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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
F= 120 Hz and F= 60 Hz
 Now, present frequency component,
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)and other frequency component.
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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:
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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[𝑛]𝑛𝑜𝑟𝑥2[𝑛]x1[n] nor x2[n] have equal energies.
(d) Neither x1[n] nor x2[n] is a finite-energy signal
Ans:
(a)
Sol: Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)T= sampling time-period = 0.25 sec
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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))
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)(a) 1000 samples/s
(b) 1500 samples/s
(c) 2000 samples/s
(d)  3000 samples/s
Ans: 
(b)
Sol:
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)From the above block diagram,
Z(t) = x(t)cos1000πt
By using modulation property of fourier transform, 
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Thus, H(ω) is a low pass filter and it will pass frequency, component of Z(ω) upt 1500π rad/sec.
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Therefore, maximum frequency component of y(t) is  
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
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))
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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 B+ B2.
The bandwidth of output will be B+ B2.
So, sampling rate will be 2(B+ 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 Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
If the input signal x(t) = cosωm(t) having spectrum
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
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: Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)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: Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
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)
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)(a) Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)

(b) Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
(c) Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
(d) Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)
Ans: (b)
Sol: Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Given that, sampling interval = 1 msec
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Therefore sampling frequency
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)after sampling new signal in frequency domain
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)Therefore, spectrum of sampled signal will be
Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE)

The document Previous Year Questions- Sampling | Signals and Systems - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Signals and Systems.
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FAQs on Previous Year Questions- Sampling - Signals and Systems - Electrical Engineering (EE)

1. What is sampling in the context of Electrical Engineering?
Ans. Sampling in Electrical Engineering refers to the process of converting a continuous-time signal into a discrete-time signal by taking samples of the continuous signal at regular intervals.
2. Why is sampling important in signal processing?
Ans. Sampling is important in signal processing as it allows for easier manipulation, processing, and transmission of signals. It also helps in reducing the complexity of the signal processing algorithms.
3. What is the Nyquist-Shannon sampling theorem?
Ans. The Nyquist-Shannon sampling theorem states that in order to accurately reconstruct a signal from its samples, the sampling rate must be at least twice the highest frequency component present in the signal.
4. What are some common methods used for sampling signals in Electrical Engineering?
Ans. Some common methods used for sampling signals in Electrical Engineering include uniform sampling, non-uniform sampling, and random sampling.
5. How does aliasing affect the sampling process in Electrical Engineering?
Ans. Aliasing in the sampling process occurs when the sampling rate is too low, resulting in the inability to accurately reconstruct the original signal. This can lead to distortion and loss of information in the sampled signal.
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