LINEAR CONTROLLINEAR CONTROL SYSTEMS SYSTEMS

Ali Karimpour

Assistant Professor

Ferdowsi University of Mashhad

2

Lecture 12

Ali Karimpour Oct 2009

Lecture 12

Time domain analysis of control systems

Topics to be covered include:

Time domain analysis. Error series. Introducing some performance criteria (ISE, ITSE,

IAE and ITAE). Introducing a prototype second order system.

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Lecture 12

Ali Karimpour Oct 2009

Some test signalsچند سیگنال تست

R

t

R

t1

R

R

t

Step input

ورودی پله

Velocity input

ورودی شیب

Acceleration input

ورودی شتاب

)()( tRutr s

RsR )(

)()( tRtutr 2)(

s

RsR

)(2

)(2

tuRt

tr 3)(

s

RsR

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Lecture 12

Ali Karimpour Oct 2009

خطا در سیستمهای کنترلError in control systems

sT

dnddj

m dessss

sssksKsG

)1).......(1)(1(

)1).......(1)(1()()(

21

21

position velocity acceleration

Type Kp Kv Ka ess ess ess

0

1

2

3

kk

R

10 0

k

k

0k

R0

0 0 k

R

0 0 0

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Lecture 12

Ali Karimpour Oct 2009

: خطاهای مختلف را برای سیستم زیر تعیین کنید1مثال

Example 1: Find the different errors in following system

+-

cer

1s

k

1)()(

s

ksKsG

ks

ksT

1)(

ksKsGKs

p

)()(lim0 k

ess

1

1

0 av KK sse

Error for step input

Errors for velocity and parabolic input

System is stable so we continue

Note that the above method doesn’t say anything about how the errors go to infinity

توجه کنید که این روش راجع به چگونگی میل خطا به .بینهایت صحبت نمی کند

0k

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Lecture 12

Ali Karimpour Oct 2009

Error seriesسری خطا

G K +

-

cer

dtrwtesRsWsEt

ee )()()()()()(0

: taround)( Expand tr

....!3

)(

!2

)()()()( 32

trtr

trtrtr

....)(!2

)()()()()()(0

2

00 dwtrdwtrdwtrte

t

e

t

e

t

e

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Lecture 12

Ali Karimpour Oct 2009

Error seriesسری خطا

....)(!2

)()()()()()(

0

2

00

dw

trdwtrdwtrte

t

e

t

e

t

e

Now consider steady value for r

....)(!3

)(!2

)()()( 3210 tr

Ctr

CtrCtrCte sssss

....)(!2

)()()()()()(

0

2

00

dw

trdwtrdwtrte e

sesess

0C2C1C

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Lecture 12

Ali Karimpour Oct 2009

Error series coefficientsضرائب سری خطا

....)(!3

)(!2

)()()( 3210 tr

Ctr

CtrCtrCte sssss

dwC e

00 )( dwC e

01 )( dwC e

0

22 )( .........

Calculation of coefficients محاسبه ضرائب

dewsW see

0

)()(

)(lim0

0 sWC es

ds

sdWC e

s

)(lim

01

ne

n

sn ds

sWdC

)(lim

0........ ........

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Lecture 12

Ali Karimpour Oct 2009

بیابید1: ضرائب سری خطا را برای سیستم مثال 2مثالExample 2: Determine the error series coefficients for system in example 1

ksWC e

s

1

1)(lim

00 20

1 )1(

)(lim

k

k

ds

sdWC e

s

........

+-

cer

1s

k

)(1

1)( sR

ks

ssE

)(sWe

32

2

02 )1(

2)(lim

k

k

ds

sWdC e

s

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Lecture 12

Ali Karimpour Oct 2009

بیابید1: سری خطا را برای ورودیهای پله و شیب در مثال 3مثال .

Example 3: Determine the error series for step and velocity inputs in example 1.

,.......)1(

2,

)1(,

1

122210 k

kC

k

kC

kC

+

-

cer

1s

k

Step:

ktr

Ctr

CtrCtrCte sssss

1

1....)(

!3)(

!2)()()( 32

10

0...)()(,1)( trtrtr sss

Ramp:

232

10 )1(1....)(

!3)(

!2)()()(

k

k

k

ttr

Ctr

CtrCtrCte sssss

0...)()(,1)(,)( trtrtrttr ssss

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Ali Karimpour Oct 2009

را مقایسه کنید3 و 1: نتایج مثالهای 4مثال

Example 4: Compare the result of example 1 and 3

+-

cer

1s

k

ktes

1

1)(

2)1(1)(

k

k

k

ttes

kess

1

1

sse

Step:

Ramp:

Example 1 Example 3

They have similar result but error series show how the error go to infinity

مشخص است که نتایج یکسان است ولی سری خطا نحوه .میل به بینهایت را نیز نشان می دهد

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Lecture 12

Ali Karimpour Oct 2009

: خطا را برای ورودی داده شده بیابید5مثال

Example 5: Determine the error for following input

,.......)1(

2,

)1(,

1

1 have we2 exampleBy

32210 k

kC

k

kC

kC

+-

cer

1s

k ttr 0sin)(

This problem can just be solve with error series!

.......cos)(,sin)(,cos)(,sin)( 03

002

0000 ttrttrttrttr ssss

tC

CtC

Ctes 03

03

0102

02

0 cos...)!3

(sin...)!2

()(

k=100

20

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Lecture 12

Ali Karimpour Oct 2009

: خطا را برای ورودی داده شده بیابید)ادامه(5مثال

Example 5: Determine the error for following input (continue)

+-

cer

1s

k

tC

CtC

Ctes 03

03

0102

02

0 cos...)3.2.1

(sin...)2.1

()(

64332210 10766.5

)101(

600,000194.0

)101(

200,0098.0

)101(

100,0099.0

101

1

CCCC

tttes 2cos)83.2.1

10766.520098.0(2sin)4

2.1

000194.00099.0()(

6

)3.622sin(02213.02cos019592.02sin010288.0)( ttttes

1002sin)( 00 kttr

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Ali Karimpour Oct 2009

. بیابید5: مقدار دقیق خطا را برای مثال 6مثال

Example 6: Determine the exact value of error in example 5.

+-

cer

1s

k 1002sin)( 00 kttr

)(1

1)( sR

ks

ssE

4

2

101

12

ss

s

jsjsssE

22101)(

4101

2002

)4)(1012(

24

jj

j

)4)(1012(

24

jj

j

jtjts ejejte 22 )0047.00098.0()0047.00098.0()(

)4.642sin(0217.02cos0196.02sin0094.0)( ttttes

jtjtt ejejete 22101 )0047.00098.0()0047.00098.0(0196.0)(

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Lecture 12

Ali Karimpour Oct 2009

. بیابید5: پاسخ سیستم را برای مثال 7مثال

Example 7: Determine the response example 5.

+-

cer

1s

k 1002sin)( 00 kttr

101

100

)(

)(

ssr

sc

[u,t]=gensig('sin',2);T1=tf(100,[1 101])lsim(T1,u,t);hold on;T2=tf(1,1)lsim(T2,u,t);

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Linear Simulation Results

Time (sec)

Am

plitu

de

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Lecture 12

Ali Karimpour Oct 2009

شاخص های عملکردیPerformance indices

Integral of the Square of the Error T

dtteISE0

2 )(

Note that these are positive values and lower value is better

Integral of the Absolute magnitude of the Error T

dtteIAE0

)(

This performance is used when the initial value of error is not very important

Integral of Time multiplied by the Squared Error

TdttteITSE

0

2 )(

Integral of Time multiplied by the Absolute Error T

dttetITAE0

)(

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Lecture 12

Ali Karimpour Oct 2009

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

s

ssC

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Lecture 12

Ali Karimpour Oct 2009

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

)1sin(

1)1cos(1)()( 2

2

2 ttetutc nntn

)1sin(

1

11)()( 2

2

tetutc n

tn

)1sin()1cos(11

1)()( 222

2tt

etutc nn

tn

)1sin(cos)1cos(sin1

1)()( 22

2tt

etutc nn

tn

1cos

222

222 1)(1)(

1)(

nn

n

nn

n

ss

s

ssC

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Lecture 12

Ali Karimpour Oct 2009

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)

Am

plitu

de13 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)

Am

plitu

de

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)

Am

plitu

de6.0,8.0,13 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)

Am

plitu

de4.0,6.0,8.0,13 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)

Am

plitu

de2.0,4.0,6.0,8.0,13 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)

Am

plitu

de0,2.0,4.0,6.0,8.0,13 n

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Lecture 12

Ali Karimpour Oct 2009

Step response پاسخ پله

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)

Am

plitu

de

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)

Am

plitu

de2,13.0 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)

Am

plitu

de3,2,13.0 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)

Am

plitu

de4,3,2,13.0 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)

Am

plitu

de28.6,4,3,2,13.0 n

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Lecture 12

Ali Karimpour Oct 2009

Specifications of a prototype second order system.

2مشخصه های یک سیستم نمونه درجه

)1sin(

1

11)()( 2

2

tetutc n

tn 1cos

22

Lecture 12

Ali Karimpour Oct 2009

Rise timeزمان صعود

:rt The time elapsed up to the instant at which the step response reaches, for the first time, the value kry. The constant kr varies from author to author, being usually either 0.9 or 1.

rt: پریQان سQزم رسQQیدن تQQا شQQده بQQرای پلQQه پاسQQخ اولین بQار بQه مقQدار

kry ثابت kr. که

یQا 0.9را معمQوال می 1 نظQر در

گیرند.

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Lecture 12

Ali Karimpour Oct 2009

Settling timeزمان نشست

:st The time elapsed until the step response enters (without leaving it afterwards) a specified deviation band, ±, around the final value. This deviation , is usually defined as a percentage of y, say 2% to 5%.

st: پریQان سQزم رسQیدن تQا شQده

بQQه پلQQه پاسQQخ حQQول ±ناحیQQه

نهQQایی مقQQدار ثابت پاسخ. که

یQا % 2را معمQوال در نظQر می 5%

گیرند.

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Lecture 12

Ali Karimpour Oct 2009

Overshootفراجهش

:pM The maximum instantaneous amount by which the step response exceeds its final value.

مQاکزیمم مقQداری کQه پاسQخ مقQQQدار از پلQQQه پاسQQQQخ نهQQQQائی

اضافه می شود.

pM:

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Lecture 12

Ali Karimpour Oct 2009

Percent Overshootدرصد فراجهش

:..OP Normalized value of overshoot. مقQQدار شQQده نرمQQال

فراجهش...: OP

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Lecture 12

Ali Karimpour Oct 2009

Peak time پیک زمان

:pt

pt:

The time at which corresponding to maximum instantaneous amount by which the step response exceeds its final value.

زمQان بQQا متنQQاطر

اولین مQQQQQQاکزیمم

پاسخ.

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Lecture 12

Ali Karimpour Oct 2009

If the closed loop system includes an RHP zero اگر سیستم حلقه بسته دارای صفرRHP باشد

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Lecture 12

Ali Karimpour Oct 2009

Undershootفرو جهش

:uM

uM:

The (absolute value of the) maximum instantaneous amount by which the step response falls below zero.

مQاکزیمم قQQدر ( مقQQداری پاسQQخ ) مطلQQق پلQه از صQفر کمQتر

می شود.

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Lecture 12

Ali Karimpour Oct 2009

Exercises

1- Repeat example 3 for parabolic input.

2 – Consider following system. +-

cer

)2(

2

n

n

ss

a) Find the step response of the system for

b) Find the rise time, settling time, overshoot, and percent overshoot.

3.0,56.12 n

+-

cer

)2(

2

n

n

ss

3 – Consider following system.

a) Find the step response of the system for

b) Find the rise time, settling time, overshoot, and percent overshoot.

9.0,56.12 n