# Sampling

 Question 1
Let $x_{1}(t)$ and $x_{2}(t)$ be two band-limited signals having bandwidth $B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{s} \mathrm{each}$. In the figure below, the Nyquist sampling frequency, in rad/s, required to sample $y(t)$, is A $20 \pi \times 10^{3}$ B $40 \pi \times 10^{3}$ C $8 \pi \times 10^{3}$ D $32 \pi \times 10^{3}$
GATE EC 2023   Signals and Systems
Question 1 Explanation:
Given that, $x_{1}(t)$ and $x_{2}(t)$ are two bandlimited signals having bandwidth $B=4 \pi \times 10^{3} \mathrm{rad} / \mathrm{sec}$. and $y(t)=x_{2}(t) \cos \left(12 \pi \times 10^{3} t\right)+x_{1}(t) \cos \left(4 \pi \times 10^{3} t\right)$ So, $\text{Nyquist Rate }=2\omega _{max}=2[16 \pi \times 10^3]=32 \pi \times 10^3\; rad/sec$
 Question 2
Consider a real-valued base-band signal x(t), band limited to $\text{10 kHz}$. The Nyquist rate for the signal $y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right )$ is
 A $\text{15 kHz}$ B $\text{30 kHz}$ C $\text{60 kHz}$ D $\text{20 kHz}$
GATE EC 2021   Signals and Systems
Question 2 Explanation:   $\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}$

 Question 3
A band limited low-pass signal x(t) of bandwidth 5 kHz is sampled at a sampling rate $f_{s}$. The signal x(t) is reconstructed using the reconstruction filter H(f) whose magnitude response is shown below: The minimum sampling rate $f_{s}$ (in kHz) for perfect reconstruction of x(t) is _______.
 A 10 B 11 C 12 D 13
GATE EC 2018   Signals and Systems
Question 3 Explanation:
Let us assume an arbitrary spectrum for x(t) as shown below: The spectrum of the sampled signal can be given as, For proper reconstruction of the signal,
\begin{aligned} f_{s}-5 & \geq 8 \\ f_{s} &\geq 8+5=13 \mathrm{kHz} \\ \text { So. } \quad f_{s(\mathrm{min})} &=13 \mathrm{kHz} \end{aligned}
 Question 4
The signal $x(t)=sin(14000\pi t)$, where t is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows :
$H(f)=\left\{\begin{matrix} 1, & |f|\leq 12 kHz\\ 0, & |f|\geq 12 kHz \end{matrix}\right.$
What is the number of sinusoids in the output and their frequencies in kHz?
 A Number = 1, frequency = 7 B Number = 3, frequencies= 2,7,11 C Number = 2, frequencies = 2, 7 D Number = 2, frequencies = 2, 11
GATE EC 2017-SET-2   Signals and Systems
Question 4 Explanation:
\begin{aligned} x(t) &=\sin (14000 \pi t) ; f_{m}=7 \mathrm{kHz} \\ f_{s} &=9000 \text { samples per second } \\ &=9 \mathrm{kHz} \end{aligned}
The spectrum of the sampled signal can be given as shown below: So, three sinusoids will be there at the output of the LPF and the frequencies of those sinusoids are 2 kHz, 7 kHz and 11 kHz.
 Question 5
Consider the signal $x(t)=cos(6\pi t)+sin(8\pi t)$, where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t)=x(2t+ 5) is
 A 8 B 12 C 16 D 32
GATE EC 2016-SET-3   Signals and Systems
Question 5 Explanation:
\begin{aligned} X(t) & =\cos (6 \pi t)+\sin (8 \pi t) \\ Y(t) & =x(2 t+5) \\ Y(t) & =\cos [6 \pi(2 t+5)+\sin (8 \pi(2 t+5)] \\ & =\cos (12 \pi t+30 \pi)+\sin (16 \pi t+40 \pi) \\ f_{m_{1}} & =6 \mathrm{Hz}_{i} \qquad f_{m 2}=8 \mathrm{Hz}\\ \text{Nyquist }&\text{sampling rate},\\ f_{s} &=2 f_{\max } \\ &=16 \text { samples/second } \end{aligned}

There are 5 questions to complete.