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COMPENSATORS
We discussed the characteristics of proportional, integral and derivative controllers, and used the
root locus method to design these controllers. We saw that the proportional controller affects the
steady-state error as well as the transient response of a closed-loop system. The function of the
integral controller is mainly to improve the steady-state errors. The derivative control has the effect
of stabilizing the closed-loop system. Now we discuss similar controllers, commonly referred as
compensators, from the point of view of their frequency response. We found that a change in the
gain K of the proportional controller causes a change in magnitude (bode) plot not the phase plot.
The phase is unaffected by the gain K. We also designed a proportional controller using the
frequency response such that it satisfies the phase margin of the open-loop system. We now
discuss the characteristics of compensators which affect the phase angle of the open-loop system.
Phase Lead Compensator( or just lead compensator)
Recall that in root locus design method, placing a zero in the left half of s-plane has the effect of
pulling the root locus to the left, thereby increasing the damping ratio. Let us investigate the effect
of a proportional-derivative controller in terms of its frequency response. For convenience let
1)( z
ssGc . The bode plots of the PD controller are shown.
Bode plots of a PD controller
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Inclusion of a proportional-derivative (PD) controller to the system adds a positive phase to the
phase response of the uncompensated system. Thus, when a system is unstable or its phase
margin is very small, a PD controller can be used to boost the phase margin, which in turn
increases the damping ratio of the closed-loop system.
A proportional controller can also be used to improve the phase margin, but using a PD controller
provides the flexibility to adjust both phase margin andc
, the gain cross-over frequency. Recall
thatc
is related to n , thereby the rise, peak and settling times, as well.
A disadvantage of the PD controller is that its gain increases with frequency. Thus, it tends to
enhance high frequency noise. To counteract this effect of a PD controller, we use a controller
given by the transfer function
1
1
,)(
ps
zs
p
zK
zpps
zsKsG
c
cc
The bode plots of a lead compensator with ,1p
zKc are as shown below.
The phase angle is always positive. This why it is called a phase lead compensator. We use it
generally to improve the dynamic response of the system. The phase is maximum for a frequency
m which is the geometric mean ofz andp. That is zpm .
To find an expression for the phase angle of the lead compensator we write
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2
2
2/1
)//()/(1
/1
)/1)(/1(
/1
/1)(
p
pzjzp
p
pjzj
pj
zjKjGc
We express
)zp/(1
p/z/tan
2
1 . The phase is maximum when zpm . Therefore,
pz
pzzp
zp
pz
m
mm
2
1tan
2
//tan
)/(1
//tan
1
1
2
1max
where
We can thus write
1
1)sin( max
The gain |)j(G| mc in dB is )log(10|/|log20)(log(20 zzpjG m
Design of Phase-Lead Compensator Using The Bode Diagram
1. Choose Kto satisfy the steady-state error requirement
2. Evaluate the phase margin ofKG(s)
3. Calculate the amount of phase margin to add,
factor)fudge(5system)teduncompensatheofPM-PMdesired( omax
4. Calculate)sin(1
)sin(1
max
max
,
p
z
5. Evaluate )log(10 . Determine the frequency where )log(10 occurs on the bode plot of
KG(s) . This is the newc
.
6. Calculate .pz,p max
and
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EXAMPLE : Consider the closed-loop system with)2(
4)(
sssG . Design a lead compensator
such that the velocity error constant 20vK , and the phase margin is at least 50o. The gain
margin should be at least 10 dB.
Solution: In order for 20)()(lim
sGssGK cos
v , we should have 202)0(lim
cosG
Thus, cc
K)0(G .10K
This means, when K=10, the compensated system will satisfy the steady-state error condition..
Draw the Bode plots of the uncompensated system. We find that the uncompensated system
phase margin is about 18o .
.
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The lead compensator must provide about 32+5 = 37o so that the compensated system phase
margin is about 50o. That is .37maxo
We find as
2152.0)38sin(1
)38sin(1)sin(1)sin(1
max
max
o
o
Now we find dB68.6)log(10
We find that -6.68 dB occurs at rad/s9 on the uncompensated bode plot. Let this be the new
gain cross-over frequency. We calculate, .18.4,4.19max pzp and
46.46
KKc
The lead controller is given by4.19
18.446.46)(
s
ssGc
Check: Find the Bode plot of the compensated system
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The step response plot shown reveals that the PO is about 20%, which is consistent with a
damping ratio of about 0.5.
EXAMPLE : Let .1
)(2s
sG Design a lead compensator such that the closed-loop system has a
damping ratio of about 0.45, and the settling time in step-response is about 4 second.
Solution: Desired phase margin is o45 . Since the settling time is about 4 sec, 1n . Using the
formula
2
)tan( PMcn , the new 0-dB frequency should be 2
)45tan(
2oc
The uncompensated system with two poles at s =0, is marginally stable ( its phase margin is zero).
Thus, we must add ooo 50545 phase.
1325.0)50sin(1
)50sin(1
o
o
The lead compensator provides a gain of dB8.8)125.0log(10)log(10 at .m
Here we choose rad/s2 cm . We now calculate the zero and pole of the compensator as
follows.
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73.0
5.51375.0
2
m
m
z
p
Now we have to find the gain K. The uncompensated system has a gain of about -12 dBat
.rad/s2c The gain Kthus should offset dB2.38.812 to make the gain at
dB.0rad/s2 betoc Thus,
.45.110 202,3
K The desired compensator has the transfer function
5.5
73.095.10
1/
1/)(
s
s
ps
zs
z
pK
ps
zsKsGc
We verify by checking the phase margin of the compensated system.
The step response of the closed-loop system is shown
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Design of Phase-Lead Compensator Using The Root Locus
Procedure
1. List the system specifications and translate them into a desired root location for the dominant
roots
2.Sketch the uncompensated root locus and determine whether the desired root locations are on it.
3. If a compensator is necessary, place the zeroz of the lead compensator directly below the
desired root location ( or to the left of the first two real poles )
4. Use the angle condition of the root locus to determine the polep of the lead compensator .5. Determine the total system gain at the desired root location, using the magnitude condition of the
root locus.
6. Repeat the steps if the design does not satisfy the specifications.
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EXAMPLE : In the control system shown, let)6)(4(
1)(
ssssG . Design a phase lead
compensator to yield 16% overshoot and settling time of 2 seconds.
Solution : For PO=16%, 504.0 . For sTs 2 , rad/s3.9683n . Therefore, the desired
poles are 4274.32 js
The root locus of the uncompensated system does not pass through 4274.32 js .
We start the design of the compensator by arbitrarily placing a zero at .2z Now we use the
angle condition to find the unknown pole.
We find ooo sss 6.406,8.594,3.120 321 . The zero chosen being
the real part of the desired poles, we get the angle os 902 . Thus, we should
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have .180)(90 321o
p
o Solving for p we geto
p 4.49 . We can solve from
p from this as
94.4)4.49tan(
4274.32
op
The gain Kis obtained by applying the angle condition of the root locus. We get
23.109))(6)(4(
4274.32
jszs
pssssK
The lead compensator transfer function is given by
94.4
)2(23.109)(
s
ssGc
We verify the results by plotting the step-response of the closed-loop system.
Not satisfactory!
We do a redesign by placing a zero at s = -4. The effect of this is canceling the pole at s = -4 .
We apply the angle and magnitude conditions to find the compensator pole and the gain.
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This results in 218.4711.87 Kp and . The compensator has the transfer function
87.11
)4(87.218)(
s
ssGc
The step-response of the closed-loop system with this compensator is as shown.
.
EXAMPLE : Let .1
)(2
ssG Design a lead compensator such that the closed-loop system has a
damping ratio of about 0.45, and the settling time in step-response is about 4 second.
Solution : First calculate the desired poles. They are
9845.111 2 jjs nn . The root locus of the uncompensated system is line
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on the imaginary axis of the s-plane, and therefore does not pass through the desired poles. We
will design a lead compensator in the formps
zsK)s(Gc
. We will place a zero at s=-2 ( an
arbitrary choice). Then we use angle conditions to find the polep.
We find 8.3K3.68p and
The step response of the closed-loop system with a compensator68.3
)1(3.8)(
s
ssGc is shown
below.
It shows excessive overshoot.
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In the absence of additional specifications, we can redesign the compensator such that its zero
cancels one of the poles at s=0. ( This is not acceptable if vK ). The redesign gives a
compensator2
94.4)(
s
ssGc . The closed-loop system step response in this case is as shown.
Phase Lag compensation
The main function of the phase lag compensator is to improve steady response without affecting
the dynamic response. A first-order lag compensator has the transfer function in the form
1
1
pz,)(
ps
zs
p
zK
ps
zsKsG
c
cc
The bode plots of a lag compensator with ,1p
zKc are as shown next.
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The Bode plots of the lag compensator has the shape of the inverted Bode plots of the lead
compensator for the reason that in lag compensator zp .
The phase angle is always negative, the reason for the name phase lag compensator. The results
are similar to the lead compensator.
Phase lag design using the root locus
Steps:
1. Draw the root locus of the uncompensated system. Locate the desired poles of the root locus
that satisfies the transient response specifications.
2. Calculate the gain at these pole locations and the corresponding error constant for the closed-
loop system.
3. Calculate the amount of increase in gain necessary to satisfy the specification. This of the
compensator.
4. Determine pole and zero of the compensator such thatp
z and nz .
EXAMPLE : Let)2(
)(
ss
KsG . Desired steady-state error to a ramp input is 0.05 and
45.0 . Design a compensator.
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Solution : With 1)( sGc , the closed-loop systems transfer function isKss
KsT
2)(
2
For 45.0 , we get .9383.4K
The steady-state error in ramp response is given byvold
s
ssKssG
e1
)(lim
1
0
. We find voldK
of the uncompensated system is 2.4691 5.2 .
We want the steady-state error in ramp response of the compensated system to be
05.01
newv
ssK
e . Therefore, .20vnewK
85.2
20
c
c
vold
vnew
p
z
K
K
Choose ( arbitrarily) 05.0z . This choice is acceptable since rad/s24.25 n . Then
00625.08
zp . Thus the new system becomes
)00625.0)(2(
)05.0(5)()(
sss
ssGsGc
We verify if the damping ratio still satisfies the specification. The closed-loop transfer function is
25.00125.500063.2
)05.0(5)(
23
sss
ssT . The poles and zero are
0.05-s:zero
0.0509-sj1.9893,-0.9777s:
Poles
The pole at 0509.0s and the zero at 05.0s practically cancel out. The damping ratio is441.0 , a very small change.
The root loci of the uncompensated and compensated systems are as shown
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The step-response of the system is as shown.
EXAMPLE : Let)20)(10(
)(
sss
KsG . The desired closed-loop 5.0 and .20vK
Design a compensator.Solution: The uncompensated system root locus is as shown. The gain of the systemcorresponding to 5.0 is 1020. The velocity error constant of the uncompensated system is
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1.5200
1020voldK . Thus, 92.3
1.5
20
c
c
vold
vnew
p
z
K
K.
If a choice of 05.0zc is made we find 0128.0pc . Thus, the lag compensator and the planttransfer function is
)0128.0)(20)(10()05.0(1020)()(
ssss
ssGsGc
We verify if the damping ratio still satisfies the specification. The closed-loop transfer function is
)()(10201
)()(1020)(
sGsG
sGsGsT
c
c
. The poles and zero are
0.0500-s:zeor
0.0504-s,j5.68733.3362-s;23.2901:
sPoles
The steady-state response of the uncompensated and compensated closed-loop systems areshown below. There is no discernable difference in the step-response. However, the velocity error
constant vK of the uncompensated system is 5.1 and the compensated system is 20.