mech261 control principles tutorial #3. block diagram transfer function consists of blocks can be...
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![Page 1: MECH261 Control Principles Tutorial #3. Block diagram Transfer Function Consists of Blocks Can be reduced](https://reader036.vdocument.in/reader036/viewer/2022081520/5697bfa51a28abf838c9783f/html5/thumbnails/1.jpg)
MECH261Control Principles
Tutorial #3
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Block diagram
Transfer Function
Consists of Blocks
Can be reduced
)(sR2G 3G1G
4G
1H
2H
)(sY
G)(sY)(sR
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Reduction techniques
2G1G21GG
2. Moving a summing point behind a block
G G
G
1G
2G21 GG
1. Combining blocks in cascade or in parallel
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5. Moving a pickoff point ahead of a block
G G
G G
G
1
G
3. Moving a summing point ahead of a block
G G
G
1
4. Moving a pickoff point behind a block
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6. Eliminating a feedback loop
G
HGH
G
1
7. Swap with two neighboring summing points
A B AB
G
1H
G
G
1
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Example 1
Find the transfer function of the following block diagrams
2G 3G1G
4G
1H
2H
)(sY)(sR
(a)
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1. Moving pickoff point A ahead of block 2G
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 block 324 GGG
1GB)(sR
21GH 2H
)(sY
)/(1 324 GGG
II
1GB)(sR C
324 GGG
2H
)(sY
21GH
4G
2G A3G 324 GGG
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4. Eliminate loop III
)(sR
)(1
)(
3242121
3241
GGGHHGG
GGGG
)(sY
)()(1
)(
)(
)()(
32413242121
3241
GGGGGGGHHGG
GGGG
sR
sYsT
)(sR1G
C
324
12
GGG
HG
)(sY324 GGG
2H
C
)(1 3242
324
GGGH
GGG
Using rule 6
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2G1G
1H 2H
)(sR )(sY
3H
(b)
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Solution:
1. Eliminate loop I
2. Moving pickoff point A behind block22
2
1 HG
G
1G
1H
)(sR )(sY
3H
BA
22
2
1 HG
G
2
221
G
HG
1G
1H
)(sR )(sY
3H
2G
2H
BA
II
I
22
2
1 HG
G
Not a feedback loop
)1
(2
2213 G
HGHH
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3. Eliminate loop II
)(sR )(sY
22
21
1 HG
GG
2
2213
)1(
G
HGHH
21211132122
21
1)(
)()(
HHGGHGHGGHG
GG
sR
sYsT
Using rule 6
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2G 4G1G
4H
2H
3H
)(sY)(sR
3G
1H
(c)
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Solution:
2G 4G1G
4H)(sY
3G
1H
2H
)(sRA B
3H4
1
G
4
1
G
I1. Moving pickoff point A behind block 4G
4
3
G
H
4
2
G
H
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2. Eliminate loop I and Simplify
II
III
443
432
1 HGG
GGG
1G)(sY
1H
B
4
2
G
H
)(sR
4
3
G
H
II
332443
432
1 HGGHGG
GGG
III
4
142
G
HGH
Not feedbackfeedback
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)(sR )(sY
4
142
G
HGH
332443
4321
1 HGGHGG
GGGG
3. Eliminate loop II & IIII
143212321443332
4321
1)(
)()(
HGGGGHGGGHGGHGG
GGGG
sR
sYsT
Using rule 6
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3G1G
1H
2H
)(sR )(sY
4G
2G A
B
(d)
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Solution:
1. Moving pickoff point A behind block 3GI
1H3
1
G
)(sY1G
1H
2H
)(sR
4G
2GA B
3
1
G
3G
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2. Eliminate loop I & Simplify
3G
1H
2G B
3
1
G
2H
32GG B
23
1 HG
H
1G)(sR )(sY
4G
3
1
G
H
23212
32
1 HGGHG
GG
II
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)(sR )(sY
12123212
321
1 HGGHGGHG
GGG
3. Eliminate loop II
12123212
3214 1)(
)()(
HGGHGGHG
GGGG
sR
sYsT
4G
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2G1G
4G
R Y
1H
3G
N
Determine the effect of R and N on Y in the following diagram
Example 2
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NTRTYYY 2121
If we set N=0, then we can get Y1:
RTYY N 101
The same, we set R=0 and Y2 is also obtained:
NTYY R 202
Thus, the output Y is given as follows:
0021 RN YYYYY
In this linear system, the output Y contains two parts, one part is related to R and the other is caused by N:
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Solution:
1. Swap the summing points A and B
2. Eliminate loop II & simplify
Y
12
2
1 HG
G
1G
4G
R
N
AB
3G
12
2131 1 HG
GGGG
R Y
N4G
II
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3. Let N=0
RHGGGGGGGHG
HGGGGGGGY
1321312112
132131211 1
12
2131 1 HG
GGGG
R Y
12
2131 1 HG
GGGG
R Y
N4G
o o
1YWe can easily get
Rewrite the diagram:
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5. Break down the summing point M:
12
421431 1 HG
GGGGGG
YN
12
2131 1 HG
GGGG
4. Let R=0, we can get:
12
2131 1 HG
GGGG
4G
YN
M
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])1(
)[(1
1
1432143142112
132131211321312112
21
NHGGGGGGGGGGHG
RHGGGGGGGHGGGGGGGHG
YYY
7. According to the principle of superposition, and can be combined together, So:
1Y 2Y
6. Eliminate above loops:
YN12
421431 1
1HG
GGGGGG
12
2131 1
1
1
HGGG
GG
NHGGGGGGGHG
HGGGGGGGGGGHGY
1321312112
14321431421122 1
1
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End