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Signal Flow Graph
1. In the signal flow graph shown in figure 𝑋2 = 𝑇𝑋1 where T, is equal to
5X1 X2
0.5
(a) 2.5
(b) 5
(c) 5.5
(d) 10
[GATE 1987: 2 Marks]
Soln. 𝑋2 = 𝑇𝑋1
𝑋2
𝑋1=
5
∆=
5
1 − 0.5= 10
Option (d)
2. For the system shown in figure the transfer function
𝐶(𝑠)
𝑅(𝑠) 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜
R(s) C(s)
S
10
S(S+1)+- +-
(a) 10
𝑆2+𝑆+10
(b) 10
𝑆2+11𝑆+10
(c) 10
𝑆2+9𝑆+10
(d) 10
𝑆2+2𝑆+10
[GATE 1987: 2 Marks]
Soln. The forward path transmittance =10
𝑆(𝑆+1)
The two closed loop are 𝐿1 =−10
𝑆(𝑆+1)
𝐿2 =−10𝑆
𝑆(𝑆 + 1)
𝐶(𝑆)
𝑅(𝑆)=
10/𝑆(𝑆 + 1)
1 − {−10
𝑆(𝑆+1)
−10
𝑆+1}
=10
𝑆(𝑆 + 1) [1 +10
𝑆(𝑆+1)+
10
(𝑆+1)]
=10 𝑆(𝑆 + 1)
𝑆(𝑆 + 1)[𝑆(𝑆 + 1) + 10 + 10𝑆]
=10
𝑆𝟐 + 𝑆 + 10𝑆 + 10=
10
𝑆2 + 11𝑆 + 10
Option (b)
3. The C/R for the signal flow graph in figure is
R CG1 G2 G3 G41 1 1
-1 -1 -1 -1
(a) 𝐺1𝐺2𝐺34
(1+𝐺1𝐺2)(1+𝐺3𝐺4)
(b) 𝐺1𝐺2𝐺3𝐺4
(1+𝐺1+𝐺2+𝐺1𝐺2)(1+𝐺3+𝐺4+𝐺3𝐺4)
(c) 𝐺1𝐺2𝐺3𝐺4
(1+𝐺1+𝐺2)(1+𝐺3+𝐺4)
(d) 𝐺1𝐺2𝐺3𝐺4
(1+𝐺1+𝐺2+𝐺3+𝐺4)
[GATE 1989: 2 Marks]
Soln. The forward path transmittance = G1 G2 G3 G4
Individual loops are, -G1, -G2, -G3, -G4.
Product of non touching loops, G1G3, G1G4, G2G3, G2G4
∆= 1 − [−𝐺1 − 𝐺2 − 𝐺3 − 𝐺4] + [𝐺1𝐺3 + 𝐺1𝐺4 + 𝐺2𝐺3 + 𝐺2𝐺4]
𝑆
𝑅=
𝐺1𝐺2𝐺3𝐺4
(1 + 𝐺1 + 𝐺2 + 𝐺3 + 𝐺4) + (𝐺1𝐺3 + 𝐺1𝐺4 + 𝐺2𝐺3 + 𝐺2𝐺4)
=𝐺1𝐺2𝐺3𝐺4
(1 + 𝐺1 + 𝐺2)(1 + 𝐺3 + 𝐺4)
Option (c)
4. In the signal flow graph of figure the gain c/r will be
r 1 2 3 4 1 c
5
-1 -1 -1
(a) 11/9
(b) 22/15
(c) 24/23
(d) 44/23
[GATE 1991: 2 Marks]
Soln. The forward path 𝑃1 = 1 × 2 × 3 × 4 = 24
The forward path 𝑃2 = 5
∆1= 1
𝐿1 = −2, 𝐿2 = −3, 𝐿3 = −4
Non touching loops → 𝐿1𝐿3
The loop 𝐿2 = −3 does not touch the path P2
So, ∆2= (1 − 𝐿2)
= 1 + 3
= 4
𝑆
𝑅=
𝑃1∆1 + 𝑃2∆2
∆
=24 × 1 + 5 × 4
1— (−2 − 3 − 4 − 5) + (−2) × (−4)
=24 + 20
(1 + 14) + 8
=44
23
Option (d)
5. In the signal flow graph of figure y/x equals
125x y
-2
(a) 3
(b) 5/2
(c) 2
(d) None of the above
[GATE 1997: 2 Marks]
Soln. Transfer function
𝑌
𝑋=
𝑃𝐾∆𝐾
∆
𝑃𝐾 = 5 × 2 × 1 = 10
∆𝐾= 1
∆𝐾= 1
∆= 1 − (−4) = 5
𝑌
𝑋=
10
5
= 2
Option (c)
6. The equivalent of the block diagram in the figure is given as
G1 G2
H
E C
F
G1 C
H/G2
E
F
(a)
G1G2
HG2
E C
F
(b)
G1
HG2
E C
F
(c)
G1G2
H/G2
E C
F
(d)
Soln.
G1 G2
H
G1 C
H/G2
G1G2
HG2
G1
HG2
G1G2
H/G2
E C
F
E
F
E C
F
E C
F
E C
F
(a)
(b)
(c)
(d)
Option (d)
7. The signal flow graph of a system is shown in the figure. The transfer
function 𝐶(𝑠)
𝑅(𝑠) of the system is
R(s)
C(s)
1 6
1
-4-3
1S S
1
-2
(a) 6
𝑆2+29𝑆+6
(b) 6𝑆
𝑆2+29𝑆+6
(c) 𝑆(𝑆+2)
𝑆2+29𝑆+6
(d) 𝑆(𝑆+27)
𝑆2+29𝑆+6
[GATE 2003: 2 Marks]
Soln. The transfer function 𝐶(𝑆)
𝑅(𝑆) of the systems?
𝐿1 =−3
𝑆, 𝐿2 = −4 ×
6
𝑆, 𝐿3 =
−2
𝑆
P1 = 1
Loops L1 and L3 are not touching the forward path
∆1= 1 − 𝐿1 − 𝐿3
= 1 +3
𝑆+
24
𝑆
=𝑆 + 27
𝑆
𝐶(𝑆)
𝑅(𝑆)= 𝐺(𝑠) =
𝑃1∆1
1 − (𝑙𝑜𝑜𝑝 𝑔𝑎𝑖𝑛) + 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑛𝑜𝑛 𝑡𝑢𝑐ℎ𝑖𝑛𝑔 𝑙𝑜𝑜𝑝𝑠
𝑆+27
𝑆
1 − (−3
𝑆
−24
𝑆
−2
𝑆) +
−2
𝑆×
−3
𝑆
𝑆+27
𝑆
1 +29
𝑆+
6
𝑆2
=𝑆(𝑆 + 27)
𝑆2 + 29𝑆 + 6
Option (d)
8. Consider the signal flow graph shown in the figure. The gain X5 / X1 is
x1 x2 x3 x4 x5a b c d
e f g
(a) 1−(𝑏𝑒+𝑐𝑓+𝑑𝑔)
𝑎𝑏𝑐
(b) 𝑏𝑒𝑑𝑔
1−(𝑏𝑒+𝑐𝑓+𝑑𝑔)
(c) 𝑎𝑏𝑐𝑑
1−(𝑏𝑒+𝑐𝑓+𝑑𝑔)+𝑏𝑒𝑑𝑔
(d) 1−(𝑏𝑒+𝑐𝑓+𝑑𝑔)+𝑏𝑒𝑑𝑔
𝑎𝑏𝑐𝑑
[GATE 2004: 2 Marks]
Soln. The forward path transmittance P1 = abcd
All the loops touch the forward path ∆1= 1
𝐿1 = 𝑏𝑒, 𝐿2 = 𝐶𝑓, 𝐿3 = 𝑑𝑔
Non touching loops are 𝐿1, 𝐿3
𝐿1𝐿3 = 𝑏𝑒 𝑑𝑔
𝑋5
𝑋1=
𝑎𝑏𝑐𝑑
1 − (𝑏𝑒 + 𝑐𝑓 + 𝑑𝑔) + 𝑏𝑒 𝑑𝑔
Option (c)
9. The transfer function Y(s) / R(s) of the system shown is
Y(s)1
S+1
1
S+1
Σ
Σ
+-
+
-
R(s)P(s)
Q(s)
Q(s)
(a) 0
(b) 1
𝑆+1
(c) 2
𝑆+1
(d) 2
𝑆+3
[GATE 2010: 1 Mark]
Soln. 𝑄(𝑠) = 𝑃(𝑠) [1
𝑆+1−
1
𝑆+1]
= 0
𝑃(𝑠) = 𝑅(𝑠) − 0
= 𝑅(𝑠)
𝑌(𝑠) =𝑃(𝑠)
𝑆 + 1
=𝑅(𝑠)
𝑆 + 1
𝑌(𝑆)
𝑅(𝑆)=
1
𝑆 + 1
Option (b)
Common Data for Question 10 and Question 11
The input-output transfer function of a plant 𝐻(𝑠) =100
𝑆(𝑆+10)2.The plant is
placed in unity negative feedback configuration as shown in the figure
below.
Σ + -r u
y
plant
10. The signal flow graph that DOES NOT model the plant transfer function
H(s) is
u 1 1/s 1/s 1/s 100 y(a)
-10 -10
u1/s 1/s1/s 100 y(b)
-100
-20
u1/s 1/s1/s 100
(c)
-100
-20
y
u1/s 1/s1/s 100
(d)
-100
y
Soln. The transfer function of a plant 𝐻(𝑠) =100
𝑆(𝑆+10)2
For the figure (a) 𝑃1 =100
𝑆3, ∆1= 1
(𝑎𝑙𝑙 𝑡ℎ𝑒 𝑙𝑜𝑜𝑝𝑠 𝑡𝑜𝑢𝑐ℎ 𝑡ℎ𝑒 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑝𝑎𝑡ℎ)
𝑌
𝑈=
100/𝑆3
1 − (−10
𝑆
−10
𝑆) +
10
𝑆×
10
𝑆
=
100
𝑆2
1 +20
𝑆+
100
𝑆2
=100
𝑆(𝑆 + 10)2
For figure (b) 𝑃1 =100
𝑆3, ∆1= 1
𝑌
𝑈=
100
𝑆3
1 − (−100
𝑆2
−20
𝑆)
=
100
𝑆3
1 +100
𝑆2+
20
𝑆
=100
𝑆(𝑆 + 10)2
For Figure (c)
𝑃1 =100
𝑆3, ∆1= 1
𝑌
𝑈=
100
𝑆3
1 − (−100
𝑆2
−20
𝑆)
=100
𝑆(𝑆 + 10)2
For Figure (d),
𝑌
𝑈=
100
𝑆3
1 − (−100
𝑆2 )=
100
𝑆3
1 +100
𝑆2
= 100 𝑆2
𝑆3(𝑆2 + 100)=
100
𝑆(𝑆2 + 100)
Which is not a transfer function of H(S)
Option (d)
11. The gain margin of the system under closed loop unity negative feedback is
𝐺(𝑠)𝐻(𝑠) =100
𝑆(𝑆+10)2
(a) 0 dB
(b) 20 dB
(c) 26 dB
(d) 46 dB
[GATE 2011: 2 Marks]
Soln. The gain margin of the system under closed loop unity negative feedback is
𝐺(𝑠)𝐻(𝑠) =100
𝑆(𝑆+10)2
∅ = −900 − 2𝑡𝑎𝑛−1 (𝜔
10)
Flow phase cross aver frequency ∅ = −1800
−180 = −900 − 2𝑡𝑎𝑛−1 (𝜔
10)
𝜔 = 10 𝑟𝑎𝑑/𝑠𝑒𝑐
𝐺(𝑗𝜔)𝐻(𝑗𝜔) =100
𝑗𝜔 (𝑗𝜔 + 10)2
|𝐺(𝑗𝜔)𝐻(𝑗𝜔)| =100
𝜔(𝜔2 + 100)
=100
10(100 + 100)=
1
20
𝐺𝑎𝑖𝑛 𝑚𝑎𝑟𝑔𝑖𝑛 (𝐺. 𝑀) =1
|𝐺(𝑗𝜔)𝐻(𝑗𝜔)|= 20
𝐺. 𝑀 𝑖𝑛 𝑑𝐵 = 20𝑙𝑜𝑔1020 = 26 𝑑𝐵
Option (c)
12. The signal flow graph for a system is given below. The transfer function
𝑌(𝑠)
𝑈(𝑠) 𝑓𝑜𝑟 𝑡ℎ𝑖𝑠 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠
U(s) Y(s)1 1
1
-4
-2
S-1
S-1
(a) 𝑠+1
5𝑠2+6𝑠+2
(b) 𝑠+1
𝑠2+6𝑠+2
(c) 𝑠+1
𝑠2+4𝑠+2
(d) 1
5𝑠2+6𝑠+2
[GATE 2013: 2 Marks]
Soln. The forward path transmittance 𝑃1 = 𝑆−1 × 𝑆1 =1
𝑆2
The forward path transmittance 𝑃2 = 𝑆−1 =1
𝑆
∆1= 1, ∆2= 1
∆= 1 − (−2𝑆−2 − 2𝑆−1 − 4𝑆−1 − 4)
= 1 +2
𝑆2+
2
𝑆+
4
𝑆+ 4
= 1 +2
𝑆2+
2
𝑆+
4
𝑆+ 4
= (5𝑆2 + 6𝑆 + 2)/𝑆2
𝑌(𝑆)
𝑈(𝑆)=
𝑃1∆1 + 𝑃2∆2
∆=
1
𝑆2+
1
𝑆
(5𝑆2 + 6𝑆 + 2)/𝑆2=
(𝑆 + 1)
5𝑆2 + 6𝑆 + 2
Option (a)
13. For the following system,
+-S
S+1 +-+ 1
S
When 𝑋1(𝑆) = 0, the transfer function 𝑌(𝑆)
𝑋2(𝑆) is
(a) 𝑆+1
𝑆2
(b) 1
𝑆+1
(c) 𝑆+2
𝑆(𝑆+1)
(d) 𝑆+1
𝑆(𝑆+2)
[GATE: 2014: 1 Mark]
Soln. With 𝑋1(𝑠) = 0, the block diagram is redrawn as
+-
S
S+1
1
S
𝑇(𝑠) =𝑌(𝑆)
𝑋2(𝑆)=
𝐺(𝑆)
1 + 𝐺(𝑆)𝐻(𝑆)
𝐺(𝑠) =1
𝑆, 𝐻(𝑠) =
𝑆
𝑆 + 1=
1/𝑆
1 +1
𝑆×
𝑆
𝑆+1
=1(𝑆 + 1)
𝑆(𝑆 + 1 + 1)=
(𝑆 + 1)
𝑆(𝑆 + 2)
Option (d)
14. Consider the following block diagram in the figure
++ ++
the transfer function 𝐶(𝑆)
𝑅(𝑆) is
(a) 𝐺1𝐺2
1+𝐺1𝐺2
(b) 𝐺1𝐺2 + 𝐺1 + 1
(c) 𝐺1𝐺2 + 𝐺2 + 1
(d) 𝐺1
1+𝐺1𝐺2
[GATE 2014: 1 Mark]
Soln. Converting the block diagram into signal flow graph as:
1
1
The forward paths
𝑃1 = 𝐺1𝐺2
𝑃2 = 𝐺2
𝑃3 = 1.1 = 1
𝐶(𝑠)
𝑅(𝑠)= 𝐺1𝐺2 + 𝐺2 + 1
Option (c)
