Chapter 7 System Compensation

(Linear Control System Design)

Instructor: Dr. Cailian Chen
 Outline

 Introduction to compensation design

 Phase Lead Compensation
 Phase Lag Compensation
 Phase Lead-lag Compensation
 PID Control

20/12/10 2
Question: What is system compensation?
Given the control plant, the procedure of controller
design to satisfy the requirement is called system
compensation.
Question: Why to compensate?
The closed-loop system has the function of self-tunning.
By selecting a particular value of the gain K, some single
performance requirement may be met.
Is it possible to meet more than one performance
requirement?
Sometimes, it is not possible.
Something new has to be done to the system in order to
make it perform as required.
1.Control system design and compensation
 Design ： Need to design the whole controller to
satisfy the system requirement.
 Compensation ： Only need to design part of the
controller with known structure.
2. Three elements for compensation
Original part of the system
Performance requirement
Compensation device
 7.1 Introduction to Compensation Design

7.1.1 Performance Requirement

1. Time domain criteria （ step response ）

 Overshoot, settling time, rising time, steady-state error
2. Frequency domain criteria
 Open-loop frequency domain criteria ：

crossover frequency, phase margin, gain margin

 Closed-loop frequency domain criteria ：

maximum value Mr , resonant frequency, bandwidth

 Frequency domain and time domain criteria
1
Resonant peak Mr 
2 1   2

Resonant frequency  r   n 1  2 2
Bandwidth b   n 1  2 2  (1  2 2 ) 2  1

Gain crossover frequency c  n 4 4  1  2 2

2
Phase margin   arctg
4 4  1  2 2


Percentage overshoot 1 2
%  e  100%
3 4
Settling time tS  ,
n n
 7.1.2 Structure of Compensator
 According to the way of compensation, the
compensator can be classified into following categories:
 + + C(s)
Original Part
- -

Compensator

(b) Feedback compensation

R(s) + Compensator + C(s)

1 Original Part
- -
Compensator
2

(c)Cascade and feedback compensation

 Compensator

+
R(s) + + C(s)
Original Part
-

(d) Feed-forward compensation

N(s)
Compensator

+ +
R(s) + + Original C(s)
Controller Part
+
-

(e) Disturbance compensation

Remark ：
 Cascade compensation and feedback
compensation are inside the feedback
loop.
 Feed-forward compensation and
disturbance compensation are outside
the feedback loop.
 7.1.3 Methods for Compensator Design
1. Frequency Response Based Method
Main idea ： By inserting the compensator, the Bode
diagram of the original system is altered to achieve
performance requirements.
Original open-loop Bode diagram ＋ Bode diagram
of compensator ＋ alteration of gain
＝ open-loop Bode diagram with compensation
2. Root Locus Based Method
Main idea: Inserting the compensator introduces new
open-loop zeros and poles to change the closed-loop root
locus to satisfy the requirement.
 7.1.2 Cascade Compensation
 Frequency response based compensation
Phase lead compensation
Phase lag compensation
Phase lead-lag compensation
 Fundmantal rule for control design ： PID control
Each requirement relates to a different region of the
frequency axis in the Bode diagram.
1.The steady-state error relates to the magnitude at low
frequency.
2.The transient response requirement relates to the gain
crossover frequency, which usually occurs at higher
frequencies.
Three design rules for cascade compensator:
1. The system is stable with satisfactory steady-state error, but
dynamic performance is not good enough.
Compensator is used to change medium and high
frequency parts to change crossover frequency and phase
margin.
2. The system is stable with satisfactory transient
performance, but the steady-state error is large.
Compensator is used to increase gain and change lower
frequency part, but keep medium and higher frequency
parts unchanged.
3. If the steady-state and transient performance are either
unsatisfactory, the compensator should be able to increase
gain of the lower frequency part and change the medium
and higher frequency parts.
 Change of Bode Diagram

L  ω L  ω L  ω

  

a)change of lower b)change of b)change of lower,

frequency part medium and higher medium and higher
frequency parts frequency parts
 7.2 Phase Lead Compensation
R1

C
1. Transfer function Ei Eo
R2
： E  s  1 1   Ts
Gc  s   o
 
Ei  s   1  Ts Passive Phase Lead Network

where

R1  R2 R1R2
  1, T  C
R2 R1  R2 1 1
 
T T
 Multiplying the transfer function by α
Lc ( )  20 lg ( T  ) 2  1  20 log (T  ) 2  1
(  1)T 
c ( )  arctg T   arctgT   arctg
1   T 2 2
dc ( ) 1  1  1
 0 , m  , m  arctg  arcsin
d T  2   1
L( )
Lc (m )  10 log  20 log a
20dB / dec 20 log a
Determination of  
 ( )
1  sin  m

1  sin m m
0    20 
Geometry mean 1 / aT
m 1/ T
2 、 Effect of phase lead compensation
L , 

40  20  20
20   40
1000 
0 142
10 23.8
100   40
2
  40
1

 
2

3

4 2
1


Example 7.1: Given a unity-feedback system with
the following open-loop transfer function
4K
G ( s) 
s ( s  2)
Please design phase lead element to satisfy the following
three requirements:
1. steady speed error constant K v  20s 1

2. phase margin   50

3. gain margin GM  10dB

 40

20 Uncompensated
0
system
-20
20
-40 G ( s) 
-60
10
0 1
10
2
10
s (0.5s  1)
-100

-120 c 0  6.17rad/s
-140   17
-160

-180
0 1 2
10 10 10

 m  50  17   33 ? Extra margin:

5o~10o
 m  50  17  5  38
   
   15
Remark: K  100
The phase lead compensator not only takes extra phase
40
margin of 33o at the uncompensated gain crossover
frequency.
However, we also have to add the magnitude part of the
compensator to the uncompensated magnitude, and
the gain crossover frequency moves to a higher value.
The ultimate phase margin is less than 33o .
It implies we need to consider extra phase margin for
determine α.
 1  sin  m 1  sin 38
   4 .2
1  sin  m 1  sin 38

1
10 lg   6.2dB m  9rad/s   c1
T
40

1 m 20

1    4.4rad/s
T 
0

1
-20

2   m   18.4rad/s -40

T -60
0 1 2
10 10 10

s  4.4 1  0.227 s
Gc  s    0.238  50

s  18.4 1  0.054s 0

-50

Complete Compensator -100

1  0.227 s
Gc  s   -150

1  0.054s -200
0 1 2
10 10 10
Comments on phase lead compensation
1 、 The slop around the gain crossover frequency is
increased. It improves the relative stability.
2 、 The closed-loop resonance peak is reduced. Also, the
overshoot is reduced.
3 、 Increase the open-loop phase margin.
4 、 The open-loop (and usually the closed-loop)
bandwidths are increased. It is beneficial for fast
response. But it may cause problems if noise exists at
the higher and unattenuated frequencies.
5 、 Take no effect on the steady-state performance.
Rules to design phase lead compensation
(1) Determine K to satisfy steady-state error
constraint
(2) Determine the uncompensated phase margin γ0
(3) estimate the phase margin  m in order to satisfy the
transient response performance constraint

(4) Determine 

(5) Calculate ωm
(6) Determine T
(7) Confirmation
 Constraints for application of lead compensation ：
 Constraint 1: The system is stable.
If it is unstable, the phase need to compensate is
too big. The noise takes severe effects on the system.
 Constraint 2: The phase cannot reduce very fast
around the gain crossover frequency.
The phase lead compensation can only provide less
than 60o extra phase margin.

20/12/10 24
 6.3 Phase Lag Compensation
R1
Transfer function:
R2
1
R2  Ei
Gc ( s )  CS Eo
C
1
R1  ( R2  )
CS
R2Cs  1 Passive phase lag network

( R1  R2 )Cs  1
 Ts  1

Ts  1
R2
T  ( R1  R2 )C ,   1
R1  R2
 L( )
1  20dB / dec
m 
T  
10 lg 

Maximum phase lag  ( ) 20 lg 

1   m 1 / T
m  arcsin 1/ T

1 
 m

The compensator has no filtering effect on the low

frequency signal, but filters high frequency noise.
The smaller β is ， the lower the noise frequency
where the noise can pass.
Comments on phase lag compensation:
1 、 Phase lag compensator is a low-pass filter. It changes the
low-frequency part to reduce gain crossover frequency. The
phase is of no consequence around the gain crossover
frequency.
2 、 Be able to amplify the magnitude of low-frequency part,
and thus reduce the steady-sate error.
3 、 The slope around gain crossover freuency is -20dB/dec.
Resonance peak is reduced, and the system is more stable.
4 、 Reduce the gain crossover frequency, and then reduce
the bandwidth. The rising time is increased. The system
response slows down.
 Rules for phase lag compensator design:
1 、 Determine K to satisfy the steady-state error ；
2 、 Draw bode diagram for umcompensated system
3 、 Find the frequency ωc1 from the uncompensated open-
loop Bode diagram where the phase margin satisfy the
performance requirement ；
4 、 Calculate the magnitude at the frequency ωc1 .
Determine β by  L c1  /20
a   10
 L ( ωc' ) 20
a  10
5 、 Determine the two break frequencies.
1 1
z   c1  p  1
 T 10 T
6. Comfirmation
Avoid introduce the maximum phase lag around the 1
gain crossover frequency c1 . So the break frequency T
is much less than c1
1 1
 c1
 T 10
The phase lag at ωc1 is
(   1)T c1
c (c1 )  arctg  T c1  arctgT c1 
1   (T c1 ) 2
10
Substitude c1T  
10
(   1)
 10(   1)
 (c1 )  arctg  arctg  arctg[0.1(   1)]
10 2
1  ( ) 100  

0.01 0.1 10

0

dB  0

 (c1 )

-20 5
-5

20 lg 

-40
-2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 010
-10

  c1  and 20 lg  versus 

 Example 6.2: Given a unity-feedback controller with
the following open-loop transfer function
K
G(s) 
s (0.1s  1)(0.2 s  1)

Please design a cascade compensator to satisfy the

following requirements ：

K v  30   40 o GM  10(dB )
c1  2.3(rad / s )

Notice: If the compensated system cannot satisfy the

requirements, we need to further alter crossover
frequency and phase of the lag compensator.  ( )
c1
Solution:
(1) K v  lim sG  s   K  30
s 0

(2) Draw the uncompensated Bode diagram

 0  38.3o

The system is unstable. The needed extra phase

compensation is 83.3o . The phase lead compensation
can not provide so big phase margin.
Two ways:
(a)Apply two cascade phase lead compensation

(b)Apply phase lag compensation

(3) For ω=2.5 rad/s, γ=45o Set ωc1=2.5

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(4) L(ωc1)=22dB

 L c1  / 20
  10  0.08

(5) 1 1
 c1  0.25
 T 10
1
 p   0.25  0.08  0.02rad/s
T
(6) 4s  1
Gc  s  
50 s  1
Confirmation:   c1   44 ,  g  6.7 rad / s, GM  10dB
o

20/12/10 33
Applicable for the following systems ：
（ 1 ） The transient performance is satisfactory, but the steady-
state performance is desired to be improved.
（ 2 ） High requirement for noise attenuation.
Drawback ： The system response is slow down.
 Comparison of phase lead and lag compensation
Phase lead compensation Phase lag compensation
Main Improve transient performance by using Improve the steady-state performance by
Idea phase lead characteristics using magnitude attenuation at the high-
frequency part
(1) Around ωc, the absolute value of slope is (1) Keep relative stability unchanged, but
reduced. Phase margin γ and gain margin reduce the steady-state error.
GM are increased. (2) Reduce ωc and then closed-loop
Effect (2) Increase the bandwidth bandwidth
(3) With bigger γ, overshoot is reduced. (3) For specific open-loop gain, γ, GM and
(4) Take no effect on the steady-state resonant peak Mr are all improved due to
performance. magnitude attenuation around ωc
(1) Broad bandwidth reduces the filtering Narrow Bandwidth increase the response
Weak for noise. time.
ness (2) For passive network implementation,
need an extra amplifier.
(1) Extra phase lead compensation is less (1) The phase lag of the uncompensated
than 550. system is fast around ωc.
(2) Require broad bandwidth and fast (2) Bandwidth and transient response are
Applic
ation
response satisfactory.
(3) No matter the noise at high-frequency (3) Require attenuation of noise
part. (4) The phase margin can be satisfied at
the low frequency.