Control chap8

CONTROL SYSTEMS
THEORY
Design Using the Graphical Tool
CHAPTER 8
STB 35103

Objectives


To use the root locus to design cascade
compensators to improve the steady-state
error.



compensator to improve the transient
response.



compensator to improve the steady-state
error and the transient response.

Introduction


In our previous chapter we learnt how to
sketch the root locus. The root locus allows
us to choose the proper loop gain (value of
K) to meet a transient response
specification (e.g. %OS).



Setting the gain at a particular value will
produce the transient response dictated by
the poles at that point on the root locus.
Thus we are limited to those response
that exist along the root locus.

Introduction


When we design a control system the two
most important things that we will
consider are:


Transient response




Settling time, percent overshoot, peak time

Steady-state error


The difference between the input and the output

Introduction – improving transient response


Example


We have a control system that produces 20% overshoot
with settling time, Ts = 20 seconds. Let’s say we want to
speed up our process to make it produce a stable output
within 10 seconds instead of 20 seconds (Ts = 10) but
still maintain 20% overshoot. Figure below shows our
root locus and the radial line.
ζ = 0.45



Based on the previous example, poles on the
same radial line will have the same
overshoot value. Pole at A has the same
overshoot as pole at B because both poles are
on the same radial line.



A point on the radial line that is further from
origin will have smaller settling time
compared to point on radial line that is close to
origin. Point at A has larger settling time
compared to point at B.



Let us assume point at A produce Ts = 20
and point at B produce Ts = 10. Our
objective is to produce transient response
with 20% overshoot and Ts = 10. So, the
best point in the s-plane to get 20% OS
and Ts =10 is at point B.
ζ = 0.45



However, based on the figure below, the root
locus only crosses point A, even if we increase
the gain the root locus will not reach point B.



One way to solve this problem is to replace the
existing system with a system that can reach
point B but this replacement is expensive.



A better method is compensate the
system with additional poles and zeros, so
that the compensated system has a root
locus that goes through the desired pole
location for some value of gain.


Disadvantage for this method is the system
order will increase and could effect our
transient response.

Introduction – improving steady-state error




Compensator can also be used to improve the
steady-state error.
In our previous chapter we varied the gain of
our system to meet the transient response
specification. When gain is varied our steadystate error will also change.




The higher the gain, the smaller the steady-state
error but the larger the percent overshoot.
Reducing the gain to reduce overshoot will increase
the steady-state error.

Introduction- Compensators


In summary, transient response is improved
with the addition of differentiation, and
steady state error is improved with the
addition of integration in the forward path.



Our focus is only for cascade compensator
for a unity feedback system.

Introduction – modes of control


There are actually three “Modes of control”.






Proportional (P)
Integral (I)
Derivative (D)

Previously we learnt to change the value of gain,
K, to get a certain transient response
specification. The mode of control for this type is
Proportional or P-controller
Proportional

Improving steady-state error via
cascade compensation


Proportional-plus-integral (PI) controller






Steady state error can be improved by placing
an open-loop pole at origin, because this will
increase the system type by one.
For example, a Type 0 system with step input
produces finite error. The error becomes zero
when we increase the type by one.
Preferably, we improve the steady-state-error
without affecting the transient response.

Type 0 has finite
error for step input

When we increase Type
0 by one into Type 1,
the error becomes zero
for step input



Example for PI controller:


Figure below is an example of a unity feedback
system with gain, K. Point A is on the root
locus which means the [sum of zeros angle] –
[sum of poles angle] = odd multiples of 180̊.



Assume the input, R(s) for this system is step
input, the steady state error for this system is
a finite value. We can reduce the steady-state
error to zero by increasing the system type.
This can be done by putting integration (poles
at s=0) at gain.



If you notice, the root locus for our
compensated system no longer goes through
point A, meaning our transient response has
changed.



In order to get the same transient response
similar to the transient response before we put
integration (poles at s=0) we must make some
changes to the compensated system to make
sure the root locus goes through point A again.



We know in root locus, the poles will always move to
zeros. Previously we introduce a pole at zero thus
changes our transient response. In order to undo the
change and still maintain the increase in system type,
we will need to put a zero very close to the
compensated pole. Both zero angle and pole angle will
cancel each other thus making the root locus resemble
the original root locus but with an increase in system
type (pole-zero cancellation)



Example 9.1


Given the system below with a step input and
57.4% overshoot, reduce the steady-state
error to zero without greatly affecting the
transient response.



Analysis (solving the proportional part)


We will first investigate the transient response and
steady-state error for this system.

First draw the poles and
zeros using the open loop
transfer function of the
unity feedback system.
Next, we need to find the
value of gain where our
system produce 57.4%
overshoot.



Next, we need to find the value of gain where
our system produce 57.4% overshoot.
ζ = 0.174

r

A
X

θ1

B
X

θ2

K = A× B × C

C
X

θ3

θζ

The gain we
found using
calculation is
K=164.6
(hint:
r = 3.987)



We are trying to improve the steady-state error so
we need to know the steady state error of our
current system. Since the input is a step, we need to
calculate the final error value for step.

1
estep (∞) =
, K p = lim KG ( s )
s →0
1+ K p

K
164.6
K p = lim
=
= 8.23
s →0 ( s + 1)( s + 2)( s + 10)
(1)(2)(10)
1
estep (∞) =
= 0.108
1 + 8.23


The value of step input is 1 but since we have error
the output that we get from the system is (1-0.108 =
0.892)



The system has three poles instead of two poles, we
need to determine whether the system is a valid
second-order approximation. We must first find the
coordinate of the system poles when gain, K =164.6
(57.4% overshoot). We can find the poles using the
closed loop transfer function of the unity feedback
system.



The closed loop transfer function, T(s) is

K
T ( s) = 3
s + 13s 2 + 32 s + 20 + K



We will substitute the value of gain where the root
locus intersect the radial line to produce 57.4 %
overshoot and get the poles of the system during this
gain. From our previous calculation we know gain, K
= 164.6, so

T ( s) =


164.6
s 3 + 13s 2 + 32s + 20 + 164.6

We will then plot the poles by factorizing the
denominator.

164.6
T (s) =
( s + 11.613) ( s + 0.693 − j3.926 ) ( s + 0.693 + j3.926 )



We will include the poles during gain, K = 164.6 in
our root locus.

The distance of the
dominant second order
poles from the jω-axis is
0.694 while the distance
of the higher order poles
is 11.61. The distance of
the higher order pole is
more than five times
the distance of the
dominant second-order
poles from the jω-axis.
Hence it is a valid 2nd
order approximation

11.61 / 0.694 = 16.729 times farther



Summary of the original (uncompensated) system is, the
system produce 57.4% overshoot when gain, K = 164.6.
The error for this system is 0.108. The system can be
approximated as a second-order system.
Parameter

Uncompensated

Percent overshoot

57.4%

Gain, K

164.6

Dominant second-order
poles

-0.693 + j3.926
-0.693 - j3.926

Higher order poles

-11.613

estep(∞)

0.108

Input =1

Output = 0.892



Solution (solving the integral part):
The question : to reduce error to zero without greatly
affecting the transient response. We reduce the error to
zero by increasing the system type. In order to increase
the system type we will need to add integration (poles at
origin). To make sure the transient response of our
original system does not change we will also put a zero
into our system very close to the compensated pole ( pole
at origin). The angular contribution of both compensated
pole and zero will cancel each other thus maintaining the
original transient response =pole-zero cancellation



We will now plot the zeros and poles using the open
loop transfer function of the compensated unity
feedback system, GOL(s)

GOL ( s )

K ( s + 0.1)

s ( s + 1) ( s + 2 ) ( s + 10 )

Poles: (0,0), (-1,0), (-2,0), (-10,0)
Zero : (-0.1, 0)



The root locus with the compensated poles and zeros

We redo the process to find
the value of gain where
radial line intersect the root
locus. (Hint: r = 3.897)
The gain found from
calculation is K = 158.2.



The gain is found using the same method as
the uncompensated system.
ζ = 0.174

r

A
X

θ1

B
X

θ2

CD
X

θ3

A× B × C × E
K=
D

θ4 θ

O

E
ζ

X

The gain we
found using
calculation is
K=158.2 (hint:
r = 3.897)
θ5



Next we find the position of poles when gain, K =
158.2. We will use the poles to check whether we can
approximate as a second-order system. The closedloop transfer function of the compensated unity
feedback system, T(s), is

T (s) =


K ( s + 0.1)
s 4 + 13s3 + 32s 2 + (10 + K ) s + 0.1K

We will then need to factorized the denominator and
plot the poles and zeros. Since the largest s value is
4, it is hard for us to do hand calculation, we need to
use software to solve for the factors.



The coordinate of the poles when gain, K = 158.2 are
plotted in the root locus



From the root locus, the higher order poles is far
from the jω-axis compared to the dominant secondorder poles. The zero is close to a close-loop pole
which means the angle contribution of both zero and
pole will cancel each other. In conclusion, the
compensated root locus can be approximated as a 2nd
order system.



The additional poles (integration) to the system has
increase the system type from Type 0 to Type 1.
Error for step input Type 1 is zero which means the
system produce no error but still maintain the same
overshoot value.



Summary result for uncompensated and
compensated system

Parameter

Uncompensated

Compensated

Percent overshoot

57.4%

57.4%

Gain, K

164.6

158.2

Dominant secondorder poles

-0.693 + j3.926
-0.693 - j3.926

-0.678 + j3.386
-0.678 - j3.386

Higher order poles

-11.613

-11.55

estep(∞)

0.108

0

Input =1

Output = 0.892

Output = 1



Based on the previous table, we can see
that our compensated system produces
zero error and at the same time the
coordinate of the dominant second order
poles and higher order poles for the
compensated an uncompensated system is
almost the same, which means the
transient response for both compensated
and uncompensated is almost the same.
The only difference is the compensated
system has zero error.

Summary - PI


Pole at A is:

a. on the root locus without compensator;
b. not on the root locus with compensator pole added;

Summary - PI


c. Pole at A is:
approximately on
the root locus
with compensator
pole and zero added
* Find angle of zero, θzc
* Find position of zero
on the real axis

Summary - PI


Closed-loop system for Example 9.1:
a. before compensation
b. after ideal integral compensation

Summary - PI


Root locus for uncompensated system

Summary - PI


Root locus for compensated system

Summary - PI


Ideal integral compensated system response and the
uncompensated system response of Example 9.1

Improving Transient Response via


Since we have solved the problem of
improving the steady-state error without
affecting the transient response, let us
now improve the transient response itself.



The objective of improving the transient
response is to design a response that has
a desirable percent overshoot and a
shorter settling time than the
uncompensated system.



Proportional-plus-derivative (PD)
controller




We can select a transient response of a system
by choosing an appropriate closed-loop pole
location on the s-plane.
If the point is on the root locus we can just
adjust the gain in order to meet the transient
response specification.



If the closed-loop poles is not on the root
locus, then we need to reshape it so that the
compensated root locus goes through the
selected closed-loop pole location.

We need to reshape the root
locus to make it go through
point B if we want to have the
transient response
specification at point B.



We can reshape the root locus by adding
poles and zeros into the forward path. One
way to speed up the original system is to add
a single zero to the forward path.




Based on Laplace theorem, the Laplace form of
derivative is s, so adding an s is adding a derivative
to the system which is why this method is called
“proportional-plus-derivative controller”.

Adding derivative will speed up our system,
(reduce the settling time, Ts)



Example for PD controller


Assume that a root locus for a system intersect
25.38% overshoot radial line at A. The settling
time, Ts, at point A is 4.26 seconds. We want
to reduce the settling time, Ts, to 1.64 seconds
but still maintain the 25.38% overshoot.

A



Point B on the radial line has the same %OS
but the settling time, Ts = 4.26s . Since we
want Ts=1.64s but maintain the %OS, we
need to move the root locus intersection from
A to B.



We can move the root locus by using a zero as
our cascade compensator. Adding a zero to our
system will change the transient response
which is why we must select the zero that will
actually move our root locus intersection
exactly at B.



Example 9.3


Given the system of Figure below, design an
ideal derivative (PD) compensator to yield a
16% overshoot, with a threefold (3 times)
reduction in settling time.



Analysis (solving the proportional part)





The question wants us to design the value of K to
yield 16% overshoot.
Additionally the question also wants us to reduce
the settling time threefold from the original
(proportional part) transient response.
First step is to draw the root locus of the
uncompensated (proportional part) system.





Use the open loop transfer function of the unity
feedback system. Plot the poles and zeros.
Find the asymptotes.
Sketch the root locus.



Next, we need to find the value of gain, K,
where the root locus intersect with the radial
line of 16% overshoot. 16% overshoot is equal
to zeta = 0.504



We calculate the value of gain using the
cosine, sine, tangent rules.
ζ = 0.504

r

A
X

θ1

The gain we
found using
calculation is
K=43.35
(hint:
r = 2.39)

B
X

θ2

K = A× B × C

C
θζ

X

θ3



Since our system is actually a 3rd order
system, we must check whether we can
approximate it as a 2nd order system or not.
This is because the equation for settling time,
Ts, is only for 2nd order system with no zeros.



We need to get the coordinate of the poles
then we will check if the distance of higher
order poles from jω-axis is five times farther
or more than the distance of the dominant
second-order poles from jω-axis. If the higher
order poles if farther then we can
approximate our system as 2nd order system.



We can get the coordinate of the poles when
gain, K = 43.35 (16% overshoot) by substitute K
value into closed loop transfer function of the
unity feedback system, T(s)

K
T (s) = 3
2
s + 10s + 24s + K
43.35
T (s) = 3
2
s + 10 s + 24 s + 43.35
43.35
T (s) =
( s + 7.59)( s + 1.205 − j 2.064)( s + 1.205 + j 2.064)



We need to know the settling time, Ts, for our
original system.

4
4
Ts =
=
= 3.320 seconds
real 1.205


The question asks us to reduce the settling
time, Ts, by threefold (divide by 3)

3.320
Ts =
= 1.107 seconds
3


We need to modify our transient response so
that our system will have Ts = 1.107 seconds.



Solution ( solving the derivative part)
We know that the new settling time for our system is Ts
= 1.107 seconds. From the new settling time we can get
the new coordinate of our dominant second order
system when Ts = 1.107 seconds. From the settling
time equation,
4
Ts =
= 1.107
ζ = 0.504
real
4
X
real =
= 3.613
1.107



θζ
-3.613 real



Now that we know the real part of the dominant
second order poles for the new settling time, Ts =
1.107, we need to calculate the imaginary part of
the coordinate.

imaginary
tan θζ =
real
imaginary = real × tan θζ

= 3.613 × tan 59.74o
= 6.193



The coordinate of our dominant second order poles
when Ts = 1.107 are

−3.613 ± j 6.193



Our system will have Ts = 1.107 and 16% OS if
the root locus goes through −3.613 + j 6.193



We need to move the root locus and make it
go through the desired compensated dominant
pole. In order to change the transient response
we must add derivative (or zero) to our
system.



We must make sure that the added zero to our
system will produce a transient response that
goes through our desired dominant second
order pole. To get the best coordinate of the
added zero we must first find the angle of our
zero in reference to the desired dominant pole



The value of the compensated zero angle can be calculated
using the fact that the [summation of zeros angle] minus
[summation of poles angle]=odd multiple of 180̊. We must use
the sine, cosine and tangent rules to calculate the angles.
ζ = 0.504

−3.613 + j 6.193

A
X

θ1

B
X

θ2

hint:
r = 7.167)

r

C
θ3 θ

ζ

O

D
θ4

X

∑ angle zeros − ∑ angle poles = 180o



We need to find the angle value of the
compensated zero, θ3
ζ = 0.504

−3.613 + j 6.193
r

A
X


θ1

hint:
r = 7.167)

B
X

θ2

C
θ3 θ

O

D
ζ

X

θ4

We must first calculate the value of angle θ1, θ2 θ4



Value of r can be calculated using
trigonometry.

6.193

r

3.613
r=
cos θζ

θζ

3.613


Using the r value, we can calculate the angle of
the compensated zero.



We need to find all angle values and calculate
the angle for compensated zero. Substitute the
values of the poles and find the angle of the
compensated zero

θ3 − (θ1 + θ 2 + θ 4 ) = 180
θ3 = 180 + (θ1 + θ 2 + θ 4 )


o

Answer: θ3 = 95.6°

From the calculation you will find that

θ3 = 180 + (68.90 + 86.46 + 120.265)
= 455.625



Based on the root locus the maximum angle is
only 180, so the value of θ3 is 455.625-360 =



We now know the angle of the compensated
zero in reference to the desired dominant pole
(Ts = 1.107). The next step is to know the
coordinate of the compensated poles.
−3.613 + j 6.193

3.613

95.6o

3.613 − σ

σ



The real part of the compensated zero
coordinate can be calculated by
6.913
= tan 180° − 95.6°
3.613 − σ
6.913
3.613 − σ =
tan 180° − 95.6°

(

(

6.913
−σ =
− 3.613
10.199
σ = 3.006


)
)

6.913
95.6º
84.4º
3.613

3.613-σ

The imaginary part of the compensated zero
coordinate is zero.

σ



Gain K can be calculated using the equation
below
ζ = 0.174

−3.613 + j 6.193
r

A
X

θ1

The gain we
found using
calculation is
K=47.75
(hint:
r = 7.167)

B
X

θ2

A× B × D
K=
C

C
θ3 θ

O

D
ζ

X

θ4



Redraw the root locus using the compensated
zero.



The coordinate of the higher order poles and
the dominant second-order poles can be
calculated using the close loop transfer
function of the compensated unity feedback
system, T(s).

K ( s + 3.006)
T (s) = 3
s + 10 s 2 + (24 + K ) s + 3.006 K
K ( s + 3.006)
=
( s + 2.775)( s + 3.613 − j 6.913)( s + 3.613 + j 6.913)



The higher order closed loop pole is not very
close with the compensated zero. Maybe the
system cannot be approximate as a 2nd order
system. We must use simulation to determine
whether the %OS is the same between
uncompensated system and compensated
system.



Once we decide the location of the
compensating zero, we can implement the
PD controller based on figure below



The transfer function of the controller is

K1 
Gc ( s ) = K 2 s + K1 = K 2  s +
÷
K2 


Improving Steady-State error and
Transient response


We can improve both steady-state error and
transient response for a system by combining the
PI controller and PD controller.



We can either improve the steady-state error first
then improve the transient response or improve
the transient response then improve the steady
state error.



PID controller that we are going to learn
improves the transient response then improves
the steady-state error.

Transient response


A PID controller is shown in figure below.



Its transfer function is

 2 K1 K 2 
K2
K2
Gc ( s ) = K1 +
+ K 3 s = K1s +
+ K3s = K3  s +
+
÷
s
s
K3 K3 


Transient response


The design technique for PID controller
consists of the following steps:
1.
2.
3.
4.

Evaluate the performance of the
uncompensated system (P-controller system).
Design the PD controller to meet the transient
response specification.
Design the PI controller to yield the required
steady-state error.
Determine the gains, K1, K2, and K3.

K1 
Gc ( s ) = K 2 s + K1 = K 2  s +
÷
K2 


Transient response


Example 9.5


Given the system below, design a PID
controller so that the system can operate with
a peak time that is two-thirds that of the
uncompensated system at 20% overshoot and
with zero steady-state error for a step input.

Transient response


Based on the PID design step, the first step is
to evaluate the performance of our system.
Sketch the root locus.
 Draw the radial line for 20% overshoot.
 Calculate the gain, K, for the system to produce 20%
overshoot and find the coordinate of the poles and
zeros during that gain.
 Determine whether we can approximate our system
as 2nd order system or not.
 Calculate the settling time, T , and the peak time, T
s
p


Ts =

4
real

π
Tp =
imaginary

Transient response


The root locus for
the
uncompensated
system
(p-controller)

Transient response


Characteristic of the uncompensated system.

Transient response


Second step is to improve in term of transient
response. We need to use PD controller. The
question requires us to reduce the peak time
to two-thirds of that of the uncompensated
system.


The peak time, Tp, for the uncompensated system is
0.297. We need to reduce it by 2/3.

2
new peak time, Tp = × 0.297
3
= 0.198

Transient response


We need to find the intersection point of the new
peak time Tp on the 20% overshoot radial line.

π
imaginary
π
π
imaginary =
=
= 15.867
Tp 0.198

Tp =

imaginary
real
imaginary
15.87
real =
=
= 8.131
tan θζ
tan(62.871)
tan θζ =

ζ = 0.456

X

intersection coordinate = - 8.13 + j15.87

θζ
real

Transient response


In order to change the transient response so that the
root locus will goes through the intersection point of
the new peak time, we must add a compensated zero
into our system.



To get the coordinate of our compensated zero we
must first get the angle of the zero in reference to
the intersection point of the new peak time.

Transient response
ζ = 0.456

−8.13 + j15.87
r

A
X

θ1

O

θ2

hint:
r = 17.831

C D

B
X

θ3

O

E
θ4

θ5 θ

X

ζ

∑ angle zeros − ∑ angle poles = (2k − 1)180o
(θ 2 + θ 4 ) − (θ1 + θ3 + θ5 ) = (2k − 1)180o

Transient response



The angle of the compensated zero is 18.37.
̊
Now that we know the angle of the
compensated zero we will then calculate the
coordinate of the compensated zero.
−8.13 + j15.87

18.37o

8.13

zc

15.87
tan18.37 =
zc − 8.13
o

15.87
zc − 8.13 =
tan18.37o
15.87
zc =
+ 8.13
o
tan18.37
= 55.92
the coordinate of our compensated zero is at
(-55.92,0)

Transient response


Sketch the root locus with the compensated
zero.

Transient response


After we design the PD controller, we design
the PI controller to reduce the steady-state
error to zero for a step input. When put an
integral (compensated pole) into our system
we must also put a zero close to the origin.



Based on the previous root locus for PD
controller, we can reduce the steady state
error to zero by putting a compensated pole
(at origin) and a zero (at 0.5).

Transient response


Next step is to sketch the root locus for the PI controller
and find the value of K for the intersection between the
root locus and the radial line for 20% overshoot.

This root locus
is also called
root locus for
PID controller
since we
combine PD
then PI
controller.

Transient response


The summary of the results

Control chap8

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