Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Signals and Systems  >  Test: Properties of LTI Systems - Electrical Engineering (EE) MCQ

Test: Properties of LTI Systems - Electrical Engineering (EE) MCQ


Test Description

20 Questions MCQ Test Signals and Systems - Test: Properties of LTI Systems

Test: Properties of LTI Systems for Electrical Engineering (EE) 2024 is part of Signals and Systems preparation. The Test: Properties of LTI Systems questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Properties of LTI Systems MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Properties of LTI Systems below.
Solutions of Test: Properties of LTI Systems questions in English are available as part of our Signals and Systems for Electrical Engineering (EE) & Test: Properties of LTI Systems solutions in Hindi for Signals and Systems course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Properties of LTI Systems | 20 questions in 15 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Signals and Systems for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Properties of LTI Systems - Question 1

 What is the rule h*(x+y) = (y+x)*h called?

Detailed Solution for Test: Properties of LTI Systems - Question 1

By definition, the commutative rule h*x=x*h.

Test: Properties of LTI Systems - Question 2

Does the system h(t) = exp([-1-2j]t) correspond to a stable system?

Detailed Solution for Test: Properties of LTI Systems - Question 2

The system corresponds to an oscillatory system, this resolving to a marginally stable system.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Properties of LTI Systems - Question 3

 Is y[n] = n*sin(n*pi/4)u[-n] a stable system?

Detailed Solution for Test: Properties of LTI Systems - Question 3

The ‘n’ term in the y[n] will dominate as it reaches to negative infinity, and hence could reach infinite values. Eventhough + infinity would not be a problem, still the resultant system would be unstable.

Test: Properties of LTI Systems - Question 4

For an LTI discrete system to be stable, the square sum of the impulse response should be

Detailed Solution for Test: Properties of LTI Systems - Question 4

If the square sum is infinite, the system is an unstable system. If it is zero, it means h(t) = 0 for all t. However, this cannot be possible. Thus, it has to be finite.

Test: Properties of LTI Systems - Question 5

What is the rule h*x = x*h called?

Detailed Solution for Test: Properties of LTI Systems - Question 5

By definition, the commutative rule h*x=x*h.

Test: Properties of LTI Systems - Question 6

 Does the system h(t) = exp(-7t) correspond to a stable system?

Detailed Solution for Test: Properties of LTI Systems - Question 6

The system corresponds to a stable system, as the Re(exp) term is negative, and hence will die down as t tends to infinity.

More precisely, if limx→cf(x)=∞limx→cf(x)=∞, then ff is upper unbounded; if limx→cf(x)=−∞limx→cf(x)=−∞, then ff is lower unbounded. (The limit in the previous statements can also be one-sided.)

One could be more precise and say that ff must be (upper/lower) unbounded in every punctured neighborhood of cc (or punctured right/left neighborhood of cc).

This essentially follows from the definition.

However, this condition is by no means sufficient. Consider

f(x)=1xsin1xf(x)=1xsin⁡1x

Then this function is unbounded in every punctured neighborhood of 00, but its limit is neither ∞∞ nor −∞−∞.

Restricting to the limit from the right or the left doesn't improve the situation.

Confusing “unbounded” with “has infinite limit” may be very dangerous.

Share

Test: Properties of LTI Systems - Question 7

What is the following expression equal to: h*(d+bd), d(t) is the delta function

Detailed Solution for Test: Properties of LTI Systems - Question 7

Apply commutative and associative rules and the convolution formula for a delta function

Test: Properties of LTI Systems - Question 8

 Is y[n] = n*sin(n*pi/4)u[-n] a causal system?

Detailed Solution for Test: Properties of LTI Systems - Question 8

The anti causal u[-n] term makes the system non causal.

Test: Properties of LTI Systems - Question 9

The system transfer function and the input if exchanged will still give the same response.

Detailed Solution for Test: Properties of LTI Systems - Question 9

By definition, the commutative rule i h*x=x*h=y. Thus, the response will be the same.

Test: Properties of LTI Systems - Question 10

In a discrete Linear Shift Invariant system the input sequence length is L, the impulse response sequence length is M, the output sequence length N is

Detailed Solution for Test: Properties of LTI Systems - Question 10

If the two discrete signals are having the length ‘L’ and ‘M’ respectively then the resultant output signal has the length as,

N = L + M – 1

The convolution of signals in one domain is equivalent to the multiplication of signals in another domain.

Test: Properties of LTI Systems - Question 11

Is y[n] = nu[n] a linear system?

Detailed Solution for Test: Properties of LTI Systems - Question 11

​​​​​

 

Test: Properties of LTI Systems - Question 12

A continuous-time system that is initially at rest is described by

where x(t) is the input voltage and y(t) is the output voltage. The impulse response of the system is

Detailed Solution for Test: Properties of LTI Systems - Question 12

Given:

Taking Laplace transform on both sides, we get
sY(s) + 3Y(s) = 2X(s)
⇒ (s + 3)Y(s) = 2X(s)

∴ Impulse response will be, h(t) = L-1(H(s)) = 2e-3tu(t)
Hence, the correct option is (C).

*Multiple options can be correct
Test: Properties of LTI Systems - Question 13

The outputs of four systems (S1, S2, S3, and S4) corresponding to the input signal sin(t), for all time t, are shown in the figure.
Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?

Detailed Solution for Test: Properties of LTI Systems - Question 13

Concept:

For an LTI system, the following results hold true:

  • The same form of the sinusoidal signal is produced at the output.
  • The only change will be in magnitude and phase.
  • ∴ If a system produces frequencies in the output other than those which are in input, then that system can’t be called as the ‘Linear shift-invariant’ system.

Analysis:

Since, LTI system does not change the frequency of sinusoidal input,
So S3 and S4 are definitely not LTI as input and output sinusoidal frequencies are different.

Test: Properties of LTI Systems - Question 14

Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system? 

Detailed Solution for Test: Properties of LTI Systems - Question 14

Concept:

Continuous-time signal: A signal of continuous amplitude and time is known as a continuous-time signal or an analog signal.

Discrete-time signal: Unlike a continuous-time signal, a discrete-time signal is not a function of a continuous argument; however, it may have been obtained by sampling from a continuous-time signal. When a discrete-time signal is obtained by sampling a sequence at uniformly spaced times, it has an associated sampling rate.

Analysis:

We will check each option to find out the incorrect one.

All the poles of the system must lie on the left side of the jω axis.

  • The system is stable if and only if all the roots lie on the left side of the S plane or jω axis. Also, we can say all the poles have a negative real part.

Example:
If ROC is Re[s]>-a, the system h(t) =e-at u(t) is causal.
Its Laplace transform 
 

f a<0, i.e., imaginary axis Re[s] =0 can be included in the ROC, the system is stable.

Zeros of the system can lie anywhere on the plane.

When zeros are in the right half of the S plane or outside the unit circle in the Z- plane, it does not cause the system to be unstable.

All the poles of the system must lie within |s|=1

  • The system is stable if and only if the ROC of the system function H(z) includes the unit circle |z|=1 and not |s|=1.
  • The system is stable in the case of rational system function H(z) if and only if all the poles of H(z) lie inside the unit circle, they must all have a magnitude smaller than unity.

All the roots of the characteristics equation must lie on the left side of the jω axis.

For stability, all the roots of the characteristics equation must lie on the left side of the s-plane, even if one root lies on the right side the system becomes unstable.

Conclusion:

So, we can say options a, b and d are satisfying property of a continuous-time casual and stable system.

Test: Properties of LTI Systems - Question 15

What is the convolution integral c(t) for a system with input x(t) and impulse response h(t), where x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2) ?

Detailed Solution for Test: Properties of LTI Systems - Question 15

Concept:

By using the impulse response of a system, convolution can be used to calculate a system's zero state response (i.e., its response when it has zero initial conditions) to an arbitrary input.  Linearity and superposition are used. 

The convolution can be defined as:

u(t)*u(t) = r(t)

Calculation:
Given signals are x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2) 
x(t) and h(t) is graphically represented by:

c(t) = x(t) * h(t)

c(t) = [u(t -1) - u(t - 3)]*[u(t) - u(t -2)]

c(t) = u(t - 1)*u(t) - u(t -1)*u(t - 2) - u(t - 3)*u(t) + u(t - 3)*u(t - 2)

c(t) = r(t - 1) - r(t - 3) - r(t - 3) + r(t + 5)

c(t) = r(t - 1) -2r(t - 3) + r(t + 5)

The output signal is represented as:


NOTES:

If the two analog signals are convolved

  1. The resultant of two signals will have a width equal to the sum of the individual width of two signals being convolved.
  2. The lower extent of the resultant signal will be equal to the sum of individual lower extents l = l1 + l2, similarly, the upper extent is the sum of individual upper extents u = u1 + u2 
  3. The area of resultant convolution is equal to the product of the area of the signals being convolved.
Test: Properties of LTI Systems - Question 16

Consider a single input single output discrete-time system with x[n] as input and y[n] as output. Where the two are related as

Which one of the following statements is true about the system?

Detailed Solution for Test: Properties of LTI Systems - Question 16

Concept:

For a system to be causal, the Present output should depend on the present or past input only.

For a system to be Stable, a Bounded input should produce a Bounded Output.

Analysis:

For 0 < n 10,

y(n) = n|x(n)|

Let x(n) is bounded, i.e. for -∞ ≤ n ≤ ∞, x(n) ≤ M, where M is finite.

So, for -∞ ≤ n ≤ ∞, |n.x(n)| will also approach a finite value. Hence in this interval y(n) is bounded. i.e. the system is stable.

Since the output y(n) is depending on the present value of the input only, the system in this interval is also causal.

For n < 0 and n > 10:

y(n) = x(n) - x(n-1).

Let input x(n) is bounded. ∴ y(n) = x(n) - x(n-1) will also be bounded, i.e. it will go to a finite value.

Hence in this interval, the system is said to be stable.

Also, because the output y(n) depends on the present and the past input values only, the system is causal as well in this interval.

∴ After considering both the intervals we can conclude that the system is both stable and causal.

Test: Properties of LTI Systems - Question 17

The signal y(t) = T{x(t)} = sin(2πt)x(t) + u(t - 2) is _______.

Detailed Solution for Test: Properties of LTI Systems - Question 17

Concept

Linearity

It is the combination of additivity and homogeneity.

Another method is by observation we can say that the given signal is linear or not.

x1(t) → y1(t)

x2(t) → y2(t)

x1(t)+x2(t) → y1(t)+y2(t)

α× x(t) → α × y(t)

Causal

A system is said to be causal if the present output depends on the present input and past values of the input but not on the future values.

x(t) = 0 for t < 0

Time invariant

A system is time-invariant if input and output characteristics do not change with time, time-varying nature is caused due to internal components.

if x(t) → y(t) then x(t - t0) → y(t - t0)

Calculation:

The given signal is multiplied with the sine function. So, the function is non-linear by observation.

Shift the signal by t0 

For Time invariance

y(t) = sin(2πt - t0)*x(t - t0) + u(t - 2 - t0)

Replace the 't' by t - t0 in the output signal

y(t) = sin(2π(t - t0))*x(t - t0) + u(t - 2 - t0)

Both the equations are the not same. So, the system given is Time-variant.

For Causality

y(t) = T{x(t)} = sin(2πt)⋆x(t) + u(t - 2)

y(1) = T{x(1)} = sin(2π×1)*x(1) + u(-1)

As the signal contains only present and past values of the input, hence the given signal is causal.

Test: Properties of LTI Systems - Question 18

Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system

Detailed Solution for Test: Properties of LTI Systems - Question 18

Causal and stable system:

  • In a causal system, the ROC of the system transfer function should be in the right half-plane and to the right of the rightmost pole.
  • In a stable LTI system, all poles of the system should be in the left half s plane and no repeated pole on the imaginary axis.In (c) option, |S| = 1, have one pole in the right-hand plane also, so it does not satisfy stable system criteria.
  • For a stable system, All the roots of the characteristic equation must be located on the left side of the jω axis
  • The system is stable if and only if the ROC of the system function H(z) includes the unit circle |z|=1 and > not |s|=1.
  • The system is stable in the case of rational system function H(z) if and only if all the poles of H(z) lie inside the unit circle, they must all have a magnitude smaller than unity.
  • All the roots of the characteristics equation must lie on the left side of the jω axis.
  • For stability, all the roots of the characteristics equation must lie on the left side of the s-plane, even if one root lies on the right side the system becomes unstable.

Conclusion:

So, we can say options a, b and d are satisfying property of a continuous-time casual and stable system.

Option c is correct for a discrete-time casual and stable LTI system.

Test: Properties of LTI Systems - Question 19

Which of the following statements about the time-invariant and time-varying control systems is INCORRECT?

Detailed Solution for Test: Properties of LTI Systems - Question 19

System

Group of components arranged in such a way that it gives the proper output. The proper output may or may not be desired output.

Eg: Fan without a regulator

Control system

It is a group of physical components arranged in such a way that it gives the desired output.

Eg: Fan with a regulator

Time invariant system

A system is called time-invariant if the shift in the input signal causes the same shift in the output signal.

Eg: y(t) = x(t+3)

  • Parameters do not change with respect to time.
  • Characteristics do not change with time.
  • If the system is linear then it is called a Linear Time-Invariant system and it can be expressed with the constant coefficient differential equation.

Time variant system

A system is said to be time-variant if the shift in the input signal does not reflect at the output, that is the different amount of shifts.

Eg: y(t) = t × x(t)

  • Parameters changes with the time. 
  • Characteristics change with time.
  • These systems are not stationary with the time during the operation of the system.
  • These systems also can be represented with the constant coefficients.
Test: Properties of LTI Systems - Question 20

What is the convolution integral c(t) for a system with input x(t) and impulse response h(t), where x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2)?

Detailed Solution for Test: Properties of LTI Systems - Question 20

Concept:

By using the impulse response of a system, convolution can be used to calculate a system's zero state response (i.e., its response when it has zero initial conditions) to an arbitrary input. Linearity and superposition are used. 
The convolution can be defined as:

Calculation:

Given signals are x(t) = u(t - 1) - u(t - 3) and h(t) = u(t) - u(t - 2) 

x(t) and h(t) is graphically represented by:

c(t) = x(t) * h(t)

c(t) = [u(t -1) - u(t - 3)]*[u(t) - u(t -2)]

c(t) = u(t - 1)*u(t) - u(t -1)*u(t - 2) - u(t - 3)*u(t) + u(t - 3)*u(t - 2)

c(t) = r(t - 1) - r(t - 3) - r(t - 3) + r(t + 5)

c(t) = r(t - 1) -2r(t - 3) + r(t + 5)

The output signal is represented as:

41 videos|52 docs|33 tests
Information about Test: Properties of LTI Systems Page
In this test you can find the Exam questions for Test: Properties of LTI Systems solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Properties of LTI Systems, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

41 videos|52 docs|33 tests
Download as PDF

Top Courses for Electrical Engineering (EE)