Lecture-3 Block Diagram Reduction - AASTMT | Homepage

Automatic Control Systems

Lecture-3Block Diagram Reduction

1

Emam FathyDepartment of Electrical and Control Engineering

email: [email protected]

Introduction• A Block Diagram is a shorthand pictorial representation of

the cause-and-effect relationship of a system.

• The interior of the rectangle represent the mathematicaloperation to be performed on the input to yield the output.

• The arrows represent the direction of information or signal flow.

dt

dx y

Introduction• The operations of addition and subtraction have a special

representation.

• The block becomes a small circle, called a summing point, withthe appropriate plus or minus sign associated with the arrowsentering 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.

Introduction

• In order to have the same signal or variable be an inputto more than one block or summing point, a takeoffpoint is used.

• This permits the signal to proceed unaltered alongseveral different paths to several destinations.

Example-1• Consider the following equations in which x1, x2, x3, are variables,

and a1, a2 are general coefficients or mathematical operators.

522113 xaxax

Canonical Form of A Feedback Control System

Characteristic Equation

• The control ratio is the closed loop transfer function of the system.

• The denominator of closed loop transfer function determines thecharacteristic equation of the system.

• Which is usually determined as:

)()(

)(

)(

)(

sHsG

sG

sR

sC

1

01 )()( sHsG

Example-4

)(sG

)(sH

1. Open loop transfer function

2. Feed Forward Transfer function

3. closed loop transfer function

4. characteristic equation

)()(

)(sG

sE

sC

)()(

)(

)(

)(

sHsG

sG

sR

sC

1

01 )()( sHsG

)()()(

)(sHsG

sE

sB

In order to analyze the system, we want to

represent multiple subsystems as a single transfer

function.

Reduction techniques

2G1G21GG

1. Combining blocks in cascade

1G

2G21 GG

2. Combining blocks in parallel (feed-forward)

3. Eliminating a feedback loop

G

H

B

A

H

OR

𝑮

𝟏 ∓ 𝑮𝑯

𝑨

𝑩 ∓ 𝑨𝑯

5. Moving a pickoff point ahead of a block

G G

G G

G

1

G

4. Moving a pickoff point behind a block

6. Moving a summing point ahead of a block

G G

G

1

7. Moving a summing point behind a block

G G

G

Note

A B AB

Example 1

R

_+

_

+1G 2G 3G

1H

2H

++

C

Example

R

_+

_

+1G 2G 3G

1H

1

2

G

H

++

C

Example

R

_+

_

+21GG 3G

1H

1

2

G

H

++

C

Example

R

_+

_

+121

21

1 HGG

GG

3G

1

2

G

H

C

Example

R

_+

_

+121

321

1 HGG

GGG

1

2

G

H

C

Example

R

_+232121

321

1 HGGHGG

GGG

C

Example

R

321232121

321

1 GGGHGGHGG

GGG

C

Example 2

Find the transfer function of the following block diagrams

2G 3G1G

4G

1H

2H

)(sY)(sR

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:

3. Moving pickoff point B behind block324 GGG

1GB

)(sR

21GH 2H

)(sY

)/(1 324 GGG

II

1GB

)(sRC

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

)(sR

1GC

324

12

GGG

HG

)(sY324 GGG

2H

C

)(1 3242

324

GGGH

GGG

2G1G

1H 2H

)(sR )(sY

3H

Example 3


Solution:

1. Eliminate loop I

2. Moving pickoff point A behind block

22

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

2G4G1G

4H

2H

3H

)(sY)(sR

3G

1H

Example 4


Solution:

2G4G1G

4H

)(sY

3G

1H

2H

)(sR

A 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

3G1G

1H

2H

)(sR )(sY

4G

2GA

B

Example 5


Solution:

1. Moving pickoff point A behind block3G

I

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

2

3

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

End of Lec 3

Example-5: Continue.

Example-6: Reduce the Block Diagram.

Example-7: Reduce the Block Diagram. (from Nise: page-242)

Example-8: Reduce the system to a single transfer function. (from Nise:page-243).

Example-9: Simplify the block diagram then obtain the close-loop transfer function C(S)/R(S). (from Ogata: Page-47)

Example-10: Multiple Input System. Determine the output C due to inputs R and U using the Superposition Method.

Example-19: Multiple-Input System. Determine the output C due to inputs R, U1 and U2 using the Superposition Method.

Example-20: Multi-Input Multi-Output System. Determine C1 and C2 due to R1 and R2.


When R1 = 0,

When R2 = 0,

Example-5: Reduce the Block Diagram to Canonical Form.


However in this example step-4 does not apply.

However in this example step-6 does not apply.

Example-6• For the system represented by the following block diagram

determine:1. Open loop transfer function


3. control ratio

4. feedback ratio

5. error ratio



8. closed loop poles and zeros if K=10.

Example-6– First we will reduce the given block diagram to canonical form

1s

K

Example-6

1s

K

ss

Ks

K

GH

G

11

1

1

Example-7• For the system represented by the following block diagram

determine:1. Open loop transfer function


3. control ratio

4. feedback ratio

5. error ratio



8. closed loop poles and zeros if K=100.

Block Diagram of Armature Controlled D.C Motor

Va

iaT

Ra La

J

c

eb

(s)IK(s)cJs

(s)V(s)K(s)IRsL

ama

abaaa


(s)E(s)K(s)IRsL abaaa


(s)IK(s)cJs ama

Block Diagram of liquid level system

11

1 qqdt

dhC

1

211

R

hhq

212

2 qqdt

dhC

2

22

R

hq


)()()( sQsQssHC 111

1

211

R

sHsHsQ

)()()(

2

22

R

sHsQ

)()( )()()( sQsQssHC 2122

11

1 qqdt

dhC

1

211

R

hhq

212

2 qqdt

dhC

2

22

R

hq

L

L

L

L


)()()( sQsQssHC 111

1

211

R

sHsHsQ

)()()(

2

22

R

sHsQ

)()( )()()( sQsQssHC 2122

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