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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.

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