linear control systems - ali karimpourkarimpor.profcms.um.ac.ir/imagesm/354/stories/lin_con/... ·...

LINEAR CONTROLLINEAR CONTROLSYSTEMSSYSTEMS

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

Lecture 20

Dr. Ali Karimpour Feb 2013

2

Lecture 20

Frequency domain chartsTopics to be covered include: Relative stability measures for minimum phase systems.

Gain margin. Phase margin.

Nichols chart or gain phase plot. Stability analysis with gain phase plot.

Bode plot. Stability analysis with Bode plot.

Step by step Bode plot construction.

Lecture 20


3

Stability margins

Stability Is Not A Yes/No Proposition

It's not enough to know that a system is stable or unstable. If a system is just barely stable, then a small gain in a system parameter could push the system over the edge, and you will often want to design systems with some margin of error.

If you're going to do that, you'll need some measure of how stable a system is. To get such measures - and there are at least two that are widely used.

Lecture 20


4

Stability margins

)(sG)(sC

)(sGc

+-

)(sR

ω=ωc

ω=ω180

G (jω) Gc(jω)

Phase Margin

We define phase margin as the phase (angle) that the frequency response

would have to change to move to the -1 point.

c : Is the gain crossover frequency

Phase margin is the most widely used measure of relative stability when working in the frequency domain.

Lecture 20


5

ω=ωc

ω=ω180

G (jω) Gc(jω)

Phase margin computation

)(sG)(sC

)(sGc

+-

)(sR

)()(180 cccm jGjG

The phase margin, is defined as follows:m

How to derive c

1)()( jGjGLet c c

Lecture 20


6

Stability margins

Phase Margin

The two different Nyquist plot above would lead to two different phase margins. The system with the frequency response with the

dashed line is less stable.

Lecture 20


7

Stability margins

Phase Margin

The two different gains shown for the Nyquist plot below would lead to the same phase margins. But do they have the same stability margin?

-1

Clearly no.

Thus we need another measure of relative stability.

Lecture 20


8

Stability margins

)(sG)(sC

)(sGc

+-

)(sR

ω=ωc

ω=ω180

G (jω) Gc(jω)

Gain Margin

We define gain margin as the gainthat the frequency response would have to increase to move to the -1

point.

180 : Is the phase crossover frequency

Gain margin is another widely used measure of relative stability when working in the frequency domain.

Lecture 20


9

ω=ωc

ω=ω180

G (jω) Gc(jω)

Gain margin computation

)(sG)(sC

)(sGc

+-

)(sR

)()(/1 180180 jGjGGM c

The gain margin, GM is defined as follows:

But gain margin is usually specified in db.

)()(/1log20 180180 jGjGGM c

How to derive 180

180)()( jGjGLet c

180 0)()(Im jGjGor c

Lecture 20


10

Stability margins

Phase and Gain Margin

-1

Same Phase Margin

But different Gain Margin

Lecture 20


11

)(sG)(sC

)(sGc

+-

)(sR

)(11

)()(11)(

)(1)(

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

sLsGsGsS

sLsL

sGsGsGsGsTsGsGsL

cc

cc

Frequency (rad/s)

||T

|| S

|| L

sMPM

Sensitivity and complementary sensitivity peak

Lecture 20


12

G (jω) Gc(jω)

Sensitivity peak

)()(11)(

jGjGjS

c

)()(11)(max

scss jGjG

jSM

1sM)(sG

)(sC)(sGc

+-

)(sR

The sensitivity peak, Ms is defined as follows:

)()(1 00 jGjG c

0

)( 01 jS

Lecture 20


13

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

-2.5

-2

-1.5

-1

-0.5

0

Nyquist Diagram

Real Axis

Imag

inar

y Ax

is

Nyquist chart (polar plot) construction )(sG

)(sC)(sGc

+-

)(sR

1 10793.2

2 12337.1

3 13882.0

4 15154.0

5 16238.0

6 17127.0

7 17920.0

8 18716.0

9 19312.0

20 22902.0

?c 6.2 ?180 7 ?GM db????PM 45

1

2

)10)(5(150)()(Let

ssssGsGc

)()( jGjGc

3

7 20

45

10793.2)101)(51(

150)1()1(

jjj

jGjGc

Lecture 20


14

-270 -225 -180 -135 -90-40

-30

-20

-10

0

10

20Nichols Chart

Open-Loop Phase (deg)

Ope

n-Lo

op G

ain

(dB)

-270 -225 -180 -135 -90-40

-30

-20

-10

0

10

20Nichols Chart

Open-Loop Phase (deg)

Ope

n-Lo

op G

ain

(dB)

)(sG)(sC

)(sGc

+-

)(sR

)10)(5(150)()(Let

ssssGsGc

)()(log20 jGjGc

db33.9

db71.2

db71.1

db29.5

db42.8

db23.11

db80.13

db18.16

db39.18

1

3

68

)()( jGjGc

1 10793.2

2 12337.1

3 13882.0

4 15154.0

5 16238.0

6 17127.0

7 17920.0

8 18716.0

9 19312.0

20 22902.0 db77.35

20

2

45

?c 6.2 ?180 7 ?GM db8.13?PM 45

Nichols chart (gain phase plot) construction

Lecture 20


15

Bode plot construction )(sG)(sC

)(sGc

+-

)(sR

)10)(5(150)()(Let

ssssGsGc

)()( jGjGc )()(log20 jGjGc

1 10793.2 db33.92 12337.1 db71.23 13882.0 db71.1

4 15154.0 db29.55 16238.0 db42.8

6 17127.0 db23.11

7 17920.0 db80.13

8 18716.0 db18.16

9 19312.0 db39.18

20 22902.0 db77.35 ?c 5.2 ?180 7 ?GM db8.13?PM 48

-80

-60

-40

-20

0

20

Mag

nitu

de (d

B)

100 101 102-270

-225

-180

-135

-90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec)

80

60

40

20

0

20

0 1

90

Bode Diagram

1003020107654321

Lecture 20


16

Step by step bode plot construction

))()(())(()()(Let

3212

21

pspspsszszsksGsGc

Step 1:

)1/)(1/)(1/()1/)(1/()()(

3212

21

321

21

pspspss

zszspppzzksGsGc

Step 2:

1/1log20

1/1log20

1/1log20)/(1log20

1/log201/log20log20)()(log20

321

2

21

pjpjpjj

zjzjkjGjGc

Step 3:

)1/

1()1/

1()1/

1(

))/(1()1/()1/()()(

321

221

pjpjpj

jzjzjkjGjGc

)1/)(1/)(1/()1/)(1/(

3212

21

pspspss

zszsk

Lecture 20


17


)1/)(1/)(1/()1/)(1/()()(Let

3212

21

pspspss

zszsksGsGc

1/1log20

1/1log20

1/1log20

)/(1log201/log201/log20log20)()(log20

321

221

pjpjpj

jzjzjkjGjGc

)1/

1()1/

1()1/

1(

))/(1()1/()1/()()(

321

221

pjpjpj

jzjzjkjGjGc

Lecture 20


18


kjGjGc log20)()(log20 kjGjGc )()(

100 101 102 103

Bode Diagram

Frequency (rad/sec)

Phas

e (d

eg)

M

agni

tude

(db)

)log(20 k

1800or Why 0 or 180?It depends to

sign of k.

Lecture 20


19(rad/sec)Frequency

10101010 3210

10

20

30

40

10

0

90

45

00 1 2 3

Phas

e (d

eg)

M

agni

tude

(db)

Step by step bode plot construction 1/log20)()(log20 1 zjjGjGc )1/()()( 1 zjjGjGc

1z

20 db / dec

20Let 1 z

Lecture 20


20


2)/(1log20)()(log20 jjGjGc ))/(1()()( 2 jjGjGc

(rad/sec)Frequency10101010 3210

40

20

0

20

80

60

90

180

2700 1 2 3

Phas

e (d

eg)

M

agni

tude

(db)

-40 db/ dec

Lecture 20


21


1/1log20)()(log20

1

pjjGjGc )

1/1()()(

1

pjjGjGc


20

10

0

10

40

30

0

45

900 1 2 3

Phas

e (d

eg)

M

agni

tude

(db)

20Let 1 p

1p

-20 db / dec

Lecture 20


22

Example 1: Derive the GM and PM of following system by use of Bode plot.


40

20

0

20

80

60

90

180

2700 1 2 3

Phas

e (d

eg)

M

agni

tude

(db)

)10)(5(500

sss

)(sC+-

)(sR

)10)(5(500)(

ssssG

Open loop transfer function is:

)110/)(15/(10

sss

Lecture 20


23

Example 1: Derive the GM and PM of following system by use of Bode plot.

)10)(5(500

sss

)(sC+-

)(sR

?c sec/5.6 rad

10PM

?180 sec/2.7 rad

dbGM 3


40

20

0

20

80

60

90

180

2700 1 2 3

Phas

e (d

eg)

M

agni

tude

(db)

Lecture 20


24

Exercises

1: Derive the gain crossover frequency, phase crossover frequency ,GM and PM of following system by use of Bode plot.

)10(1ss

)(sC200

+-

)(sR

38,,5.12: 180 mc andGMAnswer

Lecture 20


25

Exercises

2 The polar plot of an open loop system with negative unit feedback is shown.

a) Find the open loop transfer function.b) Find the closed loop transfer function.

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Nyquist Diagram

Real Axis

Imag

inar

y Ax

is

sec/18.11 rad 447.0

15025150:

)25(150: 2 ss

bss

aanswer

Lecture 20


26

Exercises

3 The Bode plot of an open loop system with negative unit feedback is shown.a) Find the open loop transfer function.b) Find the closed loop transfer function.

20020200:

)20(200: 2 ss

bss

aanswer

-80

-60

-40

-20

0

20

40

Mag

nitu

de (d

B)

100 101 102 103-180

-135

-90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec)

linear control systems - ali karimpourkarimpor.profcms.um.ac.ir/imagesm/354/stories/lin_con/... ·...

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