automatic control systems - ???????? -...

AUTOMATIC AUTOMATIC CONTROLCONTROLSYSTEMSSYSTEMS

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

Lecture 8

Ali Karimpour Nov 2012

2

Lecture 8

Time domain analysis of control systemsTopics to be covered include: Time domain analysis. Transient response of a prototype second order system Different region of S plane. Transient response of a position control system.

Lecture 8


3

Introducing a prototype second order system.2معرفی یک سیستم نمونه درجه

+-

cer)2(

2

n

n

ss

cr22

2

2 nn

n

ss

)(2

)( 22

2

sRss

sCnn

n

)2()( 22

2

nn

n

ssssC

Step response

10 if1- :are Poles 2 nn j

222

222 1)(1)(

1)(

nn

n

nn

n

ss

ss

sC

)1sin(

111)()( 2

2

tetutc n

tn 1cos

Lecture 8


4

Step response پاسخ پله

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

13 n 8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

4.0,6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

2.0,4.0,6.0,8.0,13 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

0,2.0,4.0,6.0,8.0,13 n

Lecture 8


5

Step response پاسخ پله13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

3,2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

4,3,2,13.0 n

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

28.6,4,3,2,13.0 n

Lecture 8


6

Rise time, Settling time, Overshoot, Under shoot and Peak time زمان صعود، زمان نشست، فراجهش، فروجهش و زمان پیک

Lecture 8


7

Introducing a prototype second order system. 2معرفی یک سیستم نمونه درجه

+-

cer)2(

2

n

n

ss

cr22

2

2 nn

n

ss

10 if-1- :are Poles 2 dnnn jj

n-

21 nj

d

n

jj

21

n

frequency Naturaln فرکانس طبیعی

ratio Damping نسبت میرائیfrequency damped Natural1 2 nd

فرکانس طبیعی میرا شده

factor Dampingn ضریب میرائی

Lecture 8


8

Peak time لحظه پیک

ntn 21

21

n

pt

21

n

nt

Let n=1

n=1

n=3

n=4

n=5

n=2

Lecture 8


9

Percent Overshootدرصد فراجهش

21100%..

eOP ..OP0 100%

0.100 73%0.200 53%0.300 37%0.400 25%0.500 16%0.707 4.3%

1 0%0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Per

cent

Ove

rsho

ot

Lecture 8


10

Rise time زمان صعود

rt

How can we findrise time ?

چگونه زمان صعودرا بیابیم؟

yktc r)(

ytc 9.0)(Let

nrt

5.28.0 10917.24167.01 2

nrt

Lecture 8


11

Settling time زمان نشست

st

How can we findsettling time ?

چگونه زمان نشسترا بیابیم؟

ytcy )(

%5for 2.3

nst %2for 4

n

st 7.00

Lecture 8


12

Constant natural frequency loci.

Sنواحی مختلف در صفحه Different region of S plane.

n

Constant natural damped frequency loci.

d

n

jj

21

Constant damping factor loci.

n-

Constant damping ratio loci.

cos

Lecture 8


13

مکان فرکانس طبیعی ثابتConstant natural frequency loci.

12

2 n1 n

Lecture 8


14

مکان فرکانس طبیعی میرا شده ثابتConstant natural damped frequency loci.

-1

1

-2

2

1d

2d

Lecture 8


15

Constant damping factor loci.مکان ضریب میرائی ثابت

1 n2 n

Lecture 8


16

Constant damping ratio loci.مکان نسبت میرائی ثابت

45707.0

605.0

30866.0

Lecture 8


17

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 8


18

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

ns

t

t

2.3 n

Lecture 8


19

Effect of roots loci on step responseتاثیر مکان ریشه ها بر پاسخ پله

-5 -2.5 2.5

10

5

Conjugate root is not shown ریشه مزدوج نشان داده نشده است

Lecture 8


20

Example 3: Find the poles of the following systems and its corresponding step response for k=75, 100, 200 and 1000

+-

cer)20( ss

k

kssk

srscsM

20)()()( 2

75k 15,5 21 pp

100k 10,10 21 pp

200k jpjp 1010,1010 21

1000k jpjp 3010,3010 21

Lecture 8


21

+-

cer)20( ss

k

kssk

srscsM

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)

Ampl

itude

k=75k=100k=200k=1000

Lecture 8


22

Dynamics of electromechanical systemدینامیک سیستمهاي الکترومکانیکی

Lecture 8


23

A simplified aeroplane (position control system))سیستم کنترل موقعیت(یک هواپیماي ساده شده

Position control

Lecture 8


24

Block diagram of a position control systemبلوك دیاگرام یک سیستم کنترل موقعیت

Loadr

c

cGearDC MotorPower Amplifier

with current feedback

Ks

Tacometer

Lecture 8


25

Stability بررسی پایداريc

)(sG

r

2.3614500)(~

ss

ksG)3008)(3.400(105.1)(

7

sssksG

ksssk

sGsGsM 7

7

105.1)3008)(3.400(105.1

)(1)()(

0105.112040003.3408 723 ksss

kss

72

3

105.13.340812040001

ks

ks

70

7

105.1

03.3408

105.14100593200

Stable for 0<k<273.57

kssksM

45002.3614500)( 2

045002.3612 kss

Stable for all k>0

Lecture 8


26

Analysis تجزیه تحلیلc

)(sG

r

2.3614500)(~

ss

ksG kk nn 53045002

kkn692.2

5602.3612.3612

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 8


27

Step response پاسخ پلهc

)(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)

Ampl

itude

222.2k25.7k

5.14k

274k

Lecture 8


28

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)

Ampl

itude

Velocity response cپاسخ شیب

)(sGr

2.3614500)(~

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 8


29

Analysis تجزیه تحلیلc

)(sG

r

..321 OPpppk n


0105.112040003.3408 723 ksss

)3008)(3.400(105.1)(

7

sssksG


dampedover is System303663012222.2

dampedover is System1562303022250.7

%6.42677.019218630355.14 j

%5968717.06771153178100 j

unstable is System10982.03408274 j

Lecture 8


30

Step response پاسخ پلهc

)(sG

r

)3008)(3.400(105.1)(

7

sssksG

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)

Ampl

itude

5.14k 222.2k

25.7k

100k

Lecture 8


31

Step response comparison مقایسه پاسخ پله

c

)(sGr

)3008)(3.400(105.1)(

7

sssksG

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)

Ampl

itude 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)

Ampl

itude

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)

Ampl

itude

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 8


32

Real dominant poles of transfer function.قطبهاي غالب حقیقی در تابع انتقال

Region of insignificant poles

Region of dominant

poles UnstableRegion

dD

What about D? D > 5 or 10 times of d.

Lecture 8


33

Complex dominant poles of transfer function.قطبهاي غالب مختلط در تابع انتقال


Region of dominant

poles UnstableRegion

D

What about D? D > 5 or 10 times of d.

d

Lecture 8


34

Design procedure. روش طراحی


Region of dominant

poles

D

45

45

For design purposes, such as in the pole placement design we try to put poles on: . در طراحی ها، مثلا در جاگذاري قطب سعی می کنیم قطبها در مکان زیر قرار گیرند

Lecture 8


35

Example 4: What are the dominant poles of the following transfer function.

.قطبهاي غالب در تابع انتقال زیر را بیابید: 4مثال

)16)(2(32)(

sssM

16143.0

2143.11)(

ssssC

)16)(2(32)()()(

ssssRsMsC

Step response

)()143.0143.11()( 162 tueetc tt

dominant

insignificant

insignificant

0

dominant

Lecture 8


36

Example 5: Simplify the system in example 1.

.را ساده کنید 1سیستم مثال : 5مثال

)2)(16(32)(~

sssM

)2(32)()()(~

ss

sRsMsC)2(

1616

ss

)()1616()(~ 2 tuetc t

Step response

)2(2)(~

ss

sC

)()11()(~ 2 tuetc t

)2(32

s

Step response)2)(116/(

16/32)2)(16(

32)(~

ssss

sM

)2)(16(32)(

sssM )()143.0143.11()( 162 tueetc tt

22

)2)(116/(16/32

)2)(16(32)(~

ssssssM

211

)2(2)(~

sssssC

غ ق ق

Lecture 8


37

Example 6: What are the dominant poles of the following transfer function.

.قطبهاي غالب در تابع انتقال زیر را بیابید: 6مثال

)2)(16()1.2(24.15)(

ss

ssM

169457.0

20544.01)(

ssssC

)2)(16()1.2(24.15)()()(

sss

ssRsMsC

Step response

)()9457.00544.01()( 162 tueetc tt

insignificant

0

dominant

The zero near to pole makes it insignificant

dominant insignificant

Lecture 8


38

Example 7: Discuss the dominant poles of following system.

.در مورد قطبهاي غالب سیستم زیر بحث کنید: 7مثال

)40010)((400)( 2

ssas

asM

Simplification

If (a-5) > 5 ∙ 20=100

dominant-5-a

20

)40010)(1/(400)(~2

ssas

sM

)40010(400)(~

2

sssM for a > 105

Lecture 8


39

Example 8: Compare the step response of M and its approximation for different values of k.

.بدست آورید aو تقریب آن را براي مقادیر مختلف Mپاسخ پله : 8مثال

)40010)((400)( 2

ssas

asM)40010(

400)(~2

ss

sM

step(400,[1 10 400]);hold on

a=10;step(400*a,conv([1 a],[1 10 400])

a=20;step(400*a,conv([1 a],[1 10 400]))

a=50;step(400*a,conv([1 a],[1 10 400]))

a=100;step(400*a,conv([1 a],[1 10 400]))0 0.2 0.4 0.6 0.8 1 1.2

0

0.5

1

1.5

Step Response

Time (sec)

Ampl

itude

Second orderThird order a=10Third order a=20Third order a=50Third order a=100

2nd order

a=10

a=20

a=50

a=100

Lecture 8


40

Exercises تمرینها

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

sssksG

)2.361(4500)(

ss

ksG

Let k=-300,-280,-260,……….260,280,300

Let k=-300,-280,-260,……….260,280,300

Lecture 8


41


2 Derive a suitable second order system for following system.

1 Consider following system. Find the dominant poles and insignificant poles of system.

)3)(2)(16(96)(

ssssM

)3)(2)(16(96)(

ssssM

3 Compare the step response of the system in problem 2 and step response of its second order approximation.

Lecture 8


42


5 Derive a suitable second order system for following system.

4 Consider following system. Find the dominant poles and insignificant poles of system.

)1)(3)(20)(160()05.3(3150)(

ssss

ssM

6 Compare the step response of the system in problem 5 and step response of its second order approximation.

)1)(3)(20)(160()05.3(3150)(

ssss

ssM

automatic control systems - ???????? -...

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