GATE 2023 EC Q.49: solution and answer

GATE 2023 · ECQ.49 · 2 marks · MCQ

Let $x (t) = 10 cos (10.5 W t)$ be passed through an LTI system having impulse response $h (t) = π (\frac{sin W t}{π t})^{2} cos 10 W t$ . The output of the system is

(A)
$(\frac{15 W}{4}) cos (10.5 W t)$
(B)
$(\frac{15 W}{2}) cos (10.5 W t)$
(C)
$(\frac{15 W}{8}) cos (10.5 W t)$
(D)
$(15 W) cos (10.5 W t)$

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Answer and solution

ANSWER

The output is $y (t) = (\frac{15 W}{4}) cos (10.5 W t)$ , option (A).

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

Standard pair. $\frac{sin W t}{π t} ⟷$ a rectangle of height $1$ on $∣ ω ∣ < W$ , zero elsewhere.
Multiplication property. A product in time becomes $\frac{1}{2 π} \times$ a convolution in frequency. The rectangle convolved with itself is a triangle with peak $2 W$ (the overlap length) at $ω = 0$ , falling to zero at $∣ ω ∣ = 2 W$ . So $(\frac{sin W t}{π t})^{2} ⟷ \frac{1}{2 π} (2 W) (1 - \frac{∣ ω ∣}{2 W}) = \frac{W}{π} (1 - \frac{∣ ω ∣}{2 W}), ∣ ω ∣ < 2 W$
Scale by $π$ . $g (t) = π (\frac{sin W t}{π t})^{2} ⟷ G (ω) = W (1 - \frac{∣ ω ∣}{2 W})$ for $∣ ω ∣ < 2 W$ , and $0$ otherwise.
Modulation property. $h (t) = g (t) cos 10 W t ⟷ H (ω) = \frac{1}{2} [G (ω - 10 W) + G (ω + 10 W)]$ . These are two triangles centred at $\pm 10 W$ , each extending $2 W$ on either side of its centre. Since $10 W > 2 W$ they do not overlap.
Evaluate at the input frequency $ω_{0} = 10.5 W$ . The second term needs $G (20.5 W) = 0$ , so $H (j 10.5 W) = \frac{1}{2} G (0.5 W) = \frac{1}{2} W (1 - \frac{0.5 W}{2 W}) = \frac{1}{2} \cdot \frac{3 W}{4} = \frac{3 W}{8}$
Phase. $h (t)$ is real and even, so $H (ω)$ is real; here it is also non-negative, so the phase shift is zero.
Sinusoidal steady state of an LTI system. $y (t) = 10 ∣ H (j ω_{0}) ∣ cos (ω_{0} t) = 10 \cdot \frac{3 W}{8} cos (10.5 W t) = \frac{15 W}{4} cos (10.5 W t)$ This is option (A).

KEY IDEA

A sinusoid through an LTI system comes out as the same sinusoid, scaled by $∣ H (j ω_{0}) ∣$ and shifted by $∠ H (j ω_{0})$ . Here $π (\frac{s i n W t}{π t})^{2}$ has a triangular spectrum of height $W$ reaching zero at $\pm 2 W$ . Multiplying by $cos 10 W t$ moves half of that triangle to $\pm 10 W$ . The input at $10.5 W$ lies $0.5 W$ from the centre of one copy, so $H = \frac{1}{2} \cdot W (1 - \frac{0.5 W}{2 W}) = \frac{3 W}{8}$ .

WATCH OUT FOR

Two mistakes are common. (1) Leaving out the $\frac{1}{2 π}$ in the multiplication property, or taking the triangle's peak as $W$ rather than the overlap length $2 W$ . Either error shifts $G (0)$ away from $W$ and leads to option (B) or (D). (2) Forgetting the factor $\frac{1}{2}$ from the modulation property. That gives $∣ H ∣ = \frac{3 W}{4}$ and an output of $\frac{15 W}{2} cos (10.5 W t)$ , which is option (B).

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GATE 2023 EC Q.49: solution and answer

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