block diagram representation of control systems

Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: [email protected]

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-14-15Block Diagram Representation of Control Systems

mailto:[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 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

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.

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.

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

and a1, a2 are general coefficients or mathematical operators.

522113 xaxax

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

and a1, a2 are general coefficients or mathematical operators.

522113 xaxax

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

Example-3• Draw the Block Diagrams of the following equations.

11

22

2

13

11

12

32

11

bxdtdx

dtxd

ax

dtxbdt

dxax

)(

)(

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 the characteristic equation of the system.

• Which is usually determined as:

)()()(

)()(

sHsGsG

sRsC

1

01 )()( sHsG

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

Reduction techniques

2G1G 21GG

1. Combining blocks in cascade

1G

2G21 GG

2. Combining blocks in parallel

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

Example-5: Continue.

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 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.

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

1sK

Example-6

1sK

ssK

sK

GHG

11

11

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

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.

Reduction techniques

3. Moving a summing point behind a block

G G

G

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

6. Eliminating a feedback loop

G

HGHG

1

7. Swap with two neighboring summing points

A B AB

G

1H

GG1

Example-8

R_+

_+1G 2G 3G

1H

2H

+ +

C

Example-8

R_+

_+

1G 2G 3G

1H

1

2

GH

+ +

C

Example-8

R_+

_+

21GG 3G

1H

1

2

GH

+ +

C

Example-8

R_+

_+ 21GG 3G

1H

1

2

GH

+ +

C

block diagram: reduction example

R_+

_+

121

21

1 HGGGG

3G

1

2

GH

C


R_+

_+

121

321

1 HGGGGG

1

2

GH

C


R_+

232121

321

1 HGGHGGGGG

C

Example-8

R

321232121

321

1 GGGHGGHGGGGG

C

Example 9

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)(sR C

324 GGG

2H

)(sY

21GH

4G

2GA

3G 324 GGG

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

2G1G

1H 2H

)(sR )(sY

3H

Example 10


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

3. Eliminate loop II

)(sR )(sY22

21

1 HGGG

2

2213

)1(G

HGHH

21211132122

21

1)()()(

HHGGHGHGGHGGG

sRsYsT

Using rule 6

2G 4G1G

4H

2H

3H

)(sY)(sR

3G

1H

Example 11


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

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

)(sR )(sY

4

142

GHGH

332443

4321

1 HGGHGGGGGG

3. Eliminate loop II & IIII

143212321443332

4321

1)()()(

HGGGGHGGGHGGHGGGGGG

sRsYsT

Using rule 6

3G1G

1H

2H

)(sR )(sY

4G

2GA

B

Example 12


Solution:

1. Moving pickoff point A behind block3G I

1H3

1G

)(sY1G

1H

2H

)(sR

4G

2GA B

3

1G

3G

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

)(sR )(sY12123212

321

1 HGGHGGHGGGG

3. Eliminate loop II

12123212

3214 1)(

)()(HGGHGGHG

GGGGsRsYsT

4G

Example-13: Simplify the Block Diagram.


Example-14: Reduce the Block Diagram.


Example-15: Reduce the Block Diagram. (from Nise: page-242)


Example-16: Reduce the system to a single transfer function. (from Nise:page-243).

Example-17: Simplify the block diagram then obtain the close-loop transfer function C(S)/R(S). (from Ogata: Page-47)

Example-18: 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,

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


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


(s)IK(s)cJs ama


Block Diagram of liquid level system

11

1 qqdtdh

C 1

211 R

hhq

212

2 qqdtdh

C 2

22 R

hq


)()()( 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


)()()( sQsQssHC 111 1

211 R

sHsHsQ

)()()(

2

22 R

sHsQ

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


END OF LECTURES-14-15

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block diagram representation of control systems