Feedback Control Systems ( FCS )
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Transcript of Feedback Control Systems ( FCS )
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Feedback Control Systems (FCS)
Dr. Imtiaz Hussainemail: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
Lecture-14-15Block Diagram Representation of Control Systems
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Introduction• A Block Diagram is a shorthand pictorial representation of
the cause-and-effect relationship of a system.
• The interior of the rectangle representing the block usually contains a description of or the name of the element, or the symbol for the mathematical operation to be performed on the input to yield the output.
• The arrows represent the direction of information or signal flow.
dtd
x y
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Introduction• The operations of addition and subtraction have a special
representation.
• The block becomes a small circle, called a summing point, with the appropriate plus or minus sign associated with the arrows entering the circle.
• The output is the algebraic sum of the inputs.
• Any number of inputs may enter a summing point.
• Some books put a cross in the circle.
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Introduction• In order to have the same signal or variable be an input
to more than one block or summing point, a takeoff point is used.
• This permits the signal to proceed unaltered along several different paths to several destinations.
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Example-1• Consider the following equations in which x1, x2, x3, are variables,
and a1, a2 are general coefficients or mathematical operators.
522113 xaxax
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Example-1• Consider the following equations in which x1, x2, x3, are variables,
and a1, a2 are general coefficients or mathematical operators.
522113 xaxax
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Example-2• Consider the following equations in which x1, x2,. . . , xn, are
variables, and a1, a2,. . . , an , are general coefficients or mathematical operators.
112211 nnn xaxaxax
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Example-3• Draw the Block Diagrams of the following equations.
11
22
2
13
11
12
32
11
bxdtdx
dtxd
ax
dtxbdt
dxax
)(
)(
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Canonical Form of A Feedback Control System
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Characteristic Equation• The control ratio is the closed loop transfer function of the system.
• The denominator of closed loop transfer function determines the characteristic equation of the system.
• Which is usually determined as:
)()()(
)()(
sHsGsG
sRsC
1
01 )()( sHsG
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Example-41. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. closed loop poles and zeros if K=10.
)()()()( sHsGsEsB
)()()( sGsEsC
)()()(
)()(
sHsGsG
sRsC
1
)()()()(
)()(
sHsGsHsG
sRsB
1
)()()()(
sHsGsRsE
1
1
)()()(
)()(
sHsGsG
sRsC
1
01 )()( sHsG
)(sG
)(sH
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Reduction techniques
2G1G 21GG
1. Combining blocks in cascade
1G
2G21 GG
2. Combining blocks in parallel
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Example-5: Reduce the Block Diagram to Canonical Form.
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Example-5: Continue.
However in this example step-4 does not apply.
However in this example step-6 does not apply.
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Example-6• For the system represented by the following block diagram
determine:1. Open loop transfer function2. Feed Forward Transfer function3. control ratio4. feedback ratio5. error ratio6. closed loop transfer function7. characteristic equation 8. closed loop poles and zeros if K=10.
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Example-6– First we will reduce the given block diagram to canonical form
1sK
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Example-6
1sK
ssK
sK
GHG
11
11
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Example-61. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. closed loop poles and zeros if K=10.
)()()()( sHsGsEsB
)()()( sGsEsC
)()()(
)()(
sHsGsG
sRsC
1
)()()()(
)()(
sHsGsHsG
sRsB
1
)()()()(
sHsGsRsE
1
1
)()()(
)()(
sHsGsG
sRsC
1
01 )()( sHsG
)(sG
)(sH
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Example-7• For the system represented by the following block diagram
determine:1. Open loop transfer function2. Feed Forward Transfer function3. control ratio4. feedback ratio5. error ratio6. closed loop transfer function7. characteristic equation 8. closed loop poles and zeros if K=100.
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Reduction techniques
3. Moving a summing point behind a block
G G
G
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5. Moving a pickoff point ahead of a block
G G
G G
G1
G
3. Moving a summing point ahead of a block
G G
G1
4. Moving a pickoff point behind a block
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6. Eliminating a feedback loop
G
HGHG
1
7. Swap with two neighboring summing points
A B AB
G
1H
GG1
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Example-8
R_+
_+1G 2G 3G
1H
2H
+ +
C
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Example-8
R_+
_+
1G 2G 3G
1H
1
2
GH
+ +
C
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Example-8
R_+
_+
21GG 3G
1H
1
2
GH
+ +
C
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Example-8
R_+
_+ 21GG 3G
1H
1
2
GH
+ +
C
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block diagram: reduction example
R_+
_+
121
21
1 HGGGG
3G
1
2
GH
C
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block diagram: reduction example
R_+
_+
121
321
1 HGGGGG
1
2
GH
C
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block diagram: reduction example
R_+
232121
321
1 HGGHGGGGG
C
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Example-8
R
321232121
321
1 GGGHGGHGGGGG
C
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Example 9
Find the transfer function of the following block diagrams
2G 3G1G
4G
1H
2H
)(sY)(sR
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1. Moving pickoff point A ahead of block2G
2. Eliminate loop I & simplify
324 GGG B
1G
2H
)(sY4G
2G
1H
AB3G
2G
)(sR
I
Solution:
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3. Moving pickoff point B behind block324 GGG
1GB)(sR
21GH 2H
)(sY
)/(1 324 GGG
II
1GB)(sR C
324 GGG
2H
)(sY
21GH
4G
2GA
3G 324 GGG
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4. Eliminate loop III
)(sR)(1
)(
3242121
3241
GGGHHGGGGGG
)(sY
)()(1)(
)()()(
32413242121
3241
GGGGGGGHHGGGGGG
sRsYsT
)(sR1G
C
324
12
GGGHG
)(sY324 GGG
2H
C
)(1 3242
324
GGGHGGG
Using rule 6
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2G1G
1H 2H
)(sR )(sY
3H
Example 10
Find the transfer function of the following block diagrams
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Solution:
1. Eliminate loop I
2. Moving pickoff point A behind block22
2
1 HGG
1G
1H
)(sR )(sY
3H
BA
22
2
1 HGG
2
221GHG
1G
1H
)(sR )(sY
3H
2G
2H
BA
II
I
22
2
1 HGG
Not a feedback loop
)1(2
2213 G
HGHH
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3. Eliminate loop II
)(sR )(sY22
21
1 HGGG
2
2213
)1(G
HGHH
21211132122
21
1)()()(
HHGGHGHGGHGGG
sRsYsT
Using rule 6
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2G 4G1G
4H
2H
3H
)(sY)(sR
3G
1H
Example 11
Find the transfer function of the following block diagrams
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Solution:
2G 4G1G
4H)(sY
3G
1H
2H
)(sRA B
3H4
1G
4
1G
I1. Moving pickoff point A behind block
4G
4
3
GH
4
2
GH
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2. Eliminate loop I and Simplify
II
III
443
432
1 HGGGGG
1G)(sY
1H
B
4
2
GH
)(sR
4
3
GH
II
332443
432
1 HGGHGGGGG
III
4
142
GHGH
Not feedbackfeedback
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)(sR )(sY
4
142
GHGH
332443
4321
1 HGGHGGGGGG
3. Eliminate loop II & IIII
143212321443332
4321
1)()()(
HGGGGHGGGHGGHGGGGGG
sRsYsT
Using rule 6
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3G1G
1H
2H
)(sR )(sY
4G
2GA
B
Example 12
Find the transfer function of the following block diagrams
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Solution:
1. Moving pickoff point A behind block3G I
1H3
1G
)(sY1G
1H
2H
)(sR
4G
2GA B
3
1G
3G
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2. Eliminate loop I & Simplify
3G
1H
2GB
3
1G
2H
32GG B
23
1 HGH
1G)(sR )(sY
4G3
1
GH
23212
32
1 HGGHGGG
II
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)(sR )(sY12123212
321
1 HGGHGGHGGGG
3. Eliminate loop II
12123212
3214 1)(
)()(HGGHGGHG
GGGGsRsYsT
4G
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Example-13: Simplify the Block Diagram.
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Example-13: Continue.
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Example-14: Reduce the Block Diagram.
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Example-14: Continue.
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Example-15: Reduce the Block Diagram. (from Nise: page-242)
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Example-15: Continue.
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Example-16: Reduce the system to a single transfer function. (from Nise:page-243).
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Example-17: Simplify the block diagram then obtain the close-loop transfer function C(S)/R(S). (from Ogata: Page-47)
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Example-18: Multiple Input System. Determine the output C due to inputs R and U using the Superposition Method.
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Example-18: Continue.
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Example-18: Continue.
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Example-19: Multiple-Input System. Determine the output C due to inputs R, U1 and U2 using the Superposition Method.
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Example-19: Continue.
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Example-19: Continue.
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Example-20: Multi-Input Multi-Output System. Determine C1 and C2 due to R1 and R2.
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Example-20: Continue.
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Example-20: Continue.
When R1 = 0,
When R2 = 0,
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Block Diagram of Armature Controlled D.C Motor
Va
iaT
Ra La
J
c
eb
V f=constant
(s)IK(s)cJs
(s)V(s)K(s)IRsL
ama
abaaa
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Block Diagram of Armature Controlled D.C Motor
(s)E(s)K(s)IRsL abaaa
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Block Diagram of Armature Controlled D.C Motor
(s)IK(s)cJs ama
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Block Diagram of Armature Controlled D.C Motor
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Block Diagram of liquid level system
11
1 qqdtdh
C 1
211 R
hhq
212
2 qqdtdh
C 2
22 R
hq
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Block Diagram of liquid level system
)()()( sQsQssHC 111 1
211 R
sHsHsQ
)()()(
2
22 R
sHsQ
)()( )()()( sQsQssHC 2122
11
1 qqdtdh
C 1
211 R
hhq
212
2 qqdtdh
C 2
22 R
hq
L
L
L
L
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Block Diagram of liquid level system
)()()( sQsQssHC 111 1
211 R
sHsHsQ
)()()(
2
22 R
sHsQ
)()( )()()( sQsQssHC 2122
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Block Diagram of liquid level system
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END OF LECTURES-14-15
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