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)