Previous Year Questions- Sampling

Q1: A signal x(t)= 2cos(180πt) cos(60πt) is sampled at 200 Hz and then passed through an ideal low pass filter having cut-off frequency of 100 Hz.
The maximum frequency present in the filtered signal in Hz is ____ (Round off to the nearest integer)      (2023)
(a) 95
(b) 80
(c) 110
(d) 115
Ans: (b)
Sol: Given :
x(t) = 2cos(180πt) cos(60πt)
= cos240πt + cos120πt
F₁ = 120 Hz and F₂ = 60 Hz
 Now, present frequency component,
and other frequency component.
Given cut-off frequency of LPF = 100 Hz
∴ It is pas only 60 Hz and 80 Hz frequency component.
∴ Max. frequency = 80 Hz.

Q2: Consider the two continuous-time signals defined below:
These signals are sampled with a sampling period of T = 0.25 seconds to obtain discretetime signals x₁[n] and x₂[n], respectively. Which one of the following statements is true?       (2018)
(a) The energy of x₁[n] is greater than the energy of x₂[n]
(b) The energy of  x₁[n] is greater than the energy of  x₂[n]
(c) $𝑥1\left[𝑛\right]𝑛𝑜𝑟𝑥2\left[𝑛\right]$x₁[n] nor x₂[n] have equal energies.
(d) Neither x₁[n] nor x₂[n] is a finite-energy signal
Ans: (a)
Sol: T_s = sampling time-period = 0.25 sec
Since, x₁(n) is having one more non-zero sample of amplitude '1' as compared to x₂(n). Therefore, energy of  x₁(n) greater than energy of x₂(n).

Q3: 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     (SET-2 (2017))
(a) 1000 samples/s
(b) 1500 samples/s
(c) 2000 samples/s
(d)  3000 samples/s
Ans: (b)
Sol:
From the above block diagram,
Z(t) = x(t)cos1000πt
By using modulation property of fourier transform, 
Thus, H(ω) is a low pass filter and it will pass frequency, component of Z(ω) upt 1500π rad/sec.
Therefore, maximum frequency component of y(t) is  

Q4: Let x₁(t) ↔ X₁(ω) and x₂(t) ↔ X₂(ωω) 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.        (SET-2 (2016))
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              (SET-2 (2016))
(a) 2B₁
(b) 2(B₁ + B₂)
(c) $4(𝐵1+𝐵2)$4(B₁ + B₂)
(d) ∞
Ans: (b)
Sol: Given that,
Bandwidth of X₁(ω) = B₁
Bandwidth of X₂(ω) = B₂
system has h(t) = e^-2∣t∣ and input to the system is x₁(t)⋅x₂(t)
The bandwidth of x₁(t)⋅x₂(t) is B₁ + B₂.
The bandwidth of output will be B₁ + B₂.
So, sampling rate will be 2(B₁ + B₂).

Q5: 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)      (SET-3 (2014))
(a) 10Hz
(b) 60 Hz
(c) 30 Hz
(d) 90 Hz
Ans: (c)
Sol: Given, impulse train of period 20 ms.
Then, sampling frequency
If the input signal x(t) = cosω_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

Q6: For the signal f(t) = 3sin8πt + 6sin12πt + sin14πt, the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____.         (SET-3 (2014))
(a) 7
(b) 14
(c) 18
(d) 9
Ans: (b)
Sol: Then minimum sampling frequency satisfying the nyquist criterion is 7 * 2 = 14Hz.

Q7: 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       (2013)
(a) 5 kHz
(b) 12 kHz
(c) 15 kHz
(d) 20 kHz
Ans: (a)
Sol: 
Q8: 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       (2007)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Given that, sampling interval = 1 msec
Therefore sampling frequency
after sampling new signal in frequency domain
Therefore, spectrum of sampled signal will be