GATE 2024 EE Q.20: solution and answer

GATE 2024 · EEQ.20 · 1 mark · MCQ

For the block-diagram shown in the figure, the transfer function $\frac{C ( s )}{R ( s )}$ is

GATE 2024 EE question 20: Two summing junctions in cascade followed by a single forward block G(s). — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
$\frac{G ( s )}{1 + 2 G ( s )}$
(B)
$- \frac{G ( s )}{1 + 2 G ( s )}$
(C)
$\frac{G ( s )}{1 - 2 G ( s )}$
(D)
$- \frac{G ( s )}{1 - 2 G ( s )}$

Answer and solution

ANSWER

$\frac{C ( s )}{R ( s )} = - \frac{G ( s )}{1 - 2 G ( s )}$ , option (D).

Official answer: DSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

First summing junction: $R$ enters with $+$ and the outer unity feedback $C$ enters with $-$ , so its output is $E_{1} = R - C$ .
Second summing junction: $E_{1}$ enters with a minus sign and the inner unity feedback $C$ enters with a plus sign, so $E_{2} = - E_{1} + C = - (R - C) + C = 2 C - R$ .
Forward block: $C = G E_{2} = G (2 C - R)$ .
Collect the terms in $C$ : $C - 2 GC = - GR$ , so $C (1 - 2 G) = - G R$ .
$\frac{C ( s )}{R ( s )} = - \frac{G ( s )}{1 - 2 G ( s )}$
Check with Mason's gain formula. The one forward path $R \to C$ passes through the $-$ input of the second junction, so $P_{1} = (- 1) G = - G$ . There are two loops. The inner loop goes through the $+$ input of the second junction: $L_{1} = + G$ . The outer loop goes through the $-$ input of the first junction and the $-$ input of the second: $L_{2} = (- 1) (- 1) G = + G$ . Both loops share the block $G$ , so they touch each other and the forward path. Hence $Δ = 1 - (L_{1} + L_{2}) = 1 - 2 G$ and $Δ_{1} = 1$ , giving $T = \frac{P _{1} Δ _{1}}{Δ} = - \frac{G}{1 - 2 G}$ . This is option (D).

KEY IDEA

The minus sign on the link between the two summing junctions lies on both the forward path and the outer loop. It makes the forward gain negative ( $- G$ ). In the outer loop it cancels the minus at the first junction, so that loop is effectively positive ( $+ G$ ), the same as the inner positive-feedback loop. Two touching $+ G$ loops give $Δ = 1 - 2 G$ .

WATCH OUT FOR

Counting the minus sign between the junctions only once. If you apply it to the outer loop but forget it on the forward path, you get $+ G / (1 - 2 G)$ , which is option (C). If you ignore it in the outer loop and read that loop as ordinary negative feedback, the loop signs come out wrong and you land on a $1 + 2 G$ denominator (options A and B) or on a result that is none of the options. That minus has to be counted in every path and loop that passes through it.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 10 Block diagrams & SFG questions, with notes

GATE 2024 EE Q.20: solution and answer

Answer and solution

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