# Sampling

 Question 1
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)
 A 95 B 80 C 110 D 115
GATE EE 2023   Signals and Systems
Question 1 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 $=80 \mathrm{~Hz}$.
 Question 2
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?
 A The energy of $x_{1}[n]$ is greater than the energy of $x_{2}[n]$ B The energy of $x_{1}[n]$ is greater than the energy of $x_{2}[n]$ C $x_{1}[n] nor x_{2}[n]$ have equal energies. D Neither $x_{1}[n] nor x_{2}[n]$ is a finite-energy signal
GATE EE 2018   Signals and Systems
Question 2 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 $x_2(n)$.

 Question 3
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 A 1000 samples/s B 1500 samples/s C 2000 samples/s D 3000 samples/s
GATE EE 2017-SET-2   Signals and Systems
Question 3 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
 Question 4
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
 A $2B_{1}$ B $2(B_{1}+B_{2})$ C $4(B_{1}+B_{2})$ D $\infty$
GATE EE 2016-SET-2   Signals and Systems
Question 4 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 $2(B_1+B_2)$.
 Question 5
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)
 A 10Hz B 60 Hz C 30 Hz D 90 Hz
GATE EE 2014-SET-3   Signals and Systems
Question 5 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 $50-f_m=20Hz\;\Rightarrow \;f_m=30Hz$

There are 5 questions to complete.