GATE 2025 EC – Question 38
Consider a continuous-time finite-energy signal $f(t)$ whose Fourier transform vanishes outside the frequency interval $[-\omega_c,\omega_c]$, where $\omega_c$ is in rad/sec.
The signal $f(t)$ is uniformly sampled to obtain $y(t)=f(t)\,p(t)$. Here,
$$p(t)=\sum_{n=-\infty}^{\infty}\delta(t-\tau-nT_s),$$
with $\delta(t)$ being the Dirac impulse, $T_s>0$, and $\tau>0$. The sampled signal $y(t)$ is passed through an ideal lowpass filter $h(t)=\omega_cT_s\dfrac{\sin(\omega_ct)}{\pi\omega_ct}$ with cutoff frequency $\omega_c$ and passband gain $T_s$.
The output of the filter is given by ________.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (A) $f(t)$ if $T_s<\pi/\omega_c$
Explanation
$P(\omega)=\frac{2\pi}{T_s}\sum_k\delta(\omega-k\omega_s)e^{-jk\omega_s\tau}$ with $\omega_s=2\pi/T_s$, so $Y(\omega)=\frac1{T_s}\sum_kF(\omega-k\omega_s)e^{-jk\omega_s\tau}$. The $k=0$ term is $F(\omega)/T_s$ with no delay factor. The copies do not overlap when $\omega_s>2\omega_c$, i.e. $T_s<\pi/\omega_c$. The filter (gain $T_s$, cutoff $\omega_c$) then keeps only the $k=0$ term, so the output is $f(t)$ itself: the time offset $\tau$ of the sampling instants does not delay the reconstructed signal.