Test: Signals & Systems - 2 - Electronics and Communication Engineering (ECE) MCQ

# Test: Signals & Systems - 2 - Electronics and Communication Engineering (ECE) MCQ

Test Description

## 20 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Test: Signals & Systems - 2

Test: Signals & Systems - 2 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Test: Signals & Systems - 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Signals & Systems - 2 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: Signals & Systems - 2 below.
Solutions of Test: Signals & Systems - 2 questions in English are available as part of our GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) & Test: Signals & Systems - 2 solutions in Hindi for GATE ECE (Electronics) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Signals & Systems - 2 | 20 questions in 60 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: Signals & Systems - 2 - Question 1

### For a given periodic sequence x (n) = {1, 1, 0, 0} with period N = 4, the Fourier coefficient is denoted by ck. The Value of c3* is:

Detailed Solution for Test: Signals & Systems - 2 - Question 1

The Fourier Series Coefficient of a discrete time Periodic sequence is given by:

Where N = Period of the sequence.

Calculation:

Given, x (n) = {1, 1, 0, 0} and N = 4

We are asked to calculate C3*

Putting k = 3.

Test: Signals & Systems - 2 - Question 2

### Consider the following difference equation: y[n] + 3y[n – 1] + 2y[n – 2] = 2x[n] – x[n – 1] If y[-1] = 0, y[-2] = 1, x[n] = u[n], then y[n] can be represented as

Detailed Solution for Test: Signals & Systems - 2 - Question 2

y[n] + 3y[n – 1] + 2y[n – 2] = 2x[n] – x[n – 1]
By applying Z-transform,

Y[z] + 3(z-1Y(z) + y[-1]) + 2(z-2Y(z) + y[-2] + z-1y[-1]) = 2X(z) – z-1X(z) – x[-1]

By applying partial fractions,

Test: Signals & Systems - 2 - Question 3

### The Fourier series coefficients of a periodic signal x(t), with fundamental frequency  and the rest are zero. Suppose x(t) is the input to a band-pass filter with the following magnitude and phase responses. Let y(t) be the output of the filter and the Fourier series of x(t) in the trigonometric form x(t) and the steady state response of y(t) are

Detailed Solution for Test: Signals & Systems - 2 - Question 3

The steady state response of the output is

Test: Signals & Systems - 2 - Question 4

The discrete Fourier series representation for the following sequence:

Detailed Solution for Test: Signals & Systems - 2 - Question 4

The Fourier series representation of the discrete-time periodic sequence is given by:

ak = Fourier series coefficient periodic by N.

ω0 = Fundamental frequency.

Application:

Now, we can rewrite the given sequence as

Test: Signals & Systems - 2 - Question 5

For a finite support signal x[n] = r[n] (u[n] – u[n – 11]) where r[n] is the discrete-time ramp function, the energy of x[n] is

Detailed Solution for Test: Signals & Systems - 2 - Question 5

Given signal is, x[n] = r[n] (u[n] – u[n – 11])
Energy of x[n] is

Test: Signals & Systems - 2 - Question 6

The percentage of the total energy dissipated by a 1Ω resistor in the frequency band 0 < ω < 10 rad/s when the voltage across it is v(t) = e-2t u(t) is _________

Detailed Solution for Test: Signals & Systems - 2 - Question 6

Given that f(t) = v(t) = e-2t u(t)

The total energy dissipated by the resistor is

The energy in the frequency band 0 < ω < 10 is,

The percentage of the total energy is

Test: Signals & Systems - 2 - Question 7

The z-transform of a sequence x[n] is given as x(z) = 3 + 4z – 6z-1 + 2z-2

If y[n] is the first difference of x[n], then y(z) is given by

Detailed Solution for Test: Signals & Systems - 2 - Question 7

y [n] is the first difference of x [n]
y(n) = x(n) – x(n - 1)

By applying z-transform.

⇒ Y(z) = X(z) – z-1 X(z)

⇒ Y(z) = 3 + 4z – 6z-1 + 2z-2 – z-1 (3 + 4z – 6z-1 + 2z-2)

= 3 + 4z – 6z-1 + 2z-2 – 3z-1 – 4 + 6z-2 – 2z-3

= -1 + 4z – 9z-1 + 8z-2 – 2z-3

Test: Signals & Systems - 2 - Question 8

Which of the following statements is false.

Detailed Solution for Test: Signals & Systems - 2 - Question 8

1. X(s) converges to  with ROC: Re (s) > 0.
For limit  tends to zero for σ = Re(s) > 0 as the exponent term dominates and its integral from 0 to +∞ converges.

Therefore, the given statement is false.

2.  does not tend to zero as t → +∞, which means the integral from 0 to +∞ is unbounded for all s. Thus the Laplace transform cannot converge for any value of s.

Therefore, the given statement is true.

3. For  does not tend to zero for t → -∞.

For Re(s) < 0, it does not converge for t → +∞.

Hence for no value of s does the Laplace transform converge.

Therefore, the given statement is true.

4. |t| = t u(t) – t u(-t).

For t u(t), the ROC is Re(s) > 0

For t u(-t), the ROC is Re(s) < 0

Hence, there is no value of s common to both the ROCs.Therefore, the given statement is true.

Test: Signals & Systems - 2 - Question 9

The DTFT of a signal f(n) = {a, b, c, d} is F(ω). The inverse DTFT of F(ω - π) is:

Detailed Solution for Test: Signals & Systems - 2 - Question 9

If the DTFT of a signal f(n) ↔ F(ω), then from the frequency shifting property

Calculation:

Given, f(n) = {a, b, c, d}

If f(n) ↔ F(ω)

Test: Signals & Systems - 2 - Question 10

The value of

Detailed Solution for Test: Signals & Systems - 2 - Question 10

Concept:
z-transform of a discrete signal can be written as;

Calculation:

We have to calculate:

On comparing equation (1) & (2) we find that;

Where X(z) is z-transform of nu[n] which is:

Hence our required summation is

Test: Signals & Systems - 2 - Question 11

What is the signal corresponding to the following z-transform? (Where u[n] is the unit-step signal)

Detailed Solution for Test: Signals & Systems - 2 - Question 11

Concept:
The Z-transform of standard signals with their ROC are:

Application:

Test: Signals & Systems - 2 - Question 12

Let   If y(t) = x(t - 2) then what is value of Y(e) at ω = π is ________

Detailed Solution for Test: Signals & Systems - 2 - Question 12

By applying Fourier transform,

Test: Signals & Systems - 2 - Question 13

The DTFT of x[n] = 0.5n u[n] using geometric series is:

Detailed Solution for Test: Signals & Systems - 2 - Question 13

The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by:

For a signal x(n) = an u(n), the DTFT will be:

Since u[n] is zero for n < 0 and a constant 1 for n > 0, the above summation becomes:

The geometric series states that:

Equation (1) can now be written as:

Given x[n] = 0.5n u[n], i.e. a = 0.5

The DTFT of the above given sequence using Equation (2) will be:

Test: Signals & Systems - 2 - Question 14

A discrete time signal x[n] and its discrete Fourier transform X(k) related by  Another discrete series with Fourier coefficients Y(k) given as Y(k) = X(k - 4) + X(k + 4) then the corresponding sequence y[n] will be

Detailed Solution for Test: Signals & Systems - 2 - Question 14

Test: Signals & Systems - 2 - Question 15

Consider a signal y(t) = x1(t) * x2(t). If x1(t) = 2 cos 4t u(t) and x2(t) = sin 2t u(t). The value of y(t) at t = π / 2 is _______

Detailed Solution for Test: Signals & Systems - 2 - Question 15

y(t) = x1(t) * x2(t)
x1(t) = (2 cos 4t) u(t)
x2(t) = (sin 2t) u(t)
We know convolution in time domain is multiplication in frequency domain
So, y(t) = x1(t) * x2(t)
y(t) = x1(s) ⋅ x2(s)

Put s = 0 ⇒ 0 = 4B + 16D → 1)
Put s = 1 ⇒ 4 = (A + B) (5) + (C + D) (17) → 2)
Put s = 2 ⇒ 8 = (2A + B) (8) + (2C + D) (20) → 3)
Put s = 3 ⇒ 12 = (3A + B) (13) + (3C + D) (25) → 4)
Solving 1), 2), 3) and 4) we get
A = B = 1 / 3
C = D = -1 / 3

Applying inverse Laplace transform,

Test: Signals & Systems - 2 - Question 16

The total area under the function

Detailed Solution for Test: Signals & Systems - 2 - Question 16

Test: Signals & Systems - 2 - Question 17

The sequence x[n] is (0.7)n u(n) and y (n) is x (n)*x (n). Then which of the following statement(s) is/are true?

Detailed Solution for Test: Signals & Systems - 2 - Question 17

y(n) = x(n) * x(n)
By taking z-transform both sides

Y(z) = X(z) ⋅ X(z)

x(n) = (0.7)n u(n)

We know that;

Test: Signals & Systems - 2 - Question 18

Consider a system with frequency response of  and suppose that the input to the system is x(t) = e-t u(t). If the output response is y(t), the value of y(1) is _______ (answer up to two decimal places)

Detailed Solution for Test: Signals & Systems - 2 - Question 18

Frequency response of

Input to the system is x(t) = e-t u(t)

The output in frequency domain is given by

By applying partial fraction method,

By applying the inverse Fourier transform,

Test: Signals & Systems - 2 - Question 19

Let  denote the Fourier transform of the signal x[n], where:

The value of at n = 0 will be ________. (The arrow represents the origin)

Detailed Solution for Test: Signals & Systems - 2 - Question 19

Concept:
The DTFT of a discrete sequence is a complex quantity, which can be defined as:

X(e) = Real {X(e)} + Img. {X(e)}

Where the Real {X(e)} is nothing but the DTFT of the even part of the signal.
Proof:
A signal can be written as the sum of the even part and odd part, is:

Taking the Fourier transform of the even part, we get:

Application:

We are required to find F-1{Re{X(e)}} at n = 0

The inverse DTFT of Re{X(e)} is xe(n)

Given:

Test: Signals & Systems - 2 - Question 20

Which of the following systems are linear time-invariant and causal systems.

Detailed Solution for Test: Signals & Systems - 2 - Question 20

This is an ordinary differential equation with constant co-efficient, therefore it is linear and time invariant. It is casual.

2. y[n] + 2y [n - 1] = x [n + 1]

This is a difference equation with the constant co-efficient. Therefore, it is linear and time-invariant. It is non-casual since the output depends on future values of x.

3. y[n + 1 ] + 4y[n] = 3x [n + 1] – x[n]

Rewriting the above equation by shifting one unit.

Y[n] + 4y [n - 1] = 3x [n] – x[n - 1]

This is a difference equation with constant co-efficient. So, it is linear and time invariant. The output does not depend on future values of the input, so it is causal.

The co-efficient of y means that this is non-linear and it does not depend on t. so, it is time invariant. It is causal.

## GATE ECE (Electronics) Mock Test Series 2025

24 docs|263 tests
Information about Test: Signals & Systems - 2 Page
In this test you can find the Exam questions for Test: Signals & Systems - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Signals & Systems - 2, EduRev gives you an ample number of Online tests for practice

### Up next

 Test | 25 ques
 Test | 15 ques
 Test | 20 ques
 Test | 15 ques
 Test | 20 ques

## GATE ECE (Electronics) Mock Test Series 2025

24 docs|263 tests

### Up next

 Test | 25 ques
 Test | 15 ques
 Test | 20 ques
 Test | 15 ques
 Test | 20 ques