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Test: Signal Classification - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test Digital Signal Processing - Test: Signal Classification

Test: Signal Classification for Electronics and Communication Engineering (ECE) 2024 is part of Digital Signal Processing preparation. The Test: Signal Classification questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Signal Classification MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Signal Classification below.
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Test: Signal Classification - Question 1

 Which of the following is done to convert a continuous time signal into discrete time signal?

Detailed Solution for Test: Signal Classification - Question 1

Explanation: A discrete time signal can be obtained from a continuous time signal by replacing t by nT, where T is the reciprocal of the sampling rate or time interval between the adjacent values. This procedure is known as sampling.

Test: Signal Classification - Question 2

 The deflection voltage of an oscilloscope is a ‘deterministic’ signal. True or False? 

Detailed Solution for Test: Signal Classification - Question 2

Explanation: The behavior of the signal is known and can be represented by a saw tooth wave form. So, the signal is deterministic.

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Test: Signal Classification - Question 3

The even part of a signal x(t) is:

Detailed Solution for Test: Signal Classification - Question 3

Explanation: Let x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By adding the above two equations, we get
xe(t)=(1/2)*(x(t)+x(-t)).

Test: Signal Classification - Question 4

 Which of the following is the odd component of the signal x(t)=e(jt)?

Detailed Solution for Test: Signal Classification - Question 4

Explanation: Let x(t)=e(jt)
Now, xo(t)=(1/2)*(x(t)-x(-t))
=(1/2)*(e(jt) – e(-jt))
=(1/2)*(cost+jsint-cost+jsint)
=(1/2)*(2jsint)
=j*sint.

Test: Signal Classification - Question 5

For a continuous time signal x(t) to be periodic with a period T, then x(t+mT) should be equal to:

Detailed Solution for Test: Signal Classification - Question 5

Explanation: If a signal x(t) is said to be periodic with period T, then x(t+mT)=x(t) for all t and any integer m.

Test: Signal Classification - Question 6

Let x1(t) and x2(t) be periodic signals with fundamental periods T1 and T2 respectively. Which of the following must be a rational number for x(t)=x1(t)+x2(t) to be periodic? 

Detailed Solution for Test: Signal Classification - Question 6

Explanation: Let T be the period of the signal x(t)
=>x(t+T)=x1(t+mT1)+x2(t+nT2)
Thus, we must have
mT1=nT2=T
=>(T1/T2)=(k/m)= a rational number.

Test: Signal Classification - Question 7

 Let x1(t) and x2(t) be periodic signals with fundamental periods T1 and T2 respectively. Then the fundamental period of x(t)=x1(t)+x2(t) is: 

Detailed Solution for Test: Signal Classification - Question 7

Explanation: For the sum of x1(t) and x2(t) to be periodic the ratio of their periods should be a rational number, then the fundamental period is the LCM of T1 and T2.

Test: Signal Classification - Question 8

All energy signals will have an average power of:

Detailed Solution for Test: Signal Classification - Question 8

Explanation: For any energy signal, the average power should be equal to 0 i.e., P=0.

Test: Signal Classification - Question 9

x(t) or x(n) is defined to be an energy signal, if and only if the total energy content of the signal is a:

Detailed Solution for Test: Signal Classification - Question 9

Explanation: The energy signal should have total energy value that lies between 0 and infinity.

Test: Signal Classification - Question 10

What is the period of cos2t+sin3t?

Detailed Solution for Test: Signal Classification - Question 10

Explanation: Period of cos2t=(2*pi)/2=pi
Period of sin3t=(2*pi)/3
LCM of pi and (2*pi)/3 is 2*pi.

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