· pdf filethe signal flow graph of a system is shown in the figure. the transfer function...
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Signal Flow Graph
1. In the signal flow graph shown in figure ๐2 = ๐๐1 where T, is equal to
5X1 X2
0.5
(a) 2.5
(b) 5
(c) 5.5
(d) 10
[GATE 1987: 2 Marks]
Soln. ๐2 = ๐๐1
๐2
๐1=
5
โ=
5
1 โ 0.5= 10
Option (d)
2. For the system shown in figure the transfer function
๐ถ(๐ )
๐ (๐ ) ๐๐ ๐๐๐ข๐๐ ๐ก๐
R(s) C(s)
S
10
S(S+1)+- +-
(a) 10
๐2+๐+10
(b) 10
๐2+11๐+10
(c) 10
๐2+9๐+10
(d) 10
๐2+2๐+10
[GATE 1987: 2 Marks]
Soln. The forward path transmittance =10
๐(๐+1)
The two closed loop are ๐ฟ1 =โ10
๐(๐+1)
๐ฟ2 =โ10๐
๐(๐ + 1)
๐ถ(๐)
๐ (๐)=
10/๐(๐ + 1)
1 โ {โ10
๐(๐+1)
โ10
๐+1}
=10
๐(๐ + 1) [1 +10
๐(๐+1)+
10
(๐+1)]
=10 ๐(๐ + 1)
๐(๐ + 1)[๐(๐ + 1) + 10 + 10๐]
=10
๐๐ + ๐ + 10๐ + 10=
10
๐2 + 11๐ + 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) ๐บ1๐บ2๐บ34
(1+๐บ1๐บ2)(1+๐บ3๐บ4)
(b) ๐บ1๐บ2๐บ3๐บ4
(1+๐บ1+๐บ2+๐บ1๐บ2)(1+๐บ3+๐บ4+๐บ3๐บ4)
(c) ๐บ1๐บ2๐บ3๐บ4
(1+๐บ1+๐บ2)(1+๐บ3+๐บ4)
(d) ๐บ1๐บ2๐บ3๐บ4
(1+๐บ1+๐บ2+๐บ3+๐บ4)
[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
โ= 1 โ [โ๐บ1 โ ๐บ2 โ ๐บ3 โ ๐บ4] + [๐บ1๐บ3 + ๐บ1๐บ4 + ๐บ2๐บ3 + ๐บ2๐บ4]
๐
๐ =
๐บ1๐บ2๐บ3๐บ4
(1 + ๐บ1 + ๐บ2 + ๐บ3 + ๐บ4) + (๐บ1๐บ3 + ๐บ1๐บ4 + ๐บ2๐บ3 + ๐บ2๐บ4)
=๐บ1๐บ2๐บ3๐บ4
(1 + ๐บ1 + ๐บ2)(1 + ๐บ3 + ๐บ4)
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 ๐1 = 1 ร 2 ร 3 ร 4 = 24
The forward path ๐2 = 5
โ1= 1
๐ฟ1 = โ2, ๐ฟ2 = โ3, ๐ฟ3 = โ4
Non touching loops โ ๐ฟ1๐ฟ3
The loop ๐ฟ2 = โ3 does not touch the path P2
So, โ2= (1 โ ๐ฟ2)
= 1 + 3
= 4
๐
๐ =
๐1โ1 + ๐2โ2
โ
=24 ร 1 + 5 ร 4
1โ (โ2 โ 3 โ 4 โ 5) + (โ2) ร (โ4)
=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
๐
๐=
๐๐พโ๐พ
โ
๐๐พ = 5 ร 2 ร 1 = 10
โ๐พ= 1
โ๐พ= 1
โ= 1 โ (โ4) = 5
๐
๐=
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 ๐ถ(๐ )
๐ (๐ ) of the system is
R(s)
C(s)
1 6
1
-4-3
1S S
1
-2
(a) 6
๐2+29๐+6
(b) 6๐
๐2+29๐+6
(c) ๐(๐+2)
๐2+29๐+6
(d) ๐(๐+27)
๐2+29๐+6
[GATE 2003: 2 Marks]
Soln. The transfer function ๐ถ(๐)
๐ (๐) of the systems?
๐ฟ1 =โ3
๐, ๐ฟ2 = โ4 ร
6
๐, ๐ฟ3 =
โ2
๐
P1 = 1
Loops L1 and L3 are not touching the forward path
โ1= 1 โ ๐ฟ1 โ ๐ฟ3
= 1 +3
๐+
24
๐
=๐ + 27
๐
๐ถ(๐)
๐ (๐)= ๐บ(๐ ) =
๐1โ1
1 โ (๐๐๐๐ ๐๐๐๐) + ๐๐๐๐ ๐๐ ๐๐๐ ๐ก๐ข๐โ๐๐๐ ๐๐๐๐๐
๐+27
๐
1 โ (โ3
๐
โ24
๐
โ2
๐) +
โ2
๐ร
โ3
๐
๐+27
๐
1 +29
๐+
6
๐2
=๐(๐ + 27)
๐2 + 29๐ + 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) 1โ(๐๐+๐๐+๐๐)
๐๐๐
(b) ๐๐๐๐
1โ(๐๐+๐๐+๐๐)
(c) ๐๐๐๐
1โ(๐๐+๐๐+๐๐)+๐๐๐๐
(d) 1โ(๐๐+๐๐+๐๐)+๐๐๐๐
๐๐๐๐
[GATE 2004: 2 Marks]
Soln. The forward path transmittance P1 = abcd
All the loops touch the forward path โ1= 1
๐ฟ1 = ๐๐, ๐ฟ2 = ๐ถ๐, ๐ฟ3 = ๐๐
Non touching loops are ๐ฟ1, ๐ฟ3
๐ฟ1๐ฟ3 = ๐๐ ๐๐
๐5
๐1=
๐๐๐๐
1 โ (๐๐ + ๐๐ + ๐๐) + ๐๐ ๐๐
Option (c)
9. The transfer function Y(s) / R(s) of the system shown is
Y(s)1
S+1
1
S+1
ฮฃ
ฮฃ
+-
+
-
R(s)P(s)
Q(s)
Q(s)
(a) 0
(b) 1
๐+1
(c) 2
๐+1
(d) 2
๐+3
[GATE 2010: 1 Mark]
Soln. ๐(๐ ) = ๐(๐ ) [1
๐+1โ
1
๐+1]
= 0
๐(๐ ) = ๐ (๐ ) โ 0
= ๐ (๐ )
๐(๐ ) =๐(๐ )
๐ + 1
=๐ (๐ )
๐ + 1
๐(๐)
๐ (๐)=
1
๐ + 1
Option (b)
Common Data for Question 10 and Question 11
The input-output transfer function of a plant ๐ป(๐ ) =100
๐(๐+10)2.The plant is
placed in unity negative feedback configuration as shown in the figure
below.
ฮฃ + -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 ๐ป(๐ ) =100
๐(๐+10)2
For the figure (a) ๐1 =100
๐3, โ1= 1
(๐๐๐ ๐กโ๐ ๐๐๐๐๐ ๐ก๐๐ข๐โ ๐กโ๐ ๐๐๐๐ค๐๐๐ ๐๐๐กโ)
๐
๐=
100/๐3
1 โ (โ10
๐
โ10
๐) +
10
๐ร
10
๐
=
100
๐2
1 +20
๐+
100
๐2
=100
๐(๐ + 10)2
For figure (b) ๐1 =100
๐3, โ1= 1
๐
๐=
100
๐3
1 โ (โ100
๐2
โ20
๐)
=
100
๐3
1 +100
๐2+
20
๐
=100
๐(๐ + 10)2
For Figure (c)
๐1 =100
๐3, โ1= 1
๐
๐=
100
๐3
1 โ (โ100
๐2
โ20
๐)
=100
๐(๐ + 10)2
For Figure (d),
๐
๐=
100
๐3
1 โ (โ100
๐2 )=
100
๐3
1 +100
๐2
= 100 ๐2
๐3(๐2 + 100)=
100
๐(๐2 + 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
๐บ(๐ )๐ป(๐ ) =100
๐(๐+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
๐บ(๐ )๐ป(๐ ) =100
๐(๐+10)2
โ = โ900 โ 2๐ก๐๐โ1 (๐
10)
Flow phase cross aver frequency โ = โ1800
โ180 = โ900 โ 2๐ก๐๐โ1 (๐
10)
๐ = 10 ๐๐๐/๐ ๐๐
๐บ(๐๐)๐ป(๐๐) =100
๐๐ (๐๐ + 10)2
|๐บ(๐๐)๐ป(๐๐)| =100
๐(๐2 + 100)
=100
10(100 + 100)=
1
20
๐บ๐๐๐ ๐๐๐๐๐๐ (๐บ. ๐) =1
|๐บ(๐๐)๐ป(๐๐)|= 20
๐บ. ๐ ๐๐ ๐๐ต = 20๐๐๐1020 = 26 ๐๐ต
Option (c)
12. The signal flow graph for a system is given below. The transfer function
๐(๐ )
๐(๐ ) ๐๐๐ ๐กโ๐๐ ๐ ๐ฆ๐ ๐ก๐๐ ๐๐
U(s) Y(s)1 1
1
-4
-2
S-1
S-1
(a) ๐ +1
5๐ 2+6๐ +2
(b) ๐ +1
๐ 2+6๐ +2
(c) ๐ +1
๐ 2+4๐ +2
(d) 1
5๐ 2+6๐ +2
[GATE 2013: 2 Marks]
Soln. The forward path transmittance ๐1 = ๐โ1 ร ๐1 =1
๐2
The forward path transmittance ๐2 = ๐โ1 =1
๐
โ1= 1, โ2= 1
โ= 1 โ (โ2๐โ2 โ 2๐โ1 โ 4๐โ1 โ 4)
= 1 +2
๐2+
2
๐+
4
๐+ 4
= 1 +2
๐2+
2
๐+
4
๐+ 4
= (5๐2 + 6๐ + 2)/๐2
๐(๐)
๐(๐)=
๐1โ1 + ๐2โ2
โ=
1
๐2+
1
๐
(5๐2 + 6๐ + 2)/๐2=
(๐ + 1)
5๐2 + 6๐ + 2
Option (a)
13. For the following system,
+-S
S+1 +-+ 1
S
When ๐1(๐) = 0, the transfer function ๐(๐)
๐2(๐) is
(a) ๐+1
๐2
(b) 1
๐+1
(c) ๐+2
๐(๐+1)
(d) ๐+1
๐(๐+2)
[GATE: 2014: 1 Mark]
Soln. With ๐1(๐ ) = 0, the block diagram is redrawn as
+-
S
S+1
1
S
๐(๐ ) =๐(๐)
๐2(๐)=
๐บ(๐)
1 + ๐บ(๐)๐ป(๐)
๐บ(๐ ) =1
๐, ๐ป(๐ ) =
๐
๐ + 1=
1/๐
1 +1
๐ร
๐
๐+1
=1(๐ + 1)
๐(๐ + 1 + 1)=
(๐ + 1)
๐(๐ + 2)
Option (d)
14. Consider the following block diagram in the figure
++ ++
the transfer function ๐ถ(๐)
๐ (๐) is
(a) ๐บ1๐บ2
1+๐บ1๐บ2
(b) ๐บ1๐บ2 + ๐บ1 + 1
(c) ๐บ1๐บ2 + ๐บ2 + 1
(d) ๐บ1
1+๐บ1๐บ2
[GATE 2014: 1 Mark]
Soln. Converting the block diagram into signal flow graph as:
1
1
The forward paths
๐1 = ๐บ1๐บ2
๐2 = ๐บ2
๐3 = 1.1 = 1
๐ถ(๐ )
๐ (๐ )= ๐บ1๐บ2 + ๐บ2 + 1
Option (c)