Q61GATE 2002MCQ

In the figure, $m(t)=\frac{2\sin 2\pi t}{t}$ , $s(t)=\cos 200\pi t$ and $n(t)=\frac{\sin 199\pi t}{t}$ . The output y(t) will be

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📖 Explanation

\begin{aligned} m(t) \times s(t) &=\frac{2 \sin 2 \pi t}{t} \times \cos 200 \pi t \\ &=\frac{1}{t}(\sin 202 \pi t-\sin 198 \pi t) \end{aligned}

The signal at the input of the LPF will be, $x(t)=[m(t) s(t)+n(t)] s(t)$ $=\frac{1}{t}(\sin 202 \pi t-\sin 198 \pi t+\sin 199 \pi t) \cos 200 \pi t$ $=\frac{1}{2 t}(\sin 402 \pi t+\sin 399 \pi t-\sin 398 \pi t)$ $\quad+\frac{1}{2 t}(\sin 2 \pi t+\sin 2 \pi t-\sin \pi t)$ After passing $\frac{1}{2 t}(\sin 402 \pi t+\sin 399 \pi t-\sin 398 \pi t)$ term through LPF with cut-off frequency of 1 Hz , it results in a component of $\frac{1}{2 t}(\sin 2 \pi t)$ at the output. So, after passing x(t) through the LPF, we get,

\begin{aligned} y(t) &=\frac{1}{2 t}(\sin 2 \pi t)+\frac{1}{2 t}(\sin 2 \pi t+\sin 2 \pi t-\sin \pi t) \\ &=\frac{\sin 2 \pi t}{t}+\frac{1}{2 t}(2 \sin 0.5 \pi t \cos 1.5 \pi t) \\ y(t) &=\frac{\sin 2 \pi t}{t}+\frac{\sin 0.5 \pi t}{t} \cos 1.5 \pi t \end{aligned}

Q62GATE 2001MCQ

A band limited signal is sampled at the Nyquist rate. The signal can be recovered by passing the samples through

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📖 Explanation

According to the Nyquist-Shannon sampling theorem, a signal sampled at the Nyquist rate ( $f_s = 2f_m$ ) or higher can be perfectly reconstructed without aliasing by passing the discrete samples through an ideal low-pass filter with a cutoff frequency exactly at $f_m$ . The ideal low-pass filter acts as an interpolator, convolving the discrete impulse train with a sinc function to rebuild the exact continuous-time waveform.