7/28/2019 Design of Compensators Using Root Locus Diagram
1/20
3/9/2013 1
AUTOMATIC CONTROL
THIRD POWER
Prepared by
Dr. Helmy El-Zoghby
2013
3/9/2013 2
Design of compensators
using root locus diagram
7/28/2019 Design of Compensators Using Root Locus Diagram
2/20
3/9/2013 3
A compensator (filter or controller) is a circuit whichadded to the control system to improve the system
performance.
Different types of controllers1. Two Position Controller2. Proportional Controller
3. Integral Controller4. Differential Controller
5. PI Controller6. PD Controller7. PID-Controller
8.Lead compensator9.Lag compensator
10.Lag-lead compensator11.Intelligent compensators (fuzzy , neural , genetic , .)
3/9/2013 4
Benefits of compensators
Lead compensator (as PD-controller) is used to:
- improve transient response
- improve stability
Lag compensator ( as PI-controller) is used to:- reduce or eliminate steady-state error
Lag- Lead compensator ( as PID-controller)
- used to improve all issues.
7/28/2019 Design of Compensators Using Root Locus Diagram
3/20
3/9/2013 5
Note:
For a closed loop system:
By placing the closed-loop poles at the desired location
-if the new closed-loop poles lies on the old root locus , then k
is tuned or using P-controller (amplifier).
-if the new closed-loop poles not lies on the old root locus ,then reshape the root locus by adding poles/zeros
(compensator) to system.
3/9/2013 6
Effect of adding poles on control system response
7/28/2019 Design of Compensators Using Root Locus Diagram
4/20
3/9/2013 7
Effect of adding zeros on control system response
3/9/2013 8
Lag and lead compensators concept
)(s
)z(sK)(G ccc
cps
+
+=
The controller transfer function
7/28/2019 Design of Compensators Using Root Locus Diagram
5/20
3/9/2013 9
Lead compensator
3/9/2013 10
Lag compensator
7/28/2019 Design of Compensators Using Root Locus Diagram
6/20
3/9/2013 11
Electronic circuit of lead or lag compensator
)(s
)z(s
)1
(
)1
(
.)(
)()(G c
22
11
23
14c
cc
i
o
ps
s
s
ss K
CR
CRCRCR
vv
+
+=
+
+
==
For lead R1C1> R2C2
For lag R1C1< R2C2
3/9/2013 12
Lead compensator design example
Radar tracking system
7/28/2019 Design of Compensators Using Root Locus Diagram
7/20
3/9/2013 13
draw the uncompensated system root locus and find the damping ratio and
natural frequency at k=4
design a suitable compensator using three different methods to rise the natural
frequency to 4 rad/sec at the same damping ratio
draw the compensated root locus
draw the designed controller electronic circuit
check the steady state error after adding the controller
draw the transient response before and after compensation.
3/9/2013 14
solution
For the uncompensated system
5.0)cos(
60,2then
4ppK
0
n
22
==
==
==
and
7/28/2019 Design of Compensators Using Root Locus Diagram
8/20
3/9/2013 15
030210180
)90120(0180
polesallofzerosallof180
==
++=
+= o
Angle of deficiency is:
3/9/2013 16
1-Design of compensator using minimum steady-
state error method
7/28/2019 Design of Compensators Using Root Locus Diagram
9/20
3/9/2013 17
Selection of the controller pole and zero
3/9/2013 18
The compensated system T.F is :
4.68*4/ppK c21c == czp
7/28/2019 Design of Compensators Using Root Locus Diagram
10/20
3/9/2013 19
Steady-state error before and after compensation at
unit ramp input
%202.01
02.5)2(
4
)4.5(
)9.2(68.4.lim
)()()(limk
kv
0
0
new
v
===
=++
+=
=
=
=
newnew
ss
s
sc
e
sss
ss
sHsGssG
%505.01
2)2(
4.lim)()(limk
kv
00v
===
=+
====
ess
ss ssssHssG
Before compensation
After compensation
3/9/2013 20
Note : If the input is unit step :
%01
)2(
4lim
k1
k
p
0p
==
=+
=
+
=
e ss
sss
%01
)2(4
)4.5()9.2(68.4lim
k1
k
p
0newp
==
=++
+=
+
=
new
new
ss
s
e
ssss
Before compensation
After compensation
7/28/2019 Design of Compensators Using Root Locus Diagram
11/20
3/9/2013 21
Root locus before and after compensation
3/9/2013 22
Compensator electronic circuit
)1
(
)1
(
.)4.5(
)9.2(68.4)(G
22
11
23
14c
CR
CRCRCR
s
s
s
ss
+
+
=+
+=
=
===
ktake
uftake
R
CCCR CR
10
10,68.4
3
21
23
14
=== kkkthen RRR 5.18,5.34,8.46 214
7/28/2019 Design of Compensators Using Root Locus Diagram
12/20
3/9/2013 23
2-Design of compensator using zero-pole cancellation
method
)(s
)z(s)(G cc
c
c ps K
+
+=
Since the Angle of deficiency =300
and
-Select the controller zero equal to the nearest pole
of the system to the desired point to cancel thispole
-Find the controller pole from angle condition
-Find kc from magnitude condition
3/9/2013 24
Take Zc=2 and from angle condition Pc=4
From magnitude condition
44/4*44/ppK c1c ===
)4(s
)2(s4
)(s
)z(s)(G cc
+
+=
+
+=
c
c ps K
%505.01
,2)2(
4.limk
kv0v
====+
== esssss
s
%2525.01
,4)2(
4
)4(
)2(4.limk
kv0
new
v ====++
+=
= new
new
sss essss
s
xxx o
p
30
jw
Scale 1:2
-4 -1 0-2
Steady-state error before design
The controller T.F become :
Steady-state error after design
7/28/2019 Design of Compensators Using Root Locus Diagram
13/20
3/9/2013 25
Design of compensator using PD-controller
)()(. zkkk
kkk cdd
p
ddpsssFT +=+=+=
RR
k
k
p
dRC
1
2=
=
PD-controller block diagramPD-controller electronic circuit
PD-controller transfer function
3/9/2013 26
Since the Angle of deficiency =300
-the angle of controller zero must equal to 300
-draw a line of 300 with real axis from point (p)
-Find the controller zero from angle condition
-Find kd
from magnitude condition
)()(. zkkk
kkk cdd
p
ddpsssFT +=+=+=
7/28/2019 Design of Compensators Using Root Locus Diagram
14/20
3/9/2013 27
xxo
p
30
jw
Scale 1:2
Zc=7.8-1 0-2
Compensated and
uncompensated root locus
3/9/2013 28
==
==
==
====
kthenktakeR RR
R
6.41
6.4then
k60Rthen10ufCtake
C*R0.66.3*3.5/4*4z4/ppK
21
1
2
p
c21d
k
From angle condition
8.7==kk
zd
p
c
From magnitude conditionPD-controller electronic circuit
7/28/2019 Design of Compensators Using Root Locus Diagram
15/20
3/9/2013 29
%505.01
,2)2(
4.limk
kv0v
====+
== esssss
s
%9.10109.01
,2.9)2(
4).6.06.4.(limk
kv0
new
v ====+
+== new
new
sss essss
Steady-state error before design
The controller T.F become :
Steady-state error after design
ssFT kk dp 6.06.4. +=+=
3/9/2013 30
Using zero-pole cancellation method try to solve this problem
and simulate by matlab
A-If it required to design a lead
controller to get overshoot=2%
and peak time equal 1 sec
Check these values by calculations
Lead angle =96 degree , wd=3.14 ,
wn=5, zeta=.779 ,beta= 38.
0 1 2 3 4 5 6 7 8 9 1 00
0.2
0.4
0.6
0.8
1
1.2
1.4
Time
res
p
o
n
c
e
7/28/2019 Design of Compensators Using Root Locus Diagram
16/20
3/9/2013 31
B-at 1 sec peak time and 10% overshoot
Check these values by calculationszeta=.57 , wd=3.14 , wn=3.89 , lead angle
=35, beta=55
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Time
response
32
Proportional-Integral controller (lag compensator)
Time output equation is :
+= dttetetp kk ip )()()(
where Kp is the proportional constant and Ki is the integral constant.
The block diagram of the PI controller
The PI - controller electronic
circuit.
7/28/2019 Design of Compensators Using Root Locus Diagram
17/20
3/9/2013 33
Design of lag compensator (PI-controller)
The PI-controller transfer function is :
( )pz
kkk
k
kkkk
c
c
c
p
i
p
ipi
p
s
s
s
s
s
s
sFT
+
+=
+
=
+
=+=.
The pole of the controller at s=0
k
kp
is =The zero of the controller at
pczc
x
p
3/9/2013 34
PI-controller design steps
The zero of the controller at s=-ki/kp is obtained from angle condition
For the following open loop transfer function design a controller to satisfy 0.8
damping ratio , natural frequency of 5 rad/sec , and elimination of steady-state
error
The controller gain constant kp is obtained from magnitude condition
Example
The pole of the controller at s=0
)7)(3(
1)()(
++=
sssHsG
7/28/2019 Design of Compensators Using Root Locus Diagram
18/20
3/9/2013 35
011
n37)8.0(cos)(cos,5 ====
The steady-state error is zero , PI controller is used
( )pz
kkk
k
kkkk
c
c
c
p
i
p
ipi
p
s
s
s
s
s
s
sFT
+
+=
+
=
+
=+=.
27)45108(0180
polesallofzerosallof180
=++=
+= o
Angle of deficiency is:
Lag compensator is required
3/9/2013 36
From angle condition
From magnitude condition
50*5.2
then
20*1/ppK
ki
c21c
==
===
k
K
p
cpzp
( )s
s
s
s
skk
kGp
i
pc
5.220)(
+=
+
=
5.2==kkZ p
ic
The controller transfer function is
7/28/2019 Design of Compensators Using Root Locus Diagram
19/20
3/9/2013 37
PI-controller electronic circuit
kthen
k
take
R
R
RRk p
20
1
20
2
1
1
2
=
=
==
kthen
ufC
take
R
RCk i
2
10
501
=
=
==
3/9/2013 38
XX XO
p
Pc=0Zc=-2.5-3-7
27
45 108
Scale 1:1
Re
Jw
37
Wn=5
7/28/2019 Design of Compensators Using Root Locus Diagram
20/20
3/9/2013 39
Thank You&
Any Questions?