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Control Theory II

Part one : Linear Control System

Ch.1 Frequency Response AnalysisCh.2 Bode diagramCh.3 Bode DesignCh.4 Nyquist Stability CriterionCh.5 Nichols Method

Part Two : Non-Linear Control System

Ch.6 Describing FunctionCh.7 Phase Plane Analysis

References :

1. Ogata, K. Modern Control Engineering 1996

2. Saeed, H. Syed Automatic Control Systems 2004

3. Raven, F. Automatic Control Engineering 1981

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Ch.1 Frequency Response Analysis

Design of communication systems depends upon the frequency response sincemost of the signals are to be processed are either sinusoidal or composed of

sinusoidal components.Stable linear time invariant system with sinusoidal input t X t x sin)( ,

then the steady state output has the same frequency as the input but with differentamplitude and certain phase shift )sin()( t Y t y .

Example 1 : Consider the system

with t X t x sin)( .Determine the y ss(t).solution: put js

as )90sin(1

small astsin)tan-t(sin1

)(y

T-tanand 1

1-

22ss

22

t

K X T T

XK t

T

K G

Since the phase angle is negative, this network called phase lag network .

Example 2 : Consider the system with transfer function

2

1

1

1

)(

T s

T s

sG

Determine whether this network is a lead or lag network?Solution:

For sinusoidal input t X t x sin)(

)1()1(

1

1

)(21

12

2

1

jT T jT T

T j

T j

jG

21

11

2221

2212

tantan

1

1

T T

T T

T T G

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Thus the steady state output is :

)tantansin(1

1)( 2

11

1

2221

2212

T T t

T T

T T X t y ss

if T1 > T2 then >0 which means we have lead network if T1 < T2 then

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Bode diagram

2.1 Bode Plots of Elementary Functions:In the Bode plot, the magnitude of the open loop transfer function in decibels (db)

and the phase angle are separately plotted as a function of frequency (). In general, the

numerator and denominator of any transfer function have four basic types of factors:

(1) a constant K: The constant K has a magnitude |k| and plotted as a horizontalline with magnitude (db) k log20K

The phase angle of a constant K is:=0o for positive k

=-180 o for negative k

(2) pole or zero at the origin n j )( :

(a) pole of order 1 at the origin (1/j ):

log 201

log 201

j jBode plot is a straight line with

slop=-20 db/decade, crossing the point =1 rad/sec. The phase angle of (1/j ) is -90 o.

(b) For zero of order 1 at the origin )( j :

log20 j and phase angle=+90 o

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(3) Simple pole or zero nT j )1( :a. Simple pole 1)1( T j :

at-90

T1

at45

0at0tan-angle phase T

10atdb-20

T1

atdb0

Tlog 20Gthen1,Tfrequencyhighat

01log 20Gthen1,Tfrequencylow

1log 20

o

1-

2 2

o

oT The

at

T G

(b) Simple zero :)1( T j

at90 T1

at45

0at0tan-angle phase T

10atdb-20

T1

atdb0

Tlog 20Gthen1,Tfrequencyhighat

32log 20G T1

at

01log 20Gthen1,Tfrequencylow

1log 20

o

1-

2 2

o

oT The

db

at

T G

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(4) Quadratic Factors : :])()(21[ 12

nn

j j

For >1 the quadratic factor can be factored into two first order factors with realroots. For 10 the quadratic factor have two conjugate complex roots, therefore;

180-

-90

0at01

/2-tan

frequencyhighatlog40-

frequency lowat0)2

()-(1log 20

n

2

21-

n

2

n2

2

n

n

nG

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2.2 Relative Stability:The gain crossover frequency ( i): is the frequency at which 1)( jGH . The

phase margin PM is that amount of additional phase lag at gain crossover frequency to bring the system to the verge of instability.

PM= 180 o + arg GHThe phase crossover frequency ( ) is that frequency at which the phase angle of

GH is -180 o. The GM is the reciprocal of the magnitude of the open loop transfer function at

db )( log 20)(

1

jGH jGH

GM

If gain crossover frequency is less than the phase crossover frequency, the systemis stable. But if phase crossover frequency is less than the gain crossover frequency, thesystem is unstable.

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Example 1: Consider the Bode plot for the open loop transfer function

)1.01)(6.01(5

)(sss

jG

Solution:

Put s=j then)1.01)(6.01(

5)(

j j j jG

1.0tan6.0tan90

)1.0(1log20)6.0(1log20log205log2011

22

oo

G

log 5=20 x 0.698=14 dbCorner frequencies at: 1.67, 10 rad/sec

1 2 3 4 5 10

o -127 -152 -168 -179 -188 -219.5

num=5;den=conv([0.6 1 0],[0.1 1]);

bode (num,den),gridSince positive PM & positive GM, the system is stable.

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Example:)02.01)(1.01)(4.01(

10)(

ssssG

02.0tan1.0tan4.0tan

)02.0(1log20)1.0(1log20)4.0(1log2010log20111

222

o

G

Corner frequencies at: 2.5, 10, and 50 rad/sec.Since positive PM & positive GM, the system is stable.

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Example: Find the transfer function of the given Bode diagram

Solution:1. First line having a slope of a +20 db/dec, therefore there is a term S in

numerator.2. At =1, the slope changes to zero, it means there is a term (1+s) in denominator.3. At =10, the slope changes to -20 db/dec, this indicates there is a term (1+s/10) in

denominator.4. 20 log k = 6, k=1.99=2

)10)(1(20

)10/1)(1(2

)(sss

sss

sG

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Ch.3 Bode Design3.1 Lead Compensation:

Consider a lead compensator having the following transfer function

T s

T s

k

Ts

Tsk sG

c

c

1

1

10

1

1)(

Where is called the attenuation factor of the lead compensator. It has a zero at s= -1/Tand a pole at s=-1/ T.

The minimum value of is about 0.05, this means that the maximum phase lead anglethat may be produced by a lead compensator is about 65 o.

Example: Consider the system with open loop transfer function)2(

4)(

sssG . It is

desired to designed a compensator for the system so that the static velocity error constantk v=20 sec -1, the PM is at least 50 o, and GM is at least 10 db.Solution:1. Gain adjustment to meet the required k v

10

2)2(

4Ts1

Ts1 s limGs lim c

k

k ss

k Gk v

2. plot the Bode diagram:

)2(

40)(

)2(40

)()(

1

1

j j

jG

ssskGsG

1 2 5 10 20 50 100

o -116 -135 -158 -169 -174 -178 -179

From the Bode plot : PM=17 o and GM=+ db3. Since the required PM at least 50 o thenm=(50-17)+(5 to 12)

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Addition of 5 o to 12 o is to compensate the shift in gain crossover frequency.m = 50-17+5=38 o

4. Determine the attenuation factor :

24.038sin138sin-1

sin1sin-1

or

11

)sin(

m

m

m

then determine the frequency of the uncompensated system G 1(s) where

dbT j

T jG

T

2.649.01

24.0

111

111

From plot at |G 1|=-6.2 db the c = 9 rad/sec5. determine the corner frequency of the lead compensator:

Zero at 4.419x24.01

1

cT T

Pole at 4.1824.0

91 1

cT T

The lead compensator thus determined;

sk

ss

k sG ccc 054.010.227s1

4.18

41.4)(

6. 7.4124.0

10

k k c

ss

ss

sG c 054.01227.01

104.18

41.47.41)(

The open loop transfer function of the compensated system is:

)2(4

4.1841.4 7.41)()(

sssssGsG c

7. Cheek the PM, GM and k v values:PM=50 o , GM=+ db , System type 1 then k v= 20.

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H.W.: Design a cascade compensator for the system whose transfer function is:

)001.01)(1.01()(

sssk

sG

PM 45o and k v=1000 sec -1

Answer:()001.01)(1.01(

1000)(1

ssssG , From Bode plot: PM=0

o

Take m=(45-0)+5 = 50 o , =0.13247 , dbGT

75.81

11

At -8.75 db, c =17 o rad/sec. T = 0.00214, lead compensator thus determined;

ssG c 00214.01

0.016s1

51.71

)( , then the compensated system has PM=45 o

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3.2 Lag Compensator:Consider a lag compensator with the following transfer

function:

1)( 1

1

11

)(

T

s

T s

k Ts

Tsk sG ccc

Lag compensator has a zero at s=-1/T and a pole at s=-1/ T.

The lag compensator is essentially a low pass filter.

Example: Consider the system with open loop transfer

function)15.0)(1(

1)(

ssssG . It is desired to designed a compensator for the

system so that the static velocity error constant k v=5 sec-1, the PM is at least 40 o, and GM

is at least 10 db.Solution:

1. Gain adjustment:

5Ts1

Ts1 s limGs lim 1c

k

k GGk v

2. Plot the Bode diagram of )15.0)(1(

5)(1 sss

sG .

From the plot: PM=-20 o and GM=-8 db {unstable system}Addition of a lag compensator modifies the phase curve of the Bode plot.

3. From Bode plot, the PM=40 o corresponding to =0.7 rad/sec. To avoid overlylarge time constants for a lag compensator, we shall choose =1/T=0.1 rad/sec(Note too far below 0.7 rad/sec).

m=40+12=52 0

At -128 o the frequency =0.5 rad/sec and |G 1|=-20 db.

4. To bring the magnitude curve down to 0 db at this new gain cross over frequency=0.5 rad/sec

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10 log2020

1log20)5.0(1

G

The other corner frequency 01.01010

11 xT

rad/sec

1001

101

1001101

10 )(s

sk

ss

k sG ccc

5. since the k=5 and =10, then k c=k/ =5/10=0.5

)1001)(15.0)(1()101(5

)()(ssss

ssGsG c

Then PM=40 o, GM=11 db, and k v = 5 sec -1.

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3.3 Lag-Lead Compensator:Consider the lag-lead compensator given by:

Term LagTerm Lead2

2

1

1

1 and 1 rewhe)1

1

( )

1

( )(

T s

T s

T s

T s

k sG cc

we frequently choose =. From Bode plot with T 2=10T 1, kc=1, and =10, it can be seenthat:for 0< < 1 the compensator acts as a lag compensator while 1< < 2 the compensator acts as a lead compensator and

21

11

T T

the lag-lead compensator has the form:

1 )

1)((

)1

)(1

(k )

11

( )1

1( )(

21

21c

2

2

1

1

T s

T s

T sT s

sT sT

sT

sT k sG cc

Example: Consider the system with open loop transfer function)2)(1(

1)(

ssssG .

It is desired to designed a compensator for the system so that the static velocity error constant k v=10 sec -1, the PM is at least 50 o, and GM is at least 10 db.Solution:

1. Gain adjustment:

20

102/)2)(1(

G s limGs lim cc

k

k sss

k Gk v

2. Draw Bode plot with k=20. From plot, PM=-32 o {unstable}.3. Design the phase lag portion of compensator:

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o jG 180)( at =1.5 rad/sec. We choose =1.5 rad/sec as a gain crossover frequency. Also choose

2

1T

(zero of the phase lag portion) then the new gain

crossover frequency at =0.15 rad/sec.

1

1)sin(

m . Note that at =10, m =54.9

o

Then let =10 and 015.01

2T rad/sec (pole of the phase lag portion). The

transfer function of the phase lag portion is:

ss

ss

7.66167.61

10015.015.0

4. Design of the phase lead portion of the compensator:At =1.5 rad/sec, G(j1.5)=13 db. From point (1.5 rad/sec,-13 db) draw a straightline of slop +20 db/dec. The intersection of this line with 0 db line and -20 dblines determine the corner frequencies of the phase lead portion. i.e. =0.7 and=7 rad/sec.

ss

ss

143.0143.11

101

77.0

5. Combining the transfer functions: )

7.66167.61

)(143.0143.11

()015.015.0

)(7

7.0()(

ss

ss

ss

ss

sG c

the open loop transfer function :

)5.01)(1)(7.661)(143.01()67.61)(43.11(10

)2)(1)(015.0)(7()15.0)(7.0(20

)()(

sssssss

ssssss

sGsG c

From Bode plot: PM=50 o , GM=16 db and k v = 10 sec -1 .