GATE 2025 EC Q.38: solution and answer

GATE 2025 · ECQ.38 · 2 marks · MCQ

Consider a continuous-time finite-energy signal $f (t)$ whose Fourier transform vanishes outside the frequency interval $[- ω_{c}, ω_{c}]$ , where $ω_{c}$ is in rad/sec.

The signal $f (t)$ is uniformly sampled to obtain $y (t) = f (t) p (t)$ . Here,
$p (t) = n = - \infty \sum \infty δ (t - τ - n T_{s}),$
with $δ (t)$ being the Dirac impulse, $T_{s} > 0$ , and $τ > 0$ . The sampled signal $y (t)$ is passed through an ideal lowpass filter $h (t) = ω_{c} T_{s} \frac{sin ( ω _{c} t )}{π ω _{c} t}$ with cutoff frequency $ω_{c}$ and passband gain $T_{s}$ .

The output of the filter is given by ________.

(A)
$f (t)$ if $T_{s} < π / ω_{c}$
(B)
$f (t - τ)$ if $T_{s} < π / ω_{c}$
(C)
$f (t - τ)$ if $T_{s} < 2 π / ω_{c}$
(D)
$T_{s} f (t)$ if $T_{s} < 2 π / ω_{c}$

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Answer and solution

ANSWER

The filter output is $f (t)$ provided $T_{s} < π / ω_{c}$ . This is option (A).

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

What the sampler produces. By the sifting property, $y (t) = \sum_{n} f (τ + n T_{s}) δ (t - τ - n T_{s})$ . Each impulse carries the true value of $f$ at the instant where it sits, so the offset $τ$ only moves the sampling grid. It does not delay the signal.
Spectrum of the shifted impulse train. Let $ω_{s} = 2 π / T_{s}$ . The unshifted train $\sum_{n} δ (t - n T_{s})$ has transform $ω_{s} \sum_{k} δ (ω - k ω_{s})$ . A shift by $τ$ multiplies this by $e^{- j ω τ}$ , and at each impulse $ω = k ω_{s}$ , so $P (ω) = \frac{2 π}{T _{s}} k = - \infty \sum \infty e^{- j k ω_{s} τ} δ (ω - k ω_{s}) .$
Multiplication theorem ( $Y = \frac{1}{2 π} F * P$ ): $Y (ω) = \frac{1}{T _{s}} k = - \infty \sum \infty e^{- j k ω_{s} τ} F (ω - k ω_{s}) .$ The $k = 0$ copy is $\frac{1}{T _{s}} F (ω)$ and has no phase factor, because $e^{0} = 1$ . The offset $τ$ only rotates the phase of the shifted copies ( $k \neq = 0$ ).
The filter. $h (t) = ω_{c} T_{s} \frac{sin ( ω _{c} t )}{π ω _{c} t} = T_{s} \frac{sin ( ω _{c} t )}{π t}$ . Since $\frac{sin ( ω _{c} t )}{π t}$ transforms to $1$ for $∣ ω ∣ < ω_{c}$ and $0$ elsewhere, $H (ω) = T_{s}$ for $∣ ω ∣ < ω_{c}$ and $0$ elsewhere. This is an ideal LPF with gain $T_{s}$ .
No-aliasing condition (Nyquist). The $k = \pm 1$ copies occupy $[ω_{s} - ω_{c}, ω_{s} + ω_{c}]$ and its mirror image. The copies with $∣ k ∣ \geq 2$ sit even farther out. For every shifted copy to stay clear of the passband $∣ ω ∣ < ω_{c}$ we need $ω_{s} - ω_{c} > ω_{c}$ , which gives $ω_{s} > 2 ω_{c} \Rightarrow \frac{2 π}{T _{s}} > 2 ω_{c} \Rightarrow T_{s} < \frac{π}{ω _{c}} .$
Output. Under that condition only the $k = 0$ term passes: $X_{o u t} (ω) = T_{s} \cdot \frac{1}{T _{s}} F (ω) = F (ω) \Rightarrow x_{o u t} (t) = f (t) .$ The gain $T_{s}$ cancels the $1/ T_{s}$ , and there is no delay.
Answer. The output is $f (t)$ if $T_{s} < π / ω_{c}$ , which is option (A). (B) wrongly adds a delay $τ$ . (C) and (D) use $T_{s} < 2 π / ω_{c}$ , which only gives $ω_{s} > ω_{c}$ and lets the $k = \pm 1$ copies alias into the passband. (D) also keeps a stray gain $T_{s}$ .

KEY IDEA

Shifting the sampling grid by $τ$ multiplies the $k$ -th spectral copy by $e^{- j k ω_{s} τ}$ . The baseband copy ( $k = 0$ ) gets $e^{0} = 1$ , so the offset has no effect on what the lowpass filter recovers. The sampled spectrum carries a factor $1/ T_{s}$ , which the passband gain $T_{s}$ cancels. Nyquist ( $ω_{s} > 2 ω_{c}$ ) gives $T_{s} < π / ω_{c}$ .

WATCH OUT FOR

Picking (B) on the belief that the offset $τ$ delays the output. Here $f$ is multiplied by the shifted train, so each impulse carries $f$ 's true value at its own instant, and the only $τ$ -dependent phase lands on the $k \neq = 0$ copies, which the filter removes. Other traps: writing the Nyquist condition as $ω_{s} > ω_{c}$ , which gives $2 π / ω_{c}$ as in (C) and (D), and forgetting that the gain $T_{s}$ cancels the $1/ T_{s}$ from sampling, as in (D).

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GATE 2025 EC Q.38: solution and answer

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