1 k (N/m) f(t) m x(t) C (Ns/m) f d (t) LINEAR CONTROL x(t) is the output variable and is measured by...

1

k (N/m)

f(t)

m x(t)

C (Ns/m)

fd(t)

LINEAR CONTROLLINEAR CONTROL

x(t) is the output variable and is measured by a displacement sensor. The output of a sensor is always an electrical output mainly voltage.

In the Figure, f(t) is the actuating force that excites the system in order to produce the desired output x(t). On the other hand, fd(t) is the disturbance that works against actuating force and prevents the system to reach the desired output.

2

Actuators:

Actuators are used to drive the engineering systems. DC motors, servo motors, hydraulic and pneumatic system elements are the basic type of actuating elements. The inputs of the actuators are also voltage. One can write a transfer function in order to model the behaviour of an actuator.

Gact

V1(Force, Moment, etc.)

)(

)(

1 sV

sFGact

(Voltage)

3

Sensors:

Sensors are the basic components of a control system that are used to measure the output/outputs of the system in order to obtain a feedback signal. The output of a sensor is mainly a voltage signal. One can write a transfer function in order to relate the input and output of the sensor. The input of a sensor is a physical event, temperature, pressure, displacement, velocity, acceleration, magnetic flux, rate of the fluid flow etc.)

Gsens

Displacement,pressure, temperature V2

(Voltage)

)(

)(2

sT

sVGsens

4

kcsmssG

21

)(

-Fd

Gsens

V2

+Gact

V1

+F(t) x(t)

100 N/Volt 0.1 Volt/m

OPEN-LOOP CONTROLOPEN-LOOP CONTROL

2211

VGkcsms

FGV sensdact

100actG 1.0sensG

5

221 1.01

100 Vkcsms

FV d

)(1.0

)(10

)(2122 sF

kcsmssV

kcsmssV d

Let’s take Fd=0

01.0

)(10

)(2122

kcsmssV

kcsmssV

)(10

)( 122 sVkcsms

sV

6

Objective: The desired steady-state value of x(t) is 0.05 m. In this case, the sensor output voltage is calculated as

VoltxV 005.005.01.02

The input voltage which is sufficient to obtain the output voltage, V2 is calculated as

ssss Vkcsms

V 12210

m=200 kgc=300 Ns/mk=5000 N/m

The steady-state behaviour doesn’t include time variations and the system considered as static, then

ssss Vk

V 1210

ssV15000

10005.0 VoltV ss 5.21

7

)(10

)( 122 sVkcsms

sV

ssssV

5.2

5000300200

10)(

22

ssssV

5.2

50003002002001

102001

)(2

2

255.1

125.0)(

22

sss

sV

8

255.1

125.0)(

22

sss

sV

255.1125.025

125.0125.025

)(2

2

sss

sV

255.1

25005.0)(

22

sss

sV

252 nsradn /5

5.12 n

15.052

5.1

x

9

)2()(

22

2

nn

n

ssssG

)1sin(

1

11)( 2

2

tetg n

tn )(cos 1

)15.0(cos15.015sin

15.01

11005.0)( 125*15.0

22 tetV t

42.1943.4sin00506.0005.0)( 75.02 tetV t

1.0

)()( 2 tV

tx

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9x 10

-3

Time (Second)

V2(

t) (

Vol

t)

V2(t)

V2ss=0.005 Volt

The Figure shows that, desired x(t) value is achieved if the external disturbance Fd is zero.

If Fd is not zero!

10

Let’s take Fd=80 N.

)(5000300200

1.0)(

5000300200

10)(

2122 sFss

sVss

sV d

sssssssV

80

5000300200

1.05.2

5000300200

10)(

222

ssssV

17

5000300200

1)(

22

ssssV

17

50003002002001

12001

)(2

2

255.1

250034.0

255.1085.025

085.0085.025

255.1

085.0)(

2222

sssssssss

sV

11

42.1943.4sin003439.00034.0)( 75.02 tetV t

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9x 10

-3

Time (second)

V2(t)

Fd=0

Fd=80 N

12

Can we obtain desired x(t) even if Fd is not zero. This is the main concern of a control process. This can be achieved via the impementation of a closed-loop control.

Vref

kcsmssG

21

)(

-Fd

Gsens

V2+

Gact

V1

+F(t) x(t)

100 N/Volt 0.1 Volt/m

+ Gcont

-

E(Error)

Feedback

2221

VGkcsms

FGGVV sensdactcontref

13

2221

VGkcsms

FGGVV sensdactcontref

kcsmsGGGVV

kcsmsGF

kcsmsGGGV sensactcontsensdsensactcontref 22222

111

kcsmsGGGV

kcsmsGF

kcsmsGGGV sensactcontsensdsensactcontref 2222

11

11

d

sensactcont

sens

ref

sensactcont

sensactcont

F

kcsmsGGG

kcsmsG

V

kcsmsGGG

kcsmsGGG

V

2

2

2

2

21

1

1

11

1

d

cont

ref

cont

cont

F

kcsmsG

kcsmsV

kcsmsG

kcsmsG

V

2

2

2

2

21

101

11.0

1101

110

14

d

cont

ref

cont

cont

F

kcsmsG

kcsmsV

kcsmsG

kcsmsG

V

2

2

2

2

21

101

11.0

1101

110

dcont

refcont

cont FGkcsms

VGkcsms

GV

10

1.0

10

10222

The form of the transfer function of the controller is the most important parameter in a control application. Gcont can be Proportional, Derivative, Integral or a suitable combination of these three choices.

sT

sTKG d

ipcont

11 i

pi T

KK

dpd TKK

15

Kp is the proportional constont, Ki is the integral constant and Kd is the derivative constant.

dcont

refcont

cont FGkcsms

VGkcsms

GV

10

1.0

10

10222

For proportional control, Gcont=Kp

)(10

1.0)(

10

10)(

222 sFKkcsms

sVKkcsms

KsV d

pref

p

p

Vref=0.005 Volt, Fd=80 N

Let’s take Kp=1000

sssssssV

80

1000*105000300200

1.0005.0

1000*105000300200

1000*10)(

222

16

sssssssV

80

1000*105000300200

1.0005.0

1000*105000300200

1000*10)(

222

sssssssV

1

15000300200

81

15000300200

50)(

222

sssssssV

1

15000300200200

1

8200

11

15000300200200

1

50200

1

)(22

2

sssssssV

1

755.1

04.01

755.1

25.0)(

222

sssssssV

1

755.104.0

75

04.004.0

751

755.125.0

75

25.025.0

75

)(22

2

17

sssssssV

1

755.1

75000533.0

1

755.1

7500333.0)(

222

755.1

750028.0)(

22

sss

sV

sradn /66.875 0866.066.8*2

5.1

rad484.1)0866.0(cos 1

sssssssV

1

755.104.0

75

04.004.0

751

755.125.0

75

25.025.0

75

)(22

2

18

)0866.0(cos0866.0166.8sin

0866.01

0028.00028.0)( 1266.8*0866.0

22 tetV t

484.1627.8sin00281.00028.0)( 7499.02 tetV t

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6x 10

-3

Open-loop

Closed-loop, Kp=1000.

Desired

V2(t)

Time (second)

19

For Kp=10000

sssssssV

80

10000*105000300200

1.0005.0

10000*105000300200

10000*10)(

222

sssssssV

1

105000300200

81

105000300200

500)(

222

sssssssV

1

1050003002002001

82001

1

1050003002002001

5002001

)(22

2

sssssssV

1

5255.1

04.01

5255.1

5.2)(

222

sssssssV

1

5255.104.0

525

04.004.0

5251

5255.15.2

525

5.25.2

525

)(22

2

20

sssssssV

1

5255.1

5250000762.0

1

5255.1

525004762.0)(

222

5255.1

5250046858.0)(

22

sss

sV

sradn /91.22525 032.091.22*2

5.1

rad538.1)032.0(cos 1

)032.0(cos032.0191.22sin

032.01

0046858.00046858.0)( 1291.22*032.0

22 tetV t

538.1887.22sin0046882.00046858.0)( 733.02 tetV t

21

538.1887.22sin0046882.00046858.0)( 733.02 tetV t

ess

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9x 10

-3

Desired

Open-loop

Closed-loop, Kp=1000

Closed-loop, Kp=10000

V2(t)

Time (second)Kp ess

22

By MATLAB Simulink

23

dcont

refcont

cont FGkcsms

VGkcsms

GV

10

1.0

10

10222

For proportional-derivative (PD) control, Gcont=Kp+KdsKp=10000, Kd=100

dref Fsss

Vsss

sV

)1001000(105000300200

1.0

)10010000(105000300200

)10010000(10222

dref Fss

Vss

sV

1050001300200

1.0

1050001300200

1000100000222

ssssss

sV

80

1050001300200

1.0005.0

1050001300200

1000100000222

ssssss

sV

1

1050001300200

81

1050001300200

5500222

24

ssssss

sV

1

1050001300200

81

1050001300200

5500222

ssssss

sV

1

10500013002002001

82001

1

10500013002002001

55002001

222

5255.6

04.0

5255.6

5.2025.0222

ssssss

sV

5255.604.0

525

04.004.0

525

5255.6025.0

525

5.2025.0025.0

525

222

ssssss

sV

5255.6

5250000762.0

5255.6

5250052500004762.0

222

ssssss

sV

25

PD Control by Simulink

Reference Voltage:0.005 Volt

Disturbance:-80 N

26

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8x 10

-3

Desired

Time (second)

V2(t)

PD Control by Simulink

27

Gc

PID Control by Simulink

Controller

28

Effect of the PID parameters on the output

P Control: Kp=1000

PD Control: Kp=1000, Kd=1000

PI Control: Kp=1000, KI=1000

PID Control: Kp=1000, KI=1000, Kd=1000

Desired

Output

29

Effect of the PID parameters on the output

30

STABILITY OF LINEAR CONTROL SYSTEMS

Among the many forms of performance specifications used in design, the most important requirement is that the system be stable. An unstable system is generally considered to be useless.

The stability of a control system is directly related to the location of the roots of the charactersitic equation D(s).

iii js

s-plane

Unstable region

Unstable region

Stable region

Stable region

0

j

31

Example 1:

100159

12 ss

V1(s) V2(s)

Consider an open-loop control system

159

100

)(

)()(

21

2

sssV

sVsG

)(159

100)( 122 sV

sssV

D(s)

32

D(s)=s2+9s+15

by MATLAB

>>a=[1 9 15];

>> roots (a)

s1=-2.2087

s2= -6.7913

The two roots of D(s) are in the left hand side of the s-plane. It can be easily said that the open-loop system is stable.

Response of the system to a unit step input can be obtained by MATLAB as

>>ns=[100];

>>ds=[1 9 15],

>>step(ns, ds) 0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

7Step Response

Time (sec)

Am

plitu

de

Time (s)

V2(

t)V2ss=6.66

s-plane

-2.2087-6.7913

33

Example 2:

100753

12 ss

V1(s) V2(s)

753

100

)(

)()(

21

2

sssV

sVsG

)(753

100)( 122 sV

sssV

D(s)

>>a=[1 3 75];>> roots (a)

s1=-1.5+8.53is2=-1.5-8.53i

34


>>ns=[100];

>>ds=[1 3 75],

>>step(ns, ds)

-1.5+8.53 i

-1.5-8.53 i

s-plane

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5Step Response

Time (sec)

Am

plitu

de

V2ss=1.33

35

10015

12 s

V1(s) V2(s)

Example 3:

15

100

)(

)()(

21

2

ssV

sVsG

)(15

100)( 122 sV

ssV

D(s)

>>a=[1 0 15];

>> roots (a)

s1=+3.8730i

s2= -3.8730i

The real part of the roots are zero. No roots on the right-half s-plane and system is said to be marginally stable.

36


>>ns=[100];

>>ds=[1 0 15],

>>step(ns, ds)

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14Step Response

Time (sec)

Am

plitu

deV

2(t)

Time (s)

3.873 i

-3.873 i

s-plane

37

Example 4:

100162

12 s

V1(s) V2(s)

162

100

)(

)()(

21

2

ssV

sVsG

)(162

100)( 122 sV

ssV

D(s)

>>a=[1 -2 16];

>> roots (a)

s1=1+3.8730i

s2= 1-3.8730i

The real part of the roots are positive. Two roots are on the right-half s-plane and system is said to be unstable.

38


>>ns=[100];

>>ds=[1 -2 16],

>>step(ns, ds)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-1000

-800

-600

-400

-200

0

200

400Step Response

Time (sec)

Am

plitu

deV

2(t)

Time (s)

1+3.873 i

1-3.873 i

s-plane

Unstable

39

Example 4:Consider a closed-loop control system

)()(

)(2 sHsR

sVfunctionTransfer

Mason Formula:

LoopsClosed

PathsForwardsH

1)(

V2(s)K

595

323

sss

sR(s)+

-

System

Controller

40

K595

323

sss

sR(s)+

-

V2(s)

2232595

3V

sss

sKVR

595

31

595

323223 sss

sKV

sss

sKR

)53()9(5

3

595

)3(1

595

)3(

23

23

232

KsKss

KKs

sss

sKsss

sK

R

V

41

K595

323

sss

sR(s)+

-

V2(s)

By using Mason formula

)53()9(5

3

595

31

595

3

23

23

232

KsKss

KKs

sss

sK

sss

sK

R

V

Forward Path

Loop

Is the closed-loop control system stable? For what values of K, the system is said to be stable?

D(s)

42

Routh-Hurwitz criterion: This criterion is an algebraic method that provides information about the stability of a linear time invariant system that has a characteristic equation with constant coefficients. The criterion tests whether any of the roots of characteristic equation lie in the right-half s-plane. The number of roots that lie on the j axis and in the right half plane is also indicated.

)53()9(5)( 23 KsKsssD

s3 1 K+9

s2 5 3K+5

s1 0

s0 3K+5 0

5

402

5

53455

5

)53(*1)9(*5

KKKKK

5

402 K

53

5)53(*1)9(*5

0*5)53(*5

)53(*1)9(*5

KKK

KKK

All terms must have the same sign. The number of positive roots is equal to the change in the sign.

43

K+9 >0 , K>-9

3K+5 > 0, K> -1.66

0.4K+8>0, 0.4K>-8, K>-8/0.4, K>-20

K>-1.66 System is stable for all positive K values.

For K=10; Rsss

sV

35195

3010232


>>ns=[10 30];

>>ds=[1 5 19 35],

>>step(ns, ds)

s1= -1.1333 + 3.3941is2= -1.1333 - 3.3941is3= -2.7334

s-plane

-2.7334

-1.1333+3.3941i

-1.1333-3.3941i

44

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

Response to a unit step input, V2(t)

V2(

t)

Time (s)

Stable

45

Example 1:

K 500100

2043

sss

ssR(s)+

-

V2(s)

204500100

204

500100

2041

500100

204

)(

)(3

3

32

ssKsss

ssK

sss

ssK

sss

ssK

sR

sV

KKsKssss

KsKKs

sR

sV

802450000600

8024

)(

)(2345

22

D(s)

46

KKsKsssssD 802450000600)( 2345

K>0, 24K>0, 80K>0 K>0

s5 1 50000 24K

s4 600 K 80K

s3 0

s2 0

s1

s0

600

*150000*600 K

600

80*124*600 KK

60010*3

60014320

600*600

10*3

7

7

K

KK

K

Kx

KK

7

2

103

21408000K80

K

KKxx

21408000600

1440010113256.3102.7 21116

K80

600

*1103 7 Kx

77

1030600

*1103xK

Kx

77

2101408.20

103

21408000xK

Kx

KK

010510162.2 1172 xKxK 75 101386.21034.2 xKx

0K

75 101386.21034.2 xKx

470 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5x 10

9 Step Response

Time (sec)

Am

plitu

deK=3x107

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

0.5

1

1.5

2

2.5Step Response

Time (sec)

Am

plitu

de

K=2.1386x107

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8Step Response

Time (sec)

Am

plitu

de

K=5x105

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.5

0

0.5

1

1.5

2Step Response

Time (sec)

Am

plit

ude

K=2.34x105

0 0.5 1 1.5 2 2.5 3 3.5 4-20

-15

-10

-5

0

5

10

15

20

25Step Response

Time (sec)

Am

plitu

de

K=1x105sec/6.10 rad

sec/59.188 rad

Unstable

Unstable

Stable

48

Example 2:

Ksss 20030

123

R(s)+

-

V2(s)

Ksss

K

sssK

sssK

sR

sV

20030

20030

11

20030

1

)(

)(23

23

232

s3 1 200

s2 30 K

s1 0

s0

30

*1200*30 K

30

6000 K

306000

0*30*30

6000

K

KK

K

K<6000

K>0 0 < K< 6000 Kcritical=6000

49

By MATLAB,

>>K=2000;

>>ns=[K];

>>ds=[1 30 200 K];

>>step(ns,ds)

0 0.5 1 1.5 2 2.50

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

deV

2(t)

Time (sec)

K=2000

Stable

50

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plitu

de

By MATLAB,

>>K=6000;

>>ns=[K];

>>ds=[1 30 200 K];

>>step(ns,ds)V

2(t)

Time (sec)

K=6000

D(s)=s3+30s2+200s+6000

>>ds=[1 30 200 6000];

>>roots(ds)

p1=-30.0000 p2= 0.0000 +14.1421ip3=0.0000 -14.1421i

sec/1421.14 rad

Marginally stable

51

By MATLAB,

>>K=8000;

>>ns=[K];

>>ds=[1 30 200 K];

>>step(ns,ds)

0 0.5 1 1.5 2 2.5 3-8

-6

-4

-2

0

2

4

6

8

10

12Step Response

Time (sec)

Am

plitu

de

K=6000

V2(

t)

Time (sec)

Unstable

52

ZIEGLER-NICHOLS DESIGN

The coefficients of the controller can be chosen according to Kcritical and critical which are calculated from Routh tabulation. Ziegler and Nichols recommended below relationships while determining the controller coefficients.

2* criticalcritical T

criticalcriticalT

2

sT

sTKsG d

ipc

11)(

For P control criticalp KK *5.0

For PI control criticalp KK *45.0 criticali TT *83.0

For PIDcontrol criticalp KK *6.0 criticali TT *5.0

criticald TT *125.0

53

Example 2:

Gc(s)

sss 20030

123

R(s)+

-

V2(s)

6000criticalK sec/1421.14 radcritical

.sec444.01421.14

22

critical

criticalT

3000*5.0)(,)( criticalcpc KsGKsGFor

s

KKsG

sT

KKsGFor critical

criticalci

ppc 444.0*83.0

*45.0*45.0)(,)(

ssGc

6.73262700)(

54

sTKsT

KKsGFor dp

i

ppc )(

sKs

KKsG critical

criticalcriticalc 444.0*125.0**6.0

444.0*5.0

*6.0*6.0)(

ss

sGc 8.1992.16216

3600)(

55

APPLICATION OF THE MASON FORMULAAPPLICATION OF THE MASON FORMULA

LoopsClosed

PathsForwardsH

1)(

K85

12

ss

s

K1

+

- -

+R(s) Y(s)

122

2

85

1

85

11

85

1

)(

)()(

Kss

s

ss

sK

ss

sK

sR

sYsH

56

122

2

85

1

85

11

85

1

)(

)()(

Kss

s

ss

sK

ss

sK

sR

sYsH

122

2

85

1

85

11

85

1

)(

)()(

Kss

s

ss

sK

ss

sK

sR

sYsH

112 85)(

)()(

KsKKKsss

KKs

sR

sYsH

112 85)(

)()(

KKsKKs

KsK

sR

sYsH

57

Example:

K85

12

ss

s

K1

+

- -

+R(s) Y(s)

N(s)

+

s

122

22

85

1

85

11

85

1)()(

85

1

)(

Kss

ss

ss

sK

ss

ssNsR

ss

sK

sY

58

122

22

85

1

85

11

)(85

1)(

85

1

)(

Kss

ss

ss

sK

sNss

ssR

ss

sK

sY

)()(1

1)(

)(1)(

1

)(1)()(

12

112

112

sNKsKKKs

ssR

KsKKKs

KKssY

KsKKsKs

sNssRKKssY

With respect to R(s) With respect to N(s)

59

)()()()( sEsGsEsR

)(1)()( sGsEsR

)()(1

1)( sR

sGsE

STEADY STATE ERRORSSTEADY STATE ERRORS

K+R(s) )(sGpY(s)E(s)

-

)(sG

The steady-state error value can be found for different inputs, R(s).

60

Step Input:Step Input:

t(s)

A

R(t) R(t)=Au(t)

s

AsR )(

s

A

sGsE

)(1

1)(

s

A

sGssEse

ssss )(1

1lim)(lim

00

)(lim1)(1lim

00 sG

A

sG

Ae

ss

ss

)(lim0

sGKs

s

(Step input error coefficient)

sss K

Ae

1The steady-state error for a step input with magnitude A.

61

t(s)

R(t)=At

Ramp Input:Ramp Input:

R(t)

A=Tan()

2)(

s

AsR

s

A

sGs

A

sGssEse

sssss )(1

1lim

)(1

1lim)(lim

0200

)(lim)(lim0)(lim

000 ssG

A

ssG

A

sGss

Ae

sss

ss

)(lim0

sGsKs

r

Ramp input error coefficient

rss K

Ae

62

Error coefficients and steady-state errors are valid for stable systems. Unity feedback systems are considered.

Example: Find the step input and ramp input error coefficients and steady-state errors for the closed loop system whose forward path is given as

)853(

5)(

2

sss

ssG

0

5

)80*50*3(0

50)(lim

20sGK

ss

01

1

1

1

sss K

e There is no error for step input, A=1.

8

5

80*50*3

50

)853(

5lim)(lim

2200

sss

ssssGK

ssr

63

6.15

8

8511

r

ss Ke System has steady-state error for

ramp input, A=1.

64

ROOT-LOCUS MethodROOT-LOCUS Method

K+R(s) )(sGpY(s)E(s)

-

Consider a control system,

)(1

)(

)(

)()(

sKG

sKG

sR

sYsH

p

p

)(

)()(

sD

sNsG

p

pp

)()(

)(

)(

)(1

)(

)(

)(

)()(

sNKsD

sKN

sD

sNK

sD

sNK

sR

sYsH

pp

p

p

p

p

p

65

)()(

)(

)(

)()(

sNKsD

sKN

sR

sYsH

pp

p

According to stability, the roots of the denominator of H(s) should lie on the left side of s-plane. Root Locus method gives the locations of the roots of the denominator with respect to varying controller gains, K.

Root-Locus Plot:Root-Locus Plot:

The root-locus plot of a closed loop system can be obtained easily by using the forward path transfer function of a closed loop control system.

Example: Find the root locations of a closd loop cntrol system whose process transfer function is

)84)(5(

1)(

2

ssss

ssGp

Zeros: Np(s)=0

s+1=0, s=-1

m = (number of zeros)=1

66

Poles, Dp(s)=0, n=Number of poles = 4.

p1=0

p2=-5

p3=-2+2i

p4=-2-2i

n-m=4-1=3

Plotting the root-locus :1. Determine the zeros and poles of the Gp(s)

2. Determine the angles of asymptotes,

3. Determine the intersection of asypmtotes,

4. Determine the breakaway points.

For the given system, zeros and poles are calculated as

z1=-1

p1=0

p2=-5

p3=-2+2i

p4=-2-2i

n-m=4-1=3

67

Angles of asymptotes:

10,

12

mntokformn

kk

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2-10

-8

-6

-4

-2

0

2

4

6

8

10Root Locus

Real Axis

Imag

inar

y A

xis

Root-locus by MATLAB

Poles of Gp(s)

Zeros of Gp(s)

>>ns=[1 1];

>>ds=[1 9 28 40 0]

>>rlocus(pay,payda)

68

o60314

10*20

o1803

3

14

11*21

o3003

5

14

12*22

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2-10

-8

-6

-4

-2

0

2

4

6

8

10Root Locus

Real Axis

Imag

inar

y A

xis

Root-locus by MATLAB

Intersection of asymptotes:

mn

ZerosPoles

3

)1()22()22()5(0

ii

66.23

8

3

19

60o180o

300o

-2.66

K=0 K=0

K=0

K=0

K

K

KKc

Root-Locus curve is Root-Locus curve is symmetrical with respect to symmetrical with respect to the real axis. the real axis.

Critical K value can be found by Routh tabulation.

69

Breakaway points:

A Breakaway point is the intersection point of the root-locus curve on the real axis. These points can be found by,

0)(

ds

sdGp)(

)()(

sD

sNsG

p

pp

0)(

)()(

)()(

)(2

sD

sNds

sdDsD

ds

sdN

ds

sdG

p

pp

pp

p

1)(

,1)( ds

sdNssN

pp

4056273)(

40289)(

23

234

sssds

sdD

sssssD

p

p

70

1)(

,1)( ds

sdNssN

pp

4056273)(

40289)(

23

234

sssds

sdD

sssssD

p

p

0)(

)()(

)()(

)(2

sD

sNds

sdDsD

ds

sdN

ds

sdG

p

pp

pp

p

0)1(405627340289*1 23234 ssssssss

0)405655212( 234 sssss1=-7.1447

s2=-2.36

s3= -0.4955 + 0.9687i

s4= -0.4955 - 0.9687i

There are no breakaway points because the root-locus curve doesn’t intersect the real axis.

71

Example: Plot the Root-Locus graph of the Closed-loop control system whose process transfer function is given as

)22)(6)(5(

3)(

2

sssss

ssGp

K+R(s) )(sGp

Y(s)

-

72

Example: Plot the Root-Locus graph of the Closed-loop control system whose process transfer function is given as

)22)(6)(5(

3)(

2

sssss

ssGp

Solution:

Zeros and Poles of Gp(s) are found,

s+3=0 s=-3, Number of zeros m=1

s1=0, s2=-5, s3=-6, s4=-1+1i, s5=-1-1i, Number of poles=5

Angles of the Asymptotes

1012

mntofromk

mn

kk

K+R(s) )(sGp

Y(s)

-

Process Transfer Function

73

o

o

o

o

k

k

k

k

3154

7

15

13*2

2254

5

15

12*2

1354

3

15

11*2

45415

10*2

3

2

1

0

5.24

10

15

)3()11()11()6()5(0

ii

mn

ZerosPoles

Angles of the Asymptotes

Intersection point of the Asymptotes

The Breakaway point is calculated as,

0)(

)()(

)()(

)(2

sD

sNds

sdDsD

ds

sdN

ds

sdG

p

pp

pp

p

74

Np(s)=s+3

Dp(s)=s5+13s4+ 54s3+ 82s2+ 60s

60164162525)(

1)(

234

ssssds

sdDds

sdN

p

p

0)(

)()(

)()(

2

sD

sNds

sdDsD

ds

sdN

p

pp

pp

0

)(

)3(60164162525*12

234

sD

sssss60s 82s 54s 13ss

p

2345

0180- s 492-568s-264s-54s- 4s- 2345

0.4677i - 0.6560- s

0.4677i + 0.6560- s

1.2040i - 3.3311- s

1.2040i + 3.3311- s

-5.5257s

5

4

3

2

1

Breakaway

point

75

Root Locus by MATLAB,

>>ns=[1 3];

>>ds=[1 13 54 82 60 0];

>>rlocus(ns,ds)

-15 -10 -5 0 5 10-10

-8

-6

-4

-2

0

2

4

6

8

10Root Locus

Real Axis

Imag

inar

y A

xis

-2.5

45O

135O

Intersection point

K

K

K

K

Breakaway Point

-5.53

Real

Imaginary

K=0K=0

K=0

K=0

K=0

Asymptote

76

BODE PLOTSBODE PLOTSThe bode plot gives information about the frequency response of a system.

Consider a system whose input is a harmonic function

22

2

2)(

nn

n

sssG

)cos()( tAtf

F(s) Y(s)

Y(s)=G(s) F(s)

The response y(t) can be written as,

tiGAty cos)()(

22)(

s

sAsF A

A Cos(t)

A Sin(t)

+A

-A

+A-A

2

2

77

>> w=0:0.001:30;

>> s=i*w;

>>wn=1;

>>ksi=0;

>>gs=wn^2./(s.^2+2*ksi*wn+wn^2);

>>plot(w,abs(gs))

The response of the system to a harmonic input, A*cos(t), can be found for different excitation frequencies, . For harmonic response replace s by i.

Example: Find the response of the system (G(s)) , to a harmonic excitation given as f(t)=3*Cos (5*t). Take n=1 rad/sec, =0.3.

Solution: A = 3, s=5 i,

22

2

525 nn

n

iiiG

0.0427iG )5(

radi 1416.3)5(

1416.351281.0)(),5()5(3)( tCostytCosiGty

78

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

(rad/sec)

|G(i)|

=0

=0.1

=0.5

=0.3

=0.707=1

Frequency Response of The System, Frequency Response of The System, nn=1 rad/sec.=1 rad/sec.

n

Rezonance

Rezonance

221 nrRezonance Frequency

79

-80

-60

-40

-20

0

20

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-180

-135

-90

-45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

>>wn=1;>>ksi=0.3;>>ns=[wn^2];>>ds=[1 2*ksi*wn wn^2];>>bode(ns,ds)

BODE PLOT by MATLABBODE PLOT by MATLAB

80

BODE PLOTBODE PLOT

K+R(s) )(sGpY(s)

-

)(sG

)(1

)()(

sG

sGsH

1+G(s)=0 G(s)=-1

-1

G(s) plane

=

Gain MarginGm

Phase Marginm

G(i)

Log10

Log10-180

0Gm

m

Log10|G(i)|

81

-150

-100

-50

0

50

Mag

nitu

de (

dB)

101

102

103

104

105

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

Gm=32.2 dB

m=76.91

)3000()400(

100000000)(

ssssG

Example: Find the Gain and Phase margins of the system whose transfer function is given below,

>>ns=[100000000];

>>ds=[1 3400 1200000 0];

>>bode(ns,ds)

82

EXAMPLESEXAMPLES

E1: Draw the approximate Root-Locus graph of a control system given below.

-+

R YK )25.155)(6(

84)(

2

sss

ssGp

Solution: Find Zeros ans Poles of the Gp(s)

2084 1 zs m=1

isiss 35.2,35.2,6 321 n=3

Angles of the Asymptotes:

10,

12

mntokformn

kk

o

o

k

k

2702

3

13

11*2

90213

10*2

1

0

83

Intersection of the Asymptotes with real axis

5.42

9

13

)2()35.2()35.2()6(

ii

mn

ZerosPoles

Real part of the roots goes to -4.5 as K goes to infinity.

-6

270

-2.5

-3i

270

3i

-4.5

K=0

K=0

K=0

Real

ImaginaryK

K

There is no breakaway point.

The system remains stable for All positive K values.

84

E2. The forward path transfer function of a closed-loop control system is given as.

)345(

95)(

2tKsss

ssG

Find the Kt value which makes the ramp input error coefficient Kr=2. Find the steady state errors for unit step and unit ramp input for this Kt value.

ttsr KKs

ssGsK3

9

*30*40*5

90*5)(lim2

20

Kt=1.5

)5.445(

95)(

2

sss

ssG

5.40*40*50

90*5)(lim

20sGK

ss

01

1

1

1

sss K

inputstepunitfore

5.02

11

vss K

inputrampunitfore

System has no steady-state error for unit step input.

System has steady-state error for unit ramp input.

85

E3. The root-locus graph of a closed-loop control system whose denominator is D(s)=s3+2s2+(6+K)s+(3K+1) is given below. Find the coefficients (Kp, Kı, Kd) of a PID controller which will be employed in this control system.

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5-10

-8

-6

-4

-2

0

2

4

6

8

10Root Locus

Real Axis

Imag

inar

y A

xis

K

K

K

0-3 0.5

s= 4.3211 i

s= -4.3211 i

Solution: First of all, find Kcritical by Routh tabulation.

86

s3 1 6+K

s2 2 3K+1

s1

s0

D(s)=s3+2s2+(6+K)s+(3K+1)

2

11

2

)13(*1)6(*2

KKK

2

11 K13

211

0*2)13(*2

11

KK

KK

13 K11011 KK

3

1013 KK

K<11

Kcritical=11.sec454.1

3211.4

22

critical

criticalT

sKs

KKsG critical


454.1*5.0

*6.0*6.0)(

sT

sTKsG d

ipc

11)( PID

87

sKs

KKsG critical


454.1*5.0

*6.0*6.0)(

ss

sGc 454.1*125.0*11*6.0454.1*5.0

11*6.011*6.0)(

ss

sGc 2.1078.9

6.6)(

Kp=6.6, KI= 9.078, Kd=1.2

1 k (N/m) f(t) m x(t) C (Ns/m) f d (t) LINEAR CONTROL x(t) is the output variable and is measured by...

Documents

Transcript of 1 k (N/m) f(t) m x(t) C (Ns/m) f d (t) LINEAR CONTROL x(t) is the output variable and is measured by...