INC341
Design Using Graphical Tool
(continue)
Lecture 10
INC 341
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Improving Both Steady-State Error
and Transient Response
• PI, Lag improve steady-state error
• PD, Lead improve transient response
• PID, Lead-lag improve both
(PID = Proportional plus Intergal plus
Derivative controller)
INC 341
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PID Controller
K1s  K 2  K 3 s 2
K2
Gc ( s)  K1 
 K3s 

s
s
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K 3 (s 2 
K1
K
s 2)
K3
K3
s
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PID controller design
1. Evaluate the performance of the
uncompensated system
2. Design PD controller to meet transient
response specifications
3. Simulate and Test, redesign if necessary
4. Design PI controller to get required steadystate error
5. Find K constant of PID
6. Simulate and Test, redesign if necessary
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Example
Design PID controller so that the system can operate
with a peak time that is 2/3 of uncompensated system,
at 20% OS, and steady-state error of 0 for a step input
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Step 1
• %OS = 20%  damping ratio = 0.456
 Ѳ = 62.87
• Search along ther line to find a point of
180 degree (-5.415±j10.57)
• Find a correspoding K=121.51
• Then find the peak time
Tp 
INC 341



 0.297sec
 d 10.57
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Step 2
• Decrease peak time by a factor of 2/3  get
imaginary point of a compensator pole:
d 

Tp


(2 / 3)(0.297)
 15.867
• To keep a damping ratio constant, real part of the
pole will be at

d

tan(62.87 )
 8.13
• The compensator poles will be at -8.13±j15.867
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• Sum of the angles from uncompensated poles
and zeros to the test point (-8.13±j15.867) is
-198.37
• The contribution angle for the compensator zero
is then 180-198.371 = 18.37
15.87
 tan(18.37 )
zc  8.13
zc  55.92
ดPD controller is (s+55.92)
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Step 3
• Simulate the PD compensated system to
see if it reduces peak time and improves
ss error
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Step 4
• design PI compensator (one pole at origin
and a zero near origin; at -0.5 in this
example)
s  0 .5
GPI ( s ) 
s
• Find a new point along the 0.456 damping
ratio line (-7.516±j14.67), with an associate
gain of 4.6
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INC 341
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Step 5
• Evaluate K1, K2, K3 of PID controller
K ( s  55.92)( s  0.5)
s
4.6( s  55.92)( s  0.5)

s
G PID ( s ) 
4.6( s 2  56.42s  27.96)

s
• Compare to
K1 s  K 2  K 3 s 2
Gc ( s ) 
s
K1 = 259.5, K2 = 128.6, K3 = 4.6
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Step 6
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Lead-Lag Compensator Design
Same procedures as in designing PID:
– Begin with designing lead compensator to get
the desired transient response
– design lag compensator to improve steadystate error
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Example
Design lead-lag compensator so that the system can
operate with 20% OS, twofold reduction in settling time, and
tenfold improvement in steady-state error for a ramp input
INC 341
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Step 1
• %OS = 20%  damping ratio = 0.456
 Ѳ = 62.87
• Search along ther line to find a point of
180 degree (-1.794±j3.501)
• Find a correspoding K=192.1
• Then find the settling time
Ts 
INC 341
4


4
 2.230sec
1.794
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Step 2
• Decrease settling time by a factor of 2  get a
real part of a compensator pole:

4
4

 3.588
Ts (1 / 2)(2.230)
• To keep a damping ratio constant, imaginary
part of the pole will be at
d  3.588tan(62.87 )  7.003
• The compensator poles will be at -3.588±j7.003
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• Select the compensator zero at -6 to coincide
with the open-loop pole
• Sum of the angles from uncompensated poles
and zeros to the test point (-3.588±j7.003) is
-164.65
• The contribution angle for the compensator zero
is then 180-164.65 = 15.35
7.003

pc  3.588
 tan(15.35 )
pc  29.1
Lead compensator is
( s  6)
( s  29.1)
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Then find a new K at the design point (K=1977)
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Step 3
Simulate the lead compensated system
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Step 4
• Originally the uncompensated system has
the transfer function:
K
G (s) 
, K  192.1
s ( s  6)( s  10)
INC 341
192.1
Kv 
 3.201
6 10
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• After adding the lead compensator, the system
has changed to
G LC ( s ) 
1977
s ( s  10)( s  29.1)
• Static error constant, Kv, is then 6.794 (lead
compensator has improved ss error by a factor
of 6.794/3.201=2.122)
• So the lag compensator must be designed to
improve ss error by a factor of 10/2.122=4.713
Glag ( s ) 
INC 341
(s  zc ) zc
,
 4.713
( s  pc ) pc
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Step 5
• Pick a pole at 0.01, then the associated zero will
be at 0.04713
Glag ( s ) 
( s  0.04713)
( s  0.01)
• Lag-lead compensator
Glead lag ( s ) 
( s  6)( s  0.04713)
( s  29.101)( s  0.01)
• Lag-lead compensatated open loop system
G LLC ( s ) 
INC 341
K ( s  0.04713)
s ( s  10)(s  29.101)( s  0.01)
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Step 6
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Conclusions
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INC 341
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Feedback Compensation
Put a compensator in the feedback path
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Tachometer
Popular feedback compensator, rate sensor
Tachometer generates a voltage output proportional
to input rational speed
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rate feedback
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Example
Design a feedback
compensator
to decrease settling time
by a factor of 4 and
keep a constant %OS of
20
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Step 1
%OS = 20%  damping ratio = 0.456  Ѳ = 62.87
Search along the daping ratio line to get a summation
of angle of 180 degrees at -1.809±j3.531
Find the corresponding K from the magnitude rule
K  1.8092  3.5312 3.1912  3.5312 13.1912  3.5312
 257.841
settling time
INC 341
4
4
Ts  
 2.21 sec
 1.809
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Step 2
Reduce the Settling time by a factor of 4
4
4
 
 7.236
Ts (1 / 4)(2.21)
d  7.236tan(62.87 )  14.123
A new location of poles is at -7.236±j14.123
INC 341
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At dominant pole -7.236±j14.123, KG(s)H(s) has a net
angle of = -277.33  needs an additional angle from zero
of 277.33-180 = 97.33
14.123
 tan(180  97.33 )
7.236 zc
zc  5.42
Find the corresponding K to the pole at -7.236+j14.123
using the magnitude rule:
K ( s  5.42)
s( s  5)(s  15)
K = 256.819
INC 341
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K1 K f  256.819
1
 5.42
Kf
K f  0.185
K1  1388.211
Feedback block is 0.185(s+5.42)
INC 341
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Physical System Realization
PI Compensator
R2

1 
s R C 
R2 
2

C ( s)  
R1
s
C
R1
Vi(s)
INC 341
Vo(s)
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Lag Compensator
R1
Vi(s)
Vo(s)
R2
C
INC 341
1
s
R2
R2C
Gc ( s ) 
1
R1  R2 s 
( R1  R2 )C
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PD Compensator
R2
C
Vi(s)
R1
INC 341
Vo(s)

1 
Gc ( s)   R2C  s 

R
C
1


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Lead Compensator
R1
Vi(s)
Vo(s)
C
R2
1
s
R1C
Gc ( s ) 
1
1
s

R1C R2C
s  zc

, zc  pc
s  pc
INC 341
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PID Compensator
R2
C2
R1
Vi(s)
Vo(s)
C1
1 

 R C 

R
C
Gc ( s)    2  1   R2C1s  1 2 
s 
 R1 C2 


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