GATE 2023 EE Q.12: solution and answer

GATE 2023 · EEQ.12 · 1 mark · MCQ

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

GATE 2023 EE question 12: Input R(s) enters from the left on a horizontal line that branches to two blocks. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
$\frac{2 s + 3}{s + 1}$
(B)
$\frac{3 s + 2}{s - 1}$
(C)
$\frac{s + 1}{3 s + 2}$
(D)
$\frac{3 s + 2}{s + 1}$

Answer and solution

ANSWER

$\frac{Y ( s )}{R ( s )} = \frac{3 s + 2}{s - 1}$ , option (B).

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

Call the output of the first summing junction $X (s)$ . Both of its inputs carry a $+$ sign, so $X = 2 R + 1 \cdot Y$ . The feedback adds to the input, so this is a positive-feedback loop.
The second summing junction adds the integrator output and the feedforward term: $Y = \frac{X}{s} + 3 R$ .
Substitute for $X$ : $Y = \frac{2 R + Y}{s} + 3 R$ .
Collect the $Y$ terms on one side: $Y (1 - \frac{1}{s}) = R (\frac{2}{s} + 3)$ .
Multiply both sides by $s$ , which gives $Y (s - 1) = R (3 s + 2)$ , so $\frac{Y ( s )}{R ( s )} = \frac{3 s + 2}{s - 1}$
Check with Mason's gain formula. There is one loop, $L_{1} = \frac{1}{s} \cdot 1 = + \frac{1}{s}$ , so $Δ = 1 - \frac{1}{s}$ . The forward paths are $P_{1} = \frac{2}{s}$ (through the integrator) and $P_{2} = 3$ (direct). Both paths pass through the second summing junction, which lies on the loop, so both touch it and $Δ_{1} = Δ_{2} = 1$ . Then $T = \frac{\frac{2}{s} + 3}{1 - \frac{1}{s}} = \frac{3 s + 2}{s - 1}$ .
Checking the options: as $s \to \infty$ the integrator path dies out and only the direct gain $3$ is left, so $T \to 3$ . That rules out (A) and (C). The positive feedback around the integrator puts the closed-loop pole at $s = + 1$ , and of (B) and (D) only (B) has that pole. The answer is (B).

KEY IDEA

Name the output of each summing junction and write one equation per junction using the signs printed on the diagram, then solve for $Y / R$ . In Mason's formula, both forward paths end at the summing junction that sits on the only loop, so $Δ_{1} = Δ_{2} = 1$ . Because the feedback is positive, $Δ = 1 - \frac{1}{s}$ , which puts the pole at $s = + 1$ .

WATCH OUT FOR

Reaching for the negative-feedback formula $\frac{G}{1 + G H}$ out of habit, even though both signs at the first junction are $+$ . That gives $\frac{3 s + 2}{s + 1}$ , which is the trap option (D). Swapping which gain goes to the integrator path as well gives $\frac{2 s + 3}{s + 1}$ , option (A). Another slip is treating the direct path of gain $3$ as not touching the loop and using $Δ_{2} = 1 - \frac{1}{s}$ . That gives $\frac{3 s - 1}{s - 1}$ , which is not one of the options.

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

GATE 2023 EE Q.12: solution and answer

Answer and solution

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