Chapter 10 Control Systems Design by Frequency Response[1]

9/17/2001

1

EEE 352 Automatic Control Systems

Prof. Dr. Ahmet Uçar

© Dr. Ahmet Uçar EEE 352 Chapter 10 1

Chapter 10: Frequency Response Design Method

G(j) R(j) Y(j)

H(j)

E (j) Gc(j)


G(s)R(s) Y(s)

H(s)

E(s)

)()()( tytyty sst

In the time response of the stable system, the most important part is the transient-response performance.

)(tyss

the transient response

(t0 t tf).

the steady state response

(tf < t ).

)(ty

0

)(tyt

t0ts=tf t

)()()(1

)()( sR

sGsG

sGsY

The frequency-response approach, however, gives us the transient-response performance. only indirectly manner.

Frequency-Response Approach to Control System Design

9/17/2001

2


G(j) R(j) Y(j)

H(j)

E (j)

The transient-response performance in Bode diagramThe transient-response performance is specified in terms ofa) the phase margin, PM, gain margin, GM, resonant peak magnitude .(they give a rough estimate of the system damping ratio, ).b) the gain crossover frequency, gcf, resonant frequency, r, bandwidth, BW.(they give a rough estimate of the speed of transient response, 1/ = 1/n).

Although the correlation between the transient response and frequency response is indirect,the frequency domain specifications can be conveniently met in the Bode diagram approach.



The steady state error performance in Bode diagramThe steady state performance is specified in terms ofa) static error constants, Kp, Kv and Ka (they give the steady-state accuracy) can be readfrom Bode diagram directly.

G(j) R(j) Y(j)

H(j)

E (j)


9/17/2001

3


)()()( tytyty sst

Although the correlation between the transient response and frequency response is indirect, the frequency domain specifications can be conveniently met in the Bode diagram approach.

)(tyss

The transient response:•Damping ratio, , Speed of transient

response, 1/=1/n.(t0 t tf).

The steady state response:

Static error constants,Kp, Kv and Ka

(tf < t ).

)(ty

0

)(tyt

t0 ts=tft

)()()(1

)()( sR

sGsG

sGsY

The transient-response performance and The steady state error performance in Bodediagram.

G(s)R(s) Y(s)

H(s)

E(s)



Information Obtainable from Open-Loop Frequency Response.

I. Region: The low frequency region (the region far below the gain crossover frequency)of the locus indicates the steady-state behaviour of the closed-loop system.

II. Region: The medium-frequency region (the region near the -1 + j0 point) of the locusindicates relative stability.

III. Region: The high frequency region (the region far above the gain crossoverfrequency) indicates the complexity of the system.

deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjGlog

0

|G(j

)H(j

)|[d

B]

gc

I. regionII. region

III. region

-20 dB/d

G(j) R(j) Y(j)

H(j)

E (j)


9/17/2001

4


The advantage of use Bode Diagram to design compensatorsIf the open-loop gain, K, is varied, the magnitude curve is shifted up or down withoutchanging the slope of the curve, and the phase curve remains the same.

0180)()()2(

1|)()(|)1(0)()(1)(

jHjKG

jHjKGjHjKGjP

G(j) R(j) Y(j)

H(j)

K

)()(1

)(

)(

)(

jHjKG

jKG

jR

jY



)5)(1()()(

)()(1)(

sss

KsHsG

sHsGsP

0180)()()2(

1|)()(|)1(

jHjKG

jHjKG

G(j) R(j) Y(j)

H(j)

The advantage of use Bode Diagram to design compensatorsWhen the open-loop gain K is varied, the Bode diagram is;


9/17/2001

5


The advantage of use Bode Diagram to design compensatorsThe compensator is simply modified the original Bode diagram for the frequency range ofthe compensator only and thus plotting the complete Bode diagram is a simple matter.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

0180)()()()2(

1|)()()(|)1(0)()()(1)(

jHjGjG

jHjGjGjHjGjGjP

c

c

c

For design purposes, therefore, it is best to work with the Bode diagram.



40dB

20

0

-20

-401 2 4 6 8 10 20 40 60 80 100

Gc(j)G(j)H(j)

G(j)H(j)

The advantage of use Bode Diagram to design compensatorsWhen the compensator is designed, it simply modified the original Bode diagram for thefrequency range of the compensator only.

0180)()()()2(

1|)()()(|)1(

0)()()(1)(

jHjGjG

jHjGjG

jHjGjGjP

c

c

c

G(j) R(j) Y(j)

H(j)

Gc(j)

For design purposes it is best to work with the Bode diagram.


9/17/2001

6


Step 1:A common approach to the Bode diagram is first adjusting the open-loop gain, K, sothat the requirement on the steady-state accuracy (Kp, Kv, Ka) is met. Then the magnitude and phase curves of the uncompensated system (when the open-loop gain is just adjusted) is plotted.

)()(1

)(

)(

)(

sHsKG

sKG

sR

sY

dB01log20 10

deg180)()(

1)()(1)()(

0)()(1)(

0

jHjKG

jHjKGsHsKG

sHsKGsP

js

deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjG

G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

K

Common Approach to Design Compensator with Bode diagram


Step 2: If the desired specifications on the phase margin, PM, and gain margin , GM, are notsatisfied, then a suitable compensator, Gc(s), that will reshape the open-loop transferfunction is determined. Finally, if there are any other requirements to be met, we try to satisfy them, unlesssome of them are mutually contradictory.

)()()(1

)()(

)(

)(

sHsGsG

sGsG

sR

sY

c

c

dB01log20 10

deg180)()()(

1)()()(1)()()(

0)()()(1)(

0

jHjGsG

jHjGsGsHsGsG

sHsGsGsP

c

c

jsc

c

deg180)()()()2(

0)()()(log20)1(

0

10

jHjGjG

dBjHjGjG

c

c

G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

Gc(s)

Compensator

Common Approach to Design Compensator with Bode diagram

9/17/2001

7


Requirements on Open-Loop Frequency Response.In many practical cases, compensation (controller) is essentially a compromise betweensteady-state accuracy and relative stability.To have a high value of the velocity error constant and yet satisfactory relative stability,it necessary to reshape the open-loop frequency-response curve.The gain in the low-frequency region should be large enough, and near the gaincrossover frequency, the slope of the log-magnitude curve in the Bode diagram shouldbe -20 dB/d. This slope should extend over a sufficiently wide frequency band to assurea proper phase margin. For the high-frequency region, the gain should be attenuated asrapidly as possible to minimize the effects of noise.

deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjGlog 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. Regionindicates the steady state behaviour

II. Regionindicates relative stability

III. Regionindicates the complexity of the system

G(j) R(j) Y(j)

H(j)

E (j)



deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjG

The Desired Performance Interims of Frequency Response (in Bode Diagram).The desired frequency response performance of the systems are generally given interims ofa) PM; 300 PM 600 and GM; GM 6 dBb) The static error constants Kp, Kv, Ka.c) At the gain crossover frequency, gc, the slope of the log-magnitude curve in the Bodediagram is -20 dB/decade,d) The system band wide BW frequency has to be sufficiently large enough to containsystem operation frequency.

G(j) R(j) Y(j)

H(j)

E (j)

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. Regionindicates the steady state behaviour

II. Regionindicates relative stability



9/17/2001

8


Improve the performance of I. Region:

Phase lag compensator is designed within the low frequency region (the region far belowthe gain crossover frequency) of the locus to improve the steady-state behaviour of theclosed-loop system.

)1(,)/(1

/1

1

1)(

Ts

TsK

Ts

TsKsG ccc Design parameters are; Kc, , T

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. RegionPhase Lag

II. RegionPhase Lead




Improve the performance of II. Region:

Phase lead compensator is designed within the medium-frequency region (the region nearthe -1 + j0 point) of the locus to improve relative stability.

Design parameters are; Kc, , T)10(,)/(1

/1

1

1)(

Ts

TsK

Ts

TsKsG ccc

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. RegionPhase Lag




9/17/2001

9


Improve the performance of I. and II. Regions:

Phase lag-lead compensator is designed to improve the steady-state behaviour andrelative stability of the closed-loop system.

Design parameters are; Kc, , , T1, T2

1,,

)1

(

)1

(

)(

)1

(

)1(

)1(

)1(

)1(

)(

)(

2

2

2

1

1

2

2

1

1

Ts

TsT

Ts

Ts

KsT

sT

sT

sTK

sE

sUcc

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. RegionPhase Lag





The effects of the compensator (controller ) III. Region:

It is important to note that the designed compensator or controller should not reduce theslope in III. Region.

The phase lead, phase lag, and phase lag-lead compensators do not change slope of theactual system in III region. However PD, PI and PID controller change the slope of IIIregion by adding 20 dB/d . Therefore the system becomes sensitive to the high orderdynamics.

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

I. RegionPhase Lag




9/17/2001

10


Basic Characteristics of Lead, Lag, and Lag-Lead Compensation.

Lead compensation essentially yields an appreciable improvement in transient responseand a small change in steady-state accuracy. It may accentuate high-frequency noiseeffects.

Lag compensation, on the other hand, yields an appreciable improvement in steady-stateaccuracy at the expense of increasing the transient-response time. Lag compensation willsuppress the effects of high-frequency noise signals.

Lag-lead compensation combines the characteristics of both lead compensation and lagcompensation.

The use of a lead or lag compensator raises the order of the system by 1 (unlesscancellation occurs between the zero of the compensator and a pole of the uncompensatedopen-loop transfer function).

The use of a lag-lead compensator raises the order of the system by 2 (unless cancellationoccurs between zero(s) of the lag-lead compensator and pole(s) of the uncompensatedopen-loop transfer function), which means that the system becomes more complex and it ismore difficult to control the transient response behaviour.

The particular situation determines the type of compensation to be used.



The above system may be unstable or stable but does not perform the desiredperformance then Bode diagram will be used to design the suitable compensator(controller) Gc(s) to satisfy the desired performance.

0180)()()2(

1|)()(|)1(

jHjG

jHjG

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator(Controller)

U(s)

0180)()()()2(

1|)()()(|)1(

jHjGjG

jHjGjG

c

c

The Desired Performancea) PM; 300 PM 600 and GM; GM 6 dBb) The static error constants Kp, Kv, Ka.c) At the new gain crossover frequency, gc, the slope of the log-magnitude curve in theBode diagram is -20 dB/decade,d) The system band wide BW frequency has to be sufficiently large enough to containsystem operation frequency.

G(s)R(s) Y(s)

H(s)

E(s)


9/17/2001

11


Consider a lead compensator having the following transfer function:

The phase lead compensator has a zero at s = -l/T and a pole at s = -1/(T). Since 0<<1,we see that the zero is always located to the right of the pole in the complex plane. Notethat for a small value of the pole is located far to the left. The minimum value of islimited by the physical construction of the lead compensator and is usually taken to beabout 0.05. (This means that the maximum phase lead that may be produced by a leadcompensator is about c650 )

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

)10(,)/(1

/1

1

1

)(

)(

Ts

TsK

Ts

TsK

sE

sU

c

c

where is called the attenuation factor of the lead

compensator and Kc, , T are design parameters.

j

0

s-düz.

)/(1 T T/1

Frequency-Response Characteristics of the Lead Compensator


Polar plot of a lead compensator: For Kc=1 Polar plot of a lead compensator is

)10(,)/(1

/1

1

1)(

Tj

Tj

Tj

TjjGc

Im

0

=

Re 1

= 0m

Gc(j) plane)1(2

1

)1(2

1

For a given value of , at frequency = m theangle between the positive real axis and the tangentline drawn from the origin to the semicircle givesthe maximum phase-lead angle, m.

1

1

2

12

1

sin m


9/17/2001

12


The Bode diagram of the phase lead compensator when Kc=1, =0.1 and T=1 is depictedin the Figure.The corner frequencies for the lead compensator are = l/T = 1 and = -1/(T) = 10.

)10(,)/(1

/1

1

1)(

Tj

Tj

Tj

TjjGc

For a given value of the phase leadcompensator provide the maximum phaselead angle m at frequency = m.m is the geometric mean of the twocorner frequencies:

rad/s)/(1 TT/1

10-1

100

101

102

-20

-10

0

Magnitude(d

B)

10-1

100

101

102

0

20

40

60

Pha

se(d

eg)

m

)/(1 Tm

(zero) (pole)

rad/s

20 dB/decade

TTTmm

1)

1log

1(log

2

1log

The lead compensator is basically a high-pass filter. (The high frequencies are passed, butlow frequencies are attenuated.)



The primary function of the lead compensator is to reshape the frequency-response curveto provide sufficient phase-lead angle to offset the excessive phase lag associated with thecomponents of the fixed system of the system given in Figure 1:

Assume that the desired performance specifications are given in terms of phase margin,PMd, gain margin, GMd, static error constants, Kp, Kv, and Ka, and so on.Assume also that the system given in Figure 1 does not satisfy the desired frequencyperformance then the phase lead compensator Gc(s) shown in Figure 2 needs to bedesigned.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Phase lead comp

U(s)

Figure 2:)10(,)/(1

/1

1

1)(

Tj

TjK

Tj

TjKjG

c

cc

G(s)R(s) Y(s)

H(s)

E(s)

Figure 1:

Design Lead Compensation Based on Frequency-Response Approach

9/17/2001

13


The main procedure for designing a lead compensator by the frequency response approachmay be stated as follows:a) Determine gain K from the loop transfer function KG(s)H(s) to satisfy the requirementon the given static error constant and draw the Bode diagram of KG(j)H(j). Evaluatethe phase margin, PM and compare it with the desired the phase margin, PMd anddetermine the necessary phase m that will provide by the phase lead compensator. Hencethe parameter is obtained from;

1

1sin m

b) Determine the frequency where the magnitude of the uncompensated systemKG(j)H(j) is equal to -20log(1/). Select this frequency as the new gaincrossover frequency. This frequency corresponds to ngc= m = (1/ (T)) that themaximum phase shift m occurs.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lead Comp.

U(s)

)10(,)/(1

/1

1

1)(

Tj

TjK

Tj

TjKjG

c

cc

Design Lead Compensation Based on Frequency Response Approach


The design steps for phase lead compensator.

Step 1: To satisfy the requirement on the given static error constant determine gain KT;a) with the system’s variable gain K by letting Kc=1 and KT=K, if the system’s gain isconstant thenb) with KT= Kc.

Step 2: Using the gain KT, draw the Bode diagram of KT G(j)H(j) that the gain adjustedbut uncompensated system. Evaluate the phase margin.

Step 3: Determine the necessary phase-lead angle to be added to the system. Add anadditional phase 50 to 120 to the phase-lead angle required, because the addition of the leadcompensator shifts the gain crossover frequency to the right and decreases the phasemargin.

1

1sin m

)125()( 00 PMPM dm


9/17/2001

14



Step 4: Determine the frequency where the magnitude of the uncompensated systemKTG(j)H(j) is equal to -20log(1/).Select this frequency as the new gain crossover frequency, ngc. This frequencycorresponds to ngc= m = (1/ (T)) where the maximum phase shift m occurs.

dBjG

log201

log20)(?1

4. b) Reading directly from the gain diagram of Bode,

20log10|KTG(j)H(j)|=0 dB

4. a) Analytically by;

Determiningngc = m where |KTG(j)H(j)| = -20log(1/).




Step 4: Determiningngc = m where |KTG(j)H(j)| = -20log(1/).

4. c) Geometrically; by considering the asymptotes in |KTG(j)H(j)| where the sign ofgain is changed.Let new gain crossover be m =ngc and will be determined as follows;

ngcmdB

dB

22

11

),log(20

,0

112

1

12

loglog

loglog

/log

x

dBdB

xdBdB

ddBxdB

ngc

ngc log

|G(j

)H(j

)|[d

B]

1

-x dB/decade

dB1

2 =ngc

dB2


9/17/2001

15


The design steps for phase lead compensator.Step 5: Determine the corner frequencies of the phase lead compensator as follows:

Zero of lead compensator:

Pole of lead compensator:

T

1

T

1

TK

K

T

c

KK

rad/s)/(1 T/1

Magnitude(d

B)

Pha

se(d

eg)

m

(zero) (pole)

20 dB/decade

)/(1 T

Step 6: Since KT and are determined then,6. a) calculate constant K

6. b) calculate constant Kc from

If the system has variable gain K.

If the system has fixed gain.

Step 7: Drove the compensated system’s Bode diagram. Compare PM and GM with PMd

and GMd. If desired performance has not been satisfied then repeat the design process bymodifying the pole-zero location of the compensator until a satisfactory performance isobtained.



Example 10.1: Consider the system shown in Figure where the transfer function of blocksare;

The desired frequency response performance: It is desired to design a compensatorGc(s) for the system so that the static velocity error constant Kv is 20 sec-1, the phasemargin is at least 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

1)(,)2(

4)(

sH

sssG

Solution 10.1: Step 1: Since the system has fixed gain them KT = Kc. To adjust the gainKT to meet the steady-state performance specification or to provide the required staticvelocity error constant. Since the static velocity error constant Kv is 20 sec-1, we obtain;

10220

)2(

4

1

1lim)()(lim

00

TT

T

sc

sv

KK

ss

K

Ts

TsssGssGK

Design Lead Compensator Based on Bode Diagram

9/17/2001

16


Solution 10.1: Step 2: Draw a Bode diagram of KTG(j)H(j)

)2(

40

)2(

4)(1

ssssKsG T

)2

tan90(0)()( 101

jHjG

)2log20log20(40log20)()(log20 22101010110 jHjG

Magnitudes graph;

Angle graph;

Gain (dB) Angle

10-1 46 -92

10-0 25 -116

2 16 -135

101 -8 -168

G(s)R(s) Y(s)E(s)

KT

U(s)

G1(s)

H(s)0

1

110

1

180)()()2(

0|)()(|log20

1|)()(|)1(

jHjG

dBjHjG

jHjG



Solution 10.1: Step 2: Draw a Bode diagram of G1(j)H(j) and evaluate the phasemargin.

-50

0

50

Magnitu

de (

dB

)

10-1

100

101

102

-180

-135

-90

Phase (

deg)

Bode Diagram

Gm = Inf dB (at Inf rad/sec) , Pm = 18 deg (at 6.17 rad/sec)

Frequency (rad/sec)Desired PMd= 500

000 386)1850(6)( PMPM dm

PM of G1(j)H(j) is PM=180

at gc=6.2 rad/s

The necessary phase shift m;

Gain (dB) Angle

10-1 46 -92

10-0 25 -116

2 16 -135

101 -8 -168


9/17/2001

17


Solution 10.1: Step 3: Determine design parameter; For m=380 we have

24.0615.1

384.0

sin1

sin1

1

1sin

m

mm

24.0

Solution 10.1: Step 4: Determine the frequency where the magnitude of theuncompensated system G1(j)H(j) is equal to -20log(1/).From the Bode diagram;

ngcmdB

dB

,2.6)log(20

2.6,0

2

11

985.810*2.6

155.02.6

log40

log

02.6

/40log

155.0

1

m

gc

m

decdBdB

log

|G(j

)H(j

)|[d

B]

1

-x dB/decade

dB1

2 =ngc

dB2

Select this frequency as the new gain crossover frequency: m =ngc= 9 rad/s



Zero of lead compensator:

Pole of lead compensator:

Solution 9.1: Step 6: Since the system has fixed gain then Kc

The phase lead compensator dynamic is

Solution 10.1: Step 5: Determine the corner frequencies of the lead compensator asfollows:

41.424.0911

9 TT

ngc

41.41

T

4.181

ngc

T

rad/s

rad/s

7.4124.0

10

T

c

KK

)124.00(,4.18

41.47.41

)/(1

/1

1

1)(

s

s

Ts

TsK

Ts

TsKsG ccc


9/17/2001

18


Solution 10.1: Step 7: To drove the compensated system’s Bode diagram and checkingPM and GM to be sure the design is satisfactory.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

4.18

41.47.41)(

1)(,)2(

4)(

s

ssG

sHss

sG

c

)4.18

tan2

tan90(41.4

tan0)()()( 1101 jHjGjGc

))4.18(log202log20log20(

)41.4(log208.166log20)()()(log20

2210

221010

22101010

jHjGjGc

Magnitudes graph;

Angle graph;



Solution 10.1: Step 7: Compensated system’s Bode diagram.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

)4.18

tan2

tan90(41.4

tan0)()()( 1101 jHjGjGc

))4.18(log202log20log20(

)41.4(log208.166log20)()()(log20

2210

221010

22101010

jHjGjGc

Magnitudes graph;

Angle graph;

Gain (dB) Angle

10-1 46

2 17

4.4 6

ngc= 9 0 -129.6

18.4 -6

102 -35

PM = 1800 - 129.60 500


9/17/2001

19


Step 7: Compensated system Gc(j)G(j)H(j) Bode diagram;

PM =50.50, at gcf =8.89 rad/sGM = dBStatic velocity error constant: Kv=20 sec-1

1)(

)4.18(

)41.4(7.41)(

)2(

4)(

sH

s

ssG

sssG

c

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

-100

-50

0

50

Magnitu

de (

dB

)

10-1

100

101

102

103

-180

-135

-90

Phase (

deg)

Bode Diagram

Gm = Inf dB (at Inf rad/sec) , Pm = 50.5 deg (at 8.89 rad/sec)

Frequency (rad/sec)

-40 dB/d-20 dB/d

-20 dB/d

-40 dB/d



To see how the compensator improves:a) the actual system performance the Bode diagram of actual system in Figure 1 is firstobtained.

G(s)R(s) Y(s)

H(s)

E(s)

Figure 1

b) Secondly, the Bode diagram of the system with Gc(s)=KT (satisfy the steady staterequirement only) in Figure 2 is obtained.

Figure 2

G(s)R(s) Y(s)E(s)

KT

U(s)

G1(s)

H(s)

c) Finally, the Bode diagram of the compensated system in Figure 3 is obtained and theresults are examined.

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Figure 3

Improvement on the close loop system performance: Comparison

9/17/2001

20


a) The actual system G(j)H(j) Bode diagram;

1)(,)2(

4)(

sH

sssG

G(s)R(s) Y(s)

H(s)

E(s)

PM = 520, at gcf =1.6 rad/sGM = dBStatic velocity error constant: Kv =2 sec-1

10-1

100

101

102

-100

-50

0

50

Magnitude(d

B)

Frequency (rad/sec)

Bandwidth BW =2.54 rad/s

Closed loop system

-80

-60

-40

-20

0

20

40

Mag

nitu

de

(d

B)

10-1

100

101

102

-180

-135

-90

Ph

ase

(d

eg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 51.8 deg (at 1.57 rad/sec)

Frequency (rad/sec)

-20 dB/d

-40 dB/d

The close loop system performance: Actual system

10-1

100

101

102

-60

-40

-20

0

20

Ma

gni

tud

e(d

B)

Frequency (rad/sec)


b) The system with Gc(s) = KT, Bode diagram of KTG(j)H(j) ;

PM = 180, at gcf =6.2 rad/sGM = dBStatic velocity error constant: Kv=20 sec-1


Closed loop system

1)(,)2(

40)(1

sH

sssG

-50

0

50

Magnitu

de (

dB

)

10-1

100

101

102

-180

-135

-90

Phase (

deg)

Bode Diagram

Gm = Inf dB (at Inf rad/sec) , Pm = 18 deg (at 6.17 rad/sec)

Frequency (rad/sec)

-40 dB/d

-20 dB/d

G(s)R(s) Y(s)

H(s)

E(s)

Close loop system performance: The stem with Gc(s) = KT

9/17/2001

21

10-1

100

101

102

-40

-20

0

20

Ma

gnitu

de

(dB

)

Frequency (rad/sec)


c) Compensated system Gc(j)G(j)H(j) Bode diagram;

PM =50.50, at gcf =8.89 rad/sGM = dBStatic velocity error constant: Kv=20 sec-1

Bandwidth BW =15 rad/s

Closed loop system

1)(,)4.18(

)41.4(7.41)(,

)2(

4)(

sH

s

ssG

sssG c

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

-100

-50

0

50

Ma

gnitu

de

(dB

)

10-1

100

101

102

103

-180

-135

-90

Pha

se

(d

eg)

Bode Diagram

Gm = Inf dB (at Inf rad/sec) , Pm = 50.5 deg (at 8.89 rad/sec)

Frequency (rad/sec)

-40 dB/d-20 dB/d

-20 dB/d

Close loop system performance: The stem with Gc(s) = KT



Uncompensated system PM = 520, gcf =1.6 rad/sGM = dBKv =2 sec-1

BW =2.54 rad/s

Compensated system PM = 50.50, gcf = 8.89 rad/sGM = dBKv =20 sec-1

BW = 14.5 rad/s10

-110

010

110

2-20

-10

0

10

20

30

40

50

60

Mag

nitu

de

(dB

)

Frequency (rad/sec)

KT

=

Kc

=1

0

Gc(j)

G(j)H(j)

G1(j)H(j)Gc(j)G(j)H(j)

-40 dB/d

-20 dB/d

-20 dB/d

20 dB/d

-40 dB/d

The close loop system performance: Comparison

9/17/2001

22



Compensated system PM = 50.50, gcf = 8.89 rad/sGM = dBKv =20 sec-1

BW = 14.5 rad/s

1)(

)4.18(

)41.4(7.41)(

)2(

4)(

sH

s

ssG

sssG

c

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

The close loop system performance: A Comparison


Uncompensated system has performance: PM = 520 at gcf =1.6 rad/s, GM = dB,Kv =2 sec-1 and BW=2.54 rad/sed

Compensated system has performance: PM = 50.50 at gcf = 8.89 rad/s, GM = dBKv =20 sec-1 and BW = 14.5 rad/sec.

G(s)R(s) Y(s)

H(s)

E(s)

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Note that the bandwidth is approximately equal to the gain crossover frequency. The leadcompensator causes the gain crossover frequency to increase from 1.6 to 8.89. The increaserad/sec in this frequency means an increase in bandwidth. This implies an increase in thespeed of response.The phase and gain margins are seen to be approximately 500 and + dB, respectively.The compensated system therefore meets both the steady-state and the relative stabilityrequirements.


9/17/2001

23


Homework 10.1: Consider the system shown in Figure where the transfer function ofblocks are;

The desired frequency response performance: It is desired to design Phase Leadcompensator for the system so that the static velocity error constant Kv is 10 sec-1, the phasemargin is at least 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

)1(1)(,)1(

10)(

sH

sssG

Homework 10.2: Consider the system shown in Figure where the open loop transferfunction is

)2()5)(2(

)()(

sss

KsHsG

The desired frequency response performance: It is desired to design Phase Leadcompensator for the system so that the static velocity error constant Kv is 5 sec-1, the phasemargin is at least 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Design Lead Compensator Based on Bode Diagram: Homework


Homework 10.3: Consider the system shown in Figure where the transfer function ofblocks are;


The desired frequency response performance: It is desired to design Phase Leadcompensator for the system so that the static velocity error constant Kp is 8, the phase marginis at least 40°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

)4()10)(5(

)()(

ss

KsHsG

Determine the gain K such that the phase margin is 50°,(PMd = 500). What is the gainmargin in this case.

)3(1)(,1

)(),2()(2

sHs

sGsKsGc

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Design Lead Compensator Based on Bode Diagram: Homework

9/17/2001

24


Consider a lag compensator having the following transfer function:

It has a zero at s = -l/T and a pole at s = -1/(T). Since > 1, we see that the pole isalways located to the right of the zero in the complex plane. Note that for a large value of the pole is located near to origin, s=0.Designing and selecting T large enough to limit -50<c<00 phase lag produced by alag compensator is the essential.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

Design parameters are; Kc, , T

)1(,)/(1

/1

1

1

)(

)(

Ts

TsK

Ts

TsK

sE

sU

c

c

j

0 )/(1 TT/1

The pole is located to the right of the zero.

|)/(1|10|/1| TT

Frequency Characteristics of the Lag Compensator


Polar plot of a lag compensator: For Kc=1 Polar plot of a lead compensator is

Im

0 = Re

Kc

= 0Gc(j) plane

Kc

)1(,)/(1

/1

1

1)(

Tj

Tj

Tj

TjjGc

The phase-lag characteristic is of no consequence in lag compensation. Thereforelimiting the phase lag produced by a lag compensator within -50<c<00 is important.

Phase lag larger then -50<c<00 is not desired


9/17/2001

25


rad/sT/1)/(1 T

10-2

10-1

100

101

0

10

20

Magnitude(d

B)

10-2

10-1

100

101

-60

-40

-20

0

Pha

se(d

eg)

Kc= 1

=10T = 1

(zero)(pole)

Bode diagram of a lag compensator: For Kc=1, T=1 and =10

)1(,)/(1

/1

1

1)(

Tj

Tj

Tj

TjjGc

The corner frequencies of thelag compensator are at = 1/Tand = 1/(T).

The magnitude of the lag compensator becomes 10 (or 20 dB) at low frequencies andunity (or 0 dB) at high frequencies. Thus, the lag compensator is essentially a low-passfilter.

-20 dB/d0 dB/d

0 dB/d



The primary function of a lag compensator is to provide attenuation in the high frequencyrange to give a system sufficient phase margin. The phase-lag characteristic is of noconsequence in lag compensation.

Assume that the desired performance specifications are given in terms of phase margin,PMd, gain margin, GMd, static error constants, Kp, Kv, and Ka, and so on. Assume that thesystem given in Figure 1 does not satisfy the desired frequency performance then the phaselag compensator Gc(s) shown in Figure 2 needs to be designed.

R(s) E(s)G(s)

Y(s)

H(s)Figure 1:

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

Figure 2:)1(,)/(1

/1

1

1)(

Tj

Tj

Tj

TjjGc

Design Lag Compensation Based on Bode Diagram

9/17/2001

26


The main reason to phase lag compensator is obtain the necessary gain for the steady stateperformance.The frequency response approach may be stated as follows:

a) If the system gain is fixed then Kc1 selected. The necessary gain provided by .

b) If the system gain is not fixed then Kc1 selected. Considering >1 and T that isgenerally selected T= then for a small gain Kc the desired steady state behaviour isachieved.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag Comp.

U(s)

)1(,)/(1

/1

1

1)(

Tj

TjK

Tj

TjKjG

c

cc



The open-loop transfer function of the compensated system is

The design steps for designing a lag compensator.Step 1: To satisfy the requirement on the desired static error constant determine the gain forthe following lag compensated system

)1(,)/(1

/1

1

1)(

Ts

TsK

Ts

TsKsG ccc

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag Comp.

U(s)

)()(1

1)()(

1

1)()()( 1 sHsG

Ts

TssHsG

Ts

TsKsHsGsG Tc

where G1(s)=KTG(s)

a) If the system gain is not fixed let Kc =1 and KT = K;

b) If the system gain is fixed let KT = Kc;

))...()((

))...()(()(

212

1

212

1

nN

m

zsbsbspss

zsasaszsKsG

))...()((

))...()(()(

212

1

212

1

nN

m

zsbsbspss

zsasaszssG


9/17/2001

27


Evaluate the phase margin PM and compare with desired PMd.If the gain-adjusted of uncompensated system G1(j)H(j)=KTG(j)H(j) does not satisfythe specifications on the phase and gain margins (If the desired behaviour does notsatisfied by the gain KT) and if there is a limitation on the bandwide BW of the closed loopsystem then phase lag compensator needs to be designed.

The design steps for designing a lag compensator.

Step 2: Using the gain KT thus determined, draw the Bode diagram ofG1(j)H(j)=KTG(j)H(j) the gain adjusted but uncompensated system.

G(s)R(s) Y(s)

H(s)

E(s)K

U(s))()()()( 1 jHjGjHjGKT

)1(,)/(1

/1

1

1

)(

)(

Ts

TsK

Ts

TsK

sE

sUcc



The design steps for designing a lag compensator.Step 3: From Bode diagram of G1(j)H(j)=KTG(j)H(j) find the frequency point wherethe phase angle of the open-loop transfer function is equal to -1800 plus the required phasemargin, m =(PMd+50 to 120)The addition of 5" to 12" compensates for the phase lag of the lag compensator.)

Choose this frequency as the new gain crossover frequency, ngc.

log

log

0|G1(

j)H

(j

)|[d

B]

-1800

gc

pc

m

ngc


9/17/2001

28


The design steps for designing a lag compensator.

Step 4: To prevent detrimental effects of phase lag due to the lag compensator, the pole and zero of the lag compensator must be located substantially lower than the new gain crossover frequency. Therefore, choose the corner frequency = 1/T (corresponding to the zero of the lag compensator) 1 octave to 1 decade below the new gain crossover frequency, ngc.

= 1/T =1/[(one octav-one decade)]ngc , thus we will be selected = 1/T = 10-1ngc in this course.

(If the time constants of the lag compensator do not become too large, the corner frequency =1/T may be chosen 1 decade below the new gain crossover frequency.)Notice that we choose the compensator pole, = 1/(T), and zero, = 1/T, sufficiently small. Thus the phase lag occurs at the low-frequency region so that it will not affect the phase margin.



The design steps for designing a lag compensator.Step 5: Determine the attenuation necessary to bring the magnitude curve down to 0 dB at the new gain crossover frequency, ngc. Noting that this attenuation is -20 log , determinethe value of . Then the other corner frequency (corresponding to thepole of the lag compensator) is determined from = 1/(T).

log

log

0|G1(

j)H

(j

)|[d

B]

-1800

gc

pc

m

ngc

ngc

log20

log201log201

log20

Step 6: Using the value of K determined in step 1 and that of determined in step 5, calculateconstant Kc from Kc=K/.

Compensator zero; = 1/T

Compensator pole; = 1/( T) Compensator gain; Kc=K/

Step 7: Draw the Bode diagram of compensated system Gc(j)G(j)H(j) and obtain the PM and GM to compare desired PMd and GMd.


9/17/2001

29


Example 10.2: Consider the system shown in Figure where the open loop transfer functionis

The desired frequency response performance: It is desired to design a compensator forthe system so that the static velocity error constant Kv is 5 sec-1, the phase margin is at least40°,(PMd 400) and the gain margin is at least 10 dB, (GMd 10 dB).

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

1)(,)15.0)(1(

1)(

sH

ssssG

Solution 10.2: Step 1: Since the system has fixed gain them KT = Kc. To adjust the gainKT to meet the steady-state performance specification or to provide the required staticvelocity error constant. Since the static velocity error constant Kv is 5 sec-1, we obtain;

T

T

sc

sv

K

sss

K

Ts

TsssGssGK

5

)15.0)(1(1

1lim)()(lim

00



Solution 10.2: Step 2: Draw a Bode diagram of KTG1(j)H(j)

)15.0)(1(

5

)15.0)(1()(1

jjj

sss

KsG

js

T

Magnitudes graph;

Angle graph;

G(s)R(s) Y(s)E(s)

KT

U(s)

G1(s)

H(s)

)1)5.0(log201log20log20(5log20)()( 2210

221010101 jHjG

))5.0(tantan90(0)()( 1101 jHjG

Gain (dB) Angle

10-3 73 -9010-2 53 -90.810-1 33.5 -98.5710-0 10 -161.56

2 -2.04 -198.4101 -40.2 -252.9102 -100 -268.28


9/17/2001

30


Solution 10.2: Step 2: Draw a Bode diagram of G1(j)H(j) and evaluate the phasemargin.

Desired PMd= 400

Gain (dB) Angle

10-3 73 -9010-2 53 -90.810-1 33.5 -98.5710-0 10 -161.56

2 -2.04 -198.4101 -40.2 -252.9102 -100 -268.28

G1(j)H(j) ’in GM= -4.4 (pc=1.4 r/s)PM= -130 (gc=1.8 r/s)The system is unstable.

-150

-100

-50

0

50

100

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = -4.44 dB (at 1.41 rad/sec) , Pm = -13 deg (at 1.8 rad/sec)

Frequency (rad/sec)



Solution 10.2: Step 3: From Bode diyagram of G1(j)H(j)=KTG(j)H(j) find thefrequency point where the phase angle of the open-loop transfer function is equal to -1800

plus the required phase margin, m =(PMd+50 to 120)

Thus new gain crossover frequency is ngc= 0.5 rad/s

log

log

0|G1(

j)H

(j

)|[d

B]

-1800

gc

pc

m

ngc

0000 50104010 dm PM

0

00

?1

130

50180)()(

jHjG

At =ngc= 0.5 rad/s the angle is G1(j0.5)H(j0.5)= -1300.


9/17/2001

31


Solution 10.2: Step 4: Use the corner frequency = 1/T by considering = 1/T =1/[(one octave-one decade)]ngc , or = 1/T = 10-1ngc

leads to

05.0101 1

ngcT

rad/s

Step 5:Determine the attenuation -20 log necessary to bring the magnitude curve down to 0 dB ngc=0.5 rad/sec froma) Magnitude diagram orb) Magnitude equation

dB

jHjG

19

76.18)()(5.01

log

log

0|G1(

j)H

(j

)|[d

B]

-1800

gc

pc

m

ngc

ngc

=19



Solution 10.2: Step 5:

dBjHjG 1976.18)()(5.01

log

log

0|G1(

j)H

(j

)|[d

B]

-1800

gc

pc

m

ngc

ngc

=19

991.810

log201995.0

log2019

1log20

To bring the magnitude curve down to 0 dB at this new gain crossover frequency, the lagcompensator must give the necessary attenuation, which in this case is -20 dB. Hence,

and


9/17/2001

32


Solution 10.2: Step 6: The compensator gain is.

Compensator zero;

55.09

5

T

c

KK

Compensator pole;

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag compensator

U(s)

05.01

T rad/s

rad/s0055.01

T

)19(,0055.0

05.055.0

)/(1

/1

1

1)(

s

s

Ts

TsK

Ts

TsKsG ccc

1)(,)15.0)(1(

1)(

sH

ssssG



Solution 10.2: Step 7: Bode diagram of compensated system.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag compensator

U(s)

0055.0

05.055.0)(

s

ssGc

1)(,)15.0)(1(

1)(

sH

ssssG

))5.0(tantan900055.0

(tan05.0

tan0)()()( 11011

jHjGjGc

)1)5.0(log201log20log200055.0log20(

05.0log2055.0log20)()()(

2210

21010

2210

221010

jHjGjGc

Magnitudes graph;

Angle graph;


9/17/2001

33


Solution 10.2: Step 7: Magnitudes graph;

))5.0(tantan900055.0

(tan

05.0tan0)()()(

1101

1

jHjGjGc

)1)5.0(log201log20log200055.0log20(

05.0log2055.0log20)()()(

2210

21010

2210

221010

jHjGjGc

Angle graph; Gain (dB) Angle

10-3 74 -1000

0.005 53 -1300

0.01 47 -1430

0.05 23 -1340

0.1 15 -1220

ngc= 0.5 0 -1350

1 -10 -1640

2 -22 -2000

10 -60

PM = 1800 - 1350 450

At ngc = 0.5 rad/sec


10-2

100

102

-100

-50

0

50

Ma

gnit

ud

e(d

B)

Frequency (rad/sec)


Solution 10.2: Step 7 Compensated system Gc(j)G(j)H(j) Bode diagram;


Closed loop system

0055.0

05.055.0)(

1)(,)15.0)(1(

1)(

s

ssG

sHsss

sG

c

PM=450 (gc=0.48 r/s)GM=14 dB, (pc=1.37 r/s)Kv =5 sec-1

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

-150

-100

-50

0

50

100

Magnitu

de (

dB

)

10-4

10-2

100

102

-270

-225

-180

-135

-90

Phas

e (

deg)

Bode Diagram

Gm = 14.1 dB (at 1.37 rad/sec) , Pm = 45.3 deg (at 0.484 rad/sec)

Frequency (rad/sec)

-40 dB/d-20 dB/d

-20 dB/d

-40 dB/d

-60 dB/d

Closed loop system is stable and does satisfies the desired performance.


9/17/2001

34



G(s)R(s) Y(s)

H(s)

E(s)

Figure 1


Figure 2

G(s)R(s) Y(s)E(s)

KT

U(s)

G1(s)

H(s)

c) Finally, the Bode diagram of phase lag compensated system in Figure 3 is obtainedand the results are examined.

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Figure 3


10-2

100

102

-150

-100

-50

0

50

Magnitude(d

B)

Frequency (rad/sec)


a) The actual system G(j)H(j) Bode diagram;

G(s)R(s) Y(s)

H(s)

E(s)


Closed loop system

1)(,)15.0)(1(

1)(

sH

ssssG

PM = 32.60, gcf =0.74 rad/sGM = 9.45 dB, pc=1.41 rad/sKv =1 sec-1

-150

-100

-50

0

50

100

Magnitu

de (

dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Phas

e (

deg)

Bode Diagram


Frequency (rad/sec)

-20 dB/d

-40 dB/d

-60 dB/d

System is stable

The close loop system performance: The actual system

9/17/2001

35

10-2

100

102

-100

-50

0

50

Ma

gn

itu

de

(dB

)

Frequency (rad/sec)


Bandwidth BW =2.51rad/s

Closed loop system

G1(s)R(s) Y(s)

H(s)

E(s)

1)(,)15.0)(1(

5)(1

sH

ssssG

PM=-130 (gc=1.8 r/s)GM=-4.4 dB, (pc=1.4 rad/sec)Kv =5 sec-1 -50

0

50

100

Ma

gnitu

de (

dB

)

10-3

10-2

10-1

100

101

-270

-225

-180

-135

-90

Ph

ase (

deg

)

Bode Diagram

Frequency (rad/sec)

-40 dB/d

-20 dB/d

-60 dB/d

System is unstable

b) The system with Gc(s) = KT, Bode diagram of KTG(j)H(j) ;

Close loop system performance: System with Gc(s) = KT

10-2

100

102

-100

-50

0

50

Ma

gnit

ud

e(d

B)

Frequency (rad/sec)


c) Phase lag compensated system Gc(j)G(j)H(j) Bode diagram;


Closed loop system

0055.0

05.055.0)(

1)(,)15.0)(1(

1)(

s

ssG

sHsss

sG

c

PM=450 (gc=0.48 r/s)GM=14 dB, (pc=1.37 r/s)Kv =5 sec-1

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

-150

-100

-50

0

50

100

Magnitu

de (

dB

)

10-4

10-2

100

102

-270

-225

-180

-135

-90

Phas

e (

deg)

Bode Diagram


Frequency (rad/sec)

-40 dB/d-20 dB/d

-20 dB/d

-40 dB/d

-60 dB/d

Closed loop system is stable and does satisfies the desired performance.

Performance of phase lag compensated system

9/17/2001

36



Uncompensated system PM = 32.60, gcf =0.74 r/sGM = 9.45dB, pc=1.41 r/sKv =1 sec-1

BW =1.26 rad/s

Compensated system PM=450,gc=0.48 r/sGM=14 dB, pc=1.37 r/sKv =5 sec-1

BW =0.87 rad/s

10-4

10-3

10-2

10-1

100

101

-40

-20

0

20

40

60

80

Magnitude(d

B)

Frequency (rad/sec)

KT =Kc=5

Gc(j)

G(j)H(j)

G1(j)H(j)

Gc(j)G(j)H(j)

-40 dB/d-20 dB/d

-20 dB/d-20 dB/d

-60 dB/d

Gc(j)G(j)H(j)

-60 dB/d

Performance of phase lag compensated system


Uncompensated system PM = 32.60, gcf =0.74 r/sGM = 9.45dB, pc=1.41 r/sKv =1 sec-1

BW =1.26 rad/s

Compensated system PM=450,gc=0.48 r/sGM=14 dB, pc=1.37 r/sKv =5 sec-1

BW =0.87 rad/s

Bode diagrams for G1(j) (gain-adjusted but uncompensated open-loop transfer function),Gc(j) (compensator), and Gc(j)G(j) (compensated open-loop transfer function).

The close loop system performance: Comparison

9/17/2001

37


Uncompensated system has performance: PM = 32.60 at gcf =0.74 rad/sec, GM = 9.45dB atpc=1.41 rad/sec, Kv =1 sec-1, and BW =1.26 rad/sec.

Compensated system has performance: PM=450 at gc=0.48 rad/sec, GM=14 dB atpc=1.37 rad/sec, Kv =5 sec-1, and BW =0.87 rad/sec.

G(s)R(s) Y(s)

H(s)

E(s)

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Note that the new gain crossover frequency is decreased from approximately 1 to 0.5rad/sec. This means that the bandwidth of the system is reduced.

The plot of KTG(j)H(j) clearly shows that the gain-adjusted but uncompensated systemis unstable.

The addition of the phase lag compensator stabilizes the system. The resonant peak value is3 dB, or 1.4, and this peak occurs at =0.5 rad/sec.

Design Lag Compensator Based on Bode Diagram: Results


PM=450 at gc=0.48 rad/sec, GM=14 dB at pc=1.37rad/sec, Kv=5 sec-1, andBW=0.87 rad/sec

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Compensated system has BW = 0.87 rad/sec reduced from BW =1.26 rad/sec for uncompensated case and results in slower transient response.

While the frequency range for uncompensated system is 0 < 1.26 rad/sec but its0<0.87 rad/sec for uncompensated system.

Compensators designed by different methods or by different designers (even using thesame approach) may look sufficiently different.

Any of the well-designed systems, however, will give similar transient and steady-stateperformance.

The best among many alternatives may be chosen from the economic consideration thatthe time constants of the lag compensator should not be too large.

1)(,)15.0)(1(

1)(

,0055.0

05.055.0)(

sHsss

sG

s

ssGc


9/17/2001

38


Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Since the lag compensator tends to integrate the input signal, it acts approximately as aproportional-plus-integral controller. Because of this, a lag-compensated system tends tobecome less stable. To avoid this undesirable feature, the time constant T should be madesufficiently larger than the largest time constant of the system.

Conditional stability may occur when a system having saturation or limiting iscompensated by use of a lag compensator. When the saturation or limiting takes place inthe system, it reduces the effective loop gain. Then the system becomes less stable andunstable operation may even result. To avoid this, the system must be designed so that theeffect of lag compensation becomes significant only when the amplitude of the input to thesaturating element is small. (This can be done by means of minor feedback-loopcompensation.)

1)(,)15.0)(1(

1)(

,0055.0

05.055.0)(

sHsss

sG

s

ssGc



Homework 10.5: Consider the system shown in Figure where the block of transfer functionare

The desired frequency response performance: It is desired to design Phase Lagcompensator for the system so that the static velocity error constant Kv is 10 sec-1, the phasemargin is at least 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

)1(1)(,)10)(4(

1)(

sH

ssssG


)2()5)(2(

)()(

sss

KsHsG


Design Lag Compensator Based on Bode Diagram: Homework

9/17/2001

39


Homework 10.7: Consider the system shown in Figure where the block of transferfunction are

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

Homework 10.8: A robot will be operated by NASA to build a permanent lunar station. The position control system for the gripper tool is shown in Figure 1, where H(s)=1, and

Determine a Phase Lag compensator for the system so that the static velocity errorconstant Kv is 10 sec-1, the phase margin is at least 45°,(PMd 450) and the gain margin isat least 10 dB, (GMd 10 dB).

)3(1)(,)10(

1)(

2

sH

sssG

)4()125.0)(1(

5)(

ssssG


Figure 1

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Design Lag Compensator Based on Bode Diagram: Homework


Consider a lag-lead compensator ( ) having the following transfer function:

Design parameters are Kc, T1, , T2, . The term

)1,(,

)1

(

)1

(

)(

)1

(

)1(

)1(

)1(

)1()(

2

2

1

1

2

2

1

1

Ts

Ts

Ts

Ts

K

sT

sT

sT

sTKsG

c

cc

2

1

T

j

2

1

T

1T

1

1

T

Lead part Lag part

)1(,

)(

)1

(

)1(

)1(1

1

1

1

1

Ts

Ts

sT

sT produces the effect of the lead network, and the term

)1(,

)1

(

)1

(

)1(

)1(

2

2

2

2

Ts

Ts

sT

sTproduces the effect of the lag network.

Frequency Characteristics of the Lag-Lead Compensator

9/17/2001

40


If ( = ) then the lag-lead compensator have the following transfer function:

)1(,

)1

(

)1

(

)(

)1

(

)1(

)1(

)1(

)1()(

2

2

1

1

2

2

1

1

Ts

Ts

Ts

Ts

K

sT

sT

sT

sTKsG

c

cc

Design parameters are Kc, T1, T2, and the polar plot of the lag-lead compensator with Kc=1 and = becomes as;

2

1

T

j

2

1

T

1T

1

1

T

Lead part Lag part

Im

0

=

Re1 = 0

m)

11(

2

1

= 1

)1

1(2

1

Phase lag

0< <1

Phase lead1< <



Polar plot of a lag-lead compensator;For ( = ) and Kc= 1 is;

)1(,

)1

(

)1

(

)(

)1

(

)1(

)1(

)1(

)1()(

2

2

1

1

2

2

1

1

Ts

Ts

Ts

Ts

K

sT

sT

sT

sTKsG

c

cc

It can be seen that, for 0 < < 1, the compensator acts as a lag compensator,while for 1 < < it acts as a lead compensator.

Im

0

=

Re1 = 0

m)

11(

2

1

= 1

)1

1(2

1

Phase lag

0< <1

Phase lead1< <

21

1

1

TT

The necessary angle m that provided by phase lead part by designing is again; 1

1

11

11

sin

m

The frequency 1 is the frequency at whichthe phase angle is zero. It is given by;


9/17/2001

41


Bode diagram of a lag-lead compensator: Kc=1 T1=1, T2=10T1 and = =10

110-3

10-2

10-1

100

101

102

-20

-10

0

Magnitude(d

B)

10-3

10-2

10-1

100

101

102

-100

0

100

Pha

se(d

eg)

Frequency (rad/sec)

1

1

T

1T

2

1

T

2

1

T

211 /1 TT

Lead

Lag

)1(,

)1

(

)1

(

)(

)1

(

)(

2

2

1

1

Tj

Tj

Tj

Tj

KjG cc



Bode diagram of a lag-lead compensator: Kc=1 T1=1, T2=10T1 and = =10

For <<1/(T2) and >>/T1 the slope of phase lag-lead compensator is 0 dB.The corner frequencies of the lag lead compensator are at =1/T1, = /T1, =1 /T2, and = 1/(T2).

110-3

10-2

10-1

100

101

102

-20

-10

0

Magnitude(d

B)

10-3

10-2

10-1

100

101

102

-100

0

100

Pha

se(d

eg)

Frequency (rad/sec)

1

1

T

1T

2

1

T

2

1

T

211 /1 TT

Lead

Lag


9/17/2001

42


The design of a lag-lead compensator by the frequency-response approach is based on thecombination of the design techniques discussed under lead compensation and lagcompensation.

Assume that the desired performance specifications are given in terms of phase margin,PMd, gain margin, GMd, static error constants, Kp, Kv, and Ka, and so on. Assume that thesystem given in Figure 1 does not satisfy the desired frequency performance then the phaselag lead compensator Gc(s) shown in Figure 2 needs to be designed.

R(s) E(s)G(s)

Y(s)

H(s)Figure 1:

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

Figure 2:

)1(,

)1

(

)1

(

)(

)1

(

)(

2

2

1

1

Tj

Tj

Tj

Tj

KjG cc

Lag-Lead Compensation Techniques Based on the Frequency-Response Approach


The phase lead portion of the lag-lead compensator (the portion involving T1) alters thefrequency-response curve by adding phase-lead angle and increasing the phase margin atthe gain crossover frequency.

The phase-lag portion (the portion involving T2) provides attenuation near and above thegain crossover frequency and thereby allows an increase of gain at the low-frequencyrange to improve the steady-state performance.

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag Lead Comp.

U(s)

)1(,)

1(

)1

(

)(

)1

(

)1(

)1(

)1(

)1()(

2

2

1

1

2

2

1

1

T

s

Ts

Ts

Ts

KsT

sT

sT

sTKsG ccc

Lag-Lead Compensation Techniques Based on the Frequency-Response Approach

9/17/2001

43


1. a) From the system gain K, if the system gain is not fixed; KT=K where Kc=1.1. b) From the controller gain KT=Kc, if the system gain is fixed;

The design steps for designing phase lag lead compensator ( = ).Step 1: If the desired static error constant is given then the necessary gain, KT isdetermined

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Lag Lead Comp.

U(s)

Lag-Lead ( = ) Compensation Techniques Based on the Frequency Response Approach

)1(,

)1

(

)1

(

)(

)1

(

)(

2

2

1

1

Ts

Ts

Ts

Ts

KsG cc


Step 2: Using the gain KT thus determined, draw the Bode diagram ofG1(j)H(j)=KTG(j)H(j) the gain adjusted but uncompensated system.

)()()()( 1 jHjGjHjGKT

Evaluate the phase margin PM and compare with desired PMd.2. a) If |PMd-PM| < 100 then a simple gain Gc(s) = Kc is enough to satisfy the desiredperformance.2. b) If |PMd-PM| 700 then for increase closed loop BW phase lead for decrease BWphase lag is designed.2. c) If |PMd-PM| 700 then phase lag-lead compensator is designed.


G(s)R(s) Y(s)

H(s)

E(s)KT

U(s)

9/17/2001

44


The design steps for designing phase lag lead compensator ( = ).

)125( 00 dm PM

log

log

0

|G1(

j)H

(j

)|[d

B]

-1800

gc

m

pc=ngcngcT

1

2

101


Step 3: Determining the new gain cross overfrequency for the compensated system; Fromphase diagram of G1(j)H(j) the frequencyat G1(j)H(j)=-1800 is selected as newgain cross over frequency pc=ngc.

At new gain cross over frequency, ngc, leadpart of the compensator needs to provide thenecessary phase m;


The design steps for designing phase lag lead compensator ( = ).

Step 4: The corner frequency of lag part =1/T2 (corresponding to the zero) is 1octave to 1 decade below the new gain crossover frequency, ngc.

= 1/T2 =1/[(one octave-one decade)]ngc.

In this course, = 1/T2 = 10-1ngc will be selected.


log

log

0

|G1(

j)H

(j

)|[d

B]

-1800

gc

PM=m

pc=ngcngc

T 1

2

101

9/17/2001

45


The design steps for designing phase lag lead compensator ( = ).Step 5: In order to provide the necessary phase m at new gain crossover frequency ngc thecontroller parameter is determined from;

From this point all the parameters of phase lag part have been designed and

m

mm

sin1

sin1

1

1sin

The zero of phase lag part is;2

1

T

2

1

T

rad/s

rad/sThe pole of phase lag part is;

)1(,

)1

(

)1

(

)(

2

2

Ts

Ts

sGlag

The phase lag part of the compensator dynamic is;



The design steps for designing phase lag lead compensator ( = ).Step 6: The amount of the gain dB needs to be reduced from the gain part of Bodediagram at new gain crossover frequency ngc is determined. The corner frequencies of leadpart of the compensator, = 1/T1 and = /T1 will be designed to provide - dB such that:

log0

|G1(

j)H

(j

)|[d

B]

1= ngc

20 dBdB1 = -

dB2 = 02=x

1

22

11

,0

,

TdB

dB

x

ngc

20loglogloglog

/20log

2

12

ngcxngc

dBdB

decdBdB

log

log

0

|G1(

j)H

(j

)|[d

B]

-1800

gc

m

pc=ngc

20 dB-


9/17/2001

46


The design steps for designing phase lag lead compensator ( = ).Step 6: Ones x=2= /T1 is determined then T1 is obtained and

The zero of phase lead part is = 1 /T1 rad/sec

The pole of phase lead part is = /T1 rad/sec

The phase lead part of the compensator dynamic is;

)11

(,

)(

)1

(

)(

1

1

Ts

Ts

KsG cleadPhase lag-lead compensator for ( = );

)1(,

)1

(

)1

(

)(

)1

(

)1(

)1(

)1(

)1()(

2

2

1

1

2

2

1

1

Ts

Ts

Ts

Ts

K

sT

sT

sT

sTKsG

c

cc

10-3

10-2

10-1

100

101

102

-20

-10

0

Magnitude(d

B)

10-3

10-2

10-1

100

101

102

-100

0

100P

hase(d

eg)

Frequency (rad/sec)

1

1

T

1T

2

1

T

2

1

T

211 /1 TT

LeadLag



The design steps for designing phase lag lead compensator ( = ).Step 7: Draw the Bode diagram of compensated system Gc(j)G(j)H(j) and obtain thePM and GM to compare desired PMd and GMd.

G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)

log 0

|G(j

)H(j

)|[d

B]

gc

I. region

II. region

III. region

-20 dB/d


9/17/2001

47


Example 10.3: Consider the system shown in Figure where the open loop transferfunction is

The desired frequency response performance: It is desired to design a compensator forthe system so that the phase margin is at least 50°,(PMd 500), the gain margin is at least10 dB, (GMd 10 dB) and the static velocity error constant Kv is 10 sec-1,

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

Solution 10.3: Step 1: The system has fixed gain then KT = Kc.Since the static velocity error constant Kv is 10 sec-1, we obtain;

1)(,)2)(1(

1)(

sH

ssssG

202

10

)2)(1(1

1lim)()(lim

00

cT

T

T

sc

sv

KK

K

sss

K

Ts

TsssGssGK

Design Lag Lead Compensator Based on Bode Diagram


Solution 10.3: Step 2: Draw the Bode diagram of KTG1(j)H(j)

)2)(1(

20

)2)(1(

1)(1

jjj

sssKsG

js

T

Magnitudes graph;

Angle graph;

G(s)R(s) Y(s)E(s)

KTU(s)

G1(s)

H(s)

Gain (dB) Angle

10-3 80 -90

10-2 60 -90

10-1 40 -98

100 16 -161.5

1.41 11 -180

2 3.4 -198.4

2.58 0 -211

101 -34.19 -252

102 -93.98 -268

)2log201log20

log20(20log20)()(

2210

210

10101

jHjG

)2

tantan90(0)()( 11001

jHjG

Phase Lag-Lead Design Using Bode Diagram

9/17/2001

48


Solution 10.3: Step 2: Draw a Bode diagram of KTG1(j)H(j)

-100

-50

0

50

100

Magnitu

de (

dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Phase (

deg)

Bode Diagram

Gm = -10.5 dB (at 1.41 rad/sec) , Pm = -28.1 deg (at 2.43 rad/sec)

Frequency (rad/sec)

G1(j)H(j) has PM= -280 at gc=2.4 rad/s.The system is unstable.

Desired phase margin isPMd= 500

)125(78

)125()28(50

)125(

000

00

00

PHPH d

Result: Since |PMd-PM| 700

then a phase lag lead compensator needs to be designed.

pc< gc

The system is unstable.



Solution 10.3: Step 3: The phase diagram G1(j)H(j)=KTG(j)H(j) = -1800 atpc=1.4 rad/sec. We select this frequency as new gain crossover frequency;pc=ngc=1.4 rad/sec.

At = ngc=1.4 rad/s the angle is

10-2

10-1

100

101

-250

-200

-150

-100

Pha

se(d

eg

)

Frequency (rad/sec)

pc=1.4ngc= pc

-1800

m

-1250

000

00

55550

)125(

dm PM

0

4.1125)()()(

ngc

jHjGjGc

The necessary phase m that will provide by the compensator at new gain crossover frequency ngc=1.4 rad/sec is


9/17/2001

49


Solution 10.3: Step 4:The corner frequency for the zero of the lag part of the compensator is

10-2

10-1

100

101

-250

-200

-150

-100

Phas

e(d

eg)

Frequency (rad/sec)

pc=1.4ngc= pc

-1800

m

-1250

14.0101 1

2

ngc

T

14.01

2

T



Solution 10.3: Step 5: At new gain crossover frequency pc=ngc=1.4 rad/sec in order toprovide the necessary phase m the controller parameter is

From this point all the parameters of phase lag part have been determined and

The zero of phase lag part is; rad/s

rad/sThe pole of phase lag part is;

The phase lag part of the compensator dynamic is;

m

m

sin1

sin1

1010

55sin1

55sin10

0

14.01

2

T

014.01

2

T

)110(,)014.0(

)14.0(

)1

(

)1

(

)(

2

2

s

s

Ts

Ts

sGlag


9/17/2001

50


Solution 10.3: Step 6: The amount of the gain dB that needs to be reduced from the gainpart of Bode diagram at new gain crossover frequency pc=ngc=1.4 rad/sec is

11)()(41.11

ngc

jHjG

The corner frequencies of lead part of the compensator, = 1/T1 and = /T1 is designed to provide -=-11 dB such that:

log0|G

1(j

)H(j

)|[d

B]

1= ngc

20 dBdB1 = -

dB2 = 02=x

96.410*4.1

20

4.1log

11

loglog

20/11

2

12

x

xngc

dBdB

122

11

/,0

41.1,11

TdB

dB

x

ngc

decdBdB

/20log

rad/sec



Solution 10.3: Step 6: Ones x=2= /T1 isdetermined then T1 is obtained andthe corner frequencies of lead part ofthe compensator, = 1/T1

and = /T1 are as follows;

log 0

|G1(j

)H(j

)|[d

B]

1= ngc

20 dB/decadedB1 = -

dB2 = 0

2=x=4.96

log

log

0

|G1(j

)H(j

)|[d

B]

-1800

gc

m

pc=ngc

20 dB-

The zero of phase lead part is

= 1 /T1=0.5 rad/sec

and the pole of phase lead part is

= /T1=5 rad/sec

)5(

)5.0(20

)(

)1

(

)(

1

1

s

s

Ts

Ts

KsG clead


9/17/2001

51


Solution 10.3: Step 6:

The phase lag-lead compensator

( = =10) is;

)110(

)014.0(

)14.0(

)5(

)5.0(20

)1

(

)1

(

)(

)1

(

)(

2

2

1

1

s

s

s

s

Ts

Ts

Ts

Ts

KsG cc

0

5

10

15

20

Magnitu

de (

dB

)

10-3

10-2

10-1

100

101

102

-90

-45

0

45

90

Phas

e (

deg)

Bode Diagram

Frequency (rad/sec)

26.014.72/1/1 211 TT

5.01

1

T

5

1

T

014.0

1

2

T

14.01

2

T

211 /1 TTLead

Lag

20 dB

-20 dB



Solution 10.3: Step 7: Draw the Bode diagram of compensated system Gc(j)G(j)H(j)and obtain the PM and GM to compare desired PMd and GMd.

G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)

log0

|G(j

)H(j

)|[d

B]

gc

I. region(Lag part)

II. Region(Lead part) III. region

-20 dB/d

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

1)(

)2)(1(

1)(

sH

ssssG


9/17/2001

52


Solution 10.3: Step 7: Draw the Bode diagram of compensated system Gc(j)G(j)H(j)and obtain the PM and GM compare with desired PMd and GMd.

G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

1)(,)2)(1(

1)(

sH

ssssG

Magnitudes graph;

Angle graph;

)5log202log201log20014.0log20log20(

5.0log2014.0log2020log20)()(

2210

2210

210

221010

2210

221010

jHjG

)5

tan2

tantan014.0

tan90(5.0

tan14.0

tan0)()( 11110110

jHjG



G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

1)(,)2)(1(

1)(

sH

ssssG

Gain (dB) Angle

10-1 25 -135

0.14 20.2 -127

0.26 13.5 -112.7

0.5 8.07 -105.3

100 2.9 -116

1.4 0 -130

1.58 -1.48 -135

2 -4.33 -147

5 -19.7 -199

101 -35.2 -230

Gain (dB) Angle

10-3 80 -93

10-2 58.2 -121

0.014 54.1 -129

0.044 37.2 -144

10-1 25 -135

Solution 10.3: Step 7: Draw the Bode diagram of compensated system


9/17/2001

53


Solution 10.3: Step 7: Draw the Bode diagram of compensated system.

-150

-100

-50

0

50

100

150

Mag

nitu

de (

dB

)

10-4

10-3

10-2

10-1

100

101

102

103

-270

-225

-180

-135

-90

Pha

se (

de

g)

Bode Diagram


Frequency (rad/sec)



)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

1)(

)2)(1(

1)(

sH

ssssG

Solution 10.3: Step 7: Bode diagramsfor G(j)H(j) (gain adjusted butuncompensated open-loop transferfunction), Gc(j) (compensator), andGc(j)G(j)H(j) (compensated openloop transfer function).

G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)


9/17/2001

54


G(s)R(s) Y(s)

H(s)

E(s) Gc(s)

U(s)

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

1)(,)2)(1(

1)(

sH

ssssG

Solution 10.3: Step 7: Obtain the PM and GM from the compensated system Bode diagramGc(j)G(j)H(j) and compare with desired PMd and GMd.

Compensated system PM and GM:PM = 510, gc =1.37 rad/sGM = 13 dB, pg =3.53 rad/sKv =10 sec-1

Desired system PM and GM:PMd 500

GMd 10 dB, Kv =10 sec-1




G(s)R(s) Y(s)

H(s)

E(s)

Figure 1


Figure 2

G(s)R(s) Y(s)E(s)

KT

U(s)

G1(s)

H(s)

c) Finally, the Bode diagram of phase lag-lead compensated system in Figure 3 isobtained and the results are examined.

Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)

Figure 3


9/17/2001

55


Compare the system performance with phase lag lead compensated ( = ) system.a) Bode diagram of the actual system

G(s)R(s) Y(s)

H(s)

E(s)

PM = 540, gc =0.4 rad/sGM = 15 dB, pg =1.4 rad/sKv =0.5 sec-1

BW =0.76 rad/s

10-2

100

102

-150

-100

-50

0

50

Magnit

ude(d

B)

Frequency (rad/sec)

Closed loop system is stable

1)(,)2)(1(

1)(

sH

ssssG

-150

-100

-50

0

50

Mag

nitu

de

(d

B)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Ph

ase

(d

eg)

Bode DiagramGm = 15.6 dB (at 1.41 rad/sec) , Pm = 53.4 deg (at 0.446 rad/sec)

Frequency (rad/sec)

-20 dB/d

-60 dB/d

-40 dB/d

The close loop system performance: the actual system


Compare the system performance with phase lag lead compensated ( = ) system.b) The system with Gc(s) = KT, Bode diagram of KTG(j)H(j);

G(s)R(s) Y(s)

H(s)

E(s)

PM = -280, gc =2.4 rad/sGM = -10 dB, pg =1.4 rad/sKv =10 sec-1

BW =3.19 rad/s

1)(,)2)(1(

20)(1

sH

ssssG

10-2

100

102

-80

-60

-40

-20

0

20

Magnitude(d

B)

Frequency (rad/sec)

Closed loop system is unstable

Close loop system performance: System with Gc(s) = KT

9/17/2001

56


Compare the system performance with phase lag lead compensated ( = ) system.c) Bode diagrams for Gc(j)G(j)H(j) (compensated system)

G1(s)R(s) Y(s)

H(s)

E(s)

-40

-20

0

20

40

60

80

Ma

gn

itud

e (

dB

)

10-3

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Ph

ase

(d

eg)

Bode Diagram


Frequency (rad/sec)

-20 dB/d

-40 dB/d

-40 dB/d

26.014.72/1/1 211 TT

LeadLag

-20 dB

-60 dB

0 dB

-20 dB

1)(,)2)(1(

1)(1

sH

ssssG

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

PM = 510, gc =1.37 rad/sGM = 13 dB, pg =3.53 rad/sKv =10 sec-1

BW =2.46 rad/s

10-2

100

102

-100

-50

0

50

Magnitude(d

B)

Frequency (rad/sec)





-40

-20

0

20

40

60

80

Ma

gn

itud

e (

dB

)

10-3

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Ph

ase

(d

eg)

Bode Diagram


Frequency (rad/sec)

-20 dB/d

-40 dB/d

-40 dB/d

26.014.72/1/1 211 TT

LeadLag

-20 dB

-60 dB

0 dB

-20 dB

1)(,)2)(1(

1)(1

sH

ssssG

)014.0(

)14.0(

)5(

)5.0(20)(

s

s

s

ssGc

PM = 510, gc =1.37 rad/sGM = 13 dB, pg =3.53 rad/sKv =10 sec-1

BW =2.46 rad/s

10-2

100

102

-100

-50

0

50

Magnitude(d

B)

Frequency (rad/sec)


Gc(s)R(s) U(s)

H(s)

E(s)G(s)

Y(s)


9/17/2001

57



10-3

10-2

10-1

100

101

102

-20

-10

0

10

20

30

40

50

60

70

80

Magnitude(d

B)

Frequency (rad/sec)

Gc(j)

G(j)H(j)

G1(j)H(j)

-40 dB/d

-20 dB/d

0 dB/d

-60 dB/d

-20 dB/d

-40 dB/d

-20 dB/d

Uncompansated systemPM = 540, gc =0.4 rad/sGM = 15 dB, pg =1.4 rad/sKv =0.5 sec-1

BW =0.76 rad/s

Compansated systemPM = 510, gc =1.37 rad/sGM = 13 dB, pg =3.53 rad/sKv =10 sec-1

BW =2.46 rad/s



Homework 10.9: Consider the system shown in Figure where the blocks transfer functionare;

The desired frequency response performance: It is desired to design a compensator forthe system so that the static velocity error constant Kv is 40 sec-1, the phase margin is atleast 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)

)1(1)(,)20)(1(

10)(

sH

ssssG


The desired frequency response performance: It is desired to design Phase Lag-Leadcompensator for the system so that the static velocity error constant Kv is 10 sec-1, the phasemargin is at least 40°,(PMd 100) and the gain margin is at least 10 dB, (GMd 10 dB).

)2()4)(1(

)()(

sss

KsHsG

Phase Lag-Lead Design Using Bode Diagram: Homework

9/17/2001

58


Homework 10.11: Consider the system shown in Figure where the blocks transferfunction are;

The desired frequency response performance: It is desired to design a compensator forthe system so that the static velocity error constant Kv is 40 sec-1, the phase margin is atleast 50°,(PMd 500) and the gain margin is at least 10 dB, (GMd 10 dB).

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Compensator

U(s)


The desired frequency response performance: It is desired to design Phase Lag-Leadcompensator for the system so that the static velocity error constant Kv is 20 sec-1, the phasemargin is at least 45°,(PMd 450) and the gain margin is at least 8 dB, (GMd 8 dB).

)3(1)(,)10)(5(

100)(

sH

ssssG

)4()5)(1(

1)()(

ssssHsG

Phase Lag-Lead Design Using Bode Diagram: Homework


This part of the lecture has presented detailed procedures for designing lead, lag, and lag-lead compensators by the use of conventional design methods (use of the root-locus orfrequency-response methods). Generally simple examples have been considered.

It is shown that the design of a compensator to satisfy the given specifications (in terms ofthe phase margin and gain margin) can be carried out in the Bode diagram in a simple andstraightforward manner.

It is noted that not every system can be compensated with a lead, a lag, or a lag-leadcompensator. In some cases compensators with complex poles and zeros may be used. Forsystems that cannot be designed by conventional design methods (use of the root-locus orfrequency-response methods), the pole-placement method may be used.

In a given design problem if both conventional design methods and the pole-placementmethod can be used, conventional methods (root-locus or frequency-response methods)usually result in a lower-order stable compensator. Note that a satisfactory design of acompensator for a complex system may require a creative application of all available designmethods.

Frequency-Response Approach to Control System Design: Concluding Comments

9/17/2001

59


1. Lead compensation achieves the desired result through the merits of its phase leadcontribution, whereas lag compensation accomplishes the result through the merits of itsattenuation property at high frequencies.

2. Lead compensation is commonly used for improving stability margins. Leadcompensation yields a higher gain crossover frequency than is possible with lagcompensation. The higher gain crossover frequency means a larger bandwidth. A largebandwidth means reduction in the settling time. The bandwidth of a system with leadcompensation is always greater than that with lag compensation. Therefore, if a largebandwidth or fast response is desired, lead compensation should be employed. If,however, noise signals are present, then a large bandwidth may not be desirable, since itmakes the system more susceptible to noise signals because of an increase in the high-frequency gain.

Comparison of Phase Lead, Lag, and Lag-Lead Compensation


3. Lead compensation requires an additional increase in gain to offset the attenuationinherent in the lead network. This means that lead compensation will require a larger gainthan that required by lag compensation. A larger gain, in most cases, implies larger space,greater weight, and higher cost.

4. The lead compensation may generate large signals in the system. Such large signals arenot desirable because they will cause saturation in the system.

5. Lag compensation reduces the system gain at higher frequencies without reducing thesystem gain at lower frequencies. Since the system bandwidth is reduced, the system hasa slower speed to respond. Because of the reduced high-frequency gain, the total systemgain can be increased, and thereby low-frequency gain can be increased and the steady-state accuracy can be improved. Also, any high frequency noises involved in the systemcan be attenuated.


9/17/2001

60


6. Lag compensation will introduce a pole-zero combination near the origin that willgenerate a long tail with small amplitude in the transient response.

7. If both fast responses and good static accuracy are desired, a lag-lead compensatormay be employed. By use of the lag-lead compensator, the low-frequency gain can beincreased (which means an improvement in steady-state accuracy), while at the sametime the system bandwidth and stability margins can be increased.

8. Although a large number of practical compensation tasks can be accomplished withlead, lag, or lag-lead compensators, for complicated systems, simple compensation byuse of these compensators may not yield satisfactory results. Then, differentcompensators having different pole-zero configurations must be employed.



Phase lead Compensator: with passive elements

Ts

Ts

Ts

Ts

sE

sE

i

1

1

1

1

)(

)(0

1

1

)(

)(

21

21

1

21

2

CsRR

RR

CsR

RR

R

sE

sE

i

o

1,21

21

RR

RTCRei

R1i1

Ci2

eoR2

)1(,1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc

Electronic circuits for Phase Lead Compensator

9/17/2001

61


Phase lead Compensator: R1C1>R2C2

e1ei

+-

R2i3

C2i4

R1i1

C1i2

e1e +

-

e0

i5R3

R4i6

Phase lead Phase inverter

3

40

)(

)(

R

R

sE

sE

22

11

2

1

22

11

1

2

1

1

1

1

)(

)(

CRs

CRs

C

C

sCR

sCR

R

R

sE

sE

i

CRCR

CRTCRT

211

2211 ,

)1(,1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc

13

24

11

22

23

14

23

14

RR

RR

CR

CR

CR

CRK

CR

CRK

c

c

Electronic circuits for Phase Lead Compensator using operational amplifiers


Phase lag Compensator: with passive elements

Ts

Ts

Ts

Ts

sE

sE

i

1

11

1

1

)(

)(0

1)(

1

)(

)(

21

2

CsRR

CsR

sE

sE

i

o

1,2

212

R

RRTCRei

R1

Ceo

R2

)1(,1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc

Electronic circuits for Phase Lag Compensator

9/17/2001

62


Phase lag Compensator: R1C1<R2C2

e1ei

+-

R2i3

C2i4

R1i1

C1i2

e1e +

-

e0

i5R3

R4i6

Phase lag Phase inverter

3

40

)(

)(

R

R

sE

sE

22

11

2

1

22

11

1

2

1

1

1

1

)(

)(

CRs

CRs

C

C

sCR

sCR

R

R

sE

sE

i

CRCR

CRTCRT

211

2211 ,

)1(,1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc

13

24

23

14 ,

RR

RRK

CR

CRK

c

c

Electronic circuits for Phase Lag Compensator using operational amplifiers


Phase lag-lead Compensator: with passive elements

ei

R1

C2eo

R2

C1

)1(

,,

21

212211

222111

TT

CRCRCR

TCRTCR

2

2

1

1

2

2

1

10

1

11

1

1

1

1

)(

)(

Ts

Ts

Ts

Ts

sT

sT

sT

sT

sE

sE

i

sCRsCRsCR

sCRsCR

sE

sE

i

o

212211

2211

)1)(1(

)1)(1(

)(

)(

)1(,1

11

1

1

1

1)(

2

2

1

1

2

2

1

1

T

s

Ts

Ts

Ts

KsT

sT

sT

sTKsG ccc

Electronic circuits for Lag-Lead Compensator

9/17/2001

63


Phase lag-lead Compensator:

e1

ei+

-

R4i3

C2i4

R1

i1

C1i2

e1e +

-

e0

i5 R5

R6i6R3

R2

1

,1

,

2

42

1

31

42

31

531

642

R

RR

R

RR

RR

RR

RRR

RRRKc

22

11

23

14

242

22

11

131

53

640

1

1

1)(

1

1

1)(

)(

)(

CRs

CRs

CR

CR

sCRR

sCR

sCR

sCRR

RR

RR

sE

sE

i

)1,(,1

11

1

1

1

1)(

2

2

1

1

2

2

1

1

Ts

Ts

Ts

Ts

KsT

sT

sT

sTKsG ccc

Electronic circuits for Phase Lag-Lead Compensator using operational amplifiers


A tachometer is one of the rate feedback devices.

Another common rate feedback device is the rate gyro. Rate gyros are commonly used inaircraft autopilot systems.

Velocity feedback using a tachometer is very commonly used in positional servo systems. Itis noted that, if the system is subjected to noise signals, velocity feedback may generatesome difficulty if a particular velocity feedback scheme performs differentiation of theoutput signal. (The result is the accentuation of the noise effects.)

Feedback Compensations

9/17/2001

64

Electrical & Electronics Engineering

EEE 352 Automatic Control Systems

Control Systems Design by Frequency Response

Remarks and Questions?


Chapter 10 Control Systems Design by Frequency Response[1]

Documents

Transcript of Chapter 10 Control Systems Design by Frequency Response[1]