GATE 2025 EC Q.56: solution and answer

GATE 2025 · ECQ.56 · 2 marks · MSQ

Consider a system represented by the block diagram shown below. Which of the following signal flow graphs represent(s) this system? Choose the correct option(s).

GATE 2025 EC question 56: Block diagram: R(s) enters the + input of summing junction 1. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
Figure: $R (s) 1 n_{1}$ ; $n_{1} G_{1} n_{2}$ (straight branch); $n_{1} G_{2} n_{2}$ (upper arc); $n_{2} - 1 n_{1}$ (lower arc, feedback); $n_{2} 1 Y (s)$ .
(B)
Figure: $R (s) 1 n_{1}$ ; $n_{1} G_{1} n_{2}$ ; $n_{2} 1 n_{3}$ ; $n_{1} G_{2} n_{3}$ (upper arc that skips $n_{2}$ ); $n_{2} - 1 n_{1}$ (lower arc, feedback); $n_{3} 1 Y (s)$ .
(C)
Figure: $R (s) 1 n_{1}$ ; $n_{1} G_{1} n_{2}$ (straight branch); $n_{1} G_{2} n_{2}$ (upper arc); $n_{2} 1 n_{1}$ (lower arc, feedback gain $+ 1$ ); $n_{2} 1 Y (s)$ .
(D)
Figure: $R (s) 1 n_{1}$ ; $n_{1} G_{1} n_{2}$ (straight branch); $n_{1} G_{2} n_{2}$ (upper arc); $n_{2} 1 n_{3}$ ; $n_{3} - 1 n_{1}$ (lower arc from $n_{3}$ back to $n_{1}$ , feedback); $n_{3} 1 Y (s)$ .

Answer and solution

ANSWER

Only signal flow graph (B) represents the system. The block diagram and graph B both give $\frac{Y ( s )}{R ( s )} = \frac{G _{1} + G _{2}}{1 + G _{1}}$ . Graphs A and D give $\frac{G _{1} + G _{2}}{1 + G _{1} + G _{2}}$ , and C gives $\frac{G _{1} + G _{2}}{1 - G _{1} - G _{2}}$ .

Official answer: BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Block diagram. Let $E$ be the output of the first summer. The feedback path starts after $G_{1}$ and before the second summer, so it carries only $G_{1} E$ . That gives $E = R - G_{1} E$ , so $E = \frac{R}{1 + G _{1}}$ .
The second summer adds both branches: $Y = G_{1} E + G_{2} E = (G_{1} + G_{2}) E$ .
So the target transfer function is $\frac{Y ( s )}{R ( s )} = \frac{G _{1} + G _{2}}{1 + G _{1}}$ The system has one loop, with gain $- G_{1}$ . The $G_{2}$ branch is a feedforward path outside that loop.
(A) The forward paths are $P_{1} = G_{1}$ and $P_{2} = G_{2}$ . The $- 1$ branch leaves $n_{2}$ , where $G_{1}$ and $G_{2}$ have already merged. That makes two loops, $L_{1} = - G_{1}$ and $L_{2} = - G_{2}$ . They share $n_{1}$ and $n_{2}$ , so they touch, and $Δ = 1 + G_{1} + G_{2}$ . Both forward paths touch both loops, so $Δ_{1} = Δ_{2} = 1$ and $T_{A} = \frac{G _{1} + G _{2}}{1 + G _{1} + G _{2}}$ . A is wrong.
(B) The $- 1$ branch leaves $n_{2}$ , which carries only $G_{1} n_{1}$ . $G_{2}$ joins later, at $n_{3}$ , and nothing leads from $n_{3}$ back into the graph, so the only loop is $L_{1} = - G_{1}$ and $Δ = 1 + G_{1}$ . The forward paths $P_{1} = G_{1}$ (through $n_{1}, n_{2}, n_{3}$ ) and $P_{2} = G_{2}$ (through $n_{1}, n_{3}$ ) both pass through $n_{1}$ , so both touch the loop and $Δ_{1} = Δ_{2} = 1$ . That gives $T_{B} = \frac{G _{1} + G _{2}}{1 + G _{1}}$ , which matches. Node equations give the same result: $n_{1} = R - G_{1} n_{1}$ , so $n_{1} = \frac{R}{1 + G _{1}}$ , and $Y = n_{3} = (G_{1} + G_{2}) n_{1}$ .
(C) The paths merge before the feedback node, as in A, but the feedback gain is $+ 1$ . The loops are $+ G_{1}$ and $+ G_{2}$ , so $Δ = 1 - G_{1} - G_{2}$ and $T_{C} = \frac{G _{1} + G _{2}}{1 - G _{1} - G _{2}}$ . C is wrong.
(D) The feedback leaves $n_{3}$ , and $n_{3} = n_{2} = (G_{1} + G_{2}) n_{1}$ , so the fed-back signal again includes $G_{2}$ 's output. The loops are $- G_{1}$ and $- G_{2}$ , as in A, so $T_{D} = \frac{G _{1} + G _{2}}{1 + G _{1} + G _{2}}$ . D is wrong.
Only graph B represents the block diagram. The answer is B.

KEY IDEA

In a signal flow graph, a feedback branch must start at a node that carries exactly the signal the block diagram feeds back. Here the feedback takes $G_{1}$ 's output before $G_{2}$ 's output is added. The graph therefore needs a node that holds $G_{1} E$ alone, separate from a later node where $G_{2} E$ joins. Only graph B has one, so its only loop is $- G_{1}$ and $G_{2}$ stays a feedforward path.

WATCH OUT FOR

Treating the take-off point after $G_{1}$ and the second summing junction as one node. Graphs A and D do this. The fed-back signal then becomes $(G_{1} + G_{2}) E$ , which adds a second loop $- G_{2}$ and gives $Δ = 1 + G_{1} + G_{2}$ . A second slip is missing the sign change in C: a $+ 1$ feedback branch turns the minus signs in $Δ$ into plus signs.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 10 Block diagrams & SFG questions, with notes

GATE 2025 EC Q.56: solution and answer

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