control_engineering_tutorial_7_presentation
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Control Engineering
Steady State Error
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
Tutorial #7
So to calculate the steady state error, you need to know the following:
1. Input type (step, ramp, …)2. Is the system unity feedback or not
Y RE
)()()( tytrte )()()( sY sRsE
)(lim0
ssEes
ss
The definition of the error signal is or
then by using the final value theorem, the steady state error is
Where R is the reference input and Y is the system output
Steady State Error
Y(t)
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
1. Input type
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
2. Is the system unity feedback or not
1. Unity feedback case
G
sRsR
G
GsR
G
GsRsY sRsE
1
)()(]
11[)(
1)()()()(
G
sRsE
1
)()(
G
ssRssE e
ssss
1
)(lim)(lim
00
Applying the final value theorem, we get
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
And, for various inputs, the result of this limit is summarized in the following table
But how to determine the system type???
For unity feedbacksystems, only !!
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
System type (for a unity feedback)Y+
_G(s)
R
E
system type is obtained from the denominator of the openloop transfer function, G(s)And it equals the degree of the free S in the denominator
So, the system type can be defined to be the degree of the polynomialfor which steady state error is a non-zero finite constant.
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
2. General Case (Non-unity feedback)
)(1
1)( sR
GH
GGH sE
)(lim0
ssE es
ss
Applying the final value theorem, we get
)(1
1)(]
11[)(
1)()()()( sR
GH
GGH sR
GH
GsR
GH
GsRsY sRsE
Important Note:
For the case of non-unity feedback, you can not use the summarized table
given in slide 5
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
Problem 1
• Determine the system type and error constant (Kp, Kv or Ka).
450
)()(
410)()(
2
ssH b
ssH a
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Solution
As you can see, the system is unity feedback, hence we can use the table in
slide 5. But first we need to get the system type:
Then the open loop transfer is
,410)()( ssH a3
5)(
ssG
3
)4.0(50
3
)410(5
s
s
s
sGH
As you see, there is no free s in the denominator, hence the system is type 0.
And, by looking in the table of slide 5, you see that the system to have a
steady state error, the input has to be step input, in this case you can get theerror constant Kp. For other inputs, the system has infinite steady state error
And the error constant is
3
20
3
)4.0(50limlim
00
s
sGK
ssp
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
3
20,0
3
201
1
4103
51
11.
4103
51
1)(:
)(1
limlim
lim
00
.
0
.
p
ssss
sss
K and Typeof systemtheTherefore
ss
ss
s
se
ssRinputstepfor
sRGH
se
Another Solution
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Then the open loop transfer is
3
5)(
s
sG
)3()5.12(20
3
)450
(5
3
)450(5
2
22
2
2
ss
sss
s
ssGH
As you see, the degree of the free s in the denominator is 2, hence the
system is type 2.And by looking in the table of slide 5, you see that the system to have a
steady state error, the input have to be acceleration input, in this case youcan get the error constant Ka. For other inputs, the system has zero steady
state error
And the error constant is
3.83
3
250
)3(
)5.12(20lim
)3(
)5.12(20limlim
2
02
22
0
2
0
s
s
ss
ssGH sK
sssa
,450
)()(2
ssH b
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
3
250,2
3
250
1
3
2500
1
4503
5
11.
450
3
51
1)(:.
1
00
1
450
3
5
11.
450
3
51
1)(:
0
01
1
450
3
51
11.
450
3
51
1)(:
)(1
2203
2
0
.
3
02
2
0
.
2
2
0
2
0
.
0
.
limlim
limlim
limlim
lim
a
ssss
ssss
ssss
sss
K Typeof systemtheTherefore
ss
ss
ss
se
ssRinputaccfor
Typeof notissystemtheTherefore
sss
ss
ss
se
ssRinputrampfor
Typeof notissystemtheTherefore
ss
s
ss
se
ssRinputstepfor
sRGH
se
Another Solution
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
How to calculate the transfer function for afeedback system with disturbance?
Problem: We have 2 inputs!
1. Reference input
2. Disturbance
)(sW
)(sR
cG pG
)(sY
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
What to do??? Assume disturbance input = 0
Get T.F. for the reference input.
Then assume reference input = 0
Get T.F. for the disturbance.
The general transfer function is sum of both transfer functions!!
)(sW
)(sR
cG pG
)(sY
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
)(sW
)(sR
cG pG
)(sY
For a feed back system with disturbance
)(1
)(1
)( sW GG
GsR
GG
GGsY
pc
p
pc
pc
Here, the closed loop T.F. is
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
Problem 2
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
Problem 3
W
Y
+
_
+
+
D(s)1
s2
k
+
_
Consider the system shown in the Figure below, which represents control
of the angle of a pendulum which has no damping.
R
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
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GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy
GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics
Control Engineering ENME 503Dr. Ayman Ali El-Badawy