GATE 2022 EE Q.47: solution and answer

GATE 2022 · EEQ.47 · 2 marks · MCQ

Let an input $x (t) = 2 sin (10 π t) + 5 cos (15 π t) + 7 sin (42 π t) + 4 cos (45 π t)$ is passed through an LTI system having an impulse response,

h (t) = 2 (\frac{sin ( 10 π t )}{π t}) cos (40 π t) .

The output of the system is

(A)
$2 sin (10 π t) + 5 cos (15 π t)$
(B)
$5 cos (15 π t) + 7 sin (42 π t)$
(C)
$7 sin (42 π t) + 4 cos (45 π t)$
(D)
$2 sin (10 π t) + 4 cos (45 π t)$

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Answer and solution

ANSWER

The output is $y (t) = 7 sin (42 π t) + 4 cos (45 π t)$ , option (C). The system is an ideal band-pass filter with unity gain over $30 π < ∣ ω ∣ < 50 π rad/s$ .

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Standard pair (ideal low-pass): $\frac{sin ( W t )}{π t} ⟷ {1, 0, ∣ ω ∣ < W ∣ ω ∣ > W$ . With $W = 10 π rad/s$ , the term $g (t) = 2 \frac{sin ( 10 π t )}{π t}$ has spectrum $G (ω) = 2$ for $∣ ω ∣ < 10 π$ and $0$ outside.
Modulation property: $g (t) cos (ω_{0} t) ⟷ \frac{1}{2} [G (ω - ω_{0}) + G (ω + ω_{0})]$ with $ω_{0} = 40 π$ . Each shifted copy has height $\frac{1}{2} \times 2 = 1$ , and the two copies (centred at $\pm 40 π$ , each $20 π$ wide) do not overlap.
So $H (ω)$ is an ideal band-pass filter: $H (ω) = {1, 0, 30 π < ∣ ω ∣ < 50 π otherwise$ (passband $40 π \pm 10 π rad/s$ , i.e. $15 Hz$ to $25 Hz$ , with unity gain). Since $h (t)$ is real and even, $H (ω)$ is real, so the filter adds zero phase.
Sinusoid through an LTI system: each input tone at $ω_{k}$ comes out scaled by $∣ H (ω_{k}) ∣$ and shifted by $∠ H (ω_{k})$ . Here $H$ is real and non-negative, so every passed tone keeps its amplitude and phase.
Check each input frequency against the passband $30 π < ∣ ω ∣ < 50 π$ : $10 π$ is blocked, $15 π$ is blocked, $42 π$ passes with gain 1, and $45 π$ passes with gain 1. No tone sits on a band edge, so there is no ambiguity.
Therefore $y (t) = 7 sin (42 π t) + 4 cos (45 π t)$ , which is option (C).

KEY IDEA

$h (t)$ is an ideal low-pass impulse response, $\frac{sin ( 10 π t )}{π t}$ , multiplied by $cos (40 π t)$ . That multiplication shifts the low-pass band to $\pm 40 π$ , so the system is an ideal band-pass filter over $30 π < ∣ ω ∣ < 50 π$ . The factor 2 cancels the $\frac{1}{2}$ from modulation, so the gain is exactly 1. An LTI system passes each sinusoid on its own, scaled by $H (ω_{k})$ , so we only need to check which input frequencies fall inside the band.

WATCH OUT FOR

The usual slip is to ignore the $cos (40 π t)$ factor and treat $h (t)$ as a low-pass filter with cutoff $10 π$ . That keeps the low tones and points to option (A), which is wrong. A second slip is to drop the $\frac{1}{2}$ of the modulation property. That gives a passband gain of 2, and the resulting $14 sin (42 π t) + 8 cos (45 π t)$ is not among the options. A third is to mix up rad/s and Hz and read the band as 30 to 50 Hz. The input tones are at 5, 7.5, 21 and 22.5 Hz, and none of those would fall inside a 30 to 50 Hz band.

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GATE 2022 EE Q.47: solution and answer

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