Ece4510 notes04

ECE4510/5510: Feedback Control Systems. 4–1
BASIC PROPERTIES OF FEEDBACK
4.1: Setting up an example to benchmark controllers
■ There are two basic types/categories of control systems:
OPEN LOOP:
Ctrlrr(t) y(t)
Disturbance
Plant
CLOSED LOOP:
Ctrlrr(t) y(t)
Disturbance
Plant
Sensor
■ This chapter of notes is concerned with comparing open-loop and
closed-loop control, and showing the potential beneﬁts (and some
pitfalls) of closed-loop (i.e., feedback) control.
■ We evaluate these two categories of controller in a number of ways:
disturbance rejection, sensitivity, dynamic tracking, steady-state error,
and stability.
DC motor speed control
■ In order to compare open- and closed-loop control, we will use an
extended example.
Lecture notes prepared by and copyright c⃝ 1998–2013, Gregory L. Plett and M. Scott Trimboli

ECE4510/ECE5510, BASIC PROPERTIES OF FEEDBACK 4–2
■ Recall equations of motion for a dc motor (pg. 2–19) but add a load
torque.
ea(t) eb(t)
ia(t) Ra La b
J
θ(t), τ(t)
Armature Load
τl, load
■ Assume that we are trying to control motor speed:
J ¨θ + b ˙θ
ke
˙θ + La
dia
dt
+ Raia
=
=
kτia + τl
ea
⎫
⎬
⎭
let output y = ˙θ,
disturbance w
△
= τl.
J ˙y + by
key + La
dia
dt
+ Raia
=
=
kτia + w
ea
⎫
⎬
⎭
s JY(s) + bY(s) = kτ Ia(s) + W(s)
keY(s) + sLa Ia(s) + Ra Ia(s) = Ea(s)
■ Solving the mechanical equation for Ia(s) gives
s JY(s) + bY(s) = kτ Ia(s) + W(s)
kτ Ia(s) = s JY(s) + bY(s) − W(s)
Ia(s) =
(s J + b) Y(s) − W(s)
kτ
.
■ Substituting into the electrical equation gives
keY(s) + sLa Ia(s) + Ra Ia(s) = Ea(s)
keY(s) + (sLa + Ra)
(s J + b) Y(s) − W(s)
kτ
= Ea(s).
■ Some algebra then yields

keY(s) + (sLa + Ra)
(s J + b) Y(s) − W(s)
kτ
= Ea(s)
kτ keY(s) + (sLa + Ra) (s J + b) Y(s) = kτ Ea(s) + (Ra + Las) W(s)
J Las2
+ (bLa + J Ra) s + (bRa + kτke) Y(s) = kτ Ea(s) + (Ra + Las) W(s).
■ Dividing both sides by bRa + kτ ke gives
J La
bRa+kτ ke
s2
+
bLa+J Ra
bRa+kτ ke
s + 1 Y(s) =
kτ
bRa+kτ ke
Ea(s) +
Ra+Las
bRa+kτ ke
W(s).
■ The left-hand-side can be factored into two parts:
J La
bRa + kτ ke
s2
+
bLa + J Ra
bRa + kτke
s + 1 = (τ1s + 1) (τ2s + 1) .
• Roughly, one of these time constants is mechanical; the other is
electrical.
■ If we assume that the mechanical time constant is much larger than
the electrical, the right-hand-side can be approximated by
kτ
bRa + kτ ke
Ea(s) +
Ra + Las
bRa + kτ ke
W(s) ≈ AEa(s) + BW(s),
where
A = kτ /(bRa + kτ ke)
B ≈ Ra/(bRa + kτke).
■ Then, we have overall relationship
(τ1s + 1)(τ2s + 1)Y(s) = AEa(s) + BW(s).
■ So,
Y(s) =
A
(τ1s + 1)(τ2s + 1)
Ea(s) +
B
(τ1s + 1)(τ2s + 1)
W(s).

4.2: Advantage of feedback: Disturbance rejection
■ We look at how the open-loop and feedback systems respond to a
step-like disturbance.
■ If ea(t) = ea · 1(t) (constant) and w(t) = w · 1(t) (constant), What is
steady state output?
■ Recall Laplace-transform final value theorem:
If a signal has a constant final value, it may be found as
yss = lim
s→0
sY(s).
Note: A signal will have a constant final value iff all of the poles of Y(s)
are strictly in the left-half s-plane, except possibly for a single pole at
s = 0.
■ For the input signals ea(t) and w(t), we have
Ea(s) =
ea
s
, W(s) =
w
s
.
■ So,
yss = lim
s→0
s
A
(τ1s + 1)(τ2s + 1)
ea
s
+
B
(τ1s + 1)(τ2s + 1)
w
s
= Aea + Bw.
■ This is the response of the open-loop system (without a controller).
■ Let’s make a simple controller for the open-loop system. The block
diagram looks like:
Ctrlrr(t) y(t)
Dist. w(t)
A
(τ1s + 1)(τ2s + 1)
B
A
ea(t)
Motor

■ We will design the controller to be a gain of Kol such that
ea(t) = Kolr(t),
■ Choose Kol so that there is no steady-state error when w = 0.
yss = AKolrss + Bwss
⇒ Kol = 1/A.
■ Is closed-loop any better? The block diagram looks like:
Ctrlrr(t) y(t)
Dist. w(t)
A
(τ1s + 1)(τ2s + 1)
B
A
ea(t)
Tachometer
1
■ Let’s make a similar controller for the closed-loop system (with the
possibility of a different value of K.)
ea(t) = Kcl(r(t) − y(t)).
■ The transfer function for the closed-loop system is:
Y(s) =
AKcl
(τ1s + 1)(τ2s + 1)
(R(s) − Y(s)) +
B
(τ1s + 1)(τ2s + 1)
W(s)
=
AKcl
(τ1s + 1)(τ2s + 1)
R(s) −
AKcl
(τ1s + 1)(τ2s + 1)
Y(s)
+
B
(τ1s + 1)(τ2s + 1)
W(s).
■ Combining Y(s) terms

Y(s) 1+
AKcl
(τ1s+1)(τ2s+1)
=
AKcl
(τ1s+1)(τ2s+1)
R(s)+
B
(τ1s+1)(τ2s+1)
W(s)
Y(s)
(τ1s+1)(τ2s+1)+AKcl
(τ1s+1)(τ2s+1)
=
AKcl
(τ1s+1)(τ2s+1)
R(s)+
B
(τ1s+1)(τ2s+1)
W(s).
■ This gives
Y(s) =
AKcl
(τ1s + 1)(τ2s + 1) + AKcl
R(s) +
B
(τ1s + 1)(τ2s + 1) + AKcl
W(s).
■ Employing the ﬁnal-value theorem for w = 0 gives
yss =
AKcl
1 + AKcl
rss
=
1
1 + 1
AKcl
rss.
■ If AKcl ≫ 1, yss ≈ rss.
■ Open-loop with load:
yss = AKolrss + Bwss = rss + Bwss
δy = Bwss.
■ Closed-loop with load:
yss =
AKcl
1 + AKcl
rss +
B
1 + AKcl
wss
δy ≈
B
1 + AKcl
wss.
which is much better than open-loop since AKcl ≫ 1.
ADVANTAGE OF FEEDBACK: Better disturbance rejection (by factor of
1 + AKcl).

4.3: Advantage of feedback: Sensitivity and dynamic tracking
■ The steady-state gain of the open-loop system is: 1.0
■ How does this change if the motor constant A changes?
A → A + δA
Gol + δGol = Kol(A + δA)
=
1
A
(A + δA)
= 1 +
δA
A
gain error
.
■ In relative terms:
δGol
Gol
=
δA
A
= 1.0
sensitivity
δA
A
.
■ Therefore, a 10 % change in A results in a 10 % change in gain.
Sensitivity=1.0.
■ Steady-state gain of closed-loop system is:
AKcl
1 + AKcl
.
Gcl + δGcl =
(A + δA)Kcl
1 + (A + δA)Kcl
.
■ From calculus (law of total differential)
δGcl =
dGcl
dA
δA
or
δGcl
Gcl
=
A
Gcl
dGcl
dA
sensitivity S
Gcl
A
δA
A
.

■ To calculate this, we ﬁrst compute
dGcl
dA
=
d
dA
AKcl
1 + AKcl
=
(1 + AKcl)Kcl − Kcl(AKcl)
(1 + AKcl)2
=
Kcl
(1 + AKcl)2
.
■ Then,
δGcl
Gcl
=
A
Gcl
dGcl
dA
δA
A
SGcl
A =
A
AKcl/(1 + AKcl)
Kcl
(1 + AKcl)2
=
1
1 + AKcl
.
ADVANTAGE OF FEEDBACK: Lower sensitivity to modeling error (by a
factor of 1 + AKcl)
Dynamic Tracking
■ Steady-state response of closed-loop better than open-loop: Better
disturbance rejection, better (lower) sensitivity.
■ What about transient response?
■ Open-loop system: Poles at roots of (τ1s + 1)(τ2s + 1)
➠ s = −1/τ1, s = −1/τ2.
■ Closed-loop system: Poles at roots of (τ1s + 1)(τ2s + 1) + AKcl.
➠ s =
−(τ1 + τ2) ± (τ1 + τ2)2 − 4τ1τ2(1 + AKcl)
2τ1τ2
.

■ FEEDBACK MOVES POLES
• System may have faster/slower response
• System may be more/less damped
• System may become unstable!!!
• Often a high gain Kcl results in instability.
• For this dc motor example, we can get step responses of the
following form:
0 2 4 6 8 10
0
0.5
1
1.5
High Kcl
Low Kcl
Time (sec)
y(t)
■ So, we see the ﬁrst potential downside of feedback—if the controller
is not well designed, it may make the plant’s response worse than it
was to begin with.
■ Designing controllers is a main focus of the rest of this course (and of
follow-on courses). It’s not a trivial task.
■ But, we can get a really good start toward improving the dynamic
response of the closed-loop system with a very simple controller
■ We look next at the PID controller, then return to exploring (potential)
advantages of feedback.

4.4: Proportional-integral-derivative (PID) control (a)
■ General control setup:
D(s)R(s)
U(s)
Y(s)
Disturbance
G(s)
■ Need to design controller D(s).
■ One option is PID (Proportional Integral Derivative) control design.
• Extremely popular. 90+
% of all ﬁelded controllers are PID.
• Doesn’t mean that they are great, just popular.
■ We just saw proportional control where u(t) = Kpe(t), or D(s) = Kp.
■ Proportional control tends to increase speed of response, but:
• Can allow non-zero steady-state error.
• Can result in larger transient overshoot.
• May not eliminate a constant disturbance.
■ Integral control, where D(s) =
Ki
s
, can eliminate steady-state error,
• But, transient response can get worse, and
• Stability margins can get worse.
■ Derivative control, where D(s) = Kds, can reduce oscillations in
dynamic response, but
• Steady-state error can get worse.
■ In the next sections, we look at each of these controllers separately,
then consider how to use them together.

Proportional control
■ Proportional controllers compute the control effort such that
u(t) = Kp(r(t) − y(t)) = Kpe(t) . . . D(s) = Kp.
IDEA: For plants with positive gain, if e(t) = r(t) − y(t) > 0, then I’m not
“trying hard enough.” Multiply error by (positive) Kp to “try harder.”
■ Also, if e(t) < 0, then I’ve tried too hard already. Multiply (negative)
error by (positive) gain Kp to try to pull response back.
EXAMPLE: Determine behavior of closed-loop poles for the dc motor.
Y(s)
R(s)
=
AKp
(τ1s + 1)(τ2s + 1) + AKp
.
■ Poles are roots of (τ1s + 1)(τ2s + 1) + AKp.
■ Without feedback, Kp → 0.
s1 = −1/τ1, s2 = −1/τ2.
■ With feedback,
Y(s)
R(s)
=
AKp
(τ1s + 1)(τ2s + 1) + AKp
=
AKp
τ1τ2s2 + (τ1 + τ2) s + 1 + AKp
.
■ Solving for root locations gives
s1, s2 =
−(τ1 + τ2) ± (τ1 + τ2)2 − 4τ1τ2(1 + AKp)
2τ1τ2
.
■ We can plot the locations of the poles (a “root locus” plot)
parametrically as Kp changes

Kp = 0Kp = 0
−
1
τ1
−
1
τ2
Kp =
(τ1 − τ2)2
4τ1τ2 A
−
τ1 + τ2
2τ1τ2
R(s)
I(s)
■ For 0 < Kp <
(τ1 − τ2)2
4τ1τ2 A
, poles move horizontally toward each other
along the real axis.
• Rise time gets faster since dominant pole moves farther from
origin, natural frequency increases.
• Settling time gets faster since real part of dominant pole moves
farther from origin.
• Damping remains same (no overshoot).
■ For Kp >
(τ1 − τ2)2
4τ1τ2 A
, the poles gain imaginary part.
• Settling time remains same since real part of pole locations is
unchanged.
• Rise time decreases since natural frequency increases.
• Overshoot increases since damping ratio decreases.
■ For systems having more poles than this example, increasing Kp
often leads to instability.
■ How do we improve accuracy, but keep stability?

4.5: Proportional-integral-derivative (PID) control (b)
Integral and proportional-integral control
■ Pure integral controllers compute the control effort such that:
u(t) =
Kp
Ti
t
0
e(τ) dτ, D(s) =
Kp
Ti s
.
• Ti = “Integral time” = time for output = Kp with input e(t) = 1(t).
• An alternate formulation has
u(t) = Ki
t
0
e(τ) dτ, D(s) =
Ki
s
.
■ Integral feedback can give nonzero control even at points of time
when e = 0 because of “memory.”
• In many cases this can eliminate steady-state error to step-like
reference inputs and step-like disturbances.
IDEA: To avoid instability or oscillations with proportional control, the
proportional gain Kp must be kept “small.”
■ But, then when error gets small, we no longer try very hard to correct
it—leads to ﬁnite steady-state error.
■ Also, some nonlinearities (e.g., coulombic friction) can cause output
to get stuck even if control effort is nonzero.
■ Integral control can help: If we integrate the error signal, the
integrated value will grow over time if the error is “stuck”.
■ This increases the control signal u(t) until the error starts
decreasing—making the error converge to zero.
EXAMPLE: Substitute: u(t) =
Kp
Ti
t
0
(r(τ) − y(τ)) dτ into dc-motor eqs.

τ1τ2 ÿ(t) + (τ1 + τ2) ˙y(t) + y(t) = A
Kp
Ti
t
0
(r(τ) − y(τ)) dτ + Bw(t).
■ Differentiate,
τ1τ2
...
y(t) + (τ1 + τ2)ÿ(t) + ˙y(t) =
AKp
Ti
(r(t) − y(t)) + B ˙w(t)
τ1τ2
...
y(t) + (τ1 + τ2)ÿ(t) + ˙y(t) +
AKp
Ti
y(t) =
AKp
Ti
r(t) + B ˙w(t).
■ If r(t) = cst, w(t) = cst, ˙w(t) = 0,
AKp
Ti
yss =
AKp
Ti
rss ➠ no error.
■ Steady-state tracking improves, but dy-
namic response degrades, especially
after poles leave real axis.
• Very oscillatory; possibly unstable.
■ Can be improved by adding proportional
term to integral term.
Ki=0Ki=0
−
1
τ1
−
1
τ2
R(s)
I(s)
u(t) = Kpe(t) +
Kp
TI
t
0
e(τ) dτ, D(s) = Kp 1 +
1
Ti s
.
■ Poles are at the roots of
τ1τ2s3
+ (τ1 + τ2)s2
+ (1 + AKp)s +
AKp
Ti
= 0.
Two degrees of freedom.
Derivative and proportional-derivative control
■ Pure derivative controllers compute the control effort such that:
u(t) = KpTd ˙e(t), D(s) = KpTds,
where Td = “derivative time”.

■ An alternate formulation has
u(t) = Kd ˙e(t), D(s) = Kds.
IDEA: Would like to anticipate “momentum,” which mechanically is
proportional to velocity, and subtract out its predicted contribution.
■ Contribution to control effort acts as braking force when approaching
reference value quickly.
WARNING: PURE DERIVATIVE CONTROL IMPRACTICAL SINCE
DERIVATIVE MAGNIFIES SENSOR NOISE!
■ Practical version = “lead control,” which we will study later.
■ Derivative control tends to stabilize a system.
■ Does nothing to reduce constant error! If ˙e(t) = 0, then u(t) = 0, even
if e(t) ̸= 0.
■ Motor control: Poles at roots of τ1τ2s2
+ (τ1 + τ2 + AKpTd)s + 1 = 0.
• Td enters ζ term, can make damping better.
■ PD = Proportional plus derivative control where
D(s) = Kp(1 + Tds) or D(s) = Kp + Kds.
■ Root locus for dc motor, PD control.
−
1
τ1
−
1
τ2
−
1
Td
R(s)
I(s)
■ Great damping, possibly poor steady-state error.

4.6: Proportional-integral-derivative (PID) control (c)
Proportional Integral Derivative Control
■ D(s) = Kp 1 +
1
Ti s
+ Tds or D(s) = Kp + Ki
1
s
+ Kds.
■ Need ways to design parameters {Kp, Ti , Td} or {Kp, Ki, Kd}.
■ In general (i.e., not always),
Kp, Ti ↑ ⇐⇒ error ↓, stability ↓
Td ↑ ⇐⇒ stability ↑
■ For speed control problem,
u(t) = Kp (r(t) − y(t)) +
1
Ti
t
0
(r(τ) − y(τ)) dτ + Td(˙r(t) − ˙y(t)) .
(math happens). Solve for poles
τ1τ2Tis3
+ Ti((τ1 + τ2) + AKpTd)s2
+ Ti(1 + AKp)s + AKp = 0
s3
+
τ1 + τ2 + AKpTd
τ1τ2
s2
+
1 + AKp
τ1τ2
s +
AKp
τ1τ2Ti
= 0.
■ Three coefﬁcients, three parameters. We can put poles anywhere!
Complete control of dynamics in this case.
■ Entire transfer functions are:
Y(s)
W(s)
=
Ti Bs
Tiτ1τ2s3 + Ti(τ1 + τ2)s2 + Ti (1 + AKp)s + AKp
.
Y(s)
R(s)
=
AKp(Tis + 1)
Tiτ1τ2s3 + Ti(τ1 + τ2)s2 + Ti (1 + AKp)s + AKp
.
■ We can plot responses in MATLAB:

num = [TI*B 0];
den = [TI*TAU1*TAU2 TI*(TAU1+TAU2) TI*(1+A*KP) A*KP];
step(num, den)
0 0.01 0.02 0.03 0.04 0.05
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time (sec)
y(t)(rad/sec)
P
PI
PID
Step Reference Response
0 0.01 0.02 0.03 0.04 0.05
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time (sec)
y(t)(rad/sec)
P
PI
PID
Step Disturbance Response
Ziegler–Nichols tuning of PID controllers
■ “Rules-of-thumb” for selecting Kp, Ti , Td.
■ Not optimal in any sense—but often provide good performance.
METHOD I: If system has step response like this,
Slope, A/τ
A
τd τ
Y(s)
U(s)
=
Ae−τds
τs + 1
,
(ﬁrst-order system plus delay)
■ We can easily identify A, τd, τ from this step response.
■ Don’t need complex model!
■ Tuning criteria: Ripple in impulse response decays to 25% of its value
in one period of ripple

Period1
0.25
RESULTING TUNING RULES:
P PI PID
Kp =
τ
Aτd
Kp =
0.9τ
Aτd
Ti =
τd
0.3
Kp =
1.2τ
Aτd
Ti = 2τd
Td = 0.5τd
METHOD II: Conﬁgure system as
Kur(t) y(t)Plant
■ Turn up gain Ku until system produces oscillations (on stability
boundary) Ku = “ultimate gain.”
Period, Pu
1
RESULTING TUNING RULES:
P PI PID
Kp = 0.5Ku
Kp = 0.45Ku
Ti =
1
1.2
Pu
Kp = 0.6Ku
Ti = 0.5Pu
Td =
Pu
8
Practical Problem: Integrator Overload
■ Integrator in PI or PID control can cause problems.
■ For example, suppose there is saturation in the actuator.
• Error will not decrease.
• Integrator will integrate a constant error and its value will “blow up.”

■ Solution = “integrator anti-windup.” Turn off integration when actuator
saturates.
K
Ka
K
TI s
e(t) u(t) umin
umax
■ Doing this is NECESSARY in any practical implementation.
■ Omission leads to bad response, instability.
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Time (sec)
Step Response
With antiwindup
Without antiwindup
y(t)
0 2 4 6 8 10
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Time (sec)
Control Effort
With antiwindup
Without antiwindup
u(t)

4.7: Steady-state error (a)
■ System error is any difference between r(t) and y(t). Two sources:
1. Imprecise tracking of r(t).
2. Disturbance w(t) affecting the system output.
Steady-state error (w.r.t. reference input)
■ We have already seen examples of CL systems that have some
tracking error (proportional ctrl) or not (integral ctrl) to a step input.
We will formalize this concept here.
■ Start with very general control structure with “closed-loop” transfer
function T (s) that computes Y(s) from R(s):
T(s)r(t) y(t)
■ We don’t care what is inside the box T (s). It could be any block
diagram, and we may need to compute T(s) from the block diagram.
■ The error is
E(s) = R(s) − Y(s)
= R(s) − T (s)R(s)
= [1 − T(s)] R(s).
■ Assume conditions of ﬁnal value theorem are satisﬁed (i.e.,
[1 − T(s)]R(s) has poles only in LHP except perhaps for a single pole
at the origin)
ess = lim
t→∞
e(t) = lim
s→0
s[1 − T (s)]R(s).

■ When considering steady-state error to different reference inputs, we
restrict ourselves to inputs of the type
r(t) =
tk
k!
1(t); R(s) =
1
sk+1
.
■ The first few of this family are drawn below:
t
1(t)
t
t · 1(t)
t
t2
2
· 1(t)
■ As k increases, reference tracking is progressively harder. (It is easier
to track a constant reference than it is to track a moving reference.)
■ We define a concept called system type to describe the ability of the
closed-loop system to track inputs of different kinds.
• If system type = 0, constant steady-state error for step input,
infinite s.s. error for ramp or parabolic input.
• If system type = 1, no steady-state error for step input, constant
s.s. error for ramp input, infinite s.s. error for parabolic input.
• If system type = 2, no steady-state error for step or ramp inputs,
constant s.s. error for parabolic inputs.
• And so forth, for higher-order system types.
■ To find the system type in general, we must compute the following
equation for values of k = 0, 1, . . . until we calculate a finite nonzero
value for ess:
ess = lim
s→0
1 − T (s)
sk
=
⎧
⎪⎨
⎪⎩
0, type > k;
constant, type = k;
∞, type < k.

■ As an example, consider ramp responses of different system types:
t
t
Type 0 System
t
t
Type 1 System
t
t
Type 2 System
DANGER: Higher order sounds better but they are harder to stabilize and
design. Transient response may be poor.
■ Summary table for computing steady-state error for different system
types (where the limits evaluate to ﬁnite nonzero values)
Steady-state tracking errors ess for generic closed-loop T(s)
Sys. Type Step Input Ramp Input Parabola Input
Type 0 lim
s→0
(1 − T (s)) ∞ ∞
Type 1 0 lim
s→0
1 − T (s)
s
∞
Type 2 0 0 lim
s→0
1 − T (s)
s2
EXAMPLES:
(1) Consider T(s) =
s + 1
(s + 2)(s + 3)
. What is the system type?
■ Try k = 0. Evaluate
lim
s→0
[1 − T(s)] = lim
s→0
(s + 2)(s + 3) − (s + 1)
(s + 2)(s + 3)
=
5
6
̸= 0.
■ Therefore, the system type is zero, ess = 5/6 to unit step input.
(2) Consider T(s) =
s + 6
(s + 2)(s + 3)

lim
s→0
[1 − T(s)] = lim
s→0
(s + 2)(s + 3) − (s + 6)
(s + 2)(s + 3)
=
0
6
= 0.
■ Therefore the system type must be greater than zero.
lim
s→0
1 − T (s)
s
= lim
s→0
1
s
(s + 2)(s + 3) − (s + 6)
(s + 2)(s + 3)
= lim
s→0
1
s
s2
+ 4s
(s + 2)(s + 3)
=
4
6
̸= 0.
■ Therefore, the system is type 1, ess = 2/3 to unit ramp input.
(3) Consider T(s) =
5s + 6
(s + 2)(s + 3)
lim
s→0
[1 − T (s)] = lim
s→0
(s + 2)(s + 3) − (5s + 6)
(s + 2)(s + 3)
=
0
6
= 0.
Therefore, the system type must be greater than zero.
lim
s→0
1 − T (s)
s
= lim
s→0
1
s
(s + 2)(s + 3) − (5s + 6)
(s + 2)(s + 3)
= lim
s→0
1
s
s2
(s + 2)(s + 3)
=
0
6
= 0.
Therefore, the system type must be greater than one.
lim
s→0
1 − T (s)
s2
= lim
s→0
1
s2
(s + 2)(s + 3) − (5s + 6)
(s + 2)(s + 3)
= lim
s→0
1
s2
s2
(s + 2)(s + 3)
=
1
6
̸= 0.
Therefore, the system type is two, ess = 1/6 to unit parabola input.

4.8: Steady-state error w.r.t. reference input, unity feedback
WARNING: The following method is a special case of the above general
method. Always use the appropriate method for the problem at hand!
■ Unity-feedback is when the control system looks like:
Gol(s)r(t) y(t)
■ That is, the feedback loop has a gain of exactly one.
■ If we are fortunate enough to be considering a unity-feedback
scenario, the prior rules have a simpler solution.
■ But, if there are any dynamics in the feedback loop, we DO NOT have
a unity-feedback system, and must use the more general rules from
Section 4.7.
■ For unity-feedback systems, there are some important simpliﬁcations:
T (s) =
Gol(s)
1 + Gol(s)
1 − T (s) =
1 + Gol(s)
1 + Gol(s)
−
Gol(s)
1 + Gol(s)
=
1
1 + Gol(s)
.
So,
E(s) =
1
1 + Gol(s)
R(s).
■ For test inputs of the type R(s) =
1
sk+1
ess = lim
s→0
sE(s) = lim
s→0
1
[1 + Gol(s)]sk
.

■ For a system that is type 0,
ess = lim
s→0
1
1 + Gol(s)
=
1
1 + Kp
, Kp = lim
s→0
Gol(s).
ess = lim
s→0
1
1 + Gol(s)
1
s
= lim
s→0
1
sGol(s)
=
1
Kv
, Kv = lim
s→0
sGol(s).
ess = lim
s→0
1
1 + Gol(s)
1
s2
= lim
s→0
1
s2Gol(s)
=
1
Ka
, Ka = lim
s→0
s2
Gol(s).
These formulas meaningful only for unity-feedback!
Kp = lim
s→0
Gol(s). “position error constant”
Kv = lim
s→0
sGol(s). “velocity error constant”
Ka = lim
s→0
s2
Gol(s). “acceleration error constant”
Steady-state tracking errors ess for unity-feedback case only.
Type 0
1
1 + Kp
∞ ∞
Type 1 0
1
Kv
∞
Type 2 0 0
1
Ka
EXAMPLES:
(1) Consider Gol(s) =
s + 1
(s + 2)(s + 3)
Gol(0) =
1
2 · 3
=
1
6
.

■ Therefore, system type= 0, ess to unit step=
1
1 + 1/6
=
6
7
.
(s + 1)(s + 10)(s − 5)
(s2 + 3s)(s4 + s2 + 1)
Gol(0) =
1 · 10 · (−5)
0 · 1
= ∞ ➠Type > 0.
sGol(s) =
(s + 1)(s + 10)(s − 5)
(s + 3)(s4 + s2 + 1)
sGol(s)|s=0 =
1 · 10 · (−5)
3 · 1
=
−50
3
.
■ Therefore, system type= 1, ess to unit ramp=
−3
50
.
s2
+ 2s + 1
s4 + 3s3 + 2s2
Gol(0) =
1
0
= ∞ ➠Type > 0
sGol(s) =
s2
+ 2s + 1
s3 + 3s2 + 2s
sGol(s)|s=0 =
1
0
= ∞ ➠Type > 1
s2
Gol(s) =
s2
+ 2s + 1
s2 + 3s + 2
s2
Gol(s) s=0
=
1
2
➠Type = 2.
■ Therefore, system type= 2, ess to unit parabola= 2.
KEY POINT: Open-loop Gol(s) tells us about closed-loop s.s. response.
KEY POINT: For unity-feedback systems, number of poles of Gol(s) at
s = 0 is equal to the system type.

EXAMPLE: DC-motor example with proportional control, D(s) = Kp.
Ctrlrr(t) y(t)
A
(τ1s + 1)(τ2s + 1)
D(s)G(s) =
Kp A
(τ1s + 1)(τ2s + 1)
, lim
s→0
D(s)G(s) = Kp A.
■ So system is type 0, with s.s. error to step input of
1
1 + Kp A
.
■ This agrees with prior results.
EXAMPLE: DC-motor example with PI control, D(s) = Kp 1 +
1
Tis
.
D(s)G(s) =
Kp A +
Kp A
Ti s
(τ1s + 1)(τ2s + 1)
.
lim
s→0
D(s)G(s) = ∞
lim
s→0
sD(s)G(s) =
Kp A
Ti
.
■ System is type 1, with s.s. error to ramp input of
Ti
Kp A
.
EXAMPLE: DC-motor with two-integrator controller,
D(s) = Kp 1 +
1
Tis
+
1
Ti s2
.
D(s)G(s) =
Kp A +
Kp A
Ti s
+
Kp A
Ti s2
(τ1s + 1)(τ2s + 1)
lim
s→0
s2
D(s)G(s) =
Kp A
Ti
.
■ System is type 2, with s.s. error to parabolic input of
Ti
Kp A
.

4.9: Steady-state error w.r.t. disturbance
■ Recall, system error is any difference between r(t) and y(t).
■ We have just spent considerable time considering differences
between r(t) and y(t) because the system is not capable of tracking
r(t) perfectly.
• This is an issue with T(s) for the general case, or Gol(s) for the
unity-feedback case.
■ But, another source of steady-state error can be uncontrolled inputs
to the system—disturbances.
EXAMPLE: Consider a vehicle cruise-control system. We may set the
reference speed r(t) = 55 mph, but we find that the steady-state
vehicle speed yss is affected by wind and road grade (in addition to
the cruise-control system’s ability to track the reference input.)
■ We can think of a system’s overall response to both the reference
input and the disturbance input as
Y(s) = T (s)R(s) + Tw(s)W(s).
■ We find the system type with regard to the reference input by
examining T (s); similarly, we find the system type with respect to the
disturbance input by examining Tw(s).
■ Due to linearity, we can consider these two problems separately.
• When thinking about system type with respect to reference input,
we consider W(s) = 0 and follow the procedures outlined earlier.
• When thinking about system type with respect to disturbance
input, we consider R(s) = 0 and follow the procedure below.

■ We do not wish the output to have ANY disturbance term in it, so the
output error due to the disturbance is equal to whatever output is
caused by the disturbance.
ess = rss − yss = 0 − lim
s→0
sTw(s)W(s).
■ We say that the system is type 0 with respect to disturbance if it has
nonzero steady-state error when the disturbance is 1/s.
• Type 0 system (w.r.t. disturbance) has constant ess = − lim
s→0
Tw(s).
■ We say that a system is type 1 with respect to disturbance if it has
nonzero steady-state error when the disturbance is 1/s2
.
s→0
Tw(s)
s
.
■ We say that a system is type 2 with respect to disturbance if it has
nonzero steady-state error when the disturbance is 1/s3
.
s→0
Tw(s)
s2
.
■ Summary table for computing steady-state error for different system
types with respect to disturbance (where the limits evaluate to ﬁnite
nonzero values)
Steady-state errors ess due to disturbance for generic closed-loop Tw(s)
Type 0 − lim
s→0
Tw(s) −∞ −∞
Type 1 0 − lim
s→0
Tw(s)
s
−∞
Type 2 0 0 − lim
s→0
Tw(s)
s2

NOTE: There are no special cases for unity feedback when computing
system type with respect to disturbance.
■ You must always compute the closed-loop transfer function
Tw(s) = Y(s)/W(s), and then perform the tests.
EXAMPLE: Consider the motor-control problem from before.
Ctrlrr(t) y(t)
Dist. w(t)
A
(τ1s + 1)(τ2s + 1)
B
A
ea(t)
Tachometer
1
■ We’ll use a proportional controller with gain Kcl, so
ea(t) = Kcl(r(t) − y(t)).
■ When we looked at the example earlier, we found that
Y(s) =
AKcl
(τ1s + 1)(τ2s + 1) + AKcl
R(s) +
B
(τ1s + 1)(τ2s + 1) + AKcl
W(s).
■ From this equation, we gather
T (s) =
AKcl
(τ1s + 1)(τ2s + 1) + AKcl
Tw(s) =
B
(τ1s + 1)(τ2s + 1) + AKcl
.
■ Starting with system type 0, test for ﬁnite nonzero steady-state error:
ess = − lim
s→0
Tw(s) =
−B
1 + AKcl
̸= 0.
■ Therefore, the system is type 0 with respect to disturbance (it’s also
type 0 with respect to reference input, but the two system types ARE
NOT the same in general).

Ece4510 notes04

More Related Content

What's hot (20)

Similar to Ece4510 notes04 (20)

More from K. M. Shahrear Hyder (12)

Recently uploaded (20)

Ece4510 notes04