GATE 2023 EC Q.42: solution and answer

GATE 2023 · ECQ.42 · 2 marks · MCQ

In the following block diagram, $R (s)$ and $D (s)$ are two inputs. The output $Y (s)$ is expressed as $Y (s) = G_{1} (s) R (s) + G_{2} (s) D (s)$ .

$G_{1} (s)$ and $G_{2} (s)$ are given by

GATE 2023 EC question 42: Two summing junctions in cascade, then a forward block G(s) whose output is Y(s). — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$G_{1} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$ and $G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$
(B)
$G_{1} (s) = \frac{G ( s )}{1 + G ( s ) + H ( s )}$ and $G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + H ( s )}$
(C)
$G_{1} (s) = \frac{G ( s )}{1 + G ( s ) + H ( s )}$ and $G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$
(D)
$G_{1} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$ and $G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + H ( s )}$

Answer and solution

ANSWER

$G_{1} (s) = G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$ , option (A).

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

Write the signal leaving each summing junction. Junction 1 subtracts the unity feedback from the output: $E_{1} = R - Y$ .
Junction 2 adds the disturbance and subtracts the $H$ feedback: $E_{2} = E_{1} + D - H Y = R + D - Y - H Y$ .
The forward block gives $Y = G E_{2} = G (R + D - Y - H Y)$ .
Collect the $Y$ terms on the left: $Y (1 + G + G H) = G R + G D$
So $Y (s) = \frac{G}{1 + G + G H} R (s) + \frac{G}{1 + G + G H} D (s)$
Check with Mason's gain formula. There are two loops, with gains $- G$ (through the unity outer feedback) and $- G H$ (through $H$ ). Both pass through $G$ , so they touch, and $Δ = 1 + G + G H$ . From $R$ and from $D$ the only forward path is through $G$ alone, and it touches both loops, so $Δ_{1} = 1$ for each input. Both transfer functions are therefore $G / (1 + G + G H)$ .
Therefore $G_{1} (s) = G_{2} (s) = \frac{G ( s )}{1 + G ( s ) + G ( s ) H ( s )}$ , which is option (A).

KEY IDEA

Both inputs enter the loop ahead of the single forward block $G$ , and nothing but a summing junction separates them from it, so they share one forward path (gain $G$ ) and one characteristic denominator. Each feedback loop's gain is the product around the loop: $- G \cdot 1$ for the outer loop and $- G \cdot H$ for the inner one. That gives $Δ = 1 + G + G H$ , and so $G_{1} = G_{2} = G / (1 + G + G H)$ .

WATCH OUT FOR

Writing the inner loop's contribution as $H$ in place of the loop gain $G H$ . The feedback signal $H Y$ goes back through $G$ before it reaches $Y$ , so the term is $G H$ . Dropping the $G$ gives the $1 + G + H$ denominators in options (B), (C) and (D). A second slip is to assume the disturbance sees a different denominator because $D$ enters inside the outer loop. Junction 2 is fed straight from junction 1, so $D$ and the outer feedback act at effectively the same point and both loops still act on it.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 10 Block diagrams & SFG questions, with notes

GATE 2023 EC Q.42: solution and answer

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