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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Discrete Time Systems - Electrical Engineering (EE) MCQ

Test: Discrete Time Systems - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Discrete Time Systems

Test: Discrete Time Systems for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Discrete Time Systems questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Discrete Time Systems MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Discrete Time Systems below.
Solutions of Test: Discrete Time Systems questions in English are available as part of our course for Electrical Engineering (EE) & Test: Discrete Time Systems solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Discrete Time Systems | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Discrete Time Systems - Question 1
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The output signal when a signal x(n)=(0,1,2,3) is processed through an ‘Identical’ system is:

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Explanation: An identical system is a system whose output is same as the input, that is it does not perform any operation on the input and transmits it.

Test: Discrete Time Systems - Question 2
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If a signal x(n) is passed through a system to get an output signal of y(n)=x(n+1), then the signal is said to be:

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Explanation: For example, the value of the output at the time n=0 is y(0)=x(1), that is the system is advanced by one unit.

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Test: Discrete Time Systems - Question 3
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 If the output of the system is with an input of x(n) then the system will work as:

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Explanation: From the equation given, y(n)=x(n)+x(n-1)+x(n-2)+…. .This system calculates the running sum of all the past input values till the present time. So, it acts as an accumulator.

Test: Discrete Time Systems - Question 4
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The block denoted as follows is known as:
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Explanation: If the function to this block is x(n) then the output from the block will be x(n-1). So, the block is called as delay block or delay element.

Test: Discrete Time Systems - Question 5
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 The output signal when a signal x(n)=(0,1,2,3) is processed through an ‘Delay’ system is:
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Explanation: An delay system is a system whose output is same as the input, but after a delay.

Test: Discrete Time Systems - Question 6
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 The system described by the input-output equation y(n)=nx(n)+bx3(n) is a: 
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Explanation: Since the output of the system y(n) depends only on the present value of the input x(n) but not on the past or the future values of the input, the system is called as static or memory-less system.

Test: Discrete Time Systems - Question 7
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Whether the system described by the input-output equations y(n)=x(n)-x(n-1) a Time-variant system? 
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Explanation: If the input is delayed by k units then the output will be y(n,k)=x(n-k)-x(n-k-1)
If the output is delayed by k units then y(n-k)=x(n-k)-x((n-k)-1)
=>y(n,k)=y(n-k). Hence the system is time-invariant.

Test: Discrete Time Systems - Question 8
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 The system described by the input-output equations y(n)=x2(n) is a Non-linear system.
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Explanation: Given equation is y(n)=x2(n)
Let y1(n)=x12(n) and y2(n)=x22(n)
y3(n)=y1(n)+y2(n)= x12(n)+ x22(n)≠(x1(n)+x2(n))2
So the system is non-linear.

Test: Discrete Time Systems - Question 9
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If the output of the system of the system at any ‘n’ depends only the present or the past values of the inputs then the system is said to be:
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Explanation: A system is said to be causal if the output of the system is defined as the function shown below
y(n)=F[x(n),x(n-1),x(n-2),…] So, according to the conditions given in the question, the system is a causal system.

Test: Discrete Time Systems - Question 10
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 If a system do not have a bounded output for bounded input, then the system is said to be:
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Explanation: An arbitrary relaxed system is said to be BIBO stable if it has a bounded output for every value in the bounded input. So, the system given in the question is a Non-stable system.

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