Department of Electronics and Communication Engineering
}
1950203: Control
Systems
UNIT - I: Introduction to Control Problem
Presentation By
Rupa Kumar Dhanavath
Asst. Professor
Department of ECE
MLRITM
CS – Unit I
 Presentation Outline
Problems on Block Diagram Reduction.
Rupa Kumar Dhanavath@ Dept. of ECE
3
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
H2
C
_
R
+_
+
+
G1
+
G2
G3
H1
Note: We can identify from figure that, we can shift summing point before G1
Rupa Kumar Dhanavath@ Dept. of ECE
4
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Moving the summing point before of G1, we
have:
1
G1
H2
_
R
+_
+
+
+
C
G1
G2
G3
H1
Rupa Kumar Dhanavath@ Dept. of ECE
5
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Combing G1 and G2 in Cascade,we get:
2
1
R
+_
+
+
H2
G1
3 _
C
+
G1G2
G3
H1
Note: We can identify from figure that, here we are going to re-order the
summingpoints2,3toactiveafeedbackloophere.
Rupa Kumar Dhanavath@ Dept. of ECE
6
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Eliminatingthe feedback loop G1,G2 and H1
we get:
H2
G1
C
_
R
+_
+
+
G1G2
+
G3
H1
G
1  GH
Here G =G1G2 and H =H1
Rupa Kumar Dhanavath@ Dept. of ECE
7
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Eliminatingthe feedback loop G1,G2 and H1 we get:
G
1  GH
Here G =G1G2
and H =H1
G1G2

1   G1G2  H1 

G1G2
1  G1G2 H1
Rupa Kumar Dhanavath@ Dept. of ECE
8
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
H2
G1
C
_
R
+_
+
G1G2
1  G1G2 H1
G3
Combing the two blocks in Cascade, we get
Rupa Kumar Dhanavath@ Dept. of ECE
9
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Similarly eliminatingthe second feedbackloop H 2
we get:
G
1
C
_
R
+_
G1G2 G3
1  G1G2 H1
+
G
1  GH
G1G2 G3
Here G = 1  G
1G2 H 1
H2
and H =
G1
Rupa Kumar Dhanavath@ Dept. of ECE
10
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
G1G2G3
1  G1G2 H1

 G1G2G3   H 2 
1 


1

G
G
H
G

1 2
1 
1 
G
1  GH
GG G
1
2
3
Here G = 1  G
G H
1
H2
and H =
G1
2
1
G1G2G3
G1G2G3
1  G1G2 H1


1  G1G2 H1  G2G3 H 2 1  G1G2 H1  G2G3 H 2
1  G1G2 H1
Rupa Kumar Dhanavath@ Dept. of ECE
11
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
C
R
+_
G
1 G
G1G2G3
1  G1G2 H1  G2G3 H 2
G1G2G3
Here G =
1  G1G2 H1  G2G3 H 2
Rupa Kumar Dhanavath@ Dept. of ECE
Here H =1
12
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
G1G2G3
1  G1G2 H1  G2G3 H 2

G1G2G3
1
1  G1G2 H1  G2G3 H 2
G
1 G
G1G2G3

1  G1G2 H1  G2G3 H 2  G1G2G3
G1G2G3
1 G1G2 H1  G2G3 H 2  G1G2G3
R
T .F 
C
C
R
Rupa Kumar Dhanavath@ Dept. of ECE
13
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
G4
+
R(s) +
-
G1
G2
G3
-
+
+
G6
+
H1
G5
H2
Rupa Kumar Dhanavath@ Dept. of ECE
14
C(s)
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
Apply Rule 2: Blocks in Parallel
G4
+
R(s) +
-
G1
G2
G3
-
+
+
G6
+
H1
G5
H2
Rupa Kumar Dhanavath@ Dept. of ECE
15
C(s)
CS – Unit I
MLRITM
 Example-2Reducethe lengthofthefollowingblockdiagram
Apply Rule1: Blocks in series
R(s) +
+
-
G1
G2
G3+G4+G5
G6
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
16
C(s)
CS – Unit I
MLRITM
 Example-1:Reducethelengthofthe followingblockdiagram
Apply Rule 3:
R(s) +
+
-
Elimination of feedback loop
G1
G2(G3+G4+G5)
G6
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
17
C(s)
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
Apply Rule 1:
R(s) +
-
Blocks in series
G1
1 G1H1
G2(G3+G4+G5)
G6
H2
Rupa Kumar Dhanavath@ Dept. of ECE
18
C(s)
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
Apply Rule 3: Elimination of feedback loop
R(s) +
-
G1G2(G3  G 4  G5)
1 G1H1
G6
H2
Rupa Kumar Dhanavath@ Dept. of ECE
19
C(s)
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
G
1  GH
Here G =
G1G2 (G3  G4  G5 )
1  G1 H1
and H = H 2
G1G2 (G3  G4  G5 )
1  G1 H1

 G G (G3  G4  G5 ) 
1  1 2
 H2
1

G
H

1
1

G1G2 (G3  G4  G5 )
1  G1 H1

1  G1 H1  G1G2 (G3  G4  G5 ) H 2
1  G1 H1
G1G2 (G3  G4  G5 )

1  G1 H1  G1G2 H 2 (G3  G4  G5 )
Rupa Kumar Dhanavath@ Dept. of ECE
20
CS – Unit I
MLRITM
 Example-2:Reducethelengthofthe followingblockdiagram
Apply Rule 1:
R(s)
Blocks in series
G1G2(G3 G4  G5)
1G1H1G1G2H2(G3 G4  G5)
Rupa Kumar Dhanavath@ Dept. of ECE
G6
21
C(s)
MLRITM
CS – Unit I
 Example-2:Reducethelengthofthe followingblockdiagram
R(s)
G1G2G6(G3G4G5)
1G1H1G1G2H2 (G3G4G5)
Rupa Kumar Dhanavath@ Dept. of ECE
C(s)
22
MLRITM
CS – Unit I
 Example-2:Reducethelengthofthe followingblockdiagram
G1G2G6(G3G4G5)
C (s)
 1G1H1G1G2H2(G3G4G5)
R (s)
Rupa Kumar Dhanavath@ Dept. of ECE
23
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
G4
R(s) +
+
+
G1
G2
G3
+
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
24
C(s)
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
Apply Rule 1 Blocks in series
R(s) +
+
+
G1
G2
G4
G3
+
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
25
C(s)
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
Apply Rule 2 Blocks in parallel
R(s) +
+
+
G1G2
G4
G3
+
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
26
C(s)
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
Apply Rule 3 Elimination of feedback loop
R(s) +
+
+
G1G2
C(s)
G3+G4
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
27
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
G
1  GH
Here G =
G1G2
G1G2

1  G1G2 H1
and H = H1
Rupa Kumar Dhanavath@ Dept. of ECE
28
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
Apply Rule 2 Blocks in series
R(s) +
+
G1G2
1  G1G2H1
C(s)
G3+G4
H2
Rupa Kumar Dhanavath@ Dept. of ECE
29
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
Apply Rule 3 Elimination of feedback loop
R(s) +
+
G1G2(G 3  G 4)
1  G1G2H1
C(s)
H2
Rupa Kumar Dhanavath@ Dept. of ECE
30
CS – Unit I
MLRITM
 Example-3:Reducethelengthofthe followingblockdiagram
G
1  GH
Here G =

G1G2  G3  G4 
1  G1G2 H1
and H = H 2
G1G2  G3  G4 
1  G1G2 H1

 G G  G3  G4  
1  1 2
 H2
1

G
G
H
1
2
1


G1G2  G3  G4 
1  G1G2 H1

1  G1G2 H1  G1G2 H 2  G3  G4 
1  G1G2 H1
G1G2  G3  G4 

1  G1G2 H1  G1G2 H 2  G3  G4 

G1G2  G3  G4 
1  G1G2 H1  G1G2G3 H 2  G1G2G4 H 2
Rupa Kumar Dhanavath@ Dept. of ECE
31
MLRITM
CS – Unit I
 Example-3:Reducethelengthofthe followingblockdiagram
R(s)
C(s)
G1G2(G3 G4)
1 G1G2H1G1G2G3H2 G1G2G4H2
Rupa Kumar Dhanavath@ Dept. of ECE
32
MLRITM
CS – Unit I
 Example-3:Reducethelengthofthe followingblockdiagram
C(s)
G1G2(G3 G4)

R(s) 1 G1G2H1G1G2G3H2 G1G2G4H2
Rupa Kumar Dhanavath@ Dept. of ECE
33
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
G5
R(s) +
G1
-
+
G2
G3
+
+
G4
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
34
C(s)
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
Apply Rule 3
R(s) +
G1
-
G5
+
Elimination of feedback loop
G2
G3
+
+
G4
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
35
C(s)
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
Apply Rule 1
R(s) +
G1
-
G5
Blocks in series
G2
1  G2H1
G3
+
+
G4
H2
Rupa Kumar Dhanavath@ Dept. of ECE
36
C(s)
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
Apply Rule 2
Blocks in parallel
G5
R(s) +
-
G1G2G3
1  G2H1
+
+
G4
H2
Rupa Kumar Dhanavath@ Dept. of ECE
37
C(s)
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
Apply Rule 1 Blocks in series
R(s) +
-
G1G2G3
G5 
1 G2H1
G4
H2
Rupa Kumar Dhanavath@ Dept. of ECE
38
C(s)
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram
Apply Rule 3
R(s) +
-
Elimination of feedback loop
C(s)
G1G2G3
G 4(G5 
)
1 G2H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
39
CS – Unit I
MLRITM
 Example-4: Reducethelengthofthe followingblockdiagram

G1G2G3 
G4  G5 

1  G2 H1 




G1G2G3  
1   G4  G5 
  H2
1

G
H

2
1 

G
1  GH
G4 
G5 1  G2 H1   G1G2G3 

Here G =

G1G2 G3 
G4  G5 

1  G2 H1 

and H =
H2



1
1  G2 H1
G4 
G5 1  G2 H1   G1G2G3 
 H2
1  G2 H1
G4 
G5 1  G2 H1   G1G2G3 

1  G2 H1  G4 
G5 1  G2 H1   G1G2G3 
 H2
G4G5  G2G4G5 H1  G1G2G3G4
1  G2 H1  G4G5 H 2  G2G4G5 H1 H 2  G1G2G3G4 H 2
Rupa Kumar Dhanavath@ Dept. of ECE
40
MLRITM
CS – Unit I
 Example-4: Reducethelengthofthe followingblockdiagram
R(s)
G4G5G2G4G5H1G1G2G3G4
C(s)
1 G2H1 G4G5H2 G2G4G5H1H2 G1G2G3G4H2
Rupa Kumar Dhanavath@ Dept. of ECE
41
MLRITM
CS – Unit I
 Example-4: Reducethelengthofthe followingblockdiagram
C(s)
G4G5G2G4G5H1G1G2G3G4

R(s) 1G2H1G4G5H2G2G4G5H1H2 G1G2G3G4H2
Rupa Kumar Dhanavath@ Dept. of ECE
42
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
R(s) +
+
-
G1
-
+
C(s)
G2
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H2
43
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
Elimination of feedback loop
Apply Rule 3
R(s) +
+
G1
-
+
C(s)
G2
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H2
44
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
R(s) +
+
-
G1
-
G2
1  G2H 2
C(s)
H1
Rupa Kumar Dhanavath@ Dept. of ECE
45
MLRITM
CS – Unit I
 Example-5: Reducethelengthofthe followingblockdiagram
 Now Rule 1, 2 or 3 cannot be used directly.
 Use Rule 4 & interchange order of summing
so that Rule 3 can be used on G1-H1
feedback loop.
Rupa Kumar Dhanavath@ Dept. of ECE
46
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
Apply Rule 4
1
R(s) +
+
2
-
Exchange summing order
G1
-
C(s)
G2
1  G2H 2
H1
Rupa Kumar Dhanavath@ Dept. of ECE
47
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
Apply Rule 3
2 R(s) +
+
Elimination feedback loop
1
G1
-
C(s)
G2
1  G2H 2
H1
Rupa Kumar Dhanavath@ Dept. of ECE
48
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
Apply Rule 1 Bocks in series
2 R(s) +
G1
1  G1H1
Rupa Kumar Dhanavath@ Dept. of ECE
G2
1  G2H 2
C(s)
49
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
2 R(s) +
C(s)
G1G2
1  G1H1  G2H 2  G1G2H1H 2
Now which Rule will be applied…?
-------It is blocks in parallel…???
OR
-------It is feed back loop …???
Rupa Kumar Dhanavath@ Dept. of ECE
50
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
Let us re-arrange the block diagram to understand.
Apply Rule 3: Elimination of feed back loop.
R(s)
2
+
-
C(s)
G1G2
1  G1H1  G2H 2  G1G2H1H 2
Rupa Kumar Dhanavath@ Dept. of ECE
51
CS – Unit I
MLRITM
 Example-5: Reducethelengthofthe followingblockdiagram
G
1 G
Here G =
G1G2
1  G1 H1  G2 H 2  G1G2 H1 H 2

G1G2
1
1  G1 H1  G2 H 2  G1G2 H1 H 2
G1G2
1  G1 H1  G2 H 2  G1G2 H1 H 2

and H =1
Rupa Kumar Dhanavath@ Dept. of ECE
G1G2
1  G1 H1  G2 H 2  G1G2 H1 H 2  G1G2
52
MLRITM
CS – Unit I
 Example-5: Reducethelengthofthe followingblockdiagram
R(s)
G1G2
1 G1H1 G2H 2  G1G2H1H2  G1G2
Rupa Kumar Dhanavath@ Dept. of ECE
C(s)
53
MLRITM
CS – Unit I
 Example-5: Reducethelengthofthe followingblockdiagram
G1G2
C (S )

R ( S ) 1  G1 H1  G2 H 2  G1G2 H1 H 2  G1G2
Rupa Kumar Dhanavath@ Dept. of ECE
54
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
 Note-1: By corollary, from rule 4, one can split a
summing
point to two summing point and sum
in any order.
B
B
R(s) +
+
G
C(s)
-
R(s) +
+ +
G
H
Rupa Kumar Dhanavath@ Dept. of ECE
H
55
C(s)
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Simplify, by splitting second summing point as said in note 1.
H1
R(s) +
1
2
+
-
-
G1
G2
C(s)
G3
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
56
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Apply rule 3:
Elimination of feedback loop.
H1
+
R(s)
+
-
+
-
G1
G2
C(s)
G3
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
57
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Apply rule 1
+
R(s)
+
-
-
Blocks in series
G1
1  G1H1
G2
C(s)
G3
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
58
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Apply rule 3: Elimination of feedback loop
+
R(s)
+
-
-
G1G2
1  G1H1
G3
C(s)
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
59
CS – Unit I
MLRITM
 Example-6 Reducethelengthofthe followingblockdiagram
G1G2
1  G1 H1

G1G2
1
H2
1  G1 H1
G
1  GH
Here G =
G1G2
1  G1 H1
and H = H 2
G1G2
1  G1 H1

1  G1 H1  G1G2 H 2
1  G1 H1

Rupa Kumar Dhanavath@ Dept. of ECE
G1G2
1  G1 H1  G1G2 H 2
60
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Apply rule 1 Blocks in series
+
R(s)
-
G1G2
1 G1H1G1G2H2
G3
C(s)
H3
Rupa Kumar Dhanavath@ Dept. of ECE
61
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
Apply rule 3 Elimination of feedback loop
+
R(s)
-
G1G2G3
1  G1H1  G1G2H 2
C(s)
H3
Rupa Kumar Dhanavath@ Dept. of ECE
62
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
G
1  GH
Here G =

G1G2G3
1  G1 H1  G1G2 H 2
and H = H 3
G1G2G3
1  G1 H1  G1G2 H 2

G1G2G3
1
H3
1  G1 H1  G1G2 H 2
G1G2G3
1  G1 H1  G1G2 H 2

1  G1 H1  G1G2 H 2  G1G2G3 H 3
1  G1 H1  G1G2 H 2

G1G2G3
1  G1 H1  G1G2 H 2  G1G2G3 H 3
Rupa Kumar Dhanavath@ Dept. of ECE
63
CS – Unit I
MLRITM
 Example-6: Reducethelengthofthe followingblockdiagram
R(s)
G1G2G3
1 G1H1 G1G2H2  G1G2G3H3
Rupa Kumar Dhanavath@ Dept. of ECE
C(s)
64
MLRITM
CS – Unit I
 Example-6: Reducethelengthofthe followingblockdiagram
G1G 2G3
C(s)

R(s) 1  G1H1  G1G2H 2  G1G2G3H 3
Rupa Kumar Dhanavath@ Dept. of ECE
65
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
G5
Apply rule 8
Shift take off point beyond block G3.
R(s) +
G1
-
+
G2
G3
G4
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
66
+
C(s)
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
1/
G3
Apply rule 1 Blocks in series
R(s) +
G1
-
+
G2
G3
G4
G5
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
67
+
C(s)
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
Apply rule 2 Blocks in parallel
R(s) +
G1
-
+
G2G3
G5/
G3
G4
+
H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
68
+
C(s)
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
Apply rule 3 Feedback loop
R(s) +
G1
-
+
-
G2G3
G4 
G5
G3
H1
G
1  GH
H2
Rupa Kumar Dhanavath@ Dept. of ECE
69
C(s)
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
Apply rule 1 Blocks in series
R(s) +
G1
-
G2G3
1 G2G3H1
G4 
G5
G3
H2
Rupa Kumar Dhanavath@ Dept. of ECE
70
C(s)
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram


G2G3
G5 
G1 
G

 4

1

G
G
H
G
2
3
1 
3 

R(s) +
-
C(s)
G5
(G1)( G2G3 )(G 4  )
G3
1 G2G3H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
71
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram


G2G3
G5 
 G1 
G

 4

1

G
G
H
G
2
3
1 
3 


  G3G4  G5 
G2G3
 G1 


1

G
G
H
G
2
3
1 
3




G2
 G1 
  G3G4  G5 
1

G
G
H
2
3
1 

G1G2  G3G4  G5 

1  G2G3 H1
Rupa Kumar Dhanavath@ Dept. of ECE
72
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
Apply rule 3 Eliminating Feedback loop
R(s) +
-
G1G2(G 4 G 3  G 5)
1  G2G3H1
C(s)
H2
Rupa Kumar Dhanavath@ Dept. of ECE
73
CS – Unit I
MLRITM
 Example-7: Reducethelengthofthe followingblockdiagram
G1G2  G3G4  G5 
1  G2G3 H1

G G  G3G4  G5 
1 1 2
H2
1  G2G3 H1
G
1  GH
Here G =
G G  G3G4  G5 
 1 2
1  G2G3 H1
and H = H 2
G1G2  G3G4  G5 
1  G2G3 H1

1  G2G3 H1  G1G2 H 2  G3G4  G5 
1  G2G3 H1

G1G2  G3G4  G5 
1  G2G3 H1  G1G2 H 2  G3G4  G5 
Rupa Kumar Dhanavath@ Dept. of ECE
74
MLRITM
CS – Unit I
 Example-7: Reducethelengthofthe followingblockdiagram
R(s)
C(s)
G1G2(G 4 G 3  G 5)
1 G2G3H1  G1G2H 2(G 3G 4  G 5)
Rupa Kumar Dhanavath@ Dept. of ECE
75
MLRITM
CS – Unit I
 Example-7: Reducethelengthofthe followingblockdiagram
C(S)
G1G2(G 4 G 3  G 5)

R(S) 1  G2G3H1  G1G2H 2(G 3G 4  G 5)
Rupa Kumar Dhanavath@ Dept. of ECE
76
Example-8:
Apply rule 8
Shift the take off point after block G4
H2
R(s) +
G1
+
-
G2
+
G3
G4
-
-
C(s)
H3
H1
Rupa Kumar Dhanavath@ Dept. of ECE
77
Example-8:
Apply rule 1
Blocks in series
H2
R(s) +
G1
+
G2
+
1/G4
G3
G4
-
-
C(s)
H3
H1
Rupa Kumar Dhanavath@ Dept. of ECE
78
Example-8:
Apply rule 3
Eliminating Feedback loop
H2/
G4
R(s) +
G1
+
G2
+
G3G4
-
-
C(s)
H3
H1
G
1  GH
Rupa Kumar Dhanavath@ Dept. of ECE
79
Example-8:
Apply rule 1
Blocks in series
H2/
G4
R(s) +
G1
+
-
G2
-
G3G4
1  G3G4H 3
C(s)
H1
Rupa Kumar Dhanavath@ Dept. of ECE
80
Example-8:
Apply rule 3
R(s) +
Feedback loop
G1
+
H2/
G4
-
G2G3G4
1 G3G4H 3
-
C(s)
H1
Rupa Kumar Dhanavath@ Dept. of ECE
81
G
1  GH
Here G =
G2G3G4
1  G3G4 H 3
H2
and H =
G4
G2G3G4
1  G3G4 H 3

 G2G3G4   H 2 
1 


1

G
G
H
G
3 4
3 
4 


G2G3G4
1  G3G4 H 3
 1  G3G4 H 3  G2G3 H 2 


1

G
G
H
3 4
3


G2G3G4

1  G3G4 H 3  G2G3 H 2
Rupa Kumar Dhanavath@ Dept. of ECE
82
82
Example-8:
Apply rule 1
R(s) +
Blocks in series
G1
-
G2G3G4
1  G3G4H 3  G2G3H 2
C(s)
H1
Rupa Kumar Dhanavath@ Dept. of ECE
83
Example-8:
Apply rule 3
R(s) +
-
Feedback loop
G1G2G3G4
1  G3G4H 3  G2G3H 2
C(s)
H1
Rupa Kumar Dhanavath@ Dept. of ECE
84
G
1  GH
Here G =
G1G2G3G4
1  G3G4 H 3  G2G3 H 2
and H = H1
G1G2G3G4
1  G3G4 H 3  G2G3 H 2



G1G2G3G4
1 
  H1 
 1  G3G4 H 3  G2G3 H 2 

G1G2G3G4
1  G3G4 H 3  G2G3 H 2
 1  G3G4 H 3  G2G3 H 2  G1G2G3G4 H1 


1

G
G
H

G
G
H
3 4
3
2 3
2


G1G2G3G4

1  G3G4 H 3  G2G3 H 2  G1G2G3G4 H1
Rupa Kumar Dhanavath@ Dept. of ECE
85
85
Example-8:
R(s)
G1G2G3G4
1  G3G4H 3  G2G3H 2  G1G2G3G4H1
Rupa Kumar Dhanavath@ Dept. of ECE
86
C(s)
Example-8:
G1G2G3G4
C(S)

R(S) 1  G3G4H 3  G2G3H 2  G1G2G3G4H1
Rupa Kumar Dhanavath@ Dept. of ECE
87
Example-9:
Simplify, by splitting 3rd summing point as given in Note 1
G3
1
R(s) +
-
2
+
G1
-
G2
+
+
3
C(s)
G4
H2
Rupa Kumar Dhanavath@ Dept. of ECE
H1
88
Example-9:
Apply Rule 3
Elimination of Feedback loop
G3
R(s) +
+
-
G1
-
G2
+
+ +
G4
H2
H1
G
1  GH
Rupa Kumar Dhanavath@ Dept. of ECE
89
C(s)
Example-9:
Apply Rule 8
Shift take off point after block
G3
R(s) +
+
-
G1
-
G2
+
+
G4
1  G4H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
90
C(s)
Example-9:
Apply Rule 1 Blocks in series
G3/
G2
R(s) +
+
-
G1
-
G2
+
+
G4
1  G4H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
91
C(s)
Example-9:
Now which rule we have to use?
G3/
G2
R(s) +
+
-
G1G2
-
+
+
G4
1  G4H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
92
C(s)
Example-9:
Apply Rule 2 Blocks in parallel
G3
/
G2
R(s) +
+
-
G1G2
-
1 +
+
G4
1  G4H1
H2
Rupa Kumar Dhanavath@ Dept. of ECE
93
C(s)
Example-9:
Apply Rule 1
R(s)
+
Blocks in series
+
-
G1G2
-
G3  1
G2
G4
1  G4H1
H2
G4  G2  G3 
 G3


G4

1




G
1

G
H
G2 1  G4 H1 
 2

4
1 
Rupa Kumar Dhanavath@ Dept. of ECE
94
C(s)
Example-9:
Apply Rule 3
R(s) +
Elimination of Feedback Loop
+
-
G1G2
-
(G 3  G 2)G4
G 2(1  G4H1)
H2
G1G2
G

1  GH
1  G1G2 H 2
Rupa Kumar Dhanavath@ Dept. of ECE
95
C(s)
Example-9:
Apply Rule 1
R(s) +
-
Blocks in series
G1G2
1  G1G2H 2
(G 3  G 2)G4
G 2(1  G4H1)
G1G4  G2  G3 

  G4  G2  G3  
G1G2






1  G1G2 H 2  1  G4 H1 
 1  G1G2 H 2   G2 1  G4 H1  
Rupa Kumar Dhanavath@ Dept. of ECE
96
C(s)
Example-9:
Apply Rule 3
Elimination of Feedback loop
R(s) +
G1G4(G 3  G 2)
(1  G1G 2 H 2)(1  G4H1)
-
C(s)
G
1 G
Rupa Kumar Dhanavath@ Dept. of ECE
97
Example-9:
G1G4  G2  G3 
1  G1G2 H 2  1  G4 H1 

G1G4  G2  G3 
1
1  G1G2 H 2  1  G4 H1 
G
1 G
Here G =
G1G4  G2  G3 
1  G1G2 H 2  1  G4 H1 
and H =1
G1G4  G2  G3 
1  G1G2 H 2  1  G4 H1 

1  G1G2 H 2  1  G4 H1   G1G4  G2  G3 
1  G1G2 H 2  1  G4 H1 


G1G4  G2  G3 
1  G1G2 H 2  1  G4 H1   G1G4  G2  G3 
G1G4  G2  G3 
1  G4 H1  G1G2 H 2  G1G2G4 H1 H 2   G1G4  G2  G3 
Rupa Kumar Dhanavath@ Dept. of ECE
98
Example-9:
R(s)
G1G4(G 3 G2)
1 G4H1G1G2H2G1G2G4H1H2G1G4(G2G3)
Rupa Kumar Dhanavath@ Dept. of ECE
99
C(s)
Example-9:
C(s)
G1G4(G3 G2)

R(s) 1 G4H1G1G2H2 G1G2G4H1H2 G1G4(G2 G3)
Rupa Kumar Dhanavath@ Dept. of ECE
100
Example-10:
Apply rule 2
Blocks in Parallel
G4
R(s) +
G1
-
+
+
G2
G3
-
+ G5
+
H1
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
101
C(s)
Example-10:
Apply rule 3
R(s) +
-
Elimination of Feedback Loop
G1+G4
+ G5
+
G2
-
+
G3
-
C(s)
H1
H2
H3
Rupa Kumar Dhanavath@ Dept. of ECE
102
Example-10:
Apply rule 1
R(s) +
Blocks in Series
G1+G4+G5
-
G2
1 G2H1
G3
1G3H2
H3
G2 G3 (G1  G4  G5 )
1  G2 H1  1  G3 H 2 
Rupa Kumar Dhanavath@ Dept. of ECE
103
C(s)
Example-10:
Apply rule 3
R(s) +
-
Elimination of Feedback loop
G 2 G 3 ( G 1  G 4  G 5)
(1  G2H1)(1  G 3 H 2)
H3
Rupa Kumar Dhanavath@ Dept. of ECE
104
C(s)
G
1  GH

Here G =
G2G3 (G1  G4  G5 )
1  G2 H1  1  G3 H 2 

and H =
H3

G2 G3 (G1  G4  G5 )
1  G2 H1  1  G3 H 2 
 G2 G3 (G1  G4  G5 ) 
1 
 H3
H
G

1
H
G

1




2
3
1
2


G2G3 (G1  G4  G5 )
1  G2 H1  1  G3 H 2 
 1  G2 H1  1  G3 H 2   G2G3 (G1  G4  G5 ) 

 H3
1

G
H
1

G
H




2
1
3
2


G2G3 (G1  G4  G5 )
1  G2 H1  1  G3 H 2   G2G3 (G1  G4  G5 ) H 3
G2G3 (G1  G4  G5 )
1  G2 H1  G3 H 2  G2G3 H1 H 2  G2G3 H 3 (G1  G4  G5 )
Rupa Kumar Dhanavath@ Dept. of ECE
105
Example-10:
R(s)
G2G3 (G1  G4  G5 )
1  G2 H1  G3 H 2  G2G3 H1 H 2  G2G3 H 3 (G1  G4  G5 )
Rupa Kumar Dhanavath@ Dept. of ECE
106
C(s)
Example-10:
C(s)
G2G3(G1 G4  G5)

R(s) 1 G2H1 G3H2  G2G3H1H 2  G2G3H3(G1 G4  G5)
Rupa Kumar Dhanavath@ Dept. of ECE
107
Example-11:
R(s)
G1
-
+
G2
G3
-
+
+
C(s)
+
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H3
+
108
Example-11:
Apply rule 2
R(s)
G1
Blocks in Parallel
-
+
G2
G3
-
+
+
C(s)
+
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H3
+
109
Example-11:
Apply rule 3
R(s)
G1
Elimination of Feedback Loop
-
+
G2
1+G3
C(s)
+
H1
G
1 G

G2
1  G2
Rupa Kumar Dhanavath@ Dept. of ECE
H3
+
110
Example-11:
Apply rule 8
R(s)
G1
-
Shift take off point after block
+
-
G2
1 G2
1+G3
C(s)
+
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H3
+
111
Example-11:
Apply rule 1
R(s)
G1
Blocks in series
-
+
-
G2
1 G2
1+G3
C(s)
1
1  G3
+
H1
Rupa Kumar Dhanavath@ Dept. of ECE
H3
+
112
Example-11:
Apply rule 2
R(s)
G1
Blocks in Parallel
-
+
G2(1  G 3)
1  G2
-
1
1  G3
+
H1
Rupa Kumar Dhanavath@ Dept. of ECE
C(s)
H3
+
113
Example-11:
Apply rule 1
R(s)
G1
Blocks in Series
+
-
G2(1  G 3)
1  G2
-
H1
H2
C(s)
1
1  G3
 H 2 1  G3   1  H1  H 2  H 2G3  1

1 
H1  H 2 

H


1
1  G3 
1  G3
1  G3



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Example-11:
Apply rule 3
R(s)
G1
G
1  GH
Elimination of Feedback loop
-
+
G2(1  G 3)
1  G2
-
C(s)
H1(H 2  H 2 G 3  1)
1  G3
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G
1  GH
Here G =
G2 (1  G3 )
1  G2
and H =
H1  H 2  H 2G3  1
1  G3
G2 (1  G3 )
1  G2

 G (1  G3 )   H1  H 2  H 2G3  1 
1  2


1

G
1

G

2

3

G2 (1  G3 )
1  G2

 G H  H 2  H 2G3  1 
1  2 1

1

G
2




G2 (1  G3 )
1  G2  G2 H1  H 2  H 2G3  1
G2 (1  G3 )
1  G2  G2 H1 1  H 2  H 2G3 
Rupa Kumar Dhanavath@ Dept. of ECE
116
116
Example-11:
Apply rule 1
R(s)
G1
Blocks in series
G2(1  G 3)
1  G2  G2H1(1  H 2  H 2 G 3)
Rupa Kumar Dhanavath@ Dept. of ECE
C(s)
117
Example-11:
R(s)
C(s)
G1G2(1  G 3)
1  G2  G2H1(1  H 2  H 2 G 3)
Rupa Kumar Dhanavath@ Dept. of ECE
118
Example-11:
C(s)
G1G2(1  G 3)

R(s) 1  G2  G2H1(1  H 2  H 2 G 3)
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MLRITM
 End of the Session:
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120
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