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Feedback Control Systems (FCS)

Lecture-8-9
Block Diagram Representation of Control Systems
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
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, gain,
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.
x

d
dt

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 (or
pickoff) 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.

x3  a1 x1  a2 x2  5
Example-1
x3  a1 x1  a2 x2  5
Example-2
• Draw the Block Diagrams of the following equations.

(1)
( 2)

dx1 1
x2  a1
  x1dt
dt
b
x3  a1

d 2 x2
dt 2

dx1
3
 bx1
dt
Canonical Form of A Feedback Control System
Characteristic Equation
• The control ratio is the closed loop transfer function of the system.
C( s )
G( s )

R( s ) 1  G( s ) H ( s )

• The denominator of closed loop transfer function determines the
characteristic equation of the system.
• Which is usually determined as:

1  G( s )H ( s )  0
Example-3
1. Open loop transfer function

B( s )
 G( s ) H ( s )
E( s )

2. Feed Forward Transfer function

C( s )
G( s )

R( s ) 1  G( s ) H ( s )

3. control ratio

4. feedback ratio

5. error ratio

C (s)
 G (s)
E (s)
G(s )

B( s )
G( s ) H ( s )

R( s ) 1  G ( s ) H ( s )

E( s )
1

R( s ) 1  G( s ) H ( s )

6. closed loop transfer function

H (s )

C( s )
G( s )

R( s ) 1  G( s ) H ( s )

7. characteristic equation 1  G( s )H ( s )  0
8. Open loop poles and zeros if 9. closed loop poles and zeros if K=10.
Reduction techniques
1. Combining blocks in cascade

G2

G1

G1G2

2. Combining blocks in parallel

G1
G2

G1  G2
3. Eliminating a feedback loop

G

G
1  GH

H

G
H 1

G
1 G
Example-4: Reduce the Block Diagram to Canonical Form.
Example-4: Continue.
Example-5
• For the system represented by the following block diagram
determine:
1.
2.
3.
4.
5.
6.
7.
8.

Open loop transfer function
Feed Forward Transfer function
control ratio
feedback ratio
error ratio
closed loop transfer function
characteristic equation
closed loop poles and zeros if K=10.
Example-5
– First we will reduce the given block diagram to canonical form

K
s 1
Example-5
K
s 1

K
G
 s 1
K
1  GH
1
s
s 1
Example-5 (see example-3)
1. Open loop transfer function

B( s )
 G( s ) H ( s )
E( s )

2. Feed Forward Transfer function

C( s )
 G( s )
E( s )

C( s )
G( s )
3. control ratio

R( s ) 1  G( s ) H ( s )
4. feedback ratio

5. error ratio

G(s )

B( s )
G( s ) H ( s )

R( s ) 1  G ( s ) H ( s )

E( s )
1

R( s ) 1  G( s ) H ( s )

6. closed loop transfer function

C( s )
G( s )

R( s ) 1  G( s ) H ( s )

7. characteristic equation 1  G( s )H ( s )  0
8. closed loop poles and zeros if K=10.

H (s )
Example-6
• For the system represented by the following block diagram
determine:
1.
2.
3.
4.
5.
6.
7.
8.

Open loop transfer function
Feed Forward Transfer function
control ratio
feedback ratio
error ratio
closed loop transfer function
characteristic equation
closed loop poles and zeros if K=100.
Reduction techniques

4. Moving a summing point behind a block

G

G
G

5. Moving a summing point ahead a block

G

G
1
G
6. Moving a pickoff point behind a block

G

G

1
G

7. Moving a pickoff point ahead of a block

G

G
G
8. Swap with two neighboring summing points

A

B

B

A
Example-7
• Reduce the following block diagram to canonical form.

H2
_

R
+_

+

+

G1

+

H1

C
G2

G3
Example-7
H2
G1
_

R
+_

+

+

+

C
G1
H1

G2

G3
Example-7
H2
G1
_

R
+_

+

+

C

+

G1G2
H1

G3
Example-7
H2
G1
_

R
+_

+

C
+

G1G2

+

H1

G3
Example-7
H2
G1
_

R
+_

+

G1G2
1  G1G2 H1

C
G3
Example-7
H2
G1
_

R
+_

+

G1G2G3
1  G1G2 H1

C
Example-7

R
+_

G1G2G3
1  G1G2 H 1  G2G3 H 2

C
Example 8
Find the transfer function of the following block diagram

G4
R (s )

Y (s )

G1

G2

G3

H2
H1
I

G4

R(s)

B

G1

G2

A

G3

H2
H1

G2

Solution:
1. Moving pickoff point A ahead of block
2. Eliminate loop I & simplify
B

G4  G2G3

G2

Y (s )
G4

R(s)

G1

G A G
G 4  G2G3

B

Y (s )

3

2

H2

H1G2

G4  G2G3

3. Moving pickoff point B behind block

II

R(s)

G1

B

G4  G2G3
H2

H1G2

1 /(G4  G2G3 )

C

Y (s )
4. Eliminate loop III

R(s)

G1

G4 4 G2G3
G G2G3
H
1  H 2 (G4 2 G2G3 )

C

C

Y (s )

G2 H1
G4  G2G3

R(s)

G1 (G4  G2G3 )
1  G1G 2 H1  H 2 (G4  G2G3 )

G1 (G4  G2G3 )
Y (s)

R( s ) 1  G1G 2 H1  H 2 (G4  G2G3 )  G1 (G4  G2G3 )

Y (s )
Example 9
Find the transfer function of the following block diagrams

H4

R(s)

Y (s )

G1

G3

G2
H3

H2
H1

G4
Solution:
1. Moving pickoff point A behind block

G4

I

H4

R(s)

Y (s )

G1

G3

G2
H3

H2

H3
G4
H2
G4

H1

1
G4
1
G4

A

G4

B
2. Eliminate loop I and Simplify

R(s)

G2G3G4
1  G3G4 H 4

G1

II

Y (s )
B

H3
G4
H2
G4

III

H1
II

feedback

G2G3G4
1  G3G4 H 4  G2G3 H 3

III

Not feedback

H 2  G4 H 1
G4
3. Eliminate loop II & IIII

R(s)

G1G2G3G4
1  G3G4 H 4  G2G3 H 3

Y (s )

H 2  G4 H 1
G4

G1G2G3G4
Y (s)

R( s ) 1  G2G3 H 3  G3G4 H 4  G1G2G3 H 2  G1G2G3G4 H1
Example-10: Reduce the Block Diagram.
Example-10: Continue.
Example-11: Simplify the block diagram then obtain the closeloop transfer function C(S)/R(S). (from Ogata: Page-47)
Example-11: Continue.
Superposition of Multiple Inputs
Example-12: Multiple Input System. Determine the output C
due to inputs R and U using the Superposition Method.
Example-12: Continue.
Example-12: Continue.
Example-13: Multiple-Input System. Determine the output C
due to inputs R, U1 and U2 using the Superposition Method.
Example-13: Continue.
Example-13: Continue.
Example-14: Multi-Input Multi-Output System. Determine C1
and C2 due to R1 and R2.
Example-14: Continue.
Example-14: Continue.

When R1 = 0,

When R2 = 0,
Block Diagram of Armature Controlled D.C Motor
Ra

La
c

Va

ia

eb

T

J


La s  Ra I a(s)  K b(s)  Va(s)
Js  c (s)  K m I a(s)
Block Diagram of Armature Controlled D.C Motor

La s  Ra I a(s)  K b(s)  Va(s)
Block Diagram of Armature Controlled D.C Motor

Js  c (s)  K ma I a(s)
Block Diagram of Armature Controlled D.C Motor
To download this lecture visit
http://imtiazhussainkalwar.weebly.com/

END OF LECTURES-8-9

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Lecture 8-9 block-diagram_representation_of_control_systems

  • 1. Feedback Control Systems (FCS) Lecture-8-9 Block Diagram Representation of Control Systems Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/
  • 2. 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, gain, 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. x d dt y
  • 3. 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.
  • 4. Introduction • In order to have the same signal or variable be an input to more than one block or summing point, a takeoff (or pickoff) point is used. • This permits the signal to proceed unaltered along several different paths to several destinations.
  • 5. Example-1 • Consider the following equations in which 𝑥1 , 𝑥2 , 𝑥3 , are variables, and 𝑎1 , 𝑎2 are general coefficients or mathematical operators. x3  a1 x1  a2 x2  5
  • 6. Example-1 x3  a1 x1  a2 x2  5
  • 7. Example-2 • Draw the Block Diagrams of the following equations. (1) ( 2) dx1 1 x2  a1   x1dt dt b x3  a1 d 2 x2 dt 2 dx1 3  bx1 dt
  • 8. Canonical Form of A Feedback Control System
  • 9. Characteristic Equation • The control ratio is the closed loop transfer function of the system. C( s ) G( s )  R( s ) 1  G( s ) H ( s ) • The denominator of closed loop transfer function determines the characteristic equation of the system. • Which is usually determined as: 1  G( s )H ( s )  0
  • 10. Example-3 1. Open loop transfer function B( s )  G( s ) H ( s ) E( s ) 2. Feed Forward Transfer function C( s ) G( s )  R( s ) 1  G( s ) H ( s ) 3. control ratio 4. feedback ratio 5. error ratio C (s)  G (s) E (s) G(s ) B( s ) G( s ) H ( s )  R( s ) 1  G ( s ) H ( s ) E( s ) 1  R( s ) 1  G( s ) H ( s ) 6. closed loop transfer function H (s ) C( s ) G( s )  R( s ) 1  G( s ) H ( s ) 7. characteristic equation 1  G( s )H ( s )  0 8. Open loop poles and zeros if 9. closed loop poles and zeros if K=10.
  • 11. Reduction techniques 1. Combining blocks in cascade G2 G1 G1G2 2. Combining blocks in parallel G1 G2 G1  G2
  • 12. 3. Eliminating a feedback loop G G 1  GH H G H 1 G 1 G
  • 13. Example-4: Reduce the Block Diagram to Canonical Form.
  • 15. Example-5 • For the system represented by the following block diagram determine: 1. 2. 3. 4. 5. 6. 7. 8. Open loop transfer function Feed Forward Transfer function control ratio feedback ratio error ratio closed loop transfer function characteristic equation closed loop poles and zeros if K=10.
  • 16. Example-5 – First we will reduce the given block diagram to canonical form K s 1
  • 17. Example-5 K s 1 K G  s 1 K 1  GH 1 s s 1
  • 18. Example-5 (see example-3) 1. Open loop transfer function B( s )  G( s ) H ( s ) E( s ) 2. Feed Forward Transfer function C( s )  G( s ) E( s ) C( s ) G( s ) 3. control ratio  R( s ) 1  G( s ) H ( s ) 4. feedback ratio 5. error ratio G(s ) B( s ) G( s ) H ( s )  R( s ) 1  G ( s ) H ( s ) E( s ) 1  R( s ) 1  G( s ) H ( s ) 6. closed loop transfer function C( s ) G( s )  R( s ) 1  G( s ) H ( s ) 7. characteristic equation 1  G( s )H ( s )  0 8. closed loop poles and zeros if K=10. H (s )
  • 19. Example-6 • For the system represented by the following block diagram determine: 1. 2. 3. 4. 5. 6. 7. 8. Open loop transfer function Feed Forward Transfer function control ratio feedback ratio error ratio closed loop transfer function characteristic equation closed loop poles and zeros if K=100.
  • 20. Reduction techniques 4. Moving a summing point behind a block G G G 5. Moving a summing point ahead a block G G 1 G
  • 21. 6. Moving a pickoff point behind a block G G 1 G 7. Moving a pickoff point ahead of a block G G G
  • 22. 8. Swap with two neighboring summing points A B B A
  • 23. Example-7 • Reduce the following block diagram to canonical form. H2 _ R +_ + + G1 + H1 C G2 G3
  • 29. Example-7 R +_ G1G2G3 1  G1G2 H 1  G2G3 H 2 C
  • 30. Example 8 Find the transfer function of the following block diagram G4 R (s ) Y (s ) G1 G2 G3 H2 H1
  • 31. I G4 R(s) B G1 G2 A G3 H2 H1 G2 Solution: 1. Moving pickoff point A ahead of block 2. Eliminate loop I & simplify B G4  G2G3 G2 Y (s )
  • 32. G4 R(s) G1 G A G G 4  G2G3 B Y (s ) 3 2 H2 H1G2 G4  G2G3 3. Moving pickoff point B behind block II R(s) G1 B G4  G2G3 H2 H1G2 1 /(G4  G2G3 ) C Y (s )
  • 33. 4. Eliminate loop III R(s) G1 G4 4 G2G3 G G2G3 H 1  H 2 (G4 2 G2G3 ) C C Y (s ) G2 H1 G4  G2G3 R(s) G1 (G4  G2G3 ) 1  G1G 2 H1  H 2 (G4  G2G3 ) G1 (G4  G2G3 ) Y (s)  R( s ) 1  G1G 2 H1  H 2 (G4  G2G3 )  G1 (G4  G2G3 ) Y (s )
  • 34. Example 9 Find the transfer function of the following block diagrams H4 R(s) Y (s ) G1 G3 G2 H3 H2 H1 G4
  • 35. Solution: 1. Moving pickoff point A behind block G4 I H4 R(s) Y (s ) G1 G3 G2 H3 H2 H3 G4 H2 G4 H1 1 G4 1 G4 A G4 B
  • 36. 2. Eliminate loop I and Simplify R(s) G2G3G4 1  G3G4 H 4 G1 II Y (s ) B H3 G4 H2 G4 III H1 II feedback G2G3G4 1  G3G4 H 4  G2G3 H 3 III Not feedback H 2  G4 H 1 G4
  • 37. 3. Eliminate loop II & IIII R(s) G1G2G3G4 1  G3G4 H 4  G2G3 H 3 Y (s ) H 2  G4 H 1 G4 G1G2G3G4 Y (s)  R( s ) 1  G2G3 H 3  G3G4 H 4  G1G2G3 H 2  G1G2G3G4 H1
  • 38. Example-10: Reduce the Block Diagram.
  • 40. Example-11: Simplify the block diagram then obtain the closeloop transfer function C(S)/R(S). (from Ogata: Page-47)
  • 43. Example-12: Multiple Input System. Determine the output C due to inputs R and U using the Superposition Method.
  • 46. Example-13: Multiple-Input System. Determine the output C due to inputs R, U1 and U2 using the Superposition Method.
  • 49. Example-14: Multi-Input Multi-Output System. Determine C1 and C2 due to R1 and R2.
  • 51. Example-14: Continue. When R1 = 0, When R2 = 0,
  • 52. Block Diagram of Armature Controlled D.C Motor Ra La c Va ia eb T J  La s  Ra I a(s)  K b(s)  Va(s) Js  c (s)  K m I a(s)
  • 53. Block Diagram of Armature Controlled D.C Motor La s  Ra I a(s)  K b(s)  Va(s)
  • 54. Block Diagram of Armature Controlled D.C Motor Js  c (s)  K ma I a(s)
  • 55. Block Diagram of Armature Controlled D.C Motor
  • 56. To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURES-8-9