MECH261 Control Principles Tutorial #3. Block diagram Transfer Function Consists of Blocks Can be reduced.

MECH261Control Principles

Tutorial #3

Block diagram

Transfer Function

Consists of Blocks

Can be reduced

)(sR2G 3G1G

4G

1H

2H

)(sY

G)(sY)(sR

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

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

6. Eliminating a feedback loop

G

HGH

G

1

7. Swap with two neighboring summing points

A B AB

G

1H

G

G

1

Example 1

Find the transfer function of the following block diagrams

2G 3G1G

4G

1H

2H

)(sY)(sR

(a)

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:

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

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

2G1G

1H 2H

)(sR )(sY

3H

(b)

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

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

2G 4G1G

4H

2H

3H

)(sY)(sR

3G

1H

(c)

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

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

)(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

3G1G

1H

2H

)(sR )(sY

4G

2G A

B

(d)

Solution:

1. Moving pickoff point A behind block 3GI

1H3

1

G

)(sY1G

1H

2H

)(sR

4G

2GA B

3

1

G

3G

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

)(sR )(sY

12123212

321

1 HGGHGGHG

GGG

3. Eliminate loop II

12123212

3214 1)(

)()(

HGGHGGHG

GGGG

sR

sYsT

4G

2G1G

4G

R Y

1H

3G

N

Determine the effect of R and N on Y in the following diagram

Example 2

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:

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

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:

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

])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

End