E19
7
MEEN 364 Parasuram Lecture 19, 20 August 25, 2001 1 HANDOUT E.19 - EXAMPLES ON FEEDBACK CONTROL SYSTEMS Example1 Consider the system shown below. The open loop transfer function is given by . ) 1 ( ) ( + = s s K s G The closed loop transfer function is . ) 1 ( ) 1 ( 1 ) 1 ( ) ( 1 ) ( ) ( ) ( 2 K s s K K s s K s s K s s K s G s G s R s Y + + = + + = + + + = + = Example 2 Consider the system shown. a) Determine the transfer function from ‘r’ to ‘y’. b) Determine the transfer function from ‘w’ to ‘y’. To obtain the transfer function between ‘r’ to ‘y’, assume ‘w’ to be equal to zero. Therefore the block diagram reduces to,
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Transcript of E19
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
1
HANDOUT E.19 - EXAMPLES ON FEEDBACK CONTROL SYSTEMS
Example1
Consider the system shown below.
The open loop transfer function is given by
.)1(
)(+
=ssKsG
The closed loop transfer function is
.)1(
)1(1
)1()(1
)()()(
2 KssK
KssK
ssK
ssK
sGsG
sRsY
++=
++=
++
+=+
=
Example 2
Consider the system shown.a) Determine the transfer function from ‘r’ to ‘y’.b) Determine the transfer function from ‘w’ to ‘y’.
To obtain the transfer function between ‘r’ to ‘y’, assume ‘w’ to be equal to zero.Therefore the block diagram reduces to,
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
2
Hence the open loop transfer function is
.)20()(10
20)1(10)(
)( 22121
+++
=++
⋅+
=sss
sKKsss
sKKsG
The closed loop transfer function is given by
.10)1020(
)(10
)20()(10
1
)20()(10
)(1)(
)()(
1223
21
221
221
KKssssKK
ssssKK
ssssKK
sGsG
sRsY
+++++
=
++++
+++
=+
=
To obtain the transfer function between ‘w’ and ‘y’, assume ‘r’ to be equal to zero. Hencethe block diagram reduces to,
For the above block diagram the open loop transfer function is
2010
20)1(10)( 2 ++
=++
=ssss
sG
and the feedback transfer function is given by
.)( 21
ssKKsH +=
Therefore the transfer function from ‘w’ to ‘y’ is given by
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
3
.)10)1020((
10
)20(101
)20(10
)()(1)(
)()(
1223
212
2
KKssss
ssKK
ss
sssHsG
sGsWsY
++++=
+⋅
+++
++=+
=
Example 3
A unity feedback system has the open loop transfer function
.1)( 23 ssssG
++=
Find the error constants Kp, Kv and Ka for the system.
The closed loop transfer function is given as
.1
111
1
)(1)(
)()(
23
23
23
++++=
+++
++
=+
=sss
s
sssss
s
sGsG
sRsY
The transfer function for the error signal is given by
).(1
)(1
1)()()()( 23
23
23 sRssssssR
sssssRsYsRsE
++++=
++++−=−=
Therefore the steady state error for the system can be obtained by applying the final valuetheorem as
).(lim0
ssEesss >−
=
Therefore,
).(1
lim 23
23
0sR
sssssse
sss ++++=
>−
For a unit step input, we have ,1)( ssR = therefore the steady state error is
.1
101
lim11
lim 23
23
023
23
0p
ssss Ksssss
ssssssse
+==
++++=⋅
++++⋅=
>−>−
Therefore the value of Kp is given as
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
4
.∞→pK
For a ramp input, we have ,1)( 2ssR = therefore the steady state error is
.101
lim1.1
.lim 23
2
0223
23
0v
ssss Ksssss
ssssssse ==
++++=
++++=
>−>−
Hence the value of Kv is given as
.∞→vK
For a parabolic input, we have ,1)( 3ssR = therefore the steady state error is
.111
1lim1.1
.lim 230323
23
0a
ssss Kssss
ssssssse ==
++++=
++++=
>−>−
Therefore,
.1=aK
Example 4
The block diagram shown below shows a control system in which the output member ofthe system is subject to a disturbance. In the absence of a disturbance, the output is equalto the reference. Investigate the response of this system to a unit step disturbance.
Since we are interested in the response of the system to a unit step disturbance, assumethe reference input to be equal to zero. Therefore the block diagram reduces to
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
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The open loop transfer function is given by
,1)(Js
sG = and the feedback transfer function is given as
.)( KsH =
Therefore the transfer function between the disturbance and the output is given as
.1
1
1
)()(1)(
)()(
KJsJsKJs
sHsGsG
sWsY
+=
+=
+=
For a unit step disturbance, the output reduces to
.)(
111)(KJsssKJs
sY+
=⋅+
=
The steady state value of the output can be obtained by applying the final value theoremas
.11lim)(
1lim)(lim000 KKJsKJss
sssYysssss =
+=
+⋅==
>−>−>−
Therefore the system is incapable of rejecting the disturbance completely as there is asteady state offset.
Example 5
Consider a nonunity feedback system, whose open loop transfer function is given by
,)1(
1)(Tss
sG+
=
where in the feedback loop transfer function is given as
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
6
,)( hsH = with h > 0. Determine the system type with respect to the reference input.
The closed loop transfer function is given as
.1
)1(1
)1(1
)()(1)(
)()(
2 hsTsTsshTss
sHsGsG
sRsY
++=
++
+=+
=
The transfer function for the error signal is given by
).(1)(11)()()( 2
2
2 sRhsTshsTssR
hsTssYsRsE
++−++=
++−=−=
The steady state error is obtained as
).(1lim)(lim 2
2
00sR
hsTshsTsssEse
ssss ++−++⋅=⋅=
>−>−
For a step reference input, the error is
.11lim11lim 2
2
02
2
0 hh
hsTshsTs
shsTshsTsse
ssss−=
++−++=⋅
++−++⋅=
>−>−
If the system is not unity feedback then, 1≠h , therefore the system is of Type 0. If thesystem is unity feedback, i.e., h = 1, then the system is of Type 1 as the steady state errorfor the step reference input is zero.
MEEN 364 ParasuramLecture 19, 20 August 25, 2001
7
Assignment
1) Consider a unity feedback system whose open loop transfer function is given by
.)52(
)1()( 2 +++=sss
ssG
Determine the closed loop transfer function of the system.
2) Problem 4.32a from the textbook “ Feedback control of dynamic systems”, by Franklinet.al.
Recommended Reading
“Feedback Control of Dynamic Systems” 4th Edition, by Gene F. Franklin et.al – pp 201– 215, 230 – 242.
Recommended Assignment
“Feedback Control of Dynamic Systems” 4th Edition, by Gene F. Franklin et.al –problems 4.22, 4.28, 4.33, 4.35, 4.37.
