### Feedback Control Systems (FCS) - Dr. Imtiaz Hussain

```Feedback Control Systems (FCS)
Lecture-14-15
Block Diagram Representation of Control Systems
Dr. Imtiaz Hussain
email: [email protected]
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, 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
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.
x 3  a1 x1  a 2 x 2  5
Example-1
• Consider the following equations in which x1, x2, x3, are variables,
and a1, a2 are general coefficients or mathematical operators.
x 3  a1 x1  a 2 x 2  5
Example-2
• Consider the following equations in which x1, x2,. . . , xn, are
variables, and a1, a2,. . . , an , are general coefficients or
mathematical operators.
x n  a1 x1  a 2 x 2  a n 1 x n 1
Example-3
• Draw the Block Diagrams of the following equations.
(1 )
(2)
x 2  a1
dx 1
x 3  a1
d

dt
2
dt
1
b
x2
2
 x1 dt
 3
dx 1
dt
 bx 1
Canonical Form of A Feedback Control System
Characteristic Equation
• The control ratio is the closed loop transfer function of the system.
C(s)
R( s )

G(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-4
B( s )
1. Open loop transfer function
E(s)
2. Feed Forward Transfer function
C(s)
3. control ratio
R( s )
4. feedback ratio
5. error ratio

B( s )
R( s )
E(s)
R( s )

 G ( s )H ( s )
C(s)
 G(s)
E(s)
G(s)
G(s)
1  G ( s )H ( s )

G ( s )H ( s )
1  G ( s )H ( s )
H (s)
1
1  G ( s )H ( s )
6. closed loop transfer function
C(s)
R( s )

G(s)
1  G ( s )H ( s )
7. characteristic equation 1  G ( s ) H ( s )  0
8. closed loop poles and zeros if K=10.
Reduction techniques
1. Combining blocks in cascade
G2
G1
G 1G 2
2. Combining blocks in parallel
G1
G2
G1  G 2
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.
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-6
– First we will reduce the given block diagram to canonical form
K
s 1
Example-6
K
s 1
K
G
1  GH

1
s 1
K
s 1
s
Example-6
B( s )
1. Open loop transfer function
E(s)
2. Feed Forward Transfer function
C(s)
3. control ratio
R( s )
4. feedback ratio
5. error ratio

B( s )
R( s )
E(s)
R( s )

 G ( s )H ( s )
C(s)
 G(s)
E(s)
G(s)
G(s)
1  G ( s )H ( s )

G ( s )H ( s )
1  G ( s )H ( s )
H (s)
1
1  G ( s )H ( s )
6. closed loop transfer function
C(s)
R( s )

G(s)
1  G ( s )H ( s )
7. characteristic equation 1  G ( s ) H ( s )  0
8. closed loop poles and zeros if K=10.
Example-7
• 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
3. Moving a summing point behind a block
G
G
G
3. Moving a summing point ahead of a block
G
G
1
G
4. Moving a pickoff point behind a block
G
G
1
G
5. Moving a pickoff point ahead of a block
G
G
G
6. Eliminating a feedback loop
G
G
1  GH
H
G
G
1 G
H 1
7. Swap with two neighboring summing points
A
B
B
A
Example-8
H
2
_
R
+_
+
+
G1
+
H1
C
G2
G3
Example-8
H2
G1
_
R
+_
+
+
+
C
G1
H1
G2
G3
Example-8
H2
G1
_
R
+_
+
+
C
+
G 1G 2
H1
G3
Example-8
H2
G1
_
R
+_
+
C
+
G 1G 2
+
H1
G3
block diagram: reduction example
H2
G1
_
R
+_
+
G 1G 2
1  G 1G 2 H 1
C
G3
block diagram: reduction example
H2
G1
_
R
+_
+
G 1G 2 G 3
1  G 1G 2 H 1
C
block diagram: reduction example
R
+_
G 1G 2 G 3
1  G 1G 2 H 1  G 2 G 3 H 2
C
Example-8
R
G 1G 2 G 3
1  G 1G 2 H 1  G 2 G 3 H 2  G 1G 2 G 3
C
Example 9
Find the transfer function of the following block diagrams
G4
R (s )
Y (s)
G1
G2
G3
H
H1
2
I
G4
R (s)
B
G1
G2
A
Y (s)
G3
H
H1
G2
Solution:
1. Moving pickoff point A ahead of block
2. Eliminate loop I & simplify
B
G 4  G 2G 3
G2
2
G4
R (s)
G1
G A G 2GG 3
G 4
B
2
Y (s)
3
H
H 1G 2
2
G 4  G 2G 3
3. Moving pickoff point B behind block
II
R (s)
G1
B
G 4  G 2G 3
H
H 1G 2
2
1 /( G 4  G 2 G 3 )
C
Y (s)
4. Eliminate loop III
R (s)
G1
GG4 4GG2 G2 G
3 3
C
C
Y (s)
1  H 2 ( GH4 2 G 2 G 3 )
G2H1
G 4  G 2G 3
Using rule 6
G1 (G 4  G 2 G 3 )
R (s)
Y (s)
1  G 1G 2 H 1  H 2 ( G 4  G 2 G 3 )
T (s) 
Y (s)
R (s)

G1 (G 4  G 2 G 3 )
1  G 1G 2 H 1  H 2 ( G 4  G 2 G 3 )  G 1 ( G 4  G 2 G 3 )
Example 10
Find the transfer function of the following block diagrams
R (s)
G1
G2
H1
H
H3
2
Y (s)
Solution:
1. Eliminate loop I
R (s)
G1
G2
A
I
G2
B
Y (s)
B
Y (s)
1  GH2 H 2
H1
2
H3
G2
2. Moving pickoff point A behind block
R (s)
G1
H1
1  G2H 2
G2
A
1  G2H 2
1  G2H 2
G2
H3
II
H 3  H1(
1  G2H 2
Not a feedback loop
G2
)
3. Eliminate loop II
R (s)
G 1G 2
Y (s)
1  G2H 2
H3 
H 1 (1  G 2 H 2 )
G2
Using rule 6
T (s) 
Y (s)
R (s)

G 1G 2
1  G 2 H 2  G 1G 2 H 3  G 1 H 1  G 1G 2 H 1 H 2
Example 11
Find the transfer function of the following block diagrams
H
4
R (s)
Y (s)
G1
G2
G3
H3
H
2
H1
G4
Solution:
1. Moving pickoff point A behind block
G4
I
H
R (s)
4
Y (s)
G1
G2
G3
H
H3
H
2
3
1
G4 G4
1
H
2
G4 G 4
H1
A
G4
B
2. Eliminate loop I and Simplify
R (s)
II
G 2 G 3G 4
G1
Y (s)
B
1  G 3G 4 H 4
H
3
G4
H2
III
G4
H1
II
feedback
III
Not feedback
G 2 G 3G 4
H 2  G4H1
1  G 3G 4 H 4  G 2 G 3 H 3
G4
3. Eliminate loop II & IIII
R (s)
G 1G 2 G 3 G 4
Y (s)
1  G 3G 4 H 4  G 2 G 3 H 3
H 2  G4H1
G4
Using rule 6
T (s) 
Y (s)
R (s)

G 1G 2 G 3 G 4
1  G 2 G 3 H 3  G 3 G 4 H 4  G 1G 2 G 3 H 2  G 1G 2 G 3 G 4 H 1
Example 12
Find the transfer function of the following block diagrams
H
R (s)
G2
G1
H1
G4
A
2
G3
Y (s)
B
Solution:
1. Moving pickoff point A behind block
G3
H
R (s)
G2
G1
2
A
G3
1
H1
H1
G4
I
G3
1
G3
B
Y (s)
2. Eliminate loop I & Simplify
H
2
G3
G2
H1
R (s)
1
H1
G3
G3
G1
B
G 2G3
B
 H2
II
G 2G 3
1  G 2 H 1  G 2G 3 H 2
H1
G3
G4
Y (s)
3. Eliminate loop II
R (s)
Y (s)
G 1G 2 G 3
1  G 2 H 1  G 2 G 3 H 2  G 1G 2 H 1
G4
T (s) 
Y (s)
R (s)
 G4 
G 1G 2 G 3
1  G 2 H 1  G 2 G 3 H 2  G 1G 2 H 1
Example-13: Simplify the Block Diagram.
Example-13: Continue.
Example-14: Reduce the Block Diagram.
Example-14: Continue.
Example-15: Reduce the Block Diagram. (from Nise: page-242)
Example-15: Continue.
Example-16: Reduce the system to a single transfer function.
(from Nise:page-243).
Example-17: Simplify the block diagram then obtain the closeloop 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-18: Continue.
Example-18: Continue.
Example-19: Multiple-Input System. Determine the output C
due to inputs R, U1 and U2 using the Superposition Method.
Example-19: Continue.
Example-19: Continue.
Example-20: Multi-Input Multi-Output System. Determine C1
and C2 due to R1 and R2.
Example-20: Continue.
Example-20: Continue.
When R1 = 0,
When R2 = 0,
Block Diagram of Armature Controlled D.C Motor
Ra
La
c
Va
ia
eb
T
J

L a s 
 Js
R a I a (s)  K b  (s)  V a (s)
 c  (s)  K ma I a (s)
Block Diagram of Armature Controlled D.C Motor
L a s 
R a I a (s)  K b  (s)  E a (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
Block Diagram of liquid level system
q1 
h1  h 2
q2 
R1
C1
h2
R2
C2
dh 1
dt
dh 2
dt
 q  q1
 q1  q 2
Block Diagram of liquid level system
L
q1 
Q1 ( s ) 
q2 
L
h1  h 2
C1
R1
H 1( s )  H 2 ( s )
R1
C2
Q2 (s) 
H 2(s)
R2
dt
 q  q1
L
C 1 sH 1 ( s )  Q ( s )  Q1 ( s )
h2
R2
dh 1
dh 2
dt
 q1  q 2
L
C 2 sH 2 ( s )  Q1 ( s )  Q 2 ( s )
Block Diagram of liquid level system
Q1 ( s ) 
H 1( s )  H 2 ( s )
Q2 (s) 
C 1 sH 1 ( s )  Q ( s )  Q1 ( s )
R1
H 2(s)
R2
C 2 sH 2 ( s )  Q1 ( s )  Q 2 ( s )
Block Diagram of liquid level system