Download - Time Domain Analysis of 1st Order Systems
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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.
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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 )(
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Im!lse #esonse of 1st $rder %ystem
• #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
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Im!lse #esonse of 1st $rder %ystem
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)
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%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
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%te #esonse of 1st $rder %ystem
• 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-
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%te #esonse of 1st $rder %ystem
• 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
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%te #esonse of 1st $rder %ystem
• 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
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%te #esonse of 1st
order %ystem• %ystem taes five time constants to reach its
final val!e.
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%te #esonse of 1st $rder %ystem
• 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
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#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(
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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
.)( −=
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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
)(
)(
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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
)(
)(
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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 .
)( −
−=
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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 .)( −−=
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9artial Fraction 62ansion in :atlab
• If yo! want to e2and a olynomial into artial fractions !seresidue command.
;-?-
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9artial Fraction 62ansion in :atlab
• If we want to e2and following olynomial into artial fractions
86
84
2++
+−
ss
s
;-?-
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9artial Fraction 62ansion in :atlab• If yo! want to e2and a olynomial into artial fractions !se
residue command.
( )126
+
=
S ssC )(
;-8>?-