preceded by the zero-order hold: Next, using the bilinear transformation given by
- 146. iscrete-Time Control ystems by Conventional re 4-57 Digital control system. we transform G(z)into G(w): Thus, Figure 4-58 shows the e diagram of G(jv)thus obtained. Wotice t margin is -2". ft is nec v (radlsec) gure 4-58 Bode diagrarn for the system considered rn Problem A-4-12. hap. 4 Example Problems an and gain margin. By applying a conventional design technique, it can be seen that the following lead network will satisfy the requirements: The addition of this lead network modifies the ode diagram. The gain crossover frequency is shifted to v = 4. Note that the maximum phase lead 4, that this lead network can produce is 61.93", since 1 - L 4, = sin-l---6 =. sin-' 0.8824 =: 61.93" l + & At the gain crossover frequency v = 4, the phase angle of GD(jv)G(jv)becomes -191.31" i61.93" = -129.38". Thus, the phase margin is 50.62". The gain margin is found to be approximately 13 dB. ence, the given design specifications are satisfied. now transform the controlier transfer function CD(W>into GD(z). the biíinear transfsrmation we obtain Hence, the open-loop pulse transfer function becomes The static velocity error constant Kv is obtained as follows: Thus, the static velocity error constant K,, is infinity. There is no steady-state error in the ramp response. The closed-loop pulse transfer function of the system is Figure 4-59 shows the unit-step response. Notice that the zero of the digital controller at z = 0.9048 is close to the double pole at z = 1.A pole-zero pair near point z = 1 creates a long tail with srnall amplitude in the response.
- 147. 1.6 1.8 2.0 2.2 kT ísec) Figure 4-59 Plot of c(kT)versus kT for the system designed in Problem A-4-12. Consider the digital control system shown in Figure 4-60. The plant transfer fu involves a transportation lag e-'". The delay time is 5 sec, or L = 5. Th c(t)in response to a unit-step input is as shown in Figure 4-61(a). The zero to the final value in 10 sec (measured frorn t = 5 e0 t = 15) and course, choose the sampling period to be 2.5 sec, 1 sec, or another value. In example, however, to simplify our presentation, we set the sampling period at 5 S Figure 4-60 Digital control system. Chap. 4 Example Problerns and 1 (a) Desired output c(t)in response to a unit-step input; (b)plot of u ( f ) versus t. The z transform of the plant that is preceded by the zezo-order hold is Notice that there is no unstable or critically stabie pole involved in G(z). Therefore, there is no stability problem involved in this case. Let us define the closed-loop pulse transfer function as F(z): In the present case the output c(t) in the unit-step response is specified as shown in Figure 4-61(a). Since h[l - e-0.1(15-5)]= h(l - e-') = 1, we have h = 1.5820. from the deadbeat response curve shown in Figure 4-61(a), we obtain c(0) = o c(1) = o
- 148. Design of Diccrete-Time Control Systernc by Conventianal from which we get Noting that we obtain F(z) = 0.6225z-' + 0.3775~-~= 0.6225(1 + 0.6065z-l)~-~ Once F(z) is determined, the pulse transfer function of the digital controller can b obtained frorn Equation (4-80): GD(z) = G ( z ) P - F(z)l Notice that from Equation (4-48) we have 1 - F(z) = (1 - zV1)N(z) or 1 - 0.6225zp2- 0.37'95~-~= (1 - z-')IV(z) By dividing (1 - 0 . 6 2 2 5 ~ - ~- 0.3775~-~)by (1 - z-'), N ( z ) can be deterrniined follows: N ( z ) = 1 + z-' + 0.-3775~-~ Consequently, 1 - F(z) = (1 - z-')(1 + z-' + 0.37752-') and o , , 0.6225(1 + 0.6065~-')2-~ This last equation gives the pulse transfer function of the digital controller. Since must be unity at steady state, u(t),a continuous-time signal, must be constant after steady state is reached. Eet us determine U(Z): ,,, . C ( z ) 0.6225~~'+ 0.3775~-~ /I - n 3 6 7 ~ ~ ~ 2 1 hap. 4 Exarnple Problerns and Solutions Taking the inverse z transform of U(z),we find that u(k)is constant for k r 2. Thus, there are no intersampling ripples in the output after the settling time is reached. The signal u(t)versus t is plotted in Figure 4-61(b). Consider the digital control system shown in Figure 4-62. Design a digital controller GD(z)such that the closed-loop system will exhibit the minimum settlingtime with zero steady-stateerror in a unit-ramp response. The systern should not exhibit intersampling ripples at steadystate. The sampling period Tis assumed to be 1sec. After the controller is designed, investigate the response of the system to a Kronecker delta input and a unit-step input. Soliution The first step in the design is to determine the z transform of the plant that is preceded by the zero-order hold: Now define the closed-loop pulse transfer function as E(z): Notice that if G ( z )is expanded into a series in z-' then the first term will be 0 . 5 ~ ~ ' . Hence, F(z) must begin with a term in z-': where N r n and n is the order of the system. Since the system here is of the second order, n = 2. Since the input is a unit ramp, from Equation (4-48) we require that 1 - F(z) = (1 - z-')'IV(z) (4-81) Notice that G ( z ) has a critically stable double pole at z = 1. Therefore, from the stability requirement, 1 - F(z) must have a double zero at z = 1. However, the function 1 - F(z) already involves a term (1 - z-')', and therefore it satisfies the stability requirement. Since the system should not exhibit intersampling ripples at steady state, we require U ( z )to be of the following type of series in z-': U(Z)= bo + blz-' + b2zV2+ - - + bN-l~-N+l+ b ( ~ - ~+ f N - ' + z-~-' + - .) Figure 4-62 Digital control system.
- 149. lesign oC Discrete-Time Control Systems by Conventional Methods Chap. Because the plant transfer function C;,(s) involves a double integrator, b must be ze (Otherwise, the output increases parabolically, instead of linearly.) Consequently, have U ( Z )= bo + bl 2-' + ... + b ~ - l From Figure 4-62 U ( Z )can be given by For U ( z ) to be a series in 2-' with a finite nurnber of terms, F(z) rnust be by 1 + z-': F(2) = (1 + z-')Fi(z) Then U ( z )can be written as follows: U ( z ) = 2F(2) where & ( Z )is a polynomial in z-' with a finite number of terms. By cornparing Equations (4-81) and (4-82) and by making a simple analysis, see that F(z) must involve a term with at least z - ~ .Hence, we assurne ~ ( z )= al z-' + a2z-2 + a32-' This assumed form of F(z) involves the minimurn number of terms; the tra response will settle in three sampling periods. We shall now determine constants al,a2,and a3.Frorn Equation (4-81), we 1 -. al 2-1 - a22-2 - a32 - 3 = (1 - -1 2 2 1 N ( z ) If we divide the left-hand side of this last equation by (1 - z-')', the quotient is (2 - a1)z'-'.The remainder is [2(2- al) - (1 + Q ~ ) ] Z - ~- [(2- al) + a 3 ] ~ - 3 .He N ( z ) is deterinined as N ( z ) = 1 + (2 - al)z-' and the rernainder is set equal to zero: [Y(:!- al) - (1 + a 2 ) ] ~ - ~- (2 - al + a 3 ) ~ - 3= O To satisfy this last equation, we require that 2(2 - al) - (1+ a2)= 0 2 - a l + a 3 = 0 From Equation (4-K!), we have a1.Y1+ a22-2 + a3z-3 = (1 + z-')F1(z) Jf we divide the left-hand side of this last equation by 1 + 2-l, the quotient al z-' + (uz - ul)z-'. The remainder is (al - a2 + a3)z-3. F1(z)= alz-' 4 (a2- al)zW2 roblems and Ssllations and the remainder is set equal to zero: al - az + a3 = O (4-86) y solving Equations (4-84), (4-85), and (4-86) for al, a,, and a', we obtain al = 1.25, a2 = 0.5, a3 = -0.75 Wence, and $(z) = 1.252-' - 0.75z-' = 1.252-'(1 - 0.62-') and F(z) is determined as follows: F(z) = 1.252-' + O . ~ Z - ~- 0 . 7 5 ~ - ~ The digital controller Gu(z) is then determined from Equation (4-50): ith the digital controller thus designed, the system output in response to a unit-ramp input is obtained as follows: Hence, c(0) = o Notice that from Equation (4-83) we have Thus, the control signal u ( k ) becomes zero for k s 3. Consequently, there are no intersalñrpling ripples in the response at steady state. Figure 4-63 shows plots of c(k) versus k, u(k) versus k , and u(t) versus t in the unit-ramp response.
- 150. Design of Biscrete-Time ontrol Systems by v. Chap. 4 Example Problems and For the unit-step input, 1 C ( Z )= F(z)R(z)= (1.252-' + 0.52-' - 0.75~-~)- 1 - 2-' = 1 . 2 5 ~ ~ '+ 1 . 7 5 ~ - ~+ z - ~+ z - ~+ z-' + The maximurn overshoot is 75% in the unit-step response. Notice that = 1.25(1 - 0.62-')(2)(1 - z-') = 2.5 - kv'+ 1.5z-~ The control signal u(k) becomes zero for k 2 3. Gornsequently, there are no inter- sarnpling ripples after the settling time is reached. Figure 4-64(a) shows plots of c(k) versus k , u ( k )versus k , and u(t)versus t in the response to the Kronecker delta input. Figure 4-64(b) shows similar plots in the unit-step response. Notice that when the systemis designed for the ramp input the response to a stepinput is no longer deadbeat. C(z) = F(z)R(z)= F(z) = 1.252-' + 0.5z-' - 0 . 7 5 ~ - ~ Notice that U(z)in this case becomes as follows: (sed -63 Plots of c(k)versus k , u(k)versus k , and u(t)versus t in the unit ramp response of the system designed in Problem A-4-14, Next, let us investigate the response of this systemto a Kronecker dei~aI I ~ J U ~ a unit-step input. For a Kronecker delta input. The control signal u(k) becomes zero for k 2 4. Heme, there are no ir ripples after r 3 4T = 4, Figure 4-64 (a) Plots of c(k)versus k , u(k)versus k , and u(t)versus t in the response to itef-samp the ICronecírer delta input of the system designed in Problem A-4-14; (b) plots of c(k)versus k, u(k)versus k,and u(t) versus t in the unit-step response of the same system.
- 151. Design of Discrete-Time Contr the regions in the s plane shown in Figures 4-65(a) and (b). Draw t corresponding regions in the z plane. The sampling period T is assumed to be 0.3 s (The sampling frequency is w, = 2v/T = 2~10.3= 20.9 radhec.) Figure 4-65 (a) Region in the s pla bounded by constant w lines and constant CT lines; (b) region in the s plane bounded by constant 5 lines, constant o lines, and a constant c z3 + 2 . 1 ~ ~i-. 1.442 + 0.32 = O Determine whether or not any of the roots of the characteristic equation lie ou unit circle centered at the origin of the z plane. Determine the stability of the following discrete-time system: Y(z) --- z - ~ X(z) 1+ 0.52-' - 1 . 3 4 ~ - ~-k 0 . 2 4 ~ ~ ~ hob8em B-4-4 Consider the discrete-time closed-loopcontrolsystemshownin Figure 4-13. Determln the range of gain K for stability by use of the Jury stability criterion. Solve Problem B-4-4 by using the bilinear transforrnation coupled with the Wouth stability criterion. Consider the system Suppose that the input sequence {x(k)) is bounded; that is, = constant, k = 0,1,2,. .. Show that, if al1 poles of G(z) lie inside the unit circle in the z plane, then the output y(k) is also bounded; that is, ly(k)l 5 M2 = constant, k = 0,1,2,. .. S plane State the conditions for stability, instability, and critical stability in terms of the weight- ing sequence g(kT) of a linear time-invariant discrete-time control system, Consider the digital control system shown in Figure 4-66. Plot the root loci as the gain Kis varied from Oto m. Determine the criticalvalue of gain Kfor stability.The sampling period is 0.1 sec, or T = 0.1. hat value of gain K will yield a damping ratio of the closed-loop poles equal to OS? With gain M set to yield 2; = 0.5, determine the damped natural frequency wd and the number of sarnples per cycle of damped sinusoidal oscillation. Gonsider the following characteristic equation: Figure 4-66 Digital control system for Problem B-4-8. Referring to the digital control system shown in Figure 4-67, design a digital controller GD(z)such that the damping ratio 2; of the dominant closed-loop poles is 0.5 and the number of sarnples per cycle of damped sinusoidal oscillation is 8. Assume that the sampling period is 0.1 sec, or T = 0.1. Determine the static velocity error constant. Also, determine the response of the designed system to a unit-step input. Consider the control system shown in Figure 4-68. Design a suitable digital controller that includes an integral control action. Lhe design specificationsare that the damping ratio of the dominant closed-loop poles be 0.5 and that there be at least eight samples per cycle of damped sinusoidal oscillation. The sampling period is assumed to be 0.2
- 152. Figure 4-67 Digital control system for Problem B-4-9. Figure 4-68 Digital control system for Problem B-4-10. sec, or T = 0.2. After the digital controller is designed, determine the static ve1 error constant Kv. order and has a dead .timeof 5 sec. By choosing a reasonable sampling period T, d a digital PI controller such that the dominant closed-loop poles have a dampin jof 0.5and the number of samples per cycleof damped sinusoidal oscillationis 1 Design a digital proportional-plus-derivative controller for the plant whose tra function is l/sZ,as shown in Figure 4-70. It is desired that the damping ratio 5 0 dominant dosed-loop poles be 0.5 and the undamped natural frequency be 4 rad The sampling period is 0.1 sec, or T = 0.1. After the controller is designed, the mmber of samples per cycle of damped sinusoidaf oscillation. hap. 4 Prablems -70 Digital control system for Problem B-4-12. Weferring to the system considered in Problem 8-4-9, redesign the digital controller so that the static velocity error constant K, is 12 sec-', without appreciably changing the locations of the dominant closed-loop poles in the z plane. The sampling period is assumed to be 0.2 sec, or T = 0.2. After the controller is redesigned, obtain the unit-step response and unit-ramp response of the digital control system. Consider the digital control system shown in Figure 4-71. Draw a Bode diagram in the w plane. Set the gain K so that the phase margin becomes equal to 50'. K so set, determine the gain margin and the static velocity error constant KV.The sampling period is assumed to be 0.1 sec, or T = 0.1. L LDigital control system for Problem B-4-14. Using t diagram approach in the w plane, design a digital controller for the ign specifications are that the phase margin be and the static velocity error constant KVbe 20 sec-l. The sampiing period is assumed to be 0.1 sec, or T = 0.1. After the controller is designed, calculate the number of samples per cycle of damped sinusoidal oscillation. Fignre 4-72 Digital control system for Problem B-4-15.
- 153. Consider.the digital control system shown in Figure 4-73. Using the Bode diag approach in the w plane, design a digital controller such that the phase margin is the gain margin is 12dB or more, and the static velocity error constant is 5 sec-l. sampling period is assumed to be 0.1 sec, or T = 0.1. In Ghapters 3 and 4 we were concerned wieh conventional methods for the analysis Figure 4-74 Digital control system for Problem B-4-17. which are mostly time varying andlor nonlinear. Consider the digital control system shown in Figure 4-75. Design a digital contr A modern control system may have many inputs and many outputs, and these GD(z)such that the system output will exhibit a deadbeat response to a unit step i (that is, the settling time will be the minimum possible and the steady-state error be zero; also, the system output will not exhibit intersampling ripples after the set time is reached). The sampling period T is assumed to be 1sec, or T = 1. the state space can be carried out for a
- 154. ec. 5-1 lntroduction function such as the impulse function, ste function, or sinusoi state-space methods enable the engineer to include initial condi his is a very convenient and useful feature that is not ble, state vector, and state Seate, The state of a dynamic syst is the smallest set of variables (called state variables) such that the knowledge hese variables at t = to,together with e knowledge of the input for t $: to, c letely determines the behavior of the system for any time t 2 to.Note that the concept of state is by no %tis applicable to biological systems, eeonomic syste maki system. Ef at least n variables xl,xs, ...,x, ar behavior of a dynamic system (so that once the state at t = tois spe then such n variabl Note that state variables need not be quantities. Variables that do not represenh neither measurable in choosing state variables is an advantage o speaking, however, it is convenient to choos state variables, if this is possible at all, becau feedback of al1 state variables with suitable Vee$or, Hf n state variables are needed to com f a given system, then these n state variables can components of a vector x. Such a vector is called a state vector. A state vector Bs thus a vector that determines uniquely the system state x(t) for any time t z to,once the state at t = to is given and the input u(t) for t 2 tois specified. aee. The n-dimensional space whose coordinate axes consist of the XI axis, x2axis, ...,x,, axis is called a staie space. Any state can be represented b a point in the state space. ons, %nstate-space analysis we are concerned wit tYpes are involved in the m ling of dynamic systems: 1 variables, output variables, and state variables. state-space representation for a given systern is not unique, except that the numb of state variables is the same for any of the different state-space representations the same system. For time-varying (linear or nonlinear) discrete-time systems,the state e q d may be written as x(k + 1) = f [x(k),un(k), k] and the output equation as For linear time-varying discrete-time systems, t e state equation an where x(k) = n-wector (state vectsr) y(k) = m-vector u(k) = r-vector G(k) = n x n matrix (state matrix) H(k) = n x r matrix (input matrix) atrix (output matrix) D(k) = m x r matrix (direct transmission matrix) arance of the variable k in the arguments of m implies that these matrices are time vas-ying. If not appear explicitly in the matrices, they are constant. That is, if the system is time invariant, w(k + 1) = Gx(k) + Hu(k) (5-1) As in the discrete-time case, continuous-time (linear or nonlinear) systems may be represented by the following state equation and output equation: For linear time-varying continuous-time syste equation are given by If tbe system is time invariant, then the last ~ W Oe uati~nsare simplified to Figure 5-l(a) shows the block diagram representation of the discrete-ti system defined by Equations (5-1) and (5-2), and Figure 5-l(b) shows the contin- UOUS-timecontrol system defined by Equations (5-3) (5-4). Notice that the basic coarfigurations of the diserete-time and continusus- systems are the same.
- 155. iscrefre-Time Systems (a) Consider the discrete-time system described by y(k) + aly(k - 1) + aly(k - 2) + + any(k - n) = b o u ( k ) + b l u ( k - 1 ) + - - . + b n u ( k - n ) (5-5) Y(z) bo+ blz-l + . - + b, z-"-=: ~ ( z ) 1+ alz-' + . - - + anz-" (5-6) Y(z) bozn + blzn-l + a .- = + bn U(Z) zn + al zn-1+ - . + a, (5-7) (b) following representations: Controllable canonical form . Observable canonical form . Diagonal canonical form . Jordan canonical form hod. (See Problem A-5-1.) The observable canonical form can be obtained by tions of linear time- -time systems. Section 5-3 first tr nested programming rnethod. (See Problem A-5-2.) The diagonal canonical orm and Jordan canonical form may be obtained by use of the partial-fraction- procedure and by the z transform approach. Then it presents a method for com ing (zI- G)-'. Section 5-3 concludes with discussions of the solcltdon of the li time-varying discrete-time state equation. Section 5-4 deals with the pulse tran fmction matrix. Section 5-5 first treats the discretization of linear continasous-e state-space equations. Then it discusses time response between two concecu by the fdowing eqimations:
- 156. State-Space Analysic ec. 5-2 State-Space epresentationc of Díscrete-Time Systems uations (5-8) and (5-9) are the staf The state-space representation given -- - .eequation and output equation, by Equations (5-8) and (5-9) called a controllabl ical form. [For the derivatidn of Equations (5-8) a (5-9), see Problem Note that if w 1 e the order of the state variables, that is, we define ne state variables according to the fashion then the state equation can be written as follows: -al -a2 - -an Xl(k) 1 1 O - . e O O i 2 ( k ) O = O 1 . a . O O X3(k) + O O O - - S O - - ~ , i k ) ~ b~ The output equation can be given by kquations (5-10) and (5-11) are also in the controllable canonical form. The state-space representation of the discrete time system given by Equation (5-5), (5-6), or (5-7) may be put in the followin form: Tbe state-space representation given by Equations (5-12) and (5-13) is called an observable canonicalform. [For the derivation of Equations (5-12) and (5-13), see Problem A-5-2.1 Notice that the n X n state rnatrix of the state e Equation (5-12) is the transpose of that of the state equation defi (5-8). Note that if we reverse the order of the state variables, tha then the state equation and the output equation becorne as follows: i,-,(k + 1) Xn(k 4- 1) Xdk) X2(k) y ( k ) = [ 1 O S . - O O] f i n - l(k) in(k) . Equations (5-14) and (5-15) are also + bou(k) in the observable canonical form. . If the poles of the pulse transfer function given by Equation (5-5), (5-6), or (5-7) are al1distinct, then the state-space representation may be put in the diagonal canonical form as follaws:
- 157. tate-Space Analysis C rX,W i [For the derivation of Equatlons (5-16) and (5-17), see transfer function are distinct, then the state equation and out The state-space representations in the controllable canonical form, observable can ical form, and diagonal canonical form become as follows: CONTROLLABLE CANONICAL FORM: r ,J DIAGONAL CANONICAL FORM: The given pulse trancfer hnction Y(z)IU(z)can be expanded as follows: epresentationc of Discrete-Time ence, transformation. Gonsider the system defined by Let us define a new state vector X(k) by is a nonsingular matrix. [Note that, since both x(k) an k(k) are n-dimen- sional vectors, they are related to each other by Then, by substituting Equation (5-22) into ation (5-2O), we obtain Premultiplying both sides of Equation (5-23) by et us define uation (5-24) can be written as follows: imilarly, by substituting Equation (5-22) into E uation (5-21), we obtain y defining we can write Equation (5-26) as have thus shown that the state-space representation given by Equations (5-20) "nd (5-21),
- 158. n n n V V C W W W
- 159. ec. 5-3 Solving resent the solution of a discrete-time state equation by t Consider the discrete-time syste described by Equation (5-28): x(k + 1) = Gx(k)+ Hu(k) (5- Taking the z transforrn of both sides of Equation ( 5 - N ) , we get zX(z) - zx(0) = GX(z)+ HtJ(z) where X(Z)= Z[x(k)]and U(z) = Z[u(k)J.Then - G)X(z)= zx(0) + HU(z) Premultiplying both sides of this last equation by ( z - G)-', we obtain (5- Eking; the inverse z transforrn of both sides of Esuation (5-41) gives Comparing Equation (5-30) with Equation (5-42), we obtain Gk = Z-'[(zI - 6)-lz ] and k- 1 b_=k-j-1 where k = 1,2,3,. .. . Notice that the solution inverting the matrix ( z - G), which rnay be accomplisfied by analiytical meam by use of a computes routine. The solution also reauires the inverse z transfor Obtain the state transition matrix of the following discrete-time system: x(k + 1) = Gx(k)+ Hu(k) y ( k ) = Cx(k) where Then obtain the state x(k) and the output y(k) when the input u ( k ) = 1 for k O 2,. . .. Assume that the initial state is given by From Equations (5-35) and (5-43) the state transition matrix Therefoire, we flrst obtain ( z The state transition matrix ( k )is now obtained as follows: Equation (5-45) gives the state transition matrix. Next, compute x(k). The z transform of x(k) is given by ( z ) = (zlj - G)-lzw(0) + ( z Since we obtain z2 -z2 + 22 z - 1 z - 1 Hence
- 160. Thus, the state vector x(k) is given by Finally, the output y(k) is obtained as follows: where = GWn-2 + a,-i W, = GH,-, -i- a, I = Note that a,, a2,. . . a, are the coefr'icients appearing in the determinant &'e by Equation (5-47). The si's can also be given (see Problern A-5-13) by use of th trace, as follows: olvirig Discrete-Time S (The trace of an n x n matrix is t For a higher-order determina into the form given by Equation (5 Equation (5-50) lo compute the si's ent method isconvenient for computer Determine the inverse of the matrix (zI - G), where [ 0.1 0.1 o G = 0.3 -0.1 -0.2 O O -0.3 Also, obtain Gk. From Equation (5-46), we have Although the determinant /z - 61 can be expanded easily, here for demonstration purposes let us use Equation (5-50) to compute al, a2, and a3. First, notice that a, = -trG = -tr 0.3 -0.1 -0.2 = 0.3 [Y a -1.JThen, from Equation (5-49), we obtain Hence O.1 0.4 0.1 a- = -!tr GHl = -+ ([H::--o1 -:.2][0.3-0.3 O ii.2 - ~ 2 ] [:.o7 "03 -1.021 = -4 tr 0.09 0.01 0.02 = -0.04
- 161. ee. 5-3 Solving Diserete-Time State-Space Equations m$. Consider the uation (5-52) may be found easily by recursion, as follows: x(h + 1) = G(h)x(h)+ H(h)aa(h) x(h + 2) = G(h + I)x(h + 1) + A(h + l)u(h + 1) = G(h + f)G(h)x(h)+ G(h + l)H(h)u(h)+ Eet us define the state defined by Equation (5-52) a e k = h,h + 1,h + 2 , . . . . t can be seen that t e state transition ,h ) is given by the eqisation ution of Equation (5-52) becomes k j + l ) ( j ) ( ) k > h (5-55) j= h Notice that the first term on the right-hand side of Equation (5-55) is the contribu- tion o%the initial state x(h)to the current state x(k) and that the second term is the contribution of the input ~ ( h ) ,u(h + l ) ,... ,u(k - 1). Equation (5-55) can be verified easily. Referring to Equation (5-54),we have If we substitute Equation (5-56) into k j= h
- 162. me-Space Analycic Ch j= h = G(k)x(k) + H(k)u(k) Thus, we have shown that Equation (5-55) is the solution of Equation (5-52). Once we get the soiution x(k), the output equation, Equation (5-53), becom as follows: (k,J + I)H(j)n(j) + D(k)u(k), k > If @(k)is nonsingular ecf,so that the inverr;eof q ( k , exists, then the inverse of (h,k),is given as follows: = [G(k - I)G(k - 2) e G(h)]-l = ~ - l ( h ) ~ - l ( hi 1) e - ~ - y k- 1) . A sumrnary on the state transition matrix G(k - 1 ) q k - 2) . G(h), k > h f G(k) 1s nonsingular for al1 k values considered, then i ) foranyi,j,k f G(k)is singular for any value of k , then ( j , ) , f o r k > j > i A single-input-single-output discrete-time system may be modeled by a pulse tran fer function. Extension of the pulse-transfer-function concept to a multiple multiple-output discrete-time system gives us the pulse-transfer-function ma this section we shall investigate the relationship between state-space represen and representation by the pulse-transfer-function matrix. Pulse-Transfgr-Functa'o~ The state-space representation of an order linear time-invariant discrete-time system with r inputs and m outputs can given by x(k + 1) = Gw(k) + Hu(k) y(k) = Cw(k) + Dn(k) ec. 5-4 Pulse-Transfer-Function Matrix where x(k) is an n-vector, u(k) is an r-vector, (k) is an m-vector, G is an n x n matrix, His an n x r matrix, is an m x n matrix, and D is an m x r the z transforrns of Equations (5-58) and (5-59), we obtain Pdoting that the definition of the pulse transfer function calls for the assumption of zero initial conditions, here &idstate x(O) is zero. Then we obtain where (z) is called the pulse- nsfer-function matrix. transfer function rnatrix z) characterizes the in Since the inverse of matrix (z - G) can be written as ulse-transfer-function matrix (z) can be given by the equation Clearly, the poles of (z) are the zeros of lz cbaracteristic equation of the dáscrete-time syst or zn + alzn-l + a2zn-2+ . - + an-lz + a, = O where the coefficients ai depend on the elements of G. have shown that for the system defined by x(k + 1)= Gx(k) + Hu(k) the pulse-transfer-function matrix is ]in Section 5-2 we sh representations for a &en system are rela defining a new state vector %(k)by using a si x(k) = P%(k)
- 163. ,fi are related, res ~ = i e j , The pulse-transfer-function matrix F(z)for the system and (5-62) is That is, itdoes not depend on the particular state vector x(k) c representation. The characteristic equation lz transformation, since Thus, ihe eigenvalues of G are invariant under similarity transformation. In digital control of continuous-time plants, we need to convert state-space equations into discrete-time state-space equations, can be done by introducing fictitious samplers and fictitious hol continuous-time systems. ?'he error i negligibleby using a sufficientlysmall sa time constant of the system. iscretimatiom of Continuocas-Time tate-Space Equations ecause of the convergence of t Aktklk!,the series can be differentiated term by ter A3t2 ~k tk-ld -- - A + ~ 2 t+- + ... + ----+ ... dt 2! (k - l)! = t is can be proved as follows: In particular, if s = -t, then eAte-At = e-At eAt = eA(t-t) = inverse of eA'ise-"'. Sincethe inverse af e"' alwaysexists, e*' is nonsingular. s important to point out that shall next obtain the solutiorr of the continuous-time state e where x is the state ctor (n-vector), ia the input vector (r-vector), and premultiplying both sides of this last equation by e-"', we obtain ntegrating the preceding equation between O and t gives e-*tx(t) = X(O)+ /'e-*'
- 164. Equation (5-64) is the solution of Equation (5-63). Note that t state equation starting with t e initial state x(tO)is shallpresent a procedure for discr assume that the input vector n(t) Note that the sampling operation Consider the continuous-ti Hn the following analysis, esentation, we use the notation kT a (k + 1)T instead of k an (5-66) will take the forrn x((k + 1)T)= G(T)x(kT)+ W(T)u(kT) Note that the matrices G and H depend on the samplingperio are constant matrices. To determine G(T)and so that all the components of consecutive sampling instants, or u(t) = n(kT), for kT 5 t < kT + T and " O multiplying Equation (5-71) by eATand subtractingit from Equation (5-70) gives x((k + 1)T)= eATx(kT)+ eA(k+l)T[(k+1)7 e-*T , , , , JkT / - %nce from Equation (5-69) ~ ( t )= u(kT)for kT i t < kT 3- T ,we may substitut U(T) = u(kT)= constant in this last equation. [Note that u(t) mai jump t = kT + T and thus u(kT + T )rnay be different from u(kT).Such a jump in u( at r = kl' + T ,the upper Iirnit of integration, does not affect the value of the inWP in this last equation, because the integrand does not involve impulse function " * nence, we rnay wite iscretization of Continuous-Time State- where h = T - t. then Equation (5-72) becomes x((k + 1)T)= G(T)x(kT)i- (5-75) trices C and D are constant atrices and do not depend on the sampling atrix A is nonsingular, t T )given by Equation (5-74) can be simpli- fied to In the state-space approach, notice that by assuming the input vector a(t)to be constant between any two consecutive sampling instants, the discrete-ti model can be obtained simply by integrating the continuous-time state equa- tion over one sampling period. The discrete-time state equation given by Equation (5-68) is called the zero-orderhold equivalent of the continuous-time state equation given by Equation (5-66). n general, in converting the continuous-time system equationr into a discrete- ime system equation, some sort of approximation is necessary. It is important to point out that Equation (5-75) involves no approximation, provided the input vector u(t) is constant between any two consecutive sampling instants, as assumed in the derivation. Notice that for T <<1,G(T)= G(O)= eAO= . Thus, as the sampling period T becomes very small, G(T)approaches the identity matrix. Consider the continuous-timesystem given by
- 165. Eea Obtain the continuous-time state-space representation of the system. Then discre the state equation and output equation and obtain the discrete-time state-space reDr sentation of the system. Also, obtain the pulse transfer function for the syst&nby u Equation (5-60). The continuous-time state-space representation of the system is simply Now we discretize the state equation and the output equation. Referring to Eqjuatio (5-73) and (5-74), we have G(T) = e--aT r T I ..-aT H(T)= jo = a ' Hence, the discretized version of the system equations is 1- e-aT x(k + 1) = e-OTx(k) +-a u@) Y(k) = x(k) Referring to Equation (5-60), the pulse transfer function for this system is - - / a a(l - e-aTz-") This result, of course, agrees with the z transform of C(s) where it is preced by a sampler and zero-order hold [that is, where the signal u(t) is sampled and fed a zero-order hold before being applied to G(s)]: Obtaim the dascrete-time state and output equations and the pulse transfer hnct (when the sarnpling period T = 1)of the following continuous-time svstern: which may be represented in state space by the equations The desired discrete-time state equation will have the form x((k + 1)T) = G(T)x(kT) + H(T)u(kT) where matrices G(T) and B(T) are obtained from Equations (5-73) and (5-74) 6- 1 2 . iscretization of Continuaus-Time State-Space E Thus, L r 1 I h e output equation becomes - ... hen the sampiing period is 1sec, or T = 1, the discrete-time state equation and the output equation become, respectively, and - ... The pulse-transfer-function reprecentation of this system can be obtained from Equa- tion (5-60), as follows: Note that the same pulse transfer function can be obtained by taking the z transform of C(s) when it is preceded by a sampler and zero-order hold. Assuming T = 1, we obtain - - F -.
- 166. equation into then, assurning the sampling period to be 0.05 sec, we obtain G and H as E space equation x(k + 1) = Gx(k) + Wu(k) iscretization of Continuous-Time As another example, consi es the following system: Assuming that the sampling period T is 0.05 sec and wit out specifyingthe forrnat, we get the following discrete-time state equation: x(k + 1) = Gx(k)+ where matrices G and H can be found in the following computer out A = [O 1 0 0 20.601 O O O o 0 0 1 -0.4905 O O O]; B = [O;-1;0;0.5]; [G,H] = c2d(AfB,0.05)
- 167. ec. 5-6 Liapunov Stability Analycis li Consider the system discussedin Example 5-5. Obtain the discrete-time state equation and the output equation at t = kT + AT. Also, obtain the specificexpressions for the state equation and output equation when 2" = 1 sec and lin Example 5-5, the matrices G(T)and H(T) were obtained as follows: 4, the z transform solution of the discrete-time sy shall showthat th Consider the time-invariant continuous-time system defined by Eet us assume that the input u is sampled and fed to a zero-order hold u ( ~ )= u(kT)for kT T < kT + T. Referring to Equation (5-651, the solu the state equation starting with the irnitial state x(to)is G(AT)= eAAT Then we obtain n important role in &hestability analysisof control equations. There are two methods o£ stability efirst method and the second method; both apply mination of the stability of dynamic systems described by ordinary .The first method consists entirely of procedures tions of the differential equations or difference . The second rnethod, on the other han4 does ential or difference equations. Thisis the reason second method is so useful in practice. Although there are many powerful stability criteria available for control sys- s, such as the Jury stability criterion and the Routh-Hurwitz stability criteria, puter calcu'tation.) Y are limited to linear time-invariant systerns. The second method of Liaprtnzov, To obtain the state equation and the output equation at t = kT + da", where O < AT < T , we first convert G(T)to G(Air) and H(T) to H(AT) and then substitute G(ibT)and M(dT) into Equations (5-80) and (5-811, as follows: 1 '(1 - e-'''' ) 1 EOr T = 1 and d T = 0.5 we obtain the state equation and output equation as follows: [ E ]= [ :::(7:][:rl:l] + [ u : ~ P ~ Y ] u ( ~ ) y(k + O S ) = 11 0.31hlj[:~:i] + (00920)u(l)
- 168. on the other hand, is not limited to linear time-invariant systems: both linear and nonlinear systems, time invariant or time varying. find that the second method of Liapunov is indispensable for the of nonlinear systems for which exact solutionsmay be undtainable. (It is cautlone however, that although the second method of Liapunov is applicable to any n linear system, obtaining successful results may not be an easy task. imagination may be necessary to carry out the stabifity analysis o£ systems.) The second method of Liapunov is also called t the classicaltheory of that 1energy is continuall equilibrium state is reached. The second method of Liapunov is based on a generalization of t system has an asymptotically stable equilibrium state, tkien the stored system displaced within a domain of attraction decays with increasin finally assumesits minimum value at the equilibrium state. For purely mat systems, however, there is no simple way of defining an "' circumvent this difficulty, Liapunov introduced the Liapunov fu energy function. Shisidea is, however, more general than that of en widelyapplicable. In fact, anyscaiar function satisfyingthe hypothe orems (see Theorems5-1 through 5-6) can serve as a Lia efore we discuss the Liapunov function further, it is necessary to defin positive definiteness of scalar functions. ons. A scalar function V(A)is said positive definite in a region f2(which includesthe or e state space) if v ( ~ for al1 nonzero states x in the region 92 and if V( A time-varyirig function V(X,t) is said to b ve definite in a regi (which includes the origin of the state space) if it is bounded from below time-invariant positive definite function, that is, if there exists a positive de function V(x) such that V(x, t) > V(x), for a l t 2 t, for al1 t 2 to negat to be origin and at certain other states, where it is zero. sns. A scalar function V(w) is nite if -V(x) is positive semidefinite. A scalar function V(x) is said to be i ositive and negative values, no matler s~nallthe region $2 is. ec. 5-6 Liapunov Stability Analysis In this example, we give severa1scalar functions and their classificationsaccordingto the foregoing definitions. Here we assume x to be a two-dimensionalvector. 1. V(s)= x: + x: positive definite x; . V(x) = x: + ----1 + xi positive definite 3. V(x) = (xi+ x2)2 positive sernidefinite . V(x) = -x: - (xl+ x2)' negative definite V(x) = x1x2+ xi indefinite ons. The Liapunov function, a scalar function, is a positive a time derivative nctions do not include t explicitly, then we denote them by V(xl,~ 2 ,...,x ~ ) ,or V(x). Notice that ~ ( x ,t ) as ac of V(x,t) with respect to t along a solution of the system. s that V(x, t) is a decreasing function of t. A Liapunov fuai en systern. ( k r this reason, the second method of Liapunov is a more powerful tool than conventional energy ons. Note that a system whose energy E decreases on the average but not at each instant is stable, but that E is not a Liapunov function.) er in this section we shall show that in the second method of Eiapunov the vior of V(x, t) and that of its time derivative ~ ( x ,t) = dV(x,t)/dt give information about the stability of an equilibrium state without having t Note that the simplest positive definite function is of a In general, Liapunov functions may not be of a simple quadratic form. For any Liapunov function, however, the lowest-degree terms in V must be even. This can be seen as follows. If we define then in the neighborhood of the origin the Iowest-degree terms done will become dominant and we can write V(x) as If we keep the i,'S fixed, V(X1,X2, . ..,in-l,1) is a fixed quantity. For p odd, x[ can assume both positive and negative values near the origin, which means that V(x) is not positive definite. Hence, p must be even. In what lollows, we give definitions of a syste ,an equilibrium state, stability, asymptotic §tal-iility,and instability.
- 169. pace Analysis Chap. where x is a state vector (an n-vecto functions of xl,x2,. .,x,, and t. (Note that w a model to present basic materials on stabilit Ilx - xell 5 r where llx - x,ll is called the Euclidean norrn and is defined as follows: I/x - = [(x~- le)' + ( ~ 2- ~ 2 , ) ~+ .e + (xn - x,,)~]~" Let S(6) consist of al1 points such that Ilx, - xell 5 6 and let S ( € )consist of al1 points such that An equilibrium state x, of the system of Equation (5-82) is said to be s sense of Eiapunov if, corresponding to each S(€),there is an S(6) such t tories starting in S(6) do not leave S(€) as t increases indefinitely. The re 6 depends on E and, in general, also depends on h. If 6 does not depen equilibriurn state is said to be uniformly stable. Wlat we have stated here is that $ ( E ) , there must be a region S(8) such S ( E )as t iilcreases indefinitely. ec. 5-6 Liapunov Stability Analysis An equilibriumstate xeof the syste is said to be asyrnptotically stable if it is stable in t solution starting within S(6) converges, without leaving S(€),to w, as t increases indefinitely. In practice, asymptotic stability is more im ortant than mere stability. Also, since asymptotic stability is a local conce simply to establish as oes not necessarily mean that the syste largest region o£ asy e dornain of attractio ble trajectories originate nating in the domain of attraction is asym totic stability holdsfor allstates (al1 from which trajectories originate, t e equilibriurn state is said to be asymptotically stable in the large. That e equilibrium state x,of the system given by Equation (5-82) is said to be as tically stable In the large if it is stable and if every solution converges to x, as ses indefinitely. Obviously, a necessary condition Eor asymptotic stability in e is that there be only one uilibrium state in the whole state space. In control engineeringproblems, asymptotic stability in t feature. -hf the equilibrium state is not asymptotically stable in the large, then the problem becomes one of determináng the largest region of asymptotic stabiiity. This is usually very difficult. For al1practica1purposes, however, it is sufficient to deter- mine a region of asyrnptotic stability large enoraghthat no disturbance will exceed it. n equilibrium state x, is said to be unstable if for sorne real number E > O and any real number 6 > O, no atter how small, there is always a state xoin S(6) such that the trajectory starting at this state leaves S(€). graphical representation of the foregoing definitions will clarify their meanings. Let us consider the two-dimensional case. Figures 5-2(a), (b), and (c) show equilibrium states and typical trajectories corresponding to stability, asymptotic Figure 5-2 (a) Stable equilibrium state and a representative trajectory; (b) asymp- totically stable equilibrium state and a representative trajectory; (c) unstable equi- librium ctate and a representative trajectory.
- 170. ec. 5-6 Liapunov Stability Analysis V(x) = C where Tf there exists a sealar fimction V(x, t )having continuous first partial derivatives satisfying the conditions . V(x, t) is positive definite. Q(x, t) is negative definite. The conditions of this theorem may be modified as follows: entically in t 2 tofor any toand any wo =# lution starting from xoat t = to. ly stable in the large. move toward the origin. To prove stability (but not asymptotic sta- biliay) of the origin of the systern defined by Equation (5-&2),the following theorem may be applied. . Suppose a system is described by for al1t. If there exists a scalar function VCx,t) having continuoras atives and satisfying the conditions V(x, t) is positive definite. . v(x, t) is negative semidefinite. then the equilibrium state at the origin is uniformly stable. It should be noted that the negative semidefiniteness of V(x, t) [ ~ ( x ,t) 5 O along the trajectories] means that the origin is uniformly stable but not necessarily ~niformlyasymptotically stable. Hence, in this case t e system may exhibit a limit cycle operation. . If an equilibriurn state x = of a system is unstable, then there function W(x, t) tkat determin s the instability of the eqluilibrium state. We shall present a theorem on instability in t
- 171. tate-Space Analycic w = f(x, t ) where W(x, t) is positive definite in the same region. arks. A few comments are in order when the Liapunov st tions but are not necessary conditions. includes this equilibrium state, it does not necessarily mean that the are unstable outside the region 0. For a stable or asyrnptoticauy stable equilibrium state, a Liapunov fun with the required properties always exists. proaches to the investigation of the asyrnptotic stability of linear time-i systems. For exarnple, for a continuous-time systern described by the equ K = Ax it can be stated that a necessary and sufficient condition for the asymptotic of the origin of the system is that al1 eigenvalues of A have negative real p that the zeros of the characteristic polynomial have negative real parts. Sirnilarly, for a discrete-time system represented by the equation x(k + 1) = Gx(k) ec. 5-6 Liapunov systems, is algebraic and does not require factoring of the characteristicpolynomial, e-invariant systemsthe gives not just sufficient conditions, but t stability or asymptotic stabiiity. ty analysiso£linear time-invariant systems, it is aspmed that if an eigenvalue A, o£matrix A is a complex quantity then A must have A,, the complex conjugate of A,, as its eigenvalue. Thus, any complex eigenvalues of A wili appear as conjugate complex pairs. Also, in the following discussions on stability, we shall use the conjugate transpose expression, rather than the transpose expres- sion, of matrix A, since the elernents of rnatrix A may include complex conjugates. The conjugate transpsse of t is a conjugate of the transpose: Consider the following linear time-invariant system: where x is a state vector (an n-vector) and A is an n x n constant matrix. is nonsingular. Then the only equilibrium state is the origin, stability of the equilibrium state of the linear time-invariant system can be investi- gated easily with tke second method of Liapunov. For the system defined by Equation (5-84), let us choose as a possible Liapunov function is a positive definite f x is a real vector, then chosen to be a positive definite real syrnmetric matrix.) The time derivative of V(x) along any trajectory is Since V(x) was chosen to be positive definite, we require, for asyrnptotic stability, that p(x) be negative definite. Therefore, we require that
- 172. .%te-%pace Analysis Chap. ) = posltfve definite e asymptotic stability of the system of E For a test of positive definiteness of an n x n rnatrlx, we apply Sylvester criterion, which states th essary and sirfficientcon rminants of all tke successivepri r example, the following n x n are al1real, then the Hermitianmmatrix beco r- -7 p11 Pl2 ' Fin p 2 n a Pnn where pi,denotes the complex conjugate of pij. a13 the successive principal rninors are positive, Pil Pi2 - e . . F12 p22 " ' 9 . .. .. . tead of first specifying a positive definite matri is positive definite, it is c then examine whether e definite. Note that po shall strrnrnarize what Consider the system described by x = AA where x is a statevector (an n-vector) and A is an n x re co necessary and sufficient condition for the equilibrium state totically stable in the Iarge is that, giv any positive definite positive definite real symmetric) matrix (or a positive definite real symmetric) rnatrix The scalar function x*Px is a Liapunov function for this system. [Note that in r systern considered, if the equilibrium state (theorigin) is asymptoticaliy stabl it is asyrnptotically stable in the large.] wks. In applying Theorem 5-4 to the siability analysis of linear tirn ~tinerous-timesysterns, several important rernarks rnay be rnade. 5-6 Liapunov Stability Analysis ong any trajectory, then be chosen to be positive semidefinite. If an arbitrary positive definite matrix is chosen for idefinitematrix if ~ ( x )doesnot vanishidentical the matrix equation as a root of the chasacteristic equarion, and if for every sum of two roots then the elements of ined. (Note that for a stable matrix the sum Ai + hk is always nonzero.) In determining whether or not t ere exists a positive definite Hermitian or positive definite real sym , it is csnvenient to choose is the identity matrix. Then tke ele are determined from is tested for positive definiteness. Determine the stabiiity of the equilibriurn state of the following system: The system has only one equilibrium state at the orágin. into Equation (5-851, we have Noting that A is a real matrix, must be a real symmetric matrix. This last equation may then be written as follows: where we have noted thatpzl = p12and made the appropriate substitution. If the matrix B turns out to be positive definite, then x*Bw is a Eiapunov function and the origin is asymptotically stable.
- 173. where tate-Space Analysis Equation (5-86) yields the following three equations: -2p11 + 2p12 = -1 - 2 ~ 1 1- $12 +pzz = 0 -4~12- 8p22= -I Solving for the pys,we obtain p1i = g, p12 = -&, pzz = &$ Hence, By Sylvester's criterion, this matrix is positive definite: Hence, we condude that t origin of the system is asymptotically stable in the large. It is noted that a Liapunov function for thic system is = &(23~:- 14x1x2 + 11x2) and p(x) is given by Ti(,) = -x; - x,2 time systems. As in the case of continuous-time S te-time systemsbased on the second met V(x(h-7')) = V(x(k + 1)T) - V(x(kT)) Consider the discrete-time system x = n-vector L = sampling period Suppose there exists a scalar function V(x) contlnuous in x such that 1. V(X) > O for x J. O. 2,AV(x) < O for x f O, where AV(x(kT))= ~ ( x ( k+ 1)T) - V(x(kT)) = V(f(x(k~)))- ~ ( r ( k T ) ) ec. 5-6 Liapunov Stability Analysis e equilibrium state x = totically stable in the large and V(x) is a kiapunov function. Note that in this theorem c 2 may be replaced by V(x) 5 O for all x, and V(x) does not vanish i entically for any soltution sequence {x(kT)}satisfying Equatlon (4-88). V(x) need not be negative definite if it oes not vanish identicalliy on any solution sequence o£the differesrce equation. Gonslder the discrete-time system described by where w is a state vector (an n-vector) and Gis an n X n constant nonsingular rnatrix. isthe equilibrium state. by use of the second method of Liapunov. ket us choose as a possible Liapunov ftunction positive definite Herrnitian (or a positive definite real sy V(x(k)) = W(x(k + 1))- V(x(k)) Since V(x(k)) is chosen to be positive clefinite, we require, for asymptotic stability, at AV(x(k)) be negative definite. Therefore, where ) = positive definite totic stability of the discrete-ti e system of Equation (5-89), systems, it is convenient to specifyfirst definite real symmetric) matrix Q and then to see avhether or not the P rnatrix deterrnined from G*PG - E" = -Q
- 174. w(k + 1)= Gx(k) A necessary and sufficient condition for t icallystable in the large is that, given any stability of an equilibrium state of a discrete-time system obtained by continuous-time systern is equivalent to that of the original continuous Consider a continuous-time system k = Ax and the corresponding discrete-time system x((k + 1)T) = Gx(kT) where G = Ef the continuous-time system is asymptotically stable, that is, if all the eigenv of the matrix A have negative real parts, then llGnll-+O, as n-+ m and the discretized system is also asymptotically stable. This is becau are the eigenvalues of A then the e" 's are the eigenvalues of 6.(Note t if hiT is negative.) Ht should be noted here that, if a continuous is discretized,then in certain exceptional caseshi ing on the choice of the samplingperiod T. That i time system is not asymptotically stable, the equivalent discretized system to be asyrnptoticallystable if we look at the vah instants. This phenomenon occurs only at cert If the value of Tis varied, then such hidden instability shows up as See Problem A-5-15. ConéPac~on. A norrn of ra denoted by I/x/lmay be though the Iength of the vector. There are severa1different definitions of however, has the foilowing properties: ec. 5-6 Liapunov l l ~ l l= 0, for x = A function f (w) is said to be a contraction if for some set of values of Consfder the folowing di ere x is an n-vector and (x) is also an n-vector. Assurne that (x) is a contraction Then the origin of the system of Equation (5-91) is asymptotically stabie in the large, and one of its LPa V(x) = 11d1 This can be seen as follows. Since V(x) = llxll is ositdve definite and is negative definite beca is a contraction for al1 w, we fin Liapunov function, and rem 5-5 the origin of the system stable in the large. (See Consider the following system: Determine the stability of the origin of the system. . Then, referring to Equation (5-90), the Liapunov stability equation becomes is found to be positive definite, then the origin x = is asymptotically stable in the large. From Equation (5-92) we obtain the following three equations: p,, - 2p12 = -1 from which we get
- 175. Consequectly, pace Analysis Chap. = [; $1 By applying Sylvester's criterion for the positive definiteness of matrix positive definite. Hence, the equilibrium state, the origin x = . . - in the large. Note that instead of choosing to be I we could choose semidefinite rnatrix, such as AV(x) = -x:(k) O -0.5 [l -1 1;: : ] [ -:]- [;:: Y;]= -[::]By solving this last equation, we obtaln (Direct programming method) Gonsider the discrete-time system defined by Y ( z ) - bozn+ b,zn-' +-- + bn U ( Z ) Z" + alzn-1+ . - + a, Show that a state-space representation of this system may be given by hap. 5 Example Problems and ntáasn The given system can be modified to This last equation can be written as follows: (61 - al b ~ ) z - 1+ (b2 - a2b o ) ~ - ~+.+ + (b, - a, bo)z-" 1+ alz-l + a2z-2+ + a,z-" U(z> Let us define (bl - al bo)zll + (b2- a2b o ) ~ - 2+ + (b, - a, bo)z-" Y ( Z ) = 1 + alz-' + a2zW2+ e + u,,z-~ U(z> Then Equation (5-95) becomes ket us rewrite Equation (5-96) in the following form: ~ ( 4 (bi - al bo)z-' + (b2 - a2bo)z-' + + (b, - a, bo)z-, From this last equation the following two equations may be obtained: Q ( ~ )= -al;-' Q ( z ) - a2z-2Q ( z ) - - - unz-" Q ( z ) + U(z) and F(z) = (bl - al b0)z-' Q ( z )+ (b2 - a2bo)z-' Q ( z ) + .. + (b, - a, bo)z-" Q ( z ) Now we define the state variables as follows: Xn-'(2) = z - ~Q ( z ) Xn(z)= 2-l Q ( z ) Then clearly we have zX*(z)= X2(z) zXz(z) = &(z)
- 176. State-Space Analysis C In terms of difference equations, the preceding n - 1 equations become xl(k t 1) = x,(k) xz(k + 1) = x 4 k ) ~ , - ~ ( k+ 1) = x,(k) By substituting Equation (5-100) into Equation (5-98), we obtain zX,(z) = -alXn(z) - a2Xn--l(z)- . - anX1(z)+ U ( z ) which may be transformed into a difference equation: x n ( k + 1 ) = - a n x l ( k ) - a n - l x 2 ( k ) - ~ ~ ~ - a l x n ( k ) + ~ ( k )(5 Also, Equation (5-99) can be rewritten as follows: ' Y ( Z ) = (b1 - al bo)X,(z) + (bi - a,bo)Xn--l(z)+ . . + (b, - c. bo)X1(r) By use of this last equation, Equation (5-97) can be written in the form y(k) = (b, - a, bo)xi(k) + (bn-i - a,-, bo)xz(k) + ... +(bi-aibo)x,(k)+bou(k) (5 Combining Equations (5-101) and (5-102) results in the state equation giv Equation (5-93). The output equation, Equation (5-103), can be rewritten in the given by Equation (5-94). (Nested programming method) Consider thepulse transfer function system defin Y ( z ) bo + blz-' + - .+ b,z-"qZ)= -- = U(z) l + a l z - ' + . - ~ + a n z - " Show that a state-cpace representation of this system may be given as ~ L ~ o w s : - 9 -z x l ( k t 1) O O a - . O O -a, xl(k) x z ( k + l ) 1 O V . - O O xz(k) ,- . .- . . . .. .. . . . ~ , - ~ ( k+ 1) O O + - - 1 O -a2 ~ , - ~ ( k ) x n ( k + l ) O O O 1 -al x,(k)- m - -, b, - a, bo bn-i - 6,-1 bo + b2 - a, bo bi - al bo P x1( k ) x2W y(k) = [O O O 11 ! x,-1 ( k ) - xn(k) Chap. 5 Exarnpie Problerns and Sslutions aakiora Rewrite the pulse transfer Eianction as follows: Y ( z ) - boU ( z )+ zw1[alY ( z ) - bl U(z)] + z-'[a, Y ( z ) - bZ U(z)]+ - + ~ - ~ [ a ,Y ( z ) - b, U(z)]= O or Y ( z ) = boU(z)+ z-'(b1 U(z) - al Y ( z )+ z-'{b2 U ( z )- a, Y ( z ) + Z-'[b3 ~ ( z )- a 3 ~ ( z )+ . -11) (5-106) Now define the state variables as follows: X,(Z) = z-'Lbl U ( z )- al Y ( z ) + Xn-l(z)] Xn-,(z) = z-l[bZU(Z)- a, Y ( z ) + XnV2(z)] (5-107) &(z) = ~ - l [ b , - ~U(z) - Y ( z )+ Xi(z)] Xl(z) = z-'[b, U(z) - a, Y ( z ) ] Then Equation (5-106) can be written in the form Y ( z ) = boU(z) + &(z) (5-108) By substituting Equation (5-108) into Equation (5-107) and multiplying both sides of the equations by z , we obtain ~ X n f z )= Xn-~(z)- a l x n ( z )+ (bi - al bo)U(Z) zXn-i(z) = X,-~(Z)- a2Xn(z)+ (b2 - a2bo)U(z) zXz(z) = X ~ ( Z )- a,-, Xn(z) + (b,-1 - anTlbo)U(z) zX1(z)= -a, X,(z) + (b, - a, bo)U(z) Taking the inverse z transforms of the preceding n equations and writing the resulting equations in the reverse order, we obtain xi(k + 1) = -a, x,(k) + (b, - a, bo)u( k ) xz(k + 1) = xl(k) - an-lxn(k)-t- (b,-l - a,-I bo)u(k) ~ , - ~ ( k+ 1) = x,-,(k) - a2xn(k)+ (6, - a2bo)u(k) x,(k + 1) = ~ , - ~ ( k )- alxn(k)+ (bl - al bo)u(k) Also, the inverse z transform of Equation (5-108) yields y ( k ) = x, ( k ) + bo~ ( k ) Rewritingthe state equation and the output equation in the standard vector-matrixform gives Equations (5-104) and (5-105), irespectively. (Partial-fraction-expansion programming method) Consider the pulse transfer func- tion system given by
- 177. and expanded into the following form: where Equation (5-112) can be written in the form Y ( z )= boU(z)+ AU(z) + L?L U(z)+ m - +--LL. - p i z -p2 z -pn w1 Let us define the state variables as follows: 1 X,(z) = - z -p1 u(4 1 X,(z) = - z -pn U(z> Then Equation (5-114) can be rewritten as zX,(z) = plxl(z) + U ( z ) zX2(z) = p2&(z) + U(z) State-Spaw Analycis Ch Y ( z ) bozn+ bi zn-' + - - - + bn- = U ( Z ) zn + al zn-' + ...+ a, Show that the state equation and output equation can be givenin the following canonical form if al1 poles are distinct. xi(k + 1) 0 . " . k 2 ( k : 1 ) ] = Fx,(k + 1) O 0 ~ ~ : ~ ' l ] + [ ~ ] u ( k ) Pn xn( k ) x l w y(k) = [cl c2 c n l / ( k ) ] + bou(k) xn(k) Solation The system puise transfer function can be modified as follows: Y ( z ) buzn+ bl 2"' + e e + bn-- U(2) ~ " + a ~ z " - l + . . - + a , , (bl - al bo)zn-l + (b2- a2bo)zn-' + . - + (bn - anbo) = bo r (2 - pí)(z - pz) ( z - pn) Since al1 poles of the puise transfer function Y(z)/U(z)are distinct, Y(z)/U(z) Chap. 5 Exarnple Problerns and Also, Equation (5-113) can be written as Y ( z )=: bo U ( Z )+ C ~ X , ( Z )+ Q X ~ ( Z )+ + cnXn(z) (5-116) The inverse z transforms of Equations (5-115) and (5-116) become xl(k + 1) = p,xl(k) + ~ ( k ) x2(k + 1) = p2 ~ 2 ( k )+ ~ ( k ) (5-117) xn(k + 1) = pn xn( k ) + ~ ( k ) and y ( k ) = c,xi(k) + c Z X ~ ( ~ )+ + cnxn(k)+ bou(k) (5-113) Rewriting the state equation and the output equation in the form of vector-matrix equations, we obtain Equations (5-109) and (5-110). (Partial-fraction-expansion prograrnming method) Consider the pulse transfer func- tion system defined by Y ( z ) - bozn + bl zn-l + + b,-- U ( Z ) zn + alzn-1 + - m + a, Assume that the system involves a multiple pole of order m at z = p, and that all other- - poles are distinct. Show that this system may be represented by the followii output equation: g state equation and Solutisn Since the system pulse transfer function can be written in the form (bl - al bo)zn-l + (b2- a2b o ) ~ n - 2+ . - = bo + + (bn - a, bo) ( 2 - P I ) ~ ( Z- pm+l) * (2 - pn)
- 178. we obtain Let us define the first m state variables Xl(z),&(z), ... ,&(z) by the equations and the remaining n - m state variables ;%i,+l(z),Xm+2(z), 1 Xm+1( z) = --M-- Z - pm+i W ) 1 Xm+2( Z ) = ------ U(Z) 2 - Pm+2 Notice that the m state variables defined by Equation (5-123) are related each to th next by the following equations: Xm-i(z>- 1 Xm(z) - p i By tahing the inverse z transforms of al1 of Equation (5-125), the last equation Equation (5-123), and al1 of Equation (5-124), we obtain ap. 5 Exarnple Problems and Sslutions xrn+i(k + 1) = pm+ixm+i(k)+ ~ ( k ) The output equation given by Equation (5-122) can be rewritten as follows: Y ( z ) = C I X ( Z )+ ~ ~ X Z ( Z )+ .. + cmXm(z) + cm+lXrn+l(z) + cm+zXm+,(z)+ + cnXn(Z) + bo U ( Z ) By taking the inverse z transform of this last equation, we get y(k) = clxi(k) + czxz(k) + . + crnxrn(k) + crn+ixrn+,(k) + cm+2xm+2(k)+ ' ' + ~n xn(k)+ bo~ ( k ) (5-127) Rewriting Equations (5-126) and (5-127) in the standard vector matrix form, we obtain Equations (5-119) and (5-120), respectively. Using the nested programming method (refer to Problem A-5-2), obtain the state equation and output equation for the system defined by Then draw a block diagram for the system showing al1 ctate variables. sHu&ion The given pulse transfer function can be written as Y ( z )= z-'{U(z) - 4Y(z) + z-'[SU(z) - 3Y(z)]} Define The state equation can therefore be given by and the output equation becomes - -. Figure 5-3 showsthe block diagram forthe systern defined by the state-space equations. The output of each delay elernent constitutes a state variable. Obtain a state-space representation of the system shown in Figure 5-4. The sampling period Tis 1 sec.
- 179. Figure 5-3 Block diagram for the system considered in Problem A-5-5. Y olution Vde shall first obtain the z transform of the feedfonvard transfer h n 0.3679(z + 0.7181)-- (Z - 1)(z - 0.3679) diagram modification. Let us expand G(z) into partial fractions: 1 G(z) = -- 0.6321 - z -'--- 0.6321~-' z - 1 z - 0.3679 1 - Z-' 1 - 0.3679~-' Figure 5-5 shows the block diagram for the system. Let us choose the output ~f e unit delay element as a state variable, as shown in Figure 5-5. Then we obtain ap. 5 Example Problems and Solutions Figure 5-5 Modified block diagram for the system shown in Figure 5-4. from which we get x (k + 1) O 0 6 3 2 x (k) [xi(k + l)] [-1 1' ][xi(k)] + Obtain a state-space representation of the following pulse-transfer-function system: Use the partial-fraction-expansion programming method. Also, obtain the initial values of the state variables in terms of y(O), y(l), and y(2). Then draw a block diagram for the system. rihtion Because we need the initialvalues of the state variables in terms of y(O), y (l), and y (2), we slightly modify the partial-fraction-expansion programming method pre- sented in Section 5-2. Let us expand Y(z)/U(z), zY(z)/U(z), and z2Y(z)/U(z) into partial fractions as follows:
- 180. Then we have Yow iet Then the state as follows: Erom Equatio Noting that we get us define the state variables by the following equation: P -f r =? variablesX3(z),&(z), and &(z) are related to Y(z), zY(z z2Y(z) m (5--12$), we obtain (2 + 1)2Xi(z) = U(z) (z + 1)Xz(z) = U(z) (Z 2)X3(Z) = U(Z) .(z) is given by the equation Y(z) = 5Xi(Z) - j&(Z) .t5X3(Z) Consequently, we have the state-space equations as follows: The initial data are obtained by use of Equation (5-129), as follows: ?*heblock diagram for this system is shown in Figure 5-6. Obtain a state-space representation of the followingpulse-transfer-function systemsuch that the state rnatrix is diagonal: Then obtain the initial state x(0) in terrns of y(O),y(l),y(2) and u(O), u(l), u(2). Solutisn Let us first divide the nurnerators of the right-hand sides of Y(z)IU(z), zY(z)IU(z), and z2Y(z)/U(z) by the respective denominators and expand the remain- ing terms into partial fractions, as follows: Figure 5-6 Block diagram for the system considered in Problem A-5-7.
- 181. State-Space Analysis . 5 Example Broblerns an Rewriting, we have Let uc define the state variables X1(z),&(z), and %(z) as follows: Then we have Notice that Equation (5-130) can be written as zX&) = -X,(z) + U(z) zX2(z)= -2X4z) + lJ(z) zx3(z)= -3X3(Z) + u ( Z ) from which we obtain xi(k + 1) = -xi(k) + u(k) xz(k + 1) = -2x2(k) + ~ ( k ) x3(k + 1) = -3x3(k) -iu(k) The output Y ( z )Is given by In vector-matrix notation, the state space equations become O O x1(k) [ =~ 3 ( k+ 1) -2O -30 ] [ x 2 ( k ) ] + [ ~ ] u ( k )x3(k) The initial data are obtained from Equation (5-131) as follows: Y @ ) - Y (1)- ~ ( 1 )- 2u(O) y(2) - u(2) - 2u(l) + 6u(O) Figure 5-7 shows the block diagram for the present system. Figure 5-9 Block diagram for the system considered in Problem A-5-8
- 182. Let A be an n x n matrix, and let its characteristic equation be lhli - A/ = A" + al A"-' + - e + a,-~A + a, = O Show that matrix A. satisfies its characteristic equation, or that A" + alAn'-l + .e + anWiA+ a, I = (This is the Cayley-Mamilton theorem.) where Note also that Wence, we obtain This proves the Cayley-Hamilton theorem. Referring to Problem A-5-3, it has been shown that every n x n matri own characteristic equation. The characteristic equation is not, however scalar equation of least degree that A satisfies. The least-degree polyn as a root is called the minimal poíynomial. That is, the minimal polynomia matrix A is defined as the polynomial $(A) of least degree: +(A) = A" + al A"-' + +a,-~h+a,, m r n The minimal polynomial plays an important role in the computation of polyno an n x n matrix. Let us suppose that d(A), a polynomial in A, is the greatest common al1the elements of adj (A1 - A). Show that if the coefficient of the highest-de in h of d(A) is chosen as 1 then the minimal polynomial $(A) is given by xarnple Problemc and Solutionc y assumption, the greatest common divisor of the matrix adj (A1 - A) is d (A). Therefore, where the greatest common divisor of the n2efements (which are functions of A) of is unity. Since - A) adj (A1 - A) = Al - Al we obtain (5-132) from which we find that /A - Al is divisible by d(A). Let us put (5-133) Then the coefficient of the highest-degree term in A of $(A) is unity. From Equations (5-132) and (5-133), we have Note that $(A) can be written as follows: $(A) = g ( A ) W ) + where a(A) is of Iower degree than $(A). Since $(A) = . Since 4(A) is the minimal polynomial, a ( h ) must be identically zero, or 4 0 ) = g(h)rp(h) Note that because $(A) = 0 we can write and we obtain ( 4 = g w w Note that the greatest common divisor of the n2elements of g(A) = 1 Therefore, $(A) = +(A) Then, from this last equation and Equation (5-133), we obtain It is noted that the minimal polynomial $(A) of an n X n matrix A can be determined by the following procedure: B. brnn adj (A1 - A) and write the elements of adj (Al - A) as factored polynomi- als in A.
- 183. State-Space Analysis Determine d(A) as the greatest common divisor of al1 the elemen ). Choose the coefficient of the highest-degree term in A of d be 1. If there is no common divisor, d(A) = 1. 3. The minimal polynomial +(A) is then given as A has n distinct eigenvalues, then the minimal polynomia racteristic polynomial. Also, if the multiple eigenvalues linked in a Jordan chain, the minimal polynomial and the characteristic polyno identical. If, however, the multiple eigenvalues of A are not linked in a Jorda the minimal polynomial is of lower than the characteristic p 'CJsingthefollowingmatrices A S examples, verify theforegoing state about the minimal polynomial when multipie eigenvalues are involved. iatioan First, consider the matrix A. The characteristic polynomial is give A - 2 -1 - A - O A - 2 y ¡=(A-?)'(A- O -3 A - 1 Thus, the eigenvalues of A are 2, 2, and 1. Ht can be shown that the Jordan c form of A is [H d 81and the multiple eigenvalues arelinked in the Jordan chain as shown. (For the for deriving the Jordan canonical form of A, refer to Appendix A.) To determine the minirnal polynomial, let us first obtain adj ( given by (A-2)fA-1) (A+11) 4(A-2) [ : (A - 2)(A - 1) 3(A - 2) (A - 2)" Notice that there is no common divisor of al1 the elements of adj (A d(h) =: 1.Thuc, the minimal polynomial +(A) is identical with the chara nornial, s r - Al (A - 2)2(h - 1) = h3 - 5h2+ 8A - 4 A simple calculation proves that A3 - 5a2+ 8A - 41= but Fi2 - 3A + 21 f Thus, we have shown that the minimal polynornial and the characte of this matrix A are the same. hap. 5 Exarnple robfems and Solurions Next, consider the matrix . The characteristic polynomial is given by simple computation reveals that matrix has three eigenvectors, and the Jordan [H % i]Thus, the multiple eige ues are not linked. To obtain the minimaí polynomial, we first compute adj (A1 - (A - 2)(X - 1) o (A - 2)(A - 1) 3(A - 2) (A - 212 from which it is evident that ence, As a check, let us compute +( For the given matrix ,the degree of the minimal polynomiai is lower by 1than that of the characteristic polynomial. As shown here, if the multiple eigenvalues of an n x n matrix are not linked in a Jordan chain, the minimal polynomial is of lower degree than the characteristic polynomial. Show that by use of the minimal polynomial the inverse of a nonsingular matrix A can be expressed as a polynomial in with scalar coefficients as follows: where al, a2,... ,amare coefficients of the minimal polynomial Then, obtain the inverse of the following matrix A = [:3 -1:-2:] Solution Fsr a nonsingular matrix A, its minirnal polynomial $(A) can be written as follows:
- 184. Clearly, there is no common divisor d(h) of al1 elements of adj (A1 - d ( h ) = 1. Consequently, the minimal polynomial +(A) is given by the eq Thus, the minimal polynomial +(A) is the same as the characteristic polyno Since the characteristic equation is - Al = h3 + 3h2- 7A - 17 = O we obtain $(A) = h3 + 3h2 - 7A - 17 y identifying the coefficients ai of the minimal polynomial (which is the sam characteristic polynomial in this case), we have al = 3, a, = -7, a3 = -17 The inverse of W can then be obtained from Equation (5-134) as follows: 1 + alA + a,I) = -(A2 + 3 17 7 O -4 -:I+ 3[n -. 6 -4 =A[! -;;] - 6. -4. [$ -8 ;]- - -L.17 17 17 Show that the inverse of z - G can be given by the equation adj (z1 - G) (z1 - G)-' = IzP - G/ -- 1zn-' + H1 + I ~ ? J ~ z ~ - ~+ - + Hn-1 lzli - G/ ) = A m + alAm-' + . e - +a,-lA+am where a, iO. Hence, 1 = --(Am+ alikm-' + + a,-,A2 + am-IA) am Premultiplying by A-', we obtain which is Equation (5-134). Eor the given matrix A, adj (h - A) can be given as follows: hap. 5 Example Problems and Sdlatisns where and al, a2,... ,a, are the coefficients appearing in the characteristic polynomial given by - 61 = zn + al zn-' + a2znW2+ + a, Show also that al = -trG a2 = -2 trGHl 7% simplify the derivation, assume that n = 3. (The derivation can be easily extended to the case of any positive integer n.) - z 2 G + Z ~ W ~- z G H ~+ zH2- GHz The Cayley-Hamilton theorem (see Problem A-5-9) states that an n x n matrix G satisfies its own characteristic equation. Since n = 3 in the present case, the character- istic equation is - 61 = z3 + a1z2+ a2z + a3 = O and G satisfies the following equation: G3 + al gy2 + a2G 4- as Hence Equation (5-136) simplifies to z2 + H1z + H2) = (z3 + alz2+ a2z C a3)I = jz Consequently, which is Equation (5-135) when n = 3.
- 185. Wotice that since we have Notice also that Now we have Next, we shall show that al = -trG a, = -S tr GH1 a3 = -S tr GHs We shall transform into a diagonal matrix if G involves n Iinearly inde eigenvectors (where = 3 in the present case) or into a matrix in a Jordan c forrn if G involves fewer than rz linearly independent eigenvectors. That is, -1 = D = matrix in diagonal form or = matrix in a Jordan canonical form are nonsingular transformatioh matrices. Since the following derivation applies regardless of whether matrix transformed into a diagonal matrix or into a matrix in a Jordan canonical form, use the notation - ~ G T= fi where represents either a diagonal matrix or a matrix in a Jordan canonical form the case may be. In the followingwe shall first show that trG = trD tr GHI = tr fihl tr GH2 = tr fiG2 i i 1 = n + a 1 ~ ii2= nE9, + a,I al = -trD n * a, = -S ir DHI A A a3 = - 4ir DE2 tr G = tr TBT-' = tr B trGH1 = trG(G + alI) = trG2 + tralG where Then we shall show that Chap. 5 Exarnple Let us write L O O ~ 3 1 where an asterisk denotes "either O or 1." Then a3 = -p1p2p3 Notice that = 3plp2p3 = -3a3 Thus, we have shown that Consider the following oscillator system:
- 186. Noting that we have and State-Space Analysis -sin oh cos wh -. State-Space Analysis Qbtain the continuous-time state-space representation of the system. Then the system and obtain the discrete-time state-space representation. Also o pulse transfer function of the discretized system. ution From the given transfer hnction, we have y + wZy= w2u Define X l = y 1 X 2 = - j w Then we obtain the followingcontinuous-time state-space representation of the syst The discrete-time state-space representation of the systern is obtained as fo Noting that we have -1 cos wT sin wT - - -sin wT cos wT s2 + w2 s2 + w2 and 1 -sin oh cos wh 1- cos wT Hence, the discrete-time state-space representation of the oscillator systern be as follows: cos COT sin W T ] [ ~ ~ ( ~ T ) ]+ [1 - C O T ] ~ ( -sin oT cos wT x2(k7') sin wT xl(kT) Y = 1 o~[x2(ki)] The pulse transfer function of the discretized system can be obta Equation (5-60): F(z) = C(z1 - G)-'H + D Chap. 5 Example Problems and Noting that D is zero, we have s i n w~ 1 - l [1 z - cos wT -- (1 - cos wT)(z + 1) z2 - 22 cos wT + 1 Hence, m= F(r) = (1 - cos O J T ) ( ~+ 2-')z-' U(z1 1 - 22-1 cos WT+ z-' Note that the pulse transfer function obtained in this way is the same as that obtained by taking the z transform of the system that is preceded by a zero-order hold. That is, 1 1- zP1cos wT -- (1 - cos w"(l + z-')z-l 1 - 22-' cosoT + z-' Thus, we get the same expression for the pulse transfer function. The reason for this is that discretization in the state space yields the zero-arder hold equivalent of the continuous-time system. Consider the system shown in Figure 5-8(a). This system involves complex poles. Tt is stable but not asymptotically stable in the sense of Liapunov. Figure 5-8(b) shows a Figure5-8 (a) Continuous-timesystem of Problem A-5-15; (b) discretized version of the system.
- 187. discretized version of the continuous-time system. The discretized systern is also sta but not asymptotically stable. Assuming a unit-step input, show that the discretized system may exhibit hid oscillations when the sampling period T assumes a certain value. aa The unit-step response of the continuous-time systemshownin Figure5-8 1s Hence, y(t) = cosrt [Notice that the average value of the output y(t) is zero, not unity.] The respsnse y versus r is shown in Figure 5-9(a). I (c) Figure 5-9 (a) Unit-step response y(t) of the continuous-time system shown in Figure 5-8(a); (b) plot of y(kT) versus kT of the discretized svstem shown in Figure- M ( b ) when T = 4rr sec; (c) piot of y(1cT) versus kTof the discretized system when T = T sec (Kidden oscillations are shown in the diagram.) Chap. 5 Exarnple Problerns and Soluaions The pulse transfer function of the discretized system shown in Figure 5-8(b) is 7 - - - y& =#- e-" " = (1 - z-')+&] Hence, the unit-step response is obtained as follows: The responsey(kT) becomes oscillatoryif T f nn-sec (n = 1,2,3,...).For example, the response of the discretized systern when T = $71. sec becomes as folIows: Hence, A plot of y(kT) versus kT when T = n- sec is shown in Figure 5-9(b). Clearly, the response is oscillatory. If, however, the sampling period T is rr sec, or T = n-, then The response y(kT) for k = 0,1,2, . ..is constant at unity. A plot of y(kT) versus kT when T = n- is shown in Figure 5-9(c). Notice that if T = n- sec (in fact, if T = nrr sec, where n = 1,2,3,...) the unit-step response sequence stays at unity. Such a response may give us an irnpression that y(t) is constant. The actual response is not unity but oscillates between 1and -1. Thus, the output of the discretized systemwhen T = 7 ~ .sec (or when T = nrr sec, where n = 1,2,3,.. .) exhibits hidden oscillations. Note that such hidden oscillations (hidden instability) occur only at certain particular values of the sampling period T . If the value of T is varied, such hidden oscillations (hidden instability) show up in the output as explicit oscillations. Even though the double-integrator system is dynamically simple, it represents an important class of systems. An example of double-integrator systems is a sateilite attitude control system, which can be described by
- 188. Chap. 5 Exarnple Problems and Soleiitions where J is the moment of inertia, 0 is the attitude angle, u is the control torque, a v is the disturbance torque. Consider the double-integrator system in the absence of disturbance in Define 98 = y. Shen the system equation becomes y = U Obtain a continuous-time state-space representation of the system. Then sbt discrete-time equivalent. Also obtain the pulse transfer function for the discrete- system. The discrete-time equivalent of this system can be @ven by x((k + 1)T) = Gx(kT) + Hu(kT) y(kT) = Cx(kT) Matrices G and H are obtained from Equations (5-73) and (5-74). Woting tbat Mence, the discrete-time state equation and output equation become The pulse transfer function of the discrete-time system is obtained frorn E¶*a (5-60) as follows: Y(z--- = F(z) = C(z1 - 6)-'R + D F J f 7 ) Show that the following quadratic form is positive de£inite: V(X) = 10x1+ 4x2 + X: + 2x1~2- 2x2~3- 4~1x3 ntioah The quadratic form V(x) can be written as follows: Applying Sylvester's criterion, we obtain Since al1 the successive principal minors of the matrix are positive, V(x) is positive definite. Consider the system defined by Suppose that Suppose also that there exists a scalar function V(x, t ) that has continuous first partial derivatives. If V(x,t) satisfies the conditions ,t) = O and V(x, t) r a(llxll) > O for al1 nondecreasing scalar function such that negative for al1 x # and al1 t , or ~ ( x ,t) 5 t , where y is a continuous nondecreasing scaiar function such that y(0) = 0. 3. There exists a continuous nondecreasing scalar function ,Bsuch that P(O) = O and, for al1 t, V(x, t) ..=: ,B(llx/l). 4. cr(1lxll) approaches infinity as llxll increases indefinitely, or then the origin of the system,x = ,isuniformly asymptotically stable in the large. (This is Liapunov's main stability theorem.) Prove this theorem.
- 189. oliution So prove uniform asymptotic stability in the large, we need to prove following: 1. The origin is uniformly stable. 2- Every solution is uniformly bounded. . Every solution converges to the origin when t=, m uniformly in toand llaao1l S where 6 Is fixed but arbitrarily large. That is, given two real nurnbers 6 > O u. > O. there is a real nurnber T(u. 8) such that, ' I Ilxoll 2s 8 irnplies (t;xo,to)[jS ,u9 for al1 t r: to + T(p96) the solution to the given differentíal equation. Since j3 is continuous and P(O) = O, we can take a 8 ( ~ )> O such that p(6) < a any E > O. Figure 5-10 shows the curves aí(llxl/),P(l/xll),and V(x, t). Noting t (t;woyto), t) - V ( ~ O ?to) = v( (T XO,LO),7 )d? < 0, l > to IUif /lrtol/S 8, tobeing arbitrary, we have a(€) > P(8) 2 V(X",to) 2 V( for all t 2 lo. Since a is nondecreasing and positive, this implies that Kence, we have shown that for each real number E > O there is a real nurnber such that //xoj/5 6impl t;XO,to)ll :E for al1t 2 to.Thus, we have proved u stability. Next, we shall prove that 1 t;xo, to)jl--+Owhen t-pm uniforrnly in to l/xo//5 8. Let us take any O < p < 1) 11 and find a v(p) > O such that p(v) < a(p). Figure 5-10 Curves cu(l/xl/),P(llxI/),and V(w, 1). ap. 5 Example Prsblems and us denote by ~ ' ( p ,8) > O the minimum of the continuous nondecreasing function y(llxll) on the compact set v(p) 5 I/x/I5 48). Eet us (t; xo,to)ll > v over the time interval to5 t 5 tl = to + T. Then we have (t,; xo, to), tl) 5 V(xo,to) - (ti - to)el5 P(8) - TE' = O which is a contradiction. Hence, for sometIn the interval to 5 t 5 tb such as an arbitrary tz, we have Therefore , for al1 t k to + T(p, 6) stz, which proves uniform asymptotic stabiiity. Since a(llxll)+ 03 as IIxll--+m, there exists for arbitrariiy large 6 a constant €(S) such that p(8) < a(€).Moreo since ~ ( 6 )does not depend on to, the solution uniformly bounded. thus have proved uniform asymptotic stability i In z plane analysis, an n X n matrix G whose n eigenvalues are Iess than unity in magnitude is called a stable matrix. Consider an n x n Hermitian (or real symmetric) that satisfies the following matrix equation: is a positive definite n x n Herrni tric) matrix. Prove that G is a stable matrix then a rnatrix ation (5-137) is unfque and is positive definite. Prove that matrix Prove also that although the right-hand si e of this last equation is an infinite series the matrix is finite. Finally, prove that if Equation (5-137) is satisfied by positive definite ,then matrix G is a stable rnatrix. Assume that al1 eigenvalues of G are distinct and al1 eigenvectors of G are linearly independent. ohihtion ket us assume that there exist two matrices that satisfy Equation (5-137). Then and By subtracting Equation (5-139) frorn Equation (5-138), we obtain
- 190. hap. 5 Exarnple roblerns and Solutions where -- then there exists an eigenvector xi of matrix G such t Let us define the eigenvalue that as associated with the eigenvector xi to be hi. T Gxi = kixi Hence, from Equation (5-140), we obtain Equatian (5-141) implies that A;' is an eigenvalue of G*. Since IAil < l_we h /AI-'/ > 1. This contradicts the assumption that G is a stable matrix. Hence, ES.mus a zero matrix, or it is necessary that ,the solution to Equation (5-13 tion (5-137) can be given by This proves that is a finite rnatrix. Finally, we shall prove that if Equation (5-137) is satisfied by positive definite then matrix G is a stable matrix. Let us define the eigenvector associated with an eigenvalue Ai of G as xi. Then premultiplying both sides of Equation (5-137) by xT and postmultiplying both sides xi, we obtain Hence, - xi are positive-definite, we hzve jkij2 - 1< o Hence, we have proved that matrix G is a stable rnatrix. It is noted that the proofs and derivations presented here can be extended to the case where matrix G involves multiple eigenvalues and rnultiple eigenvectors. Consider the system x(k + 1) = H(x(k))x(k) Assume that there exist positive constants cl, c2,. . . C, such that either (2) max 2"lhij(x)l < 1, for al1 x j [i:l:, } Show that in either case (x)xis a contraction for al1 x and therefore the equilibrium state of the system is asymptotically stable in the large. utiorn In case 1, define the norm by 1x11 = m y {ciixil) Then
- 191. lhi,(i)l}max {cjlxj]< max {cj]ql] = 1/41 which verifies that H(x)x is a contraction. In case 2, define the norm by n IIxll = L-d cilxil Then State-Space Analysis { i j } ic j / i j< jj~ ~ 1 ~ ~ 1i j=l Cj j= 1 j- 1 = IIxII which shows that H(x)x is a contraction. Wow consider a scalar function V ( x )= I/xII.Clearly, V(x)= nite and BV(x(k))= V(x(k + 1)) - V(x(k)) = IlH(x(k))x(k)ll- Il4k)ll < 0 and Thus, dV(x) is negative definite. Hence, V ( x )= ljxlj is a Liapunov function syctem considered, and by Theorem 5-5 the origin of the cystemis asymptoticall in the large. Prove that if al1 colutions of x(k + 1) = Gx(k) where x is an n vector and G is an n x n constant matrix, tend to zero as k app infinlty, then al1 solutionc of the system x(k + 1) = Gx(k) + Hu(k) where 1E% is an n x r constant matrix, are bounded, provided that the input vect r vector, is bounded. siÉPori Since u(k)is bounded, there exists a positive constant c such that llu(k)ll<c, k = 0 , 1 , 2,... The solution of Equation (5-144) is given by k x(k) = e k x ( 0 )+ 2G k ' ~ u ( j- 1) j= 1 hap. 5 Example Problems and Hence, k Since the origin of the hornogeneous system givenby Equation (5-143) is asymptotically stable, there exist positive constants a and b (O < b < 1) such that O < 1 1 ~ 1 1 ~< abk Then Therefore, e have thus proved that Ilx(k)ll ie bounded. Consider the system defined by the equations Determine the stability of the equilibrium state. iatio~a Define the equilibrium state as Then such an equilibrium state can be determined from the following two simultaneous equations: The equilibrium state is thus (0,10). Now let us consider a new coordinate system with the origin at the equilibrium state. Define Then the system equations become
- 192. State-Space Anaiysis Chap. To determine the stability of the origin of the system in the new coordinate system, le us apply the Liapunov stability equation given by Equation (5-90): 2 0.5 .8,,p,, ;::]lo 0 . 8 - 1; ;::J= -1; a351 be a positive definite matrix having e Solving this fast equation for matrix hap. 5 Problerns Obtain a state-space representation of the system described by the equation y(k + 2) +y(k + 1) + O.l$y(k) = u(k + 1) + 2u(k) Obtain the state equation and output equation for the system shown in Figure 5-11. y applying the Sylvester criterion for positive definiteness, we find that matrix P positive definite. Therefore, the origin (equilibrium state) is not stable. The instability of the equilibrium state can, of course, be determined by transform approach. Let us first eliminate X2 from the state equation. Then we Xr(k + 2) - 2.8X1(k + 1) + 1.6Xi(k) = O The characteristic equation for the system in the z plane is z2 - 2.82 + 1.6 = O or ( z - 2)(z - 0.8) = O Hence, z = 2, z = 0.8 Since pole z = 2 is located outside the unit circle in the z plane, the origin (equilib state) is unstable. O Obtain a state-space representation of the following puise-transfer-function syste the controllable canonical form. Obtain a state-space representation of ehe following pulse-transfer-function syste the observable canonical form. Y ( z ) z - ~+ 4 ~ - ~ Obtain a state-space representation of the following pulse-transfer-functm syste the diagonal canonical form. Y ( z ) 1 + 62-' + S Z - ~ Figure 5-11 Block diagram of a control system. Obtain the state equation and output equation for the system shown in Figure 5-12. Obtain the state-space representation of the system shown in Figure 5-13. Figure 5-14 shows a block diagram of a discrete-time multiple-input-multiple-output system. Obtain state-space equations for the system by considering xl(k),r2(k),and x3(k)as shown in the diagram to be state variables. Then define new state variables such that the state matrix becomec a diagonal matrix. Obtain the state equation and output equation for the system shown in Figure 5-15. Obtain a state-space representation of the discrete-time control system shown in Figure 5-1 6.
- 193. 2 Block diagram of a control system. Block diagram of the discrete-time multiple-input-multiple-output system of Problem B-5-8. Figure 5-15 Block diagram of a control system.
- 194. tate-Space Analysis Chap. state-space representation of the scalar difference equation system y (k + n ) + al(k)y(k + n - 1) + . + a,(k)y ( k ) = bo(k)u(k+ n ) + bl(k)u(k + n - 1) + + b,(k)u(k) wbere k = 0,1,2,. .. ,may be given by y(k) = xl(k) + bo(k - n)u(k) Determine hl(k),hz(k),...,h,(k) in terms of ai(k) and b,(k), where i = 1,2, and j = O,1,...,n , Determine also the initial values of the state variables xz(O),. ..,x,(O) in terms of the input sequence u(O),u(l),...,u(n - 1)and the sequence y (O), y (1), ... ,y (n - 1). " x(zI - 6)-' = 2k-1 Z - Zk where m is the degree of the minimal polynomial of G and the Xk9$are n X n m determined from gj(c) = ~ , ( Z I ) ~ I+ gj(zz)x2+ e + gj(zm)Xm where Obtain the pulse transfer function of the system defined by the equations Rnd the pulse transfer function of the system defined by where = [ 1 O O], D = bo Obtain a state-space representation for the systemdefined by the followingpulse-trans- fer-function matrix: Consider the discrete-time state equation Obtain the state transition matrix Consider the system defined by where matrix G is a stable matrix. Obtain the steady-state values of x(k) and y(k) when u(k)is a constant vector. Consider the system defined by
- 195. State-Space Analysis where G is a stable matrix. Show that for a positive definite (or positive semidefinite) matrix m k-O can be given by Determine a Liapunov bunction V(x) for the following system: Determine the stability of the origin of the following discrete-time system: Determine the stability of the origin of the following discrete-time system: -2 Consider the system defined by the equations xl(k + 1) = xl(k) + 0.2~2(k)+ 0.4 x2(k + 1) = 0.5~1(k)- 0.5 Determine the stability of the equilibrium state. Chap n the first part of this ch state in a finite number of s to reach some arbitrary state.) The concepts of controllability and observability were introduced by R. E. h a n . They play an important role in the op 1control of multivariable systems. fact, the conditions of controllability and ability may govern t of a complete solution to an optimal contr In the second part of this chapter we d le placement design method and state observers. Note that the concept of controllability is the basis for the solutions of the pole placement problem and the concept of observability plays an important role for the design of state observen. e design method based on pole placement coupled with state observers is one of the fundamental design methods available to control engineers. Hf the system is completely state controllable, then
- 196. the desired closed-loop poles i equalion) can be selected and t be designed. The design agpro linear time-invarian locations in the z plane is pele placernent design tech the closed-loop system are state feedback, it becomes necessary to est uch estimation can be done by use of Sta etail in this chapter. on the measurement estimates al1state variabl required for feedback to that the spacing between adjacent disturbances is sufficientlywi sufficient speed. In the final state of the design process accomplished by use of the estimated state variables rather than the actual variables, which are probably not available for direct measurement. If S state variables are measurable, then we may use those available state va use estimated state variables for those not actually measurable. In the last part of this chapter we treat a servo designproblem that usesint control coupled with the pole placement technique and the state observer. Iúotet in the regulator problem we desire to transfer the no disturbance) to the origin. Hn the seno problem, we req command input. Note that the servo system must follow t the same time must solve any regulator problem. Consequ servo system we may begin with the design of a regulator S the regulator system to a servo system. A control system is said to be co control system is controllable if ever can be controlled in a finite time period by state variable is independent of the contr this state variable and therefore the syst The solution to an optimal control sidered is not controllable. systerns. Now we shall derive this condi- tion in the following. er the discrete-time control system defined by where x(kT) = state vector (n-vector) at kth samphg instant u(kT) = control signal at kth sampling instant G = n X n matrix H = n x 1matrix T = sampling period assume that u(kT) is constant for kT 5 t < (k + 1)T. The discrete-time control system given by Equation pletely state controllable or simply state contmliable if t
- 197. Pole Placernent and Observer Design Chap state controllability. j=o we obtain the rank of the controllability matrix is n, then for an arbitrary state x(nT) = there exists a sequence of unbounded control signals u(O), u(T), ...,u((n - 1 controllability, let us assume that rank[H: GHi .GnW1 quently, we have for any i rank[HiGHi - - .iG'-lH] < n complete state controllability. ec. 6-2 Controllability m defined by Equation (6-1) is completely state controllable, then ansfer any initial state to any arbitrary state in at most n sampling however, that this is true if and only if the magnitude of u magnitude of u(kT) isboun it may take more than n sa a system is defined by where x(kT) is an n-vector, u(kT) is an r-vector, G is an n x n matrix, and H is an n x r matrix, then it can be proved that the con lete state controllabil- ity is that the n x nr matrix be of rank n, or that is of rank n and u(kT) is a scalar, then it is possible to scalar equations from which a sequence (k = 0,1,2, ...,n - 1) can be uniquely det at any initial state x(0) is transferred to the desired state in n sam Note also that, if the control signal is not a scalar sequence of u(kT) is not unique. Then there exists r u(kT) to bring the initial state w(0) to a state in not more than . Consider the s where x(kT) = state vector (n-vector) at kth sampling instant u(kT) = control vector (r-vector) at kth sampling instant 6 = n X n matrix H = n x r matrix T = sampling period If the eigenvectors of Gare distinct,then it is possible to find a transformation matrix such that
- 198. Pole Placernent and Observer if the eigenvalues of G are distinct then the eigenvectors of G are ,the converse is not true. (For example, an n x n real symmetri rnatrix is an eigenvector of (i = 1,2,.. . ,a). Let us define . , Substituting Equation (6-5) into Equation (6-4), we obtain Let us define responding state variable cannot be controlled by any of the ui(kT). Hence, t condition for complete state controllability is that, if the eigenvectors of distinct, then the system is cornpletely state controllable if and onlyif no row 0 has al1zero elements. Ht is important to note pply this conditionfor C Q ~ P ~ state controllability we must put the rnatrix n Equal-ion(6-6) into daga form. If the G matrix in Equation (6-4) does not possess distinct eigenvectors, th diagonalization is impossible. In such a case, we may transform G into a J Q ~ ~ cmonicalform. If, for example, G has eigenvalues A,, A,, A,, A,, A,, h6,. . . ,h, and n - 3 distinct eigenvectors, then the Jordan canonical form of G is ec. 6-2 Controllability The 3 x 3 and 2 x 2 submatrices on the main Suppose it is possible to find a transformation matrix f we define a new state vector k by then substituting Equation (6-7) into Equation (6-4) gives he conditionsfor the complete state controllability of the systemof E may then be stated as follows: The system is completely state controllab if (1) no two Jordan blocks in (6-8) are associated wi (2) the elements o an block are not al1z correspond to distinct eigenvalues are not al1 zero. ing conditions for st f Equation (6-8) same eigenvalues. This point is elaborated as follows. Consider the followingsystem where the two Jo the same eigenvalues Al: Although every state variable is affected by u(k), this system is uncontrollable, since the rank of the controllability matsix pplying the preceding criterion for state co trollabilaty, no two of Equation (6-8) should be associated with t e same eigenvalues.
- 199. ec. 6-2 Controllabiiity The following systems are completely state controllable: where w(kT) = state vector (n-vector) at kth sampling instant u(kT) = control signal (scalar) at kth sa y(kT) = output vector (m-vector) at kth sampling instant G = n X n matrix = n X 1matrix = m X n matrix The system defined by Equations (6-9) and (6-10) is said to be completely output controllable, or simply output controllable, if it as possible to construct an uncon- strained control signal u(kT) defined over a finite number of sa hat, starting from any initial desired point (an arbitrary n what follows we derive the condition for Note that, if a system is co pletely output controllable al exists that will transfer any initial ou e in at most n sampling periods. Since
- 200. where x(kT) = state vector (n-vector) at kth sampling instant (kT) = output vector (m-vector) at kth sampling instant G = n X n matrix H = n X r matrix j=o we obtain +Du(nT) ec. 6-2 Controllability be of rank m : at the presence of matrix to establish complete output controllability. what follows we shall briefly state the conditions for complete sta ntrollability of linear time-invariant continuous-time control systems. x = state vector (n-vector) o = control vector (r-vector) y = output vector (m-vector) = m X n matrix D = m X r matrix CompleteState Controllability. sufficientconditionforcom- plete state controllability for this syste in a way similar to what was used in the case of the iscrete-time system. Here, we shall present only lete state controllability is that the n x nr be of rank n or that it contains n linearly inde ent column vectors. (This matrix is commonly called the controllability matrix for the continuous-time system.) ?'he condition for complete state controllability can also be stated in terms of transfer functions or transfer matrices. A necessary and su complete state controllability is that no cancellation occur in or transfer matrix. If cancellation occurs, the system cannot be controlled in the direction of the canceled mode. Output Controllability. s in the case of the discrete-time control system, complete state controllability is neither neces y no* sufficientfor controlling the output of a linear time-in~ariantC O ~ ~ ~ ~ U Q U S -control system. lt can be proved that the condition for complete output controllability is that the rank of the m x (n + l)r rnatrix
- 201. Pole Placement and Observer Design Cha x((k + 1)T) = Gx(kT) where x(kT) = state vector (n-vector) at kth sampling instant (kT) = output vector (m-vector) at kth sampling instant x(kT) = Gk~ ( 0 ) 6 = n X n matrix = m x n matrix for the determination of xl(0), x2(0),...,x,(O). For a completely observable systern, given we must be able to determine xl(0),x2(O),.. .,x,(O). Noting that y(kT) is an m observers. vector, the preceding n simultaneous equations yield nm equations, al1 involving x1(0),x2(0),...,x,(O). To obtain a unique set of solutionsxl(0),x2(0),...,xn(0)from is described by the equations x((k + 1)T ) = Gx(kT) + Hu(kT) then and y(kT) is investigating a necessary and sufficient con
- 202. Pole Placement and Observer efined by Equations (6-15) and (6-16), repeated here: Then Equations (6-18) an ere Al, A,, . ..,An are n distinct eigenvalues of G. Th ervable if and only if none of the colu al1 zero elernents. This is because, if the ith column of elernents, then the state variable &(O) will not appear therefore cannot be determined from ob to i(0) by the nonsingular matrix the rnatrix G involves multiple eigenvalues and ca a diagonal rnatrix, then by using a suitabletransformation G into the Jordan canonical form: is in the Jordan canonic Then Equations (6-18) and (6-19) can be written as follows: Hence, The system is completely observable if and only if (1) no two associated with the same eigenvalue, (2) none of the co row of each Jordan block consists of allzero t correspond to distinct eigenvalues consist o clarify condition 2, in Exam le 6-3, which follows, we have enclosed in e first row of eachJor The following systems are completely observable. The following systems are not completely observable: at no pole-zero cancellation be observed in the output. Show that the following system is not completely observable:
- 203. Note that the control signal ~k(kT)does not affect the complete observability of the system. To examine complete observability, we may sirnply set u(kT) = O. For t system, we have 14 -6 61 Notice that Hence, the rank of the matrix [C* i 6;" system is not completely observable. In fact,inthis systemapole-zero cancellation occursin the pulsetransfer of the system. The pulse transfer function between Xl(r) and U(z)is and the ~ u l s etransfer function between Y ( z )and Xl(z)is Clearly, the (z $. 1)factors in the numerator and denominator cancel each other. l n means that there are nonzero initial states x(0) that cannot be determi&d from t measurement of y (kT). ents. The pulse transfer function has no cancellation if and only if th mpletely state controllable and completely observable. (See Probl A-6-4.) This means that a canceled transfer function does not carry along al1 information characterizing the dynamic system.- n what follows, we S al1 examine the relations nd observability. Consider the system SI defined by equations x((k + 1)T) = @x(kT) + Ha(kT) y(kT) = Cx(k'l") where x(kT) = state vector (n-vector) at kt er(kT) = control vector (r-vector) at kth sampling inskant (kT) = output vector (m-vector) at kth sampling instant G = n X n matrix E% = n X r matrix = m X n matrix ual counterpart, which we cal1 syste k((k + 1)T) = G*X(kT) + (6-22) (kT) = H*%(kT) (6-23) where X(kT) = state vector (n-vector) at kt G(kT) = control vector (m-vector) at kth sampling instant "(kT) = output vector (r-vector) at kth sampling instant G* = conjugate transpose of G H* = conjugate transpose of H * = conjugate transpose of shall now examine an analogy between controllability and observability. This analogy is referred to as the principle of duality, due to Kalman. The principle of duality states that system SIdefined by Equations (6-20) and (6-21) is completely state controllable (observable) if and only if system S, defined ) and (6-23) is completely observable (state controllable). To let us write down the necessary and sufficient conditions for complete state controllability and complete observability for systems SI and S2, respectively. FOR SYSTEM S,: necessary and sufficient condition for complete state controlIability is that rank[HiGHi....G"-lH] = n A necessary and suffi nt condition for complete observability is that FOR SYSTEM S,:
- 204. (6-25) the continuous-time to the discrete-time case. Consider the following continuous-time control system: (6-27) This system is completely state controllable and completely observable, since the rank of the controllability matrix defined by the equations: X = Ax is 2 and the rank of the observability matrix rn = state vector (n-vector) is also 2. Notice that the eigenvalues of the state matrix are y = output vector (m-vector) AI = j, h2 = -j A = n X n matrix The discrete-time control system obtained by discretizing the continuous-time control = m X n matrix system defined by Equations (6-26) and (6-27) may be given as follows: for complete observability is that the rank of the n x nm matrix (6-29) be n. (This n X nrn matrix is commonly called the observability matrix fo where T is the sampling period. continuous-time system.) Let us show that the discretized system given by Equations (6-28) and (6-29) is completely state controllable and completely observable if and only if 2nrr Im (Al - A2) = 1 + 1=k - T T f nn, n = 1,2,3,. .. For the discrete-time control systemobtained by discretizingthe continuous-time pletely observable in the absence of sampling remains completely state co control system, we have the following controllability matrix: and completely observable after the introduction of sampling if and only if, 1- cos T cos T + 1 - 2 cos2T eigenvalue of the characteristic equation for the continuous-time control §y§ sinT -sinT+2cosTsinT relationship the rank of the observability matrix Notice ehat the rank of [H i GH]is 2 if and only if T # n?l-(where n = 1,2,3,. . .). Also, Re hi= Re A,
- 205. ole Placement and Observer Design 1 cos T O sin T is 2 if and only if T # ~ 1 7 ~ .(where n = 1,2,3,. ..). From theforegoinganalysis,weconclude that the discretized systemis co state controllable and completely observable if and only if T # nrr, wher & 3 , . * - - Note that it is alwayspossible to avoid the loss of controllability and observa by choosing a sampling period T not equai to nrr. %nthis section we shall first reviewtechniques for transforming state-space e into canonical forms. Then we shall review the invariance property of conditions for the controllability matrix and observability matrix. I crete-time state equation and output equation e shall review techniques for tr forming the state-space equations define quations (6-30) and (6-31) int following three canonical for Controllable canonical form Observable canonical form Diagonal or Sordan canonical form (Note that the diagonal canonical form is a special case of the Jor form.) It is assumed that the system defined by Equations (6-30) completely state controllable and completely observable. The svstem defined by Eauations (6-30) , 1) can be transformed into a controiable canonicalform by meam transtormation matrix whei and seful Transformations in State-Space Analysis and Desi he elements aishown in matrix are coefficients of the c - G/ = zn+ alzn-' + + a,-~z+ a, = O (For details of the derivations of the preceding two equations, see (6-37) y(k) = [b, - a, b, i bnWl- a,-, bo (6-3 8) wbere the bk7sare those coefficients appearing in t e numerator of the following pulse transfer function:
- 206. Pole Placement and Observer Note that D = = bo.T e system given by Equations (6-37) an controllable canonical form. defined by Equations (6-30) an (6-31) can be transformed into an observable canonical form by means of th transformation matrix is given by Equation (6-34). It can be shown t and where the bk'S are those coefficients appearing in the numerator of t Equations (6-30) and (6-31) become as follows: X(k + 1) = &k(k) + I%u(k) Useful Transformationc in tate-Cpace Analysis and efined by Equations (4-41) an (6-42) is in an observable canonical f the eigenvalues pi onding eigenvectors gl, Then Pn Thus, if we define then Equations (42-30) and (6-31) can be given by tht (6-44) hus, Equations (6-43) and (6-44) can be written in the form where the ai 9sand the Pi'S are constants such that aiPi is the residue at the z = pi, that is, such that aipiwill appear in the numerator of the term l/(z - p,) when the pulse transfer function is expanded into partial fractions as follows:
- 207. Pole Placement and In many cases we choose al= a2 = - .= a, = 1. [Note that the necessary a sufficient condition for the system to be completely state controllable is that ii# (E = 1,2,. .. ,n) and th t the condition for it to be completely observa piJ;O(i = 1,2,..., n). esp, of matrix G, then we choose the trans 'S are eigenvectors correspond to eigenvalues) or ge nvectors (which co d to multiple lues). (For details generalized eigenvectors, see = rnatrix in Jordan canonical form Now if we define then Equations (6-30) and (6-31) can be given as follows: es pm+l,pm+2,...,p, t are al1distinct and different frornpl and, in addition, if th (which implies that the rninirnal polynornial is identical to the characteristic po nomial), then the state-space equations in the Jordan canonical form are given follows: seful Transformations i n nalysis and Design n many cases we choose a, = = - - = a, = 1. [Note that the necessary and sufficient condition for the (i = m , m + 1,. ..,n) and m + l , m + 2,...,n).] Note that if the rank minimal polynomial is s - will have a different Jordan c (an arbitrary n x n nonsingular matrix) be a si and write Then -l H] Similarly, for the observability matrix define *1 be an arbitrary n x n onsingular matrix and write
- 208. Then rank N = rank r"J gain matrix. x(k + 1) = Gx(k) + Hu(k) where x(k) = state vector (n-vector) at kth samgling instant u(k) = control signal (scalar) at kth sampling instant @ = n x n matrix H = n X 1matrix Figure 6-1 (a) Open-loop control system; (b) closed-loop control system with assume that the magnitude of the control signal u(k) isunbounded. If t signal u(k) is chosen as where B is the state feedback gain matrix (a 1 x n matrix), then the system becornes a closed-loop control system as shown in Figure 6-P(b), an becomes Note thatwe choose matrix K such that the eigenvalues of G - HK are the desired closed-loop poles, ,u1,luz, . .,p,.
- 209. lacement and Observer sthat there are q linearly independentcolumn vectors in the controlla t us define such q linearly independent colu Also, let us choose n - q a . Vq+l i Vq+2 : ' :vn] the transformation matrix, let us - 1 ~= Then we have or here q linearly independent colum ca milton theorem to express matrices of uppose the system of Equation (6-50) is not completely state control1 e rank of the controllability matrix is less than n, or q vectors. That is, ence, Equation (6-53) may be written as follows: i Gv,+~.a i Gv,] 1; ; ..e o To simplify the notation, let us define esígn via P d e Placement [Q %]= G 2= (n - q ) X q -m matrix o o 1 Then Equation (6-53) can be written as follows: e . f GV,+~:- :Gv,] = [ Thus, Next, referring to Equation (6-54), we have (6-56) e written in terms of q uation (6-56) may be written as follows: hus, where
- 210. Pole Placement and Observer Design Chap. esign vía Pole Placernent Now consider the closed-loop system equation given by Equation (6-51). characteristic equation is Let us define and partition the matrix kto give where Kll is a 1 x q matrix and q) rnatrix. Now, 1 x n matrix can be written as follows: Then the characteristicequa which is of rank n, and where Then, referring to Equations (6-35) and (6-36), and Substituting Equations (6-55), (6-57), and (6-58) into this last equation, we obt -1H= gE =. Next we define Equation (6-59) shows that m values (closed-loop pole lo e characteristic equation lz - &: + HKI becomes as follows: locations. The desired eigenvalues of G - HM are kl, h,...,p,; any complex ues are to occur as conjugate pairs. Noting that the c aracteristic ecljuatisn o original system given by Equation (6-50) is ( = zn + alzn-l + azzn-2+ - . + a,-~z + a, = O we define a transformation matrix where = [HiGH: ... c - 1 ~ 1
- 211. Pole Placement and bserver Design Chap. The characteristic equation with the eigenvalues is given by ( Z - P ~ ) ( z ~ 2 ) ' ' ( z - ~ n ) Equating the coefficients of equal powers of z of Equations (6-63) and (6-64) obtain al = al + 6, a2 = a2+ 62 a, = a, + S, ence, frorn Equation (6-62) we = [a, S,-1 m SI] . . = [a, - a, i anl1- a,-l : :al - where the ai's and the ai 's are known coefficients and ave determined the required feedback gain matri cients and a known matrix of the system. This proves the sufficient c is, if the system defined by Equation (6-50) is cornpletely state con it is always possible to de mine the required state feedback gain arbitrary pole placement . nce, we have proved that a necessary condition for arbitrary pole placernent isthat the systembe completel labre. . The expression given by Equation (6-65) Bs not only one used for the deterrnination of the state feedback gain rnatrix K. T other expressions available. In the following, we shall present m e such ex commonly called Ackermann's formula. Gonsider the system defined by Equation (6-50). is cornpletely state controllable. y using the state f wish to place closed-loop poles at z = pl,z = ,u2,...,z = p,. That i characteristic equation to be 1 = ( Z - pl)(z - / L 2 ) " ' ( Z - p,) = zn + alzn-l + a 2 ~ n - 24 .- + anmlz + a, = O et us define G = G - H K Since the Cayley-Harnilton theorem states that 6 satisfies its own charad equation, we have 6"+ al6"-1+ a2Gn-2 + .. Ytie shall rmtilize this lase equaticdn to derive Acltermanm's formula. ec. 6-5 Design via Pole Placement Consider now the following identities: - G = G - H K G2= (G - H K ) ~= G~- G ding equations in order by a,, a,-l,. ..,a0 (where a. = 1), ng the results, we sbtain which can be written as follows: Notice that #G)= ence, Equation (6-66) may be modified to read ince the system is completely state controllable, t e controllability matrix is of rank n and its inverse exists. hen Equation (6-67) can be modified to the form Premultiplying both sides of this last equation by [O O O 11,we obtain
- 213. Pole Placement and Observer where Note that -' 1 = z2 + z + 0.16 0.16 z + l Hence, al = 1, az = 0.16 Determine a suitable state feedback gain matrix R such that the system will closed-loop poles at z = 0.5 +j0.5, z = 0.5 'j0.5 Let us first examine the rank of the controllabílitv matrix. The rank of is 2. Thus, the system is completely state controllable, and therefore arbitrary placement is possible. The characteristic equation for the desired system is Hence, al = -1, az = 0.5 We shall demonstrate four different ways to determine matrix K. Method l. From Equation (6-69), the state feedback gainmatrix K is given asfollo R = [a2- a2I al - a,] Notice that the original systemis alread n a controllable canonicalform,and theref the transformation matrix T becomes Hence, ec. 6-5 Design via Pole Placement Thus, ethod 3. From Equation (6-72), the desired state feedback gain matrix mined as follows: = [l 1][& t 2 1 - l where = (G - piP)-lEI7 i = 1,2 = [G - (0.5 +j0.5) r - 1 1 Similarly, for p2 = 0.5 - j05, we have ijz = [G - (0.5 - j0.5) Consequently, we have Hence, the desired state feedback gain matrix M is determined to be
- 214. esign via Pole Placernent
- 215. Pole Placement and esign via Pole Placement 217 If we use the state feedback u(k) = -> , . , becomes Referring to Equation (6-761, we have = 6 - &[--a, -a,-, - -a,] Consider the system given by Determine the state feedback gain matrix K such that when the control signal is given by the closed-loop system (regulator system) exhibits the deadbeat response to an initial state x(0). Assurne that the control signal u(k) is unbounded. Weferring to Equation (6-76), for the deadbeat response we have The system given by Equation (6-77) is already in the controllable canonical form. Therefore, in Equation (6-78), T = .The characteristic equation for the system given by Equation (6-77) is Thus, Consequently, Equation (6-78) becomes This gives the desired state feedback gain matrix. Let us verify that the response of this system to an arbitrary initial state x(0) is indeed the deadbeat response. Since the closed-loop state equation becornes to discrete-time control systems. here is no such thing as deadbeat r continuous-time control systems. deadbeat control, any nonzero error be driven to zero in at most n sam ng periods if the magnitude of the sca if the initial state is given by- - - u ( k ) is unbounded. The settling time depends on theWsamplingperiod, sin response settles down in at most n sampling periods. If the sampling perh chosen very small, the settling time will also be very small, which implies th control signal must have an extremely large magnitude. Othenvise, it will possible to bring the error response to zero in a short time period. In deadbeat control, the sampling period is the only design parameter. T if the deadbeat response is desired, the designer rnust choose the sampling pe carefully so that m extremely iarge control magnitude is not required k1-m operatmn of tke system. Note that Et Bs not physicallgr possible to iescrezse where a and b are arbitrary constants, then we have [ = [ ul[::::;] = [u u][:] = [a]
- 216. Pole Placement and Observer Design Thus, the state x(k) for k = 2,3,4,. .. becomes zero and the response deadbeat. a Vectw. Thus far, we en the control signal is a sc f the control signal is a vector quantity (r-vector), the response can be S because we have more freedom to choose control signals ul(k), u2(k),. speed up the response. For example, in the case of the nth-order system wi control, the deadbeat response can be achieved in at most n sampling p the case of the vector control n(k), the deadbeat response can be achieve than n sampling periods. n the case of the vector control, however, the etermination of the feedback galn matrix R becornes more cornplex. ut. Thusfar, we have consi n the regulator systern, the reference input is fiied for a long peri externa1 disturbances create nonzero states. The characteristic equation system determines the spe hich the nonzero states what follows, we shall cons e case where the system Considen: the system in Figure 6-2. The plant is described following state and output equations: x(k + 1)= Gx(k) + Hu(k) The control signal u(k) is given by u(k) = I;=,r(k) - Mx(k) By eliminating u(k) from the state equation, we have x(k + 1)= (G - WK)x(k) + HKor(k) The characteristic equation for the system is As stated earlier, if the system is completely state controlable, t gain rnatrix K can be determined to yield the desired closed-loop poles. u Figure 6-2 State feedback control system. esign via Pole Piacement It is important to point out state feedback can change the characteristic rration for the system, but in d changed. Therefore, it is necessary state is unity, or y(m) = 1. To clarify the details, Consider the system defined by x(k + 1) = Gx(k) + Hu(k) where Design a control system such that the desired closed-loop poles of the characteristic equation are at Thus, the desired characteristic polynomial is given by - G + HRI = ( z - 0.5 - j0.5)(z - 0.5 +j0.5) The state feedback gain rnatrix K can be determined as K = [0.34 -21 (See Example 6-6 for the computation for determining matrix K.) Using this K matrix, the state equation becomes where The gain constant Ko can be deterrnined in state space or can be deterrnined in the z plane usingthe pulse transfer function. In this example, we shalluse the latter approach. The pulse transfer function Y(z)IR(z) for this system is given by where
- 217. ence, Thus, lacement and Observer Design To determine gain constant Ko,we use the condition that the steady-state sutprat for the unit-step input is unity, or lirn y (lc) = lim (1- z-') Y(z) k-m 2-1 2 - 1 Ko , z= lirn--- 2-1 z z2 - z + 0.5 z - 1 Figure 6-3 (a) Block diagram of the control system designed in Exarnple 6-8; (b) simplified block diagram. State Observers MATLAB Program 6-1 num = [O O 051; = filter(num,den,r); axis(vj; gricl title('Unit-Step Response') xlaibel('kl) ylabel('y(k)'l Unit-StepResponse Io6l..a61...........,. .......... ..; .........:...........................i.............:.............: ........... o 0 .........................1.2 ............:.............:............;.............:.............:.............: : o : Unit-step response of the system shown in Figure 6-3(b).
- 218. -6 Regulator system with a state observer. c. 6-6 State Observers rnator, is a subsystem the observed s e (estimate) state variables, we ~nustbe able to obiain k - n + 1)and ~ ( k ) ,~ ( k- l), ...,~ ( k- G-l x(k + 1) = ~ ( k )+ 6-lHu(k) w(k) = G-' x(k + 1) - 6-'Hu(k) y shifting k by 1, we get variables plus some (but not all) of the measlrrable state variables is refe reduced-order state observation. x(k - 1) = G-' x(k) - G-' Hu(k - 1) uting Equation (6-81) into Equation (6-82), we obtain x(k - 1) = G-'[G-'x(k + 1) - G-'Nu(k)] - G-'Hu(k - 1) we obtain the state and output equations as follows: -C 2 x ( k + 1) - G-2Ha(k) - 6-'Hu(k - 1) x(k + 1)= Gx(k) + Hn(k) ~ ( k- 2) = G - 2 ~ ( k )- C2Hu(k - 1) - = 6-3x(k + 1) - g ~ - ~Hn(k) - GM2Hn(k - 1) - 6-'H x(k) = state vector (n-vector) u(k) = control vector (r-vector) - 6-" Mu(k) - 6-"" y(k) = output vector (m-vector) - . . . - 6-"Wu(k - n + 1) G = n X n nonsingular matrix H = n X r matrix g Equation (6-81) into Equation (6-&O), we obtain e = m X n rnatrix ~ ( k )= ~ x ( k )= CG-'X(~+ 1) - CG-'Wu(k)
- 219. Pole Placement and Observer Design imilarly, y combining the preceding n equatio Pele Placement and Observer Design ich is also equivalent to ufficient condition for k + 1) can be obtained by prernultiplying both sides of Equation (6-83) by the verse of the rnatrix given in Equation (6-84), as follows: (6-86) surement noises, however, Notice that the right-hand side of Equation (6-83) is entirely kn x(k + 1) can be determined if and only if Since rnatrix @ is nonsingular, multiplication of each row X(k + 1) = Gk(k) + Hn(k) (6-87) Equation (6-84) by 6"does not change the rank condition. is equivalent to y(k) = CX(k) (6-88) al system. If the initial conditions for the actual system defined by Equa- 9) and (6-80) and the dynamic rnodeli defined by Equations (6-87) and san m x n matrix (m > l ) ,then the inverse ofthe matrix of Equation (6-84) cannot Notice that the right-hand side of Equation (6-83) is entirely knc - x(k + 1) can be determined if and only if rank Since rnatrix @ is nonsingular, multiplication of each row of the left Equation (6-84) by 6"does not change the rank condition. Hence, Eqi is equivalent to
- 220. e(k) = x(k) - X(k) then by subtracting Equation (6-87) from Equation (6-79), x(k + 1) - %(k+ 1)= G[x(k) - X(k)] or e(k -t 1) = Ge(k) ' If matrix G is a stable matri where matrix K, serves as a weighting matrix. (This means that the dyn presence of discrepancies between the G and H matri of the actualsystem, the addition of the difference be the estimated output will help reduce the differences consists of the differencebetween the measured output and the estim is called a full-order state observer. Figure 6-7 State feedback control system. ared with the actual state ~ ( k ) .Since the owever, it is possibIe to compare f(k) = Consider the state feedback control s equations are x(k + 1) = Gx(k) + Hn(k) (6-89) = cx(k) (6-90) where x(k) = state vector (n-vector) u(k) = control vector (r-vector) (k) = output vector (m-vector) G = n x n nonsingular matrix H = n X r matrix = m X n matrix K = state feedback gain matrix (n x r ssume that the system is completely state controllable and completely observ- but x(k) is not available for direct measurement. Figure observer incorporated into the system of Figure 6-7. The observe to form the control vector u(k),or From Figure 6-23,we have her re Reis the observer feedback gain rnatrix (an n X m matrix). This last equation can be rnodified to read
- 221. and those of G* - is n. [This is the condition for complete obs f the system defined by Consider the system x(k + 1) = Gx(k)+ Hu(k) X(k + 1) = GT(k) + Hu(k) Design a full-order state observer, assuming that the system configuration is identical to that shown in Figure 6-8. The desired eigenvalues of the observer matrix are z = 0.5 +j0.5, z = 0.5 - j0.5 Now let us define the differencebetween x(k) and k(k)as the error e(k): and so the desired characteristic equation is e(k) = x(k) - T(k) (z - 0.5 - j0.5)(z - 0.5 +j0.5) = z2 - z + 0.5 = O Since the configuration of the state observer is specifiedas shown in Figure 6-8, Then Equation (6-34) becomes matrix. Lhe rank of "ihe errar vector wiU converge to zers for
- 222. Sec. 6-6 State z 0.16 + k, = O which reduces to Then the characteristic equation becomes z2 -t (1+ k2)z + kl + 0.16 = O Since the desired characteristic equation is z2 - z + 0.5 = O by comparing Equation (6-97) with this last equation, we obtain kl = 0.34, k2 = -2 Note that the dual relationship exists between the systern state equation cons Example 6-6 and that of the present system. The state feedbackgain matrix K in Example 6-6 was K = L0.34 -21. The observer feedback gain matrix K, o here is related to rnatrix K by the relationship K, = K*. have thus far discussedfull-order pr tion observers. They are prediction observers because the estimate sampling period ahead of the measurement y(k), rning the matrix K, to exist and + KeCI = O to have prescribed eigenvaliues. In wh a more general approach to determine the observe matrix Ke. Consider the system defined by x(k + 1) = Gx(k) + Hafl(k) y ( k ) = Cx(k) where x(k) = state vector (n-vector) u(k) = control vector (r-vector) y(k) = output signal (scalar) G = n Y, n nonsingular matrix = 1 X n matrix The system is assumed &obe completely state controllable an able. Thus, the inverse of "1 aliso assume that the control law to be used is where Z(k) is the observed state and K is an r system configuration is the same as that shown The state observer dynamics are given by Z(k + 1)= GZ(k) + Hu(k) + K,ty(k) - j(k)] (6-100) First, define where and where a,, a2,...,anwlare coefficients in the characteristic equation of the original state equation given by Equation (6-(as), Next, define (6-104) where $(k) is an n-vector. y use of Equation (6-104), Equations (6-98) and (6-99) can be modified to read
- 223. Pole Placemesit and Observer Notice that where
- 224. from which we get iacement and Observer Design 8, = a, - a, Then, frorn Equation (6-113), we Hence, system. Figure 6-9 Alternative representation of the observed-state feedback control sgi State Observers Oncewe selectthe desiredeigenvalues (or desired characteristicequation), the observer can be designed in a way simil placement problem. The desired chara onds at least four or five in some applications, deadbeat response rnay be desired. we wish to have deadbeat response, the desired characteristic equation uation (6-114) with Equation (6-118), we require t he expressiongiven by Equation (6-11 r the determination of the observer fe matrix K,. In what follows, we shall deriv letely observable s ned by Equations (6-98) and systern the output y( scalar. Referring to Equation (6-112), tbe characteristic equation for th ere ]E$, is an n x 1 matrix. Define Then Equation (6-120) becomes In the observer design we etermine matrix k, so that this last charac- teristic equation is identical to the desired characteristic equation for the error vector, which is zn + alzn-l -t- + anm1z+ a, =O (6-121) That is,
- 225. x(k + 1) = Gx(k) + Hu(k) wlth state feedback the desired matrix K was obtained as given by Equation (6-68), re K = [o o .. o I][H '.GWi a i cn-lH]-l +(G) ere, iri the observer design proble with state feedback u ( k ) = -&: w(k) the desired matrix &: can therefore be obtained in a fo (6-122), as follows: g; = [O O ... o i][@*; :.. :.( & y 1 e* e conjugate transpose of both sides of this Zast equation, we have we obtain Notice that, since & = Consequently, use of Equation (6-125), tke desired observer fee uation (6-124), can be rewritten as follows: where +(G) is the desired characteristic polynomial of tke error dynamics. The expression for Ke given by Equation (6-126) is commonly called Ackermann's formula for the determination of the observer fee back gain matrix Re. rediction observer is given by Equation (6-92): The obsewed-state feedback is given by Ef this last equation is substituted into the observer equation, we obtain
- 226. i = 1,2,3,... ,n zn + alzn-l + + a,-l z + a, = O (z - pl)(z - ,u2)'" ' ( Z - p,) = zn + alzn-l + * .. + an-lz + a, Consider the double integrator system given by the equations given by Equation (6-126): where +(@S)= Gn+ alGn-l + + a,-] G + a, for this system. It is desired that the error vector exhibit deadbeat response. Use the Hf the desired eigenvalues ,u1,,u2,...,pnof matrix G - Re&' are First we check the observability condition. Notice that the rank of the observer feedback gain matrix Ke may be given by the equ is 2. Hence, the system is completely observable. Next we examine the characteristic equation for the system: Comparing this characteristic equation with Note that the -7 are the ergenvecto In the speeial case wkaere we d z2 + a l z + a2 = O onse, sothat p1 = J U ~= = JL~,= O, Equation (G- following equation will give the matrix Kefor the given by i = 1,2,3,... ,n r details of the derivations of Equations (6-129) an the system is low, assume an observer feedback gain ents. Then the elements of respectively. The a,'s are the coefficients of the desire e coefficients of like powers of z s f of the desired characteristic polynomial, w (z - p1)(z - p2)- " ' ( Z - p,) = zn + alzn-l + * .. + an-lz + a, where the pi 'S are the desired eigenvalues of G - Consider the double integrator system given by the equations x(k + 1) = Gx(k) + Hu(k) where and Tis the sampling period. (See Problem 8-5-16 fo where time state-space equations for the double integrator S server configuration is the sarne as that shown in Figure 6-8, design a state observer for this system. It is desired that the error vector exhibit deadbeat response. Use the four different methods listed in the foregoing First we check the observability condition. Notice that the rank of C is 2. Hence, the system is completely observable. Next we examine the characteristic equation for the system: Comparing this characteristic equation with