Z transfrm ppt

THE z-TRANSFORM
SWATI MISHRA
1

CONTENTS
• z-transform
• Region Of Convergence
• Properties Of Region Of Convergence
• z-transform Of Common Sequence
• Properties And Theorems
• Application
• Inverse z- Transform
• z-transform Implementation Using Matlab
2

Why z-Transform ?
• The z transform is a mathematical tool commonly used for the analysis and
synthesis of discrete-time control systems.
• The z transform in discrete-time systems play a similar role as the Laplace
transform in continuous-time systems
3

z-Transform
• The z-transform is the most general concept for the transformation of discrete-time series.
• The Laplace transform is the more general concept for the transformation of continuous
time processes.
4
TRANSFORM

The Transforms
5
The Laplace transform of a function f(t):
∫
∞
−
=
0
)()( dtetfsF st
The one-sided z-transform of a function x(n):
∑
∞
=
−
=
0
)()(
n
n
znxzX
The two-sided z-transform of a function x(n):
∑
∞
−∞=
−
=
n
n
znxzX )()(

• z-transform comes about same way as CTFT (Continuous Time Fourier Transform) in
which the variable is
Laplace Transform
►CTFT is generalized to Laplace Transform by going from
j σ + j
C.T.F.T L.T
Where , s = σ + j ; σ = real part
6
ω
ω generalized to
ω ω
ω

Motivation For C.T.F.T To L.T
j σ + j
CTFT L.T
Where , s = σ + j ;
σ = real part
This enlarges the capability of the transform to handle much wider class of
signal and system .This was the motivation for going to Laplace transform
from Continuous time fourier transform.
7
ω ω
ω

• In the similar manner z transform is generalized from D.T.F.T ( Discrete Time
Fourier Transform ) .
…………………(1)
• D.T.F.T of the sequence x(n) is given by eqation no . (1)
8
∑
∞
−∞=
−
n
nj
enx ω
)()(nx
D.T.F.T
∑
∞
−∞=
−
=
n
nj
enxX ω
ω )()(

• Here we notice that variable in this D.T.F.T obtained in eq. no. 1. is
9
ωj
e−
ωj
er.
ωj
e generalized to
Magnitude =1 Here we
introduce ‘r’

• Where r may be 1
If r = 1 : D.T.F.T
If r not equal to 1 : New transform : z-transform
• In Laplace transform we worked on the j realization to .We worked
on rectangular coordinates.
• Whereas in z- transform we work in polar coordinates. We specify complex
by its distance from the origin which is ‘r’ and angle it makes with the +ve
real axis i.e ‘ ‘.
10
ω σ + jω
ωj
e
ω
Z = ejω
ω

• In eq.1 if we give a new name say ‘z’.
• If we wish to generalize D.T.F.T into a new transform then is replaced by z
• Then
• The z- transform reduces to the Fourier transform when the magnitude of the
transform variable z is unity.
11
ωj
er.
ωj
e
∑
∞
−∞=
−
=
n
nj
enxX ω
ω )()(
∑
∞
−∞=
−
=
n
n
znxzX )()(

POLES AND ZEROS
12
• When X(z) is a rational function, i.e., a ration of polynomials in z, then:
 The roots of the numerator polynomial are referred to as the zeros of X(z), and
 The roots of the denominator polynomial are referred to as the poles of X(z).
• Note that no poles of X(z) can occur within the region of convergence since the z-
transform does not converge at a pole.
• Furthermore, the region of convergence is bounded by poles.

REGION OF CONVERGENCE
13
• The z-transform of x(n) can be viewed as the Fourier transform of x(n) multiplied by
an exponential sequence r-n
, and the z-transform may converge even when the
Fourier transform does not.
• By redefining convergence, it is possible that the Fourier transform may converge
when the z-transform does not.
• For the Fourier transform to converge, the sequence must have finite energy, or:
∞<∑
∞
−∞=
−
n
n
rnx )(

CONVERGENCE, CONTINUED
14
∑
∞
−∞=
−
=
n
n
znxzX )()(
• The power series for the z-transform is called a Laurent series:
• The Laurent series, and therefore the z-transform, represents an analytic function at
every point inside the region of convergence, and therefore the z-transform and all
its derivatives must be continuous functions of z inside the region of convergence.
• In general, the Laurent series will converge in an annular region of the z-plane.

EXAMPLE
15
)()( nuanx n
=
The z-transform is given by:
∑∑
∞
=
−
∞
−∞=
−
==
0
1
)()()(
n
n
n
nn
azznuazX
Which converges to:
azfor
az
z
az
zX >
−
=
−
= −1
1
1
)(
Clearly, X(z) has a zero at z = 0 and a pole at z = a.
×
a
Region of convergence

16
• Stable System : A system is said to be stable if
∞<∑
∞
−∞=k
kh )(
Which means that a bounded input will not yield an unbounded output.
•Causal System : A causal system is one in which changes in output do not
precede changes in input. In other words,
[ ] [ ] .for)()(then
for)()(If
021
021
nnnxTnxT
nnnxnx
<=
≤=
Linear, shift-invariant systems are causal if h(n) = 0 for n < 0.

PROPERTIES OF ROC
17
• The region of convergence (ROC) for a given DT transfer function is a disk or annulus which
contains no poles. In general, the ROC is not unique, and the particular ROC in any given
case depends on whether the system is causal or anti-causal.
• If the ROC includes the unit circle, then the system is bounded-input, bounded-output
(BIBO) stable.
• If the ROC extends outward from the pole with the largest (but not infinite) magnitude,
then the system has a right-sided impulse response.

• If the ROC extends outward from the pole with the largest magnitude and there is
no pole at infinity, then the system is causal.
• If the ROC extends inward from the pole with the smallest (nonzero) magnitude,
then the system is anti-causal.
18

• If you need stability then the ROC must contain the unit circle.
• If you need a causal system then the ROC must contain infinity and
the system function will be a right-sided sequence.
19

• If you need an anti-causal system then the ROC must contain the origin and the system
function will be a left-sided sequence.
20

• If you need both, stability and causality, all the poles of the system function
must be inside the unit circle. The unique x[n] can then be found
21

z-Transform Of Common Sequence
22

23
PROPERTIES AND THEOREMS
 Multiplication by a Constant
 Linearity of the z Transform
 Multiplication by ak
 Shifting Theorem
 Complex Translation Theorem
 Initial Value Theorem
 Final Value Theorem

Multiplication by a constant
If X(z) is the z transform of x(n), then
Z [ax(n)] = a Z[x(n)] = a X(z)
where a is a constant.
To prove this, note that by definition
Z [ax(n)] =
24
∑∑
∞
=
−
∞
=
−
==
00
)()()(
n
n
n
n
zaXznxaznax

Linearity of the z transform
The z transform possesses an important property: linearity. This means
that, if f(n) and g(n) are z-transformable and α and β are scalars, then
x(n) formed by a linear combination
x(n) = αf(n) + βg(n)
has the z transform
X(z) = αF(z) + βG(z)
where F(z) and G(z) are the z transforms of f(n) and g(n), respectively.
25

26
Multiplication By Ak
If X(z) is the z transform of x(k), then the z transform of ak
x(n) can be given by X(a-1
z):
Z[ak
x(n)] = X(a-1
z)
This can be proved as follows:
Z[an
x(n)] =
= X(a-1
z)
∑∑
∞
=
−−
∞
=
−
=
0
1
0
))(()(
n
n
n
nn
zanxznxa

27
Shifting Theorem
Also call real translation theorem. If x(t) = 0 for t < 0 and x(t) has the z transform X(z), then
Z[x(t-nT)] = z-n
X(z)
and
Z[x(t+nT)] =
n = zero or a positive integer



 − ∑
−
=
−
1
0
)()(
n
k
kn
zkTxzXz

28
EXAMPLE
Q. Find the z transforms of unit-step functions that are delayed by 1 sampling period and 4
sampling periods, respectively, as shown in figure (a) and (b) below

29
SOLUTION
Using the shifting theorem ,we have
Z [1(t-T)] = z-1
Z [1(t)] =
Also,
Z [1(t-4T)] = z-4
Z [1(t)] =
(Note that z-1
represents a delay of 1 sampling period T, regardless of the value of T.)
1
1
1
1
11
1
−
−
−
−
−
=
− z
z
z
z
1
4
1
4
11
1
−
−
−
−
−
=
− z
z
z
z

30
Q. Obtain the z transform of
Solution: Referring to Equation (2. 18), we have
Z [x(k-1)] = z-1
X(z)
The z transform of ak
is
Z[ak
] =
and so
Z [f(a)] = Z[ak-1
] =
where k = 1,2,3, ....



≤
=
=
−
0,0
....,3,2,1,
)(
1
k
ka
af
k
1
1
1
−
− az
1
1
1
1
11
1
−
−
−
−
−
=
− az
z
az
z

31
Complex Translation Theorem
If x(t) has the z transform X(z), then the z transform of e-at
x(t) can be given by X(z eat
).
To prove
Z [e-at
x(t)]
Thus, we see that replacing z in X(z) by z eat
gives the z transform of e- at
x(t).
)(
))((
)(
0
0
aT
k
kaT
k
kakT
zeX
zekTx
zekTx
=
=
=
∑
∑
∞
=
−
∞
=
−−

32
EXAMPLE
Q. By using the complex translation theorem, obtain the z transforms of:
1. e-at
sin ωt and
2. e-at
cosωt, respectively,

33
Solution
1. We know that
Z [sin ωt ] =
Using the complex translation theorem
Z
21
1
cos21
sin
−−
−
+− zTz
Tz
ω
ω
221
1
cos21
sin
]sin[ −−−−
−−
−
+−
=
zeTze
Tze
te aTaT
aT
at
ω
ω
ω

34
Solution
2. We know that
Z [cos ωt ]
Using the complex translation theorem
Z[e-at
cos ωt ]
21
1
cos21
cos1
−−
−
+−
−
=
zTz
Tz
ω
ω
221
1
cos21
cos1
−−−−
−−
+−
−
=
zeTze
Tze
aTaT
aT
ω
ω

35
Initial Value Theorem
• If x(t) has the z transform X(z) and if exists, then the initial value x(0) of x(t) or x(k) is
given by
• The initial value theorem is convenient for checking z transform calculations for possible
errors. Since x(0) is usually known, a check of the initial value by can easily spot
errors in X(z), if any exist.
)(lim zX
z ∞→
)(lim)0( zX
z
x
∞→
=
)(lim zX
z ∞→

36
EXAMPLE
Q. Determine the initial value x(0) if the z transform of x(t) is given by
By using the initial value theorem, we find
Referring to Example 2-2, notice that this X(z) was the z transform of
and thus x(0) = 0, which agrees with the result obtained earlier.
)1)(1(
)1(
)( 11
1
−−−
−−
−−
−
=
zez
ze
zX T
T
0
)1)(1(
)1(
lim)0( 11
1
=
−−
−
= −−−
−−
∞→ zez
ze
x T
T
z
t
etx −
−= 1)(

37
Final Value Theorem
• The final value of x(n), that is, the value of x(n) as approaches infinity, can be given by
(2. 27)
)]()1[(lim)(lim 1
1
zXznx
zn
−
→∞→
−=

38
EXAMPLE
Q. Determine the final value of
by using the final value theorem.
• By applying the final value theorem to the given X(z), we obtain
)(∞x
0,
1
1
1
1
)( 11
>
−
−
−
= −−−
a
zez
zX aT
[ ])()1(lim)( 1
1
zXzx
z
−
→
−=∞












−
−
−
−= −−−
−
→ 11
1
1 1
1
1
1
)1(lim
zez
z aTz
1
1
1
1lim 1
1
1
=





−
−
−= −−
−
→ ze
z
aTz

APPLICATION
39
• A closed-loop (or feedback) control system is shown in Figure.
• If you can describe your plant and your controller using linear difference equations, and if
the coefficients of the equations don't change from sample to sample, then your controller
and plant are linear and shift-invariant, and you can use the z transform.

40
• Suppose xn = output of the plant at sample time n
un = command to the DAC at sample time n
a and b = constants set by the design of the plant
• You can solve the behavior equation of the plant over time.
• Furthermore you can also investigate what happens when you add feedback to
the system.
• The z transform allows you to do both of these things.

• Deals with many common feedback control problems using continuous-time control.
• Also used in sampled-time control situations to deal with linear shift-invariant difference
equations.

TRANSFER FUNCTION
• The function H(Z) is called the “Transfer Function" of the system – it shows how the input
signal is transformed into the output signal.
H(Z)=Y(Z)/X(Z)
• In Z domain, the Transfer Function of a system isn't affected by the nature of the input
signal, nor does it vary with time.
• We can predict the behavior of the motor using H(Z).
• Let's say we want to see what the motor will do if x goes from 0 to 1 at time n = 0, and
stays there forever. This is called the ‘unit step function’ and the Z-Transform of the unit
step response is H(Z)=Z/(Z-1).
• Thus we can know everything about the system behavior and avoid undesirable
situations.

SOFTWARE
• You can write software from the Z-Transform with utter ease.
• Like, if you have a Transfer Function of a system, then the software turns it into
a Z-domain equation which can then be converted into a difference
equation which in turn can be turned into a software very quickly.
• This saves the manual work and a software for a plant can be produced within
seconds.

INVERSE Z-TRANSFORM
44
The inverse z-transform can be derived by using Cauchy’s integral theorem. Start with
the z-transform
∑
∞
−∞=
−
=
n
n
znxzX )()(
Multiply both sides by zk-1
and integrate with a contour integral for which the contour
of integration encloses the origin and lies entirely within the region of convergence of
X(z):
transform.-zinversetheis)()(
2
1
2
1
)(
)(
2
1
)(
2
1
1
1
11
nxdzzzX
i
dzz
i
nx
dzznx
i
dzzzX
i
C
k
n C
kn
C n
kn
C
k
=
=
=
∫
∑ ∫
∫ ∑∫
−
∞
−∞=
−+−
∞
−∞=
−+−−
π
π
ππ

z-TRANSFORM IMPLEMENTATION USING MATLAB
45

MATLAB Code For Finding z-TRANSFORM
• MATLAB Symbolic Toolbox gives the z-transform of a function .
• MATLAB program for Z-Transform
Program Code
% z-Transform %
clc;
close all;
clear all;
syms 'n';
syms 'z';
x=input('Input the sequence to be converted');
a=symsum((x*(z^(-n))),n,0,2);
disp(‘z-Transform of the given sequence is ');
disp(a);
46

Example of Output
•Input the sequence to be converted [1 2 3 4]
•z-Transform of the given sequence is
[ 1+1/z+1/z^2, 2+2/z+2/z^2, 3+3/z+3/z^2, 4+4/z+4/z^2]
47

MATLAB Code For Finding z-inverse Transform
48

• Output obtained : 1 4 5 -6 -32
• Therefore x (0) = 1, x(1) = = 4, x(2) 5, x(3) = -6, x(4) = -32
49

MATLAB Code For Pole And Zero Plot
• Pole-zero Diagram The MATLAB function “z-plane” can display the pole-zero
• Program Code
%Plotting zeros and poles of z-transform
clc;
close all;
clear all;
disp('For plotting poles and zeros');
b=input('Input the numerator polynomial coefficients');
a=input('Input the denominator polynomial coefficients');
[b,a]=eqtflength(b,a);
[z,p,k]=tf2zp(b,a);
zplane(z,p);
disp('zeros');
disp(z);
disp('poles');
disp(p);
disp('k');
disp(k);
50

• Example of Output
For plotting poles and zeros
Input the numerator polynomial coefficients[1 2 3 4]
Input the denominator polynomial coefficients[1 2 3]
zeros
-1.6506
-0.1747 + 1.5469i
-0.1747 - 1.5469i
poles
0
-1.0000 + 1.4142i
-1.0000 - 1.4142i
k
1
51

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