Q1GATE 2023NAT

A signal $x(t)=2 \cos (180 \pi t) \cos (60 \pi t)$ is sampled at $200 \mathrm{~Hz}$ and then passed through an ideal low pass filter having cut-off frequency of $100 \mathrm{~Hz}$ . The maximum frequency present in the filtered signal in $\mathrm{Hz}$ is ____ (Round off to the nearest integer)

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📖 Explanation

Given :

\begin{aligned} x(\mathrm{t})&=2 \cos (180 \pi \mathrm{t}) \cos (60 \pi \mathrm{t}) \\ &=\cos 240 \pi \mathrm{t}+\cos 120 \pi \mathrm{t} \end{aligned}

$\mathrm{F}_{1}=120 \mathrm{~Hz}$ and $\mathrm{F}_{2}=60 \mathrm{~Hz}$ Now, present frequency component,

\begin{aligned} & =\mathrm{nf}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =\mathrm{f}_{1}, \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1}, 2 \mathrm{f}_{\mathrm{s}} \pm \mathrm{f}_{1} \\ & =120 \mathrm{~Hz}, 200 \pm 120, \ldots . \\ & =120 \mathrm{~Hz}, 80 \mathrm{~Hz}, 320 \mathrm{~Hz}, \ldots . \end{aligned}

and other frequency component.

\begin{aligned} & =n f_{s} \pm f_{2} \\ & =f_{2}, f_{s} \pm f_{2}, 2 f_{s} \pm f_{2}, \ldots \\ & =60,200 \pm 60, \ldots \\ & =60 \mathrm{~Hz}, 140,260 \mathrm{~Hz}, \ldots \end{aligned}

Given cut-off frequency of LPF $=100 \mathrm{~Hz}$ $\therefore$ It is pas only $60 \mathrm{~Hz}$ and $80 \mathrm{~Hz}$ frequency component. $\therefore$ Max. frequency $=80 \mathrm{~Hz}$ .

Q2GATE 2018MCQ

Consider the two continuous-time signals defined below:

x_{1}(t)=\left\{\begin{matrix} |t|, & -1\leq t \leq 1 \\ 0 & otherwise \end{matrix}\right.

x_{2}(t)=\left\{\begin{matrix} 1-|t|, & -1\leq t \leq 1\\ 0, & otherwise \end{matrix}\right.

These signals are sampled with a sampling period of T=0.25 seconds to obtain discretetime signals $x_{1}[n] \; and \; x_{2}[n]$ , respectively. Which one of the following statements is true?

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📖 Explanation

x_1(t)=\left\{\begin{matrix} |t|, & -1 \leq t \leq 1\\ 0, & \text{ otherwise} \end{matrix}\right.

$T_s=$ sampling time-period = 0.25 sec $x_1(n)=\{ 1,0.75,0.5,0.25,0,0.25,0.5,0.75,1\}$

x_2(t)=\left\{\begin{matrix} 1-|t|, & -1 \leq t \leq 1\\ 0, & \text{ otherwise} \end{matrix}\right.

$x_2(n)=\{ 0,0.25,0.5,0.75,1,0.75,0.5,0.25,0\}$ Since, $x_1(n)$ is having one more non-zero sample of amplitude '1' as compared to $x_2(n)$ . Therefore, energy of $x_1(n)$ greater than energy of $x_2(n)$ .

Q3GATE 2017MCQ

The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples uniquely. The required minimum sampling rate is

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📖 Explanation

From the above block diagram, $Z(t)=x(t) \cos 1000 \pi t$ By using modulation property of fourier transform, $Z(\omega )=\frac{1}{2}[X(\omega + 1000 \pi t)+X(\omega - 1000 \pi t)]$ Now, $h(t)=\frac{\sin 1500 \pi t}{\pi t}=1500 Sa(1500 \pi t)$ Thus, $H(\omega )$ is a low pass filter and it will pass frequency, component of $Z(\omega )$ upto $1500 \pi$ rad/sec. Therefore, maximum frequency component of $y(t)$ is $\omega _m =1500 \pi$ rad/sec or $f_m =750 Hz$ So, the minimum sampling rate for $y(t)$ is $f_{s\; min}=2f_m=1500Hz=1500$ samples/sec

Q4GATE 2016MCQ

Let $x_{1}(t)\leftrightarrow X_{1}(\omega ) \; and \; x_{2}(t) \leftrightarrow X_{2}(\omega \omega )$ be two signals whose Fourier Transforms are as shown in the figure below. In the figure, $h(t)=e^{-2|t|}$ denotes the impulse response. For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can be uniquely reconstructed from its samples, is

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📖 Explanation

Given that, Bandwidth of $X_1(\omega )=B_1$ Bandwidth of $X_2(\omega )=B_2$ system has $h(t)=e^{-2|t|}$ and input to the system is $x_1(t) \cdot x_2(t)$ The bandwidth of $x_1(t) \cdot x_2(t)$ is $B_1+B_2$ . The bandwidth of output will be $B_1+B_2$ . So, sampling rate will be $2(B_1+B_2)$ .

Q5GATE 2014MCQ

A sinusoid x(t) of unknown frequency is sampled by an impulse train of period 20 ms. The resulting sample train is next applied to an ideal lowpass filter with cutoff at 25 Hz. The filter output is seen to be a sinusoid of frequency 20 Hz. This means that x(t)

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📖 Explanation

Given, impulse train of period 20 ms. Then, sampling frequency $=\frac{1}{20 \times 10^{-3}}=50Hz$ If the input signal $x(t)= \cos \omega _m (t)$ having spectrum The filtered out sinusoidal signal has 20 Hz frequency, the sampling must be under sampling. The output signal which is an under sampled signal with sampling frequency 50 Hz is and $50-f_m=20Hz\;\Rightarrow \;f_m=30Hz$

Q6GATE 2014NAT

For the signal $f(t) = 3 \sin 8 \pi t + 6 \sin 12 \pi t + \sin 14 \pi t$ , the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____.

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📖 Explanation

\begin{aligned} f_{m_1} &=4Hz \\ f_{m_2} &=6Hz \\ f_{m_3} &=7Hz \end{aligned}

Then minimum sampling frequency satisfying the nyquist criterion is 7*2=14Hz.

Q7GATE 2013MCQ

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency in kHz which is not valid is

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📖 Explanation

\begin{aligned} (f_s)_{min}&=2f_m \\ (f_s)_{min}&=2 \times 5=10kHz \\ f_s&\geq 10kHz \end{aligned}

Q8GATE 2007MCQ

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be

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📖 Explanation

Given that, sampling interval =1 msec i.e. $T_s=1 \; msec=10^{-3}\; sec$ Therefore sampling frequency $f_s=\frac{1}{T_s}=\frac{1}{10^{-3}}=1 kHz$ after sampling new signal in frequency domain $U_T(f)=\frac{1}{T_s}\sum_{n=-\infty }^{\infty }U(f-nf_s)$ Therefore, spectrum of sampled signal will be

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