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Test: Spectrum Analysis Of Sampling Process - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Spectrum Analysis Of Sampling Process

Test: Spectrum Analysis Of Sampling Process for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Spectrum Analysis Of Sampling Process questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Spectrum Analysis Of Sampling Process MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Spectrum Analysis Of Sampling Process below.
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Test: Spectrum Analysis Of Sampling Process - Question 1

Statement (I): Aliasing occurs when the sampling frequency is less than twice the maximum frequency in the signal.
Statement (II): Aliasing is a reversible process.

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 1

Explanation: Aliasing is an irreversible process. Once aliasing has occurred then signal can-not be recovered back.

Test: Spectrum Analysis Of Sampling Process - Question 2

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

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 2

Explanation: fs (min) =2fm
fs (min) =2*5 =10 KHz
So, fs >=1o KHz.

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Test: Spectrum Analysis Of Sampling Process - Question 3

Let x(t) be a continuous-time, real valued signal band-limited to F Hz. The Nyquist sampling rate in Hz, For y(t) =x(0.5t) +x(t)-x(2t) is

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 3

Explanation: Expansion in time domain in compression in frequency domain and vice-versa. So, the maximum frequency component in given signal is 2F Hz. And according to sampling theorem.
Nyquist rate =2fm =4F Hz.

Test: Spectrum Analysis Of Sampling Process - Question 4

Increased pulse-width in the flat-top sampling leads to:

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 4

Explanation: As pulse width is increased, the width of the first lobe of the spectrum is decreased. Hence, increased pulse-width in the flat-top sampling, leads to attenuation of high frequencies in reproduction.

Test: Spectrum Analysis Of Sampling Process - Question 5

A bandpass sampling extends from 4-6 kHz. What is the smallest sampling frequency required to retain all the information in the signal.

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 5

Explanation: fh =6 kHz
Bandwidth = 2 kHz
Fs =4 kHz.

Test: Spectrum Analysis Of Sampling Process - Question 6

A signal represented by x(t) =5cos 400πt is sampled at a rate 300 samples/sec. The resulting samples are passed through an ideal low pass filter of cut-off frequency 150 Hz. Which of the following will be contained in the output of the LPF?

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 6

Explanation: x (t) =5cos400πt
fm =200 Hz
The output of the LPF will contain frequencies which are less than fc =150 Hz.
So, fs-fm =300-200 =100 Hz is the only component present in the output of LPF.

Test: Spectrum Analysis Of Sampling Process - Question 7

A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t). The resulting signal is passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be :

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 7

Explanation: m (t) g (t)->M (f)*G (f)
After low pass filtering with fc =1 kHz, hence the output is zero.

Test: Spectrum Analysis Of Sampling Process - Question 8

An LTI system having transfer function s2+1/s2+2s+1 and input x(t) =sin(t+1) is in steady state. The output is sampled at ws rad/s to obtain the final output {y (k)}. Which of the following is true? 

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 8

Explanation: x (t) =sin (t+1)
w = 1 rad/s
X(s) = es/s2+1
Y(s) = es/s2+2s+1
Y (∞) =0.

Test: Spectrum Analysis Of Sampling Process - Question 9

A digital measuring instrument employs a sampling rate of 100 samples/second. The sampled input x(n) is averaged using the difference equation:
Y (n) =[x (n)+x (n-1)+x(n-2)+x(n-4)/4] For a step input, the maximum time taken for the output to reach the final value after the input transition is 

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 9

Explanation: Since output y depends on input, such as no delay, delay by 1 unit, and delay by 2 unit, delay by 4 unit, so it will sum all the samples after 4 Ts (maximum delay), to get one sample of y[n].
T =40 msec.

Test: Spectrum Analysis Of Sampling Process - Question 10

The sinusoid x(t) =6cos10πt is sampled at the rate of 15 Hz and applied to ideal rectangular LPF with cut-off frequency of 10 Hz, then the output of filter contains: 

Detailed Solution for Test: Spectrum Analysis Of Sampling Process - Question 10

Explanation: Output filter is the device that is used to filter some frequencies and make some frequencies in the response and in this case it contains 10π rad/sec.

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