linear control systems - ferdowsi university of...

LINEAR CONTROL

SYSTEMS

Ali Karimpour

Associate Professor

Ferdowsi University of Mashhad

Lecture 12

Dr. Ali Karimpour Feb 2013

2

Lecture 12

Topics to be covered include:

v Different region of S plane.

u Constant natural frequency loci.

u Constant natural damped frequency loci.

u Constant damping factor loci.

u Constant damping ratio loci.

v Transient response of a position control system.

Time domain analysis of control systems

Lecture 12


3

Introducing a prototype second order system.

2هعزفی یک سیستن نوونه درجه

+

-

c e r

)2(

2

n

n

ss

c r 22

2

2 nn

n

ss

10 if-1- :are Poles 2 dnn jj

frequency Naturaln فرکانس طبیعی

ratio Damping نسبت میرائی

frequency damped Natural1 2 nd

فرکانس طبیعی میرا شده

factor Dampingn ضریب میرائی

n-

21 nj

d

n

j

j

21

n

θ

Lecture 12


4

v Constant natural frequency loci.

Sنواحی هختلف در صفحه

Different region of S plane.

n

v Constant natural damped frequency loci.

d

n

j

j

21

v Constant damping factor loci.

n-

v Constant damping ratio loci.

cos

Lecture 12


5

هکاى فزکانس طبیعی ثابت

Constant natural frequency loci.

1 2

2 n

1 n

Lecture 12


6

هکاى فزکانس طبیعی هیزا شذه ثابت

Constant natural damped frequency loci.

-1

1

-2

2

1d

2d

Lecture 12


7

Constant damping factor loci. هکاى ضزیب هیزائی ثابت

1 n2 n

Lecture 12


8

Constant damping ratio loci. هکاى نسبت هیزائی ثابت

45

707.0

60

5.0

30

866.0

Lecture 12


9

Example 1: Find the loci of a system where damping

ratio is greater then 0.707 and damping factor greater

than 2

707.0

45

2 n

Lecture 12


10

Example 2: Find the loci that the percent overshoot

is less than 16% and settling time is less than 1 sec

(according to 5% bound).

6.0

60

5.0%16.. OP

605.0

2.3

2.31

2.3

ns

n

s

t

t

2.3 n

Lecture 12


11

Effect of roots loci on step response تاثیز هکاى ریشه ها بز پاسخ پله

-5 -2.5 2.5

10

5

Conjugate root is not shown ریشه هشدوج نشاى داده نشذه است

Lecture 12


12

Example 3: Find the poles of the following systems and

its corresponding step response for k=75, 100, 200 and

1000

+

-

c e r

)20( ss

k

kss

k

sr

scsM

20)(

)()(

2

75k 15,5 21 pp

100k 10,10 21 pp

200k jpjp 1010,1010 21

1000k jpjp 3010,3010 21

Lecture 12


13

+

- c e r

)20( ss

k

kss

k

sr

scsM

20)(

)()(

2

k=75 k=100

k=200

k=1000

0 0.2 0.4 0.6 0.8 1 1.20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(sec)

0 0.5 1 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(sec)

0 0.1 0.2 0.3 0.4 0.5 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

(sec)

0 0.1 0.2 0.3 0.4 0.5 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

(sec)

0 0.5 1 1.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude

k=75

k=100

k=200

k=1000

Lecture 12


14

Dynamics of electromechanical system

دیناهیک سیستوهای الکتزوهکانیکی

Lecture 12


15

A simplified aeroplane (position control system)

(سیستن کنتزل هوقعیت)یک هواپیوای ساده شذه

Position control

Lecture 12


16

Block diagram of a position control system

بلوک دیاگزام یک سیستن کنتزل هوقعیت

Load r

c

cGear

DC Motor

Power Amplifier

with current

feedback

Ks

Tacometer

Lecture 12


17

Block diagram of a position control system

بلوک دیاگزام یک سیستن کنتزل هوقعیت

tt sJB

1 m

ikaI

s

1 cN

l

bk

sk

r

1k

rI

2k

k

tk

Load

c

Gear

DC Motor

Power Amplifier

with current

feedback

Ks

Tacometer

t V

a a sL R

1

bk

Lecture 12


18

Let ignore the velocity feedback.

.اس فیذبک سزعت اغواض کنیذ

tt sJB

1 m

ikaI

s

1 mN

c

bk

t V

sk

r

1k

rI

2k

k

a a sL R

1

tt sJB

1 m

ikaI

s

1 mN

c

bk

t V

sk

r

1k

rI

2k

k

tk

aa sLR

1

Lecture 12


19

Simplification ساده ساسی

tt sJB

1 m

ikaI

s

1 mN

c

bk

t V

sk

r

1k

rI

2k

k

a a sL R

1

c

)(sG

r

))((1

))(()(

21

1

ttaa

bi

aa

ttaa

is

sJBsLR

kk

sLR

kk

sJBsLRs

Nkkkk

sG

bittttaa

is

kksJBkksJBsLRs

Nkkkk

)())(( 21

1

Lecture 12


20

Simplification

ساده ساسی و عذد گذاری

c

)(sG

r

bittttaa

is

kksJBkksJBsLRs

NkkkksG

)())(()(

21

1

Let we ignore La با اغواض اسLa دارین

ta

bitta

ta

is

JkkR

kkBkkBRss

JkkR

Nkkkk

sG

)(

)()(

~

21

21

21

1

Lecture 12


21

Simplification

ساده ساسی و عذد گذاری

c

)(sG

r

ta

bitta

ta

is

bittttaa

is

JkkR

kkBkkBRss

JkkR

Nkkkk

sG

kksJBkksJBsLRs

NkkkksG

)(

)()(

~

)())(()(

21

21

21

1

21

1

0001.001.0

005.011.003.05

0636.0195.010 21

ml

mlaa

bsi

JJ

BBNLR

kkkkk

0002.0100

01.00001.0

015.0100

1005.0

2

2

lmt

lmt

JNJJ

BNBB

Lecture 12


22

Simplification ساده ساسی و عذد گذاری

c

)(sG

r

2.361

4500)(

~

)3008)(3.400(

105.1

12040003.3408

105.1)(

7

2

7

ss

ksG

sss

k

sss

ksG

0001.001.010005.0

1.0003.050636.0

195.010 21

mllm

aab

si

JJBB

NLRk

kkkk

0002.0100

01.00001.0

015.0100

1005.0

2

2

lmt

lmt

JNJJ

BNBB

ta

bitta

ta

is

bittttaa

is

JkkR

kkBkkBRss

JkkR

Nkkkk

sG

kksJBkksJBsLRs

NkkkksG

)(

)()(

~

)())(()(

21

21

21

1

21

1

Lecture 12


23

Stability بزرسی پایذاری c

)(sG

r

2.361

4500)(

~

ss

ksG

)3008)(3.400(

105.1)(

7

sss

ksG

ksss

k

sG

sGsM

7

7

105.1)3008)(3.400(

105.1

)(1

)()(

0105.112040003.3408 723 ksss

ks

s

72

3

105.13.3408

12040001

ks

ks

70

7

105.1

03.3408

105.14100593200

Stable for 0<k<273.57

kss

ksM

45002.361

4500)(

2

045002.3612 kss

Stable for all k>0

Lecture 12


24

Analysis تجشیه تحلیل c

)(sG

r

2.361

4500)(

~

ss

ksG kk nn 5304500

2

kkn

692.2

560

2.3612.3612

sn tOPppk ..21

dampedover is System81.12.30331222.2

damping critically is System16.1806.18025.7

unstable is System233842


022.0%3.4255707.06.1806.1805.14 j

022.0%601110163.010966.180274 j

045002.3612 kss

Lecture 12


25

Step response پاسخ پله c

)(sG

r

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

Time (sec)

Am

plit

ude

222.2k

25.7k

5.14k

274k

Lecture 12


26

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.02

0.04

0.06

0.08

0.1

0.12Step Response

Time (sec)

Am

plit

ude

Velocity response c پاسخ شیب

)(sG

r

2.361

4500)(

~

ss

ksG

222.2k

25.7k

5.14k

274k

222.2

68.27

036.0

25.7

3.90

011.0

5.14

6.180

005.0

274

3414

00029.0

k

vk

sse

Lecture 12


27

Analysis تجشیه تحلیل c

)(sG

r

..321 OPpppk n


0105.112040003.3408 723 ksss

)3008)(3.400(

105.1)(

7

sss

ksG


dampedover is System303663012222.2

dampedover is System1562303022250.7

%6.42677.019218630355.14 j

%5968717.06771153178100 j

unstable is System10982.03408274 j

Lecture 12


28

Step response پاسخ پله c

)(sG

r

)3008)(3.400(

105.1)(

7

sss

ksG

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

Time (sec)

Am

plit

ude

5.14k 222.2k

25.7k

100k

Lecture 12


29

Step response comparison هقایسه پاسخ پله

c

)(sG

r

)3008)(3.400(

105.1)(

7

sss

ksG

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plit

ude 22.2k

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Am

plit

ude

22.2k

25.7k

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plit

ude

22.2k

25.7k

5.14k

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

(sec)

22.2k

25.7k

5.14k

100k

)2.361(

4500)(

~

ss

ksG

Lecture 12


30

Exercises توزینها

1- Find the roots of following system for -301<k<301 and show

them on the s plane.

2- Find the roots of following system for -301<k<301 and show

them on the s plane.

)3008)(3.400(

105.1)(

7

sss

ksG

)2.361(

4500)(

ss

ksG

Let k=-300,-280,-260,……….260,280,300

Let k=-300,-280,-260,……….260,280,300

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