time response 1

8/6/2019 Time Response 1

http://slidepdf.com/reader/full/time-response-1 1/68

Time Response Analysis, Design

Indices

18 February 20111



System

18 February 20112



a s e me response o a con rosystem?

Time response:

d ifferentia l e uation.

18 February 20113




me response c ons s s o :

2.Forced response

18 February 20114




Natural res onse is caused b the s stem itself.

Also called transient response or homogenous

solution

Forced response is caused by the input to the

-.or particular solution.

18 February 20115




Time response PlotNatural res onse or

Steady-stateTime solution to

Transient responseresponse

differential equation

Caused by the

system itself

Caused bythe input

18 February 20116



How to obtain time response?

Schematic Time ResponseWhyrans er

Function?

M)(t f

x M &&

Kx Transfer Function

Differential Equation

K MssF s X sG

+

==21

)()()(

18 February 20117

)(t f Kx x M =+&& )(2sF KX MXs =+



ow o o a n me response romtransfer function?

Given:

,concept of poles and zeros

18 February 20118




What are oles ?First rule:

The values of Laplace transform variable,s, a cause e rans er unc on o

become infinite.

Any roots of the denominator of thetransfer function that are common to the

18 February 20119roots of numerator




Transfer FunctionFirst rule:

)3)(2)(5(

)3()(

+++

+=

sss

ssG

When s = -5, or s = -2 or s = -3, then,G(s)=infinity! Therefore, the poles

of the transfer function G(s) are -5,-2 and -3.

Second rule:

)3)(2)(5()(

+++=

ssssG Although the term (s+3) can be

cancelled out, the value -3 is stillthe poles of the transfer function

18 February 201110

G(s).




What are zeros ?First rule:

The values of Laplace transform variable,s, a cause e rans er unc on o

become zero.

Any roots of the numerator of the transfer function that are common to roots of the

18 February 201111denominator




Transfer FunctionFirst rule:

)3)(2)(5(

)3()(

+++

+=

sss

ssG

When s = -3, then, G(s)= 0.Therefore, the zero of the transfer

function G(s) is -3.

Second rule:

)3)(2)(5()(

+++=

ssssG Although the term (s+3) can be

cancelled out, the value -3 is stillthe zero of the transfer function

18 February 201112

G(s).




Given:

How to obtain the time response?1. Find C(s)

2. Expan e rans er unc on us ng par a rac onexpansion technique

3. Perform inverse La lace transform

18 February 201114



ow o o a n me response romtransfer function? –Find C(s)

Given:

)(

)()(

s R

sC

Input

Output sG ==

)()()( s RsGsC =

18 February 201115




What is R s ?

Step input: At r =)(S

As R =)(Laplace Transform

Ramp input: t t r =2

)(S

s R =Laplace Transform

Sine input: )sin( t r(t) ω = 22)( =

As RLaplace Transform

18 February 201116




Given:

R(s) unit step input implies A = 1

)()()( = s RsGsC

)2(1)(

+=s

sC

18 February 201117



ow o o a n me response romtransfer function? – Expand C(s)

Ex and C s usin Partial Fraction Ex ansionTechnique

21 +s

)5( +=ss

s

)5()2(

5)(

+

+=

++=

sss

s B

s AsC

18 February 201118



ow o o a n me response romtransfer function? – Expand C(s)

)2( +==s B A How to find A

)5(5 ++ ssss and B?

5)5(0

=+

=

→ss

s A

53

52

5

3

)(

)2(=

+=s

s B

5+ss

18 February 201119

−



How to obtain time response from transferfunction? – Inverse Laplace Transform

32

5++=ss

sC

2 Inverse Laplace 2

sTransform 5

3 3 −

t et c

532 −+=

5+s

Transform e5 55

18 February 201120



a s me response o a con rosystem?

Input pole generatesforce res onse

System pole generates

natural response

System pole generatesnatural response in theform of e-αt .

Both zero and polegenerates theam litude of time

18 February 201122

response



a s me response o a con rosystem? An Example

)(sC s R1

)( =What are poles and

zeros?

)5)(4)(2( +++ sssPoles = -2, -4 and -5Zero = -3

By inspection:

)( 4321+++≡K K K K

sC

Force

on time response

18 February 201123

response




)(sC s R1

)( =

)5)(4)(2( +++ sss

By inspection:)5()4()2(

)( 4321

++

++

++≡

s

K

s

K

s

K

s

K sC

−−−

Transform eeec 4321+++≡

–

18 February 201124




)( 4321+++≡K K K K

sC How tossss

t t t eK eK eK K t c

5

4

4

3

2

21)( −−−

+++≡

eva uate 1,K2, K3 and K4?

)5)(4)(2(

)3()(

+++

+=

ssss

ssC

11

3

)5)(4)(2(

)3(

)5)(4)(2(

)3()(

00

1 =+++

+=

+++

+==

→→ sssss

s

ssss

ssK ssC

18 February 201125




330)3()3()( 1 =

+=

+=

+==

sssK ssC

00 →→ ss

12

1

)52)(42(2

)32(

)5)(4(

)3(

)5)(4)(2(

)3)(2()()2(

222 −

=

+−+−−

+−=

++

+=

+++

++==+

−→−→ ss sss

s

ssss

ssK sC s

8

1

)54)(24)(4(

)34(

)5)(2(

)3(

)5)(4)(2(

)3)(4()()4(

44

1

−=

+−+−−

+−=

++

+=

+++

++==+

−→−→ sssss

s

ssss

ssK sC s

15

2

)45)(25(5

)35(

)4)(2(

)3(

)5)(4)(2(

)3)(5()()5( 1

−

−=

+−+−−

+−=

++

+=

+++

++==+

−−sss

s

ssss

ssK sC s

18 February 201126




t t t eK eK eK K t c

54

43

221)( −−−

+++≡

t t t eeet c 542 2113−−−

+−−≡3 1581240

18 February 201127




18 February 201128



ys e

18 February 201130



What is the first order of a control system?

One Pole Time Response

at et c−

−=1)(

Solution Using Inspection

)(ass

asC

+=

1)(1 =====aasa

ssC K

)()( 21

as

K

s

K sC

++=

00

++

→→

aasassss

1)(

)()( −===+

=+=aaaas

sC asK

18 February 201131

at eK K t c−

+= 21)( −+−→−→asass asas



What is the first order of a control system? –Performance Parameters

Time constant

63% ofFinal

T c1

=

18 February 201133



a s e rs or er o a con ro sys em –Performance Parameters

Rise Time

me o r sefrom 0.1 to 0.9

of final va lue

T T T r

11.031.2

%10%90 −=

T

aa

r

r

2.2=

−=

18 February 201134

a




Settling Time

Time to reach

2% of finalvalue

T s4=

18 February 201135




Time constant

Rise Time

aT c =

o na

value

aT r

2.2=

Time to rise from0.1 to 0.9 of finalvalue

Settling Time

4Time to reach

18 February 201136

as =2% of final value



What is the first order of a control system? –An Example

Time constant50 1

Rise Time

sec50

==a

T c50+s

s

sec50

2.22.2==a

T r

pole: pole = a = -50

By inspection the solution of

Settling Time

44t et c

501

−−=

unit step input is:

18 February 201137

50as




By experiment we get this

Find transfer function?

Step 1: Find the poleTime constant is the time

to reach 63% of final value

Final value = 0.72

63% Final value = 0.63X0.72=0.45Time constant = 0.13 sec

18 February 201139




Step 1: Find the pole

me cons an = . sec

1=c

a

T

7.713.0

11===

cT a

at K −−

General solution

a54.5)7.7(72.072.072.0 ===→= aK

a

K

54.5K

18 February 201140

7.7+=

+=

sass




)(=sG

at are t e t me constant, r se t me, sett ng t meand steady-state value?

18 February 201141



General Response of Sec ond

Order System

18 February 201142



control system?

m wi h w l

18 February 201143



control system?

Under-damped Time Response

a

acbb

2

42

2,1

−±−=σ

81

2322

)1(2)9)(1(422

2

2,1 j

+−=

±−=

−±−=

σ

σ

σ − t d

General solution

σ ±−= 81

18 February 201145

812 j−−=σ −− d

nd jω σ σ ±−=2,1

,



me response o a secon or ercontrol system?

Un-damped Time Response

a

acbb

2

42

2,1

−±−=σ

2

36

)1(2

)9)(1(400 2

2,1

j±=

−±−=σ )cos(1)( φ ω −−= t t c

General solution

18 February 2011473

3

2

1

j

j

−=

+=

σ

σ



me response o secon or ercontrol system? - Summary

Over-damped Un-damped

Poles are realand distinct

Poles areimaginary

Under-damped Critic ally-damped

Poles arereal andimaginary

Poles are realand repeated

18 February 201149



me response o secon or ercontrol system? - Summary

18 February 201150



me response o secon or ercontrol system? - Example

)cos()( φ ω σ

−=−

d t d Aet cGeneral

acbb 42

2,1

−±−=

σ

under-damped )23.13cos()( 5 φ −=− t Aet c

)1(2

)200)(1(41010 2

2,1

−±−=σ

23.1352,

2

)175(410

1

2,1

j

j

±−=

±−

=

σ

σ

18 February 201151

The poles are real and imaginary.Therefore UNDER-DAMPED.




What are poles and zeroes?-

Write the general expression of the step response?

State the nature of response?

5

)6)(3( ++ ss

18 February 201153







)7(10 +s

)20)(10( ++ ss

18 February 201154







20

14462 ++ ss

18 February 201155







2+s

92 +s

18 February 201156







5+=s

2)10( +s

18 February 201157



Time Response of Second Order

Natural Frequency

18 February 201158



me response o secon or ercontrol system?

So far, we based on the poles of the

second order system

a

ac

22,1

−−=σ

18 February 201161




We are going to learn two

quantities that will help us analyze

second order systemNatural

Fre uenc

Damping

Ratio

ω 18 February 201162



me response o secon or ercontrol system? Natural Frequency

Natural

Frequency n

a ura requency o a secon

order system is the frequency of

osc a on o e sys em w oudamping

18 February 201163



me response o secon or ercontrol system? Damping Ratio

Damping FrequencyDecaylExponentia=

Ratio (rad/sec)FrequencyNatural

18 February 201165




36 Find damping ratio

362.42++

=

sss an e na ura

frequency?

Natural Frequency

=

Damping Ratio

6=

=

n

n

ω

ω

35.02.4

.

==ξ

n

18 February 201168

time response 1

Documents