Time Domain Analysis of 1st Order Systems

8/16/2019 Time Domain Analysis of 1st Order Systems

1/19

Introduction

• For the first order system given below

13

10

+

=

ssG )(

5

3

+

=

ssG )(

151

53

+

=

s/

/

• D.C gain is 10 and time constant is 3 seconds.

• And for following system

• D.C Gain of the system is 3/5 and time constant is 1/5

seconds.


2/19

Im!lse #esonse of 1st $rder %ystem

• Consider the following 1st order system

1+Ts

K )(sC )(s R

0 t

&'t(

1

1== )()( ss R δ

1+=

Ts

K sC )(


3/19


• #e)arrange above e*!ation as

1+=

Ts

K sC )(

T s

T K sC

/

/)(

1+=

T t

eT

K

t c

/

)(

−=

• In order to reresent the resonse of the system in time domain

we need to com!te inverse +alace transform of the abovee*!ation.

at

Aeas

A

L

−−

=

+

1


4/19


T t e

T

K t c

/)( −=• If ,-3 and -s then

0 2 4 6 8 100

0.5

1

1.5

Time

c ( t )

K/T*exp(-t/T)


5/19

%te #esonse of 1st $rder %ystem

• Consider the following 1st order system

1+Ts

K

)(sC )(s R

ssU s R

1== )()(

( )1+=

Tss

K sC )(

1+−=

TsKT

sK sC )(

• In order to find o!t the inverse +alace of the above e*!ation we

need to brea it into artial fraction e2ansion

Forced Response Natural Response


6/19


• aing Inverse +alace of above e*!ation

+

−=

1

1

Ts

T

sK sC )(

T t

et uK t c /

)()( −

−=

• here !'t(-1T t

eK t c /)( −−= 1

K eK t c 63201 1

.)( =−= −

• hen t-


7/19


• If ,-10 and -1.5s then T t eK t c /)( −−= 1

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

11

Time

c ( t )

K*(1-exp(-t/T))

Unit Step Input

Step Response

1

10===

Input

output statesteadyK GainC D

.

%63


8/19


• If ,-10 and -1 3 5 4 T t eK t c /)( −−= 1

0 5 10 150

1

2

3

4

5

6

7

8

9

10

11

Time

c ( t )

K*(1-exp(-t/T))

T=3s

T=5s

T=7s

T=1s


9/19

%te #esonse of 1st

order %ystem• %ystem taes five time constants to reach its

final val!e.


10/19


• If ,-1 3 5 10 and -1 T t eK t c /)( −−= 1

0 5 10 10

1

2

3

4

5

6

7

8

9

10

11

Time

c ( t )

K*(1-exp(-t/T))

K=1

K=3

K=5

K=10


11/19

#elation etween %te and im!lse

resonse

• he ste resonse of the first order system is

• Differentiating c(t) with resect to t yields

T t T t KeK eK t c

//)( −− −=−= 1

( )T t KeK dt

d

dt

t dc /)( −−=

T t e

T

K

dt

t dc /)( −= 'im!lse resonse(


12/19

62amle71• Im!lse resonse of a 1st order system is given below.

• Find o!t

– ime constant

– D.C Gain ,

– ransfer F!nction

– %te #esonse

t et c

503

.)( −=


13/19

62amle71• he +alace ransform of Im!lse resonse of a

system is act!ally the transfer f!nction of the system.

•

herefore taing +alace ransform of the im!lseresonse given by following e*!ation.

t et c

503

.)( −=

)(..

)( sS S

sC δ ×+

=×

+

=

50

31

50

3

503

.)()(

)()(

+

==

S s RsC

ssC

δ

12

6

+

=

S s R

sC

)(

)(


14/19

62amle71• Im!lse resonse of a 1st order system is given below.

• Find o!t

– ime constant -

– D.C Gain ,-8

– ransfer F!nction

– %te #esonse

– Also Draw the %te resonse on yo!r noteboo

t et c

503

.)( −=

12

6

+

=

S s R

sC

)(

)(


15/19

62amle71• For ste resonse integrate im!lse resonse

t et c

503

.)( −=

dt edt t c t ∫=∫

− 503

.)(

C et c t s +−= − 506 .)(

• e can find o!t C if initial condition is nown e.g. cs'0(-0

C e +−= ×− 050

60 .

6=C

t s et c

50

66 .

)( −

−=


16/19

62amle71•

If initial Conditions are not nown then artial fractione2ansion is a better choice

12

6

+

=

S s R

sC

)(

)(

( )126

+

=

S ssC )(

( ) 12126

+

+=

+ s

B

s

A

S s

ss Rs R

1=)(,)( inputstepaissince

( ) 5066

12

6

.+−=

+ ssS s

t et c 5066 .)( −−=


17/19

9artial Fraction 62ansion in :atlab

• If yo! want to e2and a olynomial into artial fractions !seresidue command.

;-?-


18/19

9artial Fraction 62ansion in :atlab

• If we want to e2and following olynomial into artial fractions

86

84

2++

+−

ss

s

;-?-


19/19

9artial Fraction 62ansion in :atlab• If yo! want to e2and a olynomial into artial fractions !se

residue command.

( )126

+

=

S ssC )(

;-8>?-

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Time Domain Analysis of 1st Order Systems