Linear Time Invariant Systems MCQ Quiz - Objective Question with Answer for Linear Time Invariant Systems - Download Free PDF

Time Invariance in Systems

Definition: A system is considered time-invariant if its behavior and characteristics do not change over time. In other words, if an input signal is delayed by a certain amount of time, the output signal will be equally delayed by the same amount. Mathematically, a system is time-invariant if for any input signal x[n] and any time delay n0, the following condition holds true:

Y[n] = T{x[n]} implies Y[n - n₀] = T{x[n - n₀]}

Here, T represents the system's transformation, Y[n] is the output, and x[n] is the input.

Analysis of the Given Options:

Let's analyze each option to determine if the system is time-invariant:

Option 1: Y[n] = nx[n]

To check for time invariance, we need to examine if a time shift in the input signal results in an equivalent time shift in the output signal:

If the input is x[n], the output is Y[n] = nx[n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n0] = (n - n₀)x[n - n₀]

Since the output is not simply a time-shifted version of the original output (nx[n]), this system is not time-invariant.

Option 2: Y[n] = x[n] - x[n - 1]

For this option, we again check if a time shift in the input results in an equivalent time shift in the output:

If the input is x[n], the output is Y[n] = x[n] - x[n - 1].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[n - n₀] - x[n - n₀ - 1]

This is indeed the time-shifted version of the original output. Therefore, this system is time-invariant.

Option 3: Y[n] = x[-n]

Here, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[-n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[-(n - n₀)] = x[-n + n₀]

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Option 4: Y[n] = x[n] cos(2πf₀n)

For this option, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[n] cos(2πf₀n).

If the input is shifted by n₀, the new input is x[n - n_0], and the output should be:

Y[n - n₀] = x[n - n₀] cos(2πf₀(n - n₀))

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Conclusion:

After analyzing all the options, it is clear that:

Option 2: Y[n] = x[n] - x[n - 1] is the correct answer as it satisfies the condition for time invariance.
 

Concept

The convolution of an input signal with the impulse response of a Linear Time-Invariant (LTI) system is a fundamental concept in signal processing. This operation provides the output of the system for the given input signal.

In mathematical terms, if x(t) is the input signal and h(t) is the impulse response of the system, then the output y(t) is given by:

\(y(t) = x(t) * h(t)\)

Here, * denotes the convolution operation.

The convolution of an input signal with the impulse response of an LTI system provides the output of the system for that input. This is because the impulse response characterizes the system's behavior, and convolution integrates this behavior over the input signal to produce the output.

The response of an LTI (linear time-invariant) or LSI (linear shift-invariant) system is given by the convolution of input and impulse responses.

Explanation:

• An LTI/LSI system is characterized by two main properties: linearity and time or shift invariance. One of the fundamental principles regarding LTI systems is that their output can be determined by convolving the input signal with the system's impulse response.
• Convolution: This mathematical operation is used to describe the relationship between the input signal, the system's impulse response, and the output signal in LTI systems. The convolution integral (for continuous-time systems) or the convolution sum (for discrete-time systems) precisely calculates how the input and the system's impulse response interact to produce the system's output. Therefore, this is the correct answer.
• Correlation: Correlation is a measure of the similarity of two signals as a function of a time lag applied to one of them. Though related to convolution in its formulation, correlation is typically used for different purposes, such as signal detection or similarity measurement, not for describing the output of an LTI system in response to a given input.
• Superposition: This principle is inherent to the operation and characterization of linear systems, stating that the response caused by two or more stimuli is the sum of the responses that would have been caused by each stimulus individually. While the superposition principle is key to understanding linear systems, the specific process by which the input signal and the impulse response combine to produce the system output is convolution, not superposition itself.
• None: This option is not correct because convolution is indeed the operation that defines the response of an LTI system to a given input signal.

Given: \(\frac{d y(t)}{d t}+3 y(t)=2 x(t)\)

Taking Laplace transform on both sides, we get

sY(s) + 3Y(s) = 2X(s)

⇒ (s + 3)Y(s) = 2X(s)

⇒ \( \frac{Y(s)}{X(s)}=\frac{2}{s+3} \)

⇒ \(H(s)=\frac{2}{s+3}\)

∴ Impulse response will be, h(t) = L^-1(H(s)) = 2e^-3tu(t)

Hence, the correct option is (C).

Given h(t) is Real and Even. When sinusoidal input applied to LTI system having even impulse response, then output will also be sinusoidal.

here, \(y(t) =\left.H(W)\right|_{W=10.5 W} \cdot 10 \cos (10.5 W t) \)

let, h(t) = f(t) cos 10 Wt

where, \( f(t) =\pi\left(\frac{\sin W t}{\pi t}\right)^2\)

Now, \(H(W)=\frac{1}{2}[F(W+10 W)+F(W-10 W)]\)

∴ \(\left.H(W)\right|_{W=10.5 W} =\frac{3}{8} W \)

Hence, \(y(t) =\left(\frac{3}{8} W\right)(10 \cos 10.5 W t) \)

= \(=\frac{15}{4} W \cos 10.5 W t\)

Concept:

Linear system

The system is said to be linear if it follows Homogenous and Superposition property.

1) The system is homogenous if 

y(t) = α₁x(t) 

y(t) = α₂x(t)

2) The system follows Superposition if

y(t) = α1x(t) + α2x(t)

Time invariant system

if y(t - t₀) = x(t - t0)

and y(t') = x(t - t₀)  

Analysis:

For the given signal check linearity property

y(t) = x(t)cos(t)

1) check homogeneity

y(t) = α1x(t)cos(t)

y(t) = α2x(t)cos(t)

2) check superposition 

y(t) = α1x(t)cos(t) + α2x(t)cos(t)

The system follows both homogeneity and superposition thus the system is Linear.

check the time invariance

y(t - t0) = x(t - t0)cos(t - t0)

and y(t') = x(t - t0)cos(t)

since y(t - t0) ≠ y(t') 

System is not time invariant.

Hence option (3) is correct

Concept:

Linearity: Necessary and sufficient condition to prove the linearity of the system is that the linear system follows the laws of superposition i.e. the response of the system is the sum of the responses obtained from each input considered separately.

y{ax1[n] + bx2[t]} = a y{x1[n]} + b y{x2[n]}

Conditions to check whether the system is linear or not.

• The output should be zero for zero input
• There should not be any non-linear operator present in the system

Causality: A system is causal, if the output of the system does not depend on future inputs, but only on past input.

Time-variance check:

y (t) =  x (t2)

Shifting the input first,

y (t,k) = x(t²- k)

Shifting the output,

y (t-k) = x((t-k)2)

y(t,k) ≠  y(t-k)

Given system is time variant.

Application:

The continuous-time system described by y(t) = x(t2)

The system depends on the future inputs and hence it is non-causal.

A time shift of t0 in the input does not give the t0 shift in the output and hence it is time-variant.

As there are no non-linear operators present, it is linear.

Therefore, the given system is non-causal, linear and time-variant

Convolution sum is defined as

\(y\left[ n \right] = h\left[ n \right]*g\left[ n \right] = \mathop \sum \limits_{k = - \infty }^\infty h\left[ k \right]g\left[ {n - k} \right]\)

For causal sequence,

y[n] = h[0] g[n] + h[1] g [n-1] + h[2] g[n-2] +….

For n = 0,

y[0] = h[0] g[0] + h[1] g[-1] +…..

= h [0] g[0]            (g[-1] = g[-2] =……..0)

y[0] = h[0] g[0]

For. n =1,  y [1] = h [0] g [1] + h [1] g [0] + h [2] g [-1] + ……

y [1] = h [0] g [1] + h [1] g [0] +....

\(\frac{1}{2} = g\left[ 1 \right] + \frac{1}{{2}}g\left[ 0 \right]\)

Now, \(g\left[ 0 \right] = \frac{{y\left[ 0 \right]}}{{h\left[ 0 \right]}} = \frac{1}{1} = 1\)

g [1] = 0.5 - 0.5 = 0.

Explanation:

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.

Conclusion:

Option 1 is not correct.

Concept:

A linear time-invariant system is given as:

y(t) = x(t) ⊕ h(t), i.e.

\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)h\left( {t - \tau } \right)d\tau \) 

Application:

Given x(t) = 5 (Constant) as shown:

Also, the impulse response of filter h(t) is given as:

The output will be the convolution of the input and the impulse response, i.e.

\(y\left( t \right) = \;\mathop \smallint \nolimits_0^3 3.\left( 5 \right).dT\)

y(t) = 45

∴ The steady-state value of output is 45

Explanation:

The Characteristics equation is an essential tool for analyzing different system parameters

• It is obtained by equating the denominator of the closed-loop transfer function to 0
• It helps in the determination of system stability
• Also provides information about the system bandwidth, speed, response time, etc. 

From solving the characteristics equation we could infer that

• The roots of the characteristic equation give the location of the Closed-loop poles of the system
• These roots if strictly lie on the left half of the S-plane, then the system is stable
• If these roots have at least one root on the imaginary axis or on the right half of S-plane then the system is unstable
• The system transfer function solely determines the characteristic equation
• It is independent of the input applied
• Therefore system stability is independent of input

Hence statement a and c are correct,

Statement d is incorrect

If the system is practically realizable (stable) then

• For bounded input, it must give bounded output
• If finite magnitude input is applied then the system must also produce finite magnitude output
• The above is known as (BIBO) stable

Hence statement b is correct

Therefore statements a b and c are correct

The correct answer is option 3.

NOTE:

It is to be noted here that stability or instability is the characteristic property of the control system and thus depends on the closed-loop poles of the system.

Therefore, we can say that stability is a factor of the system which is independent of the input of the system. However, the steady-state output of the system is dependent on the poles of the applied input.

Concept:

Some important convolution results are

• u(t) * u(t) = t u(t)
• e­^-at u(t) * e­^-at u(t) = t e­^-at u(t)
• e­^-t u(t) * u(t) = (1- e­^-t) u(t)
• e­^-at u(t) * e­^-bt u(t) \(= \frac{1}{{\left( {b - a} \right)}}{e^{ - at}} - {e^{ - bt}}\;u\left( t \right)\) 
• \(t\;{\rm{u}}\left( {\rm{t}} \right){\rm{*u}}\left( {\rm{t}} \right) = \frac{{{{\rm{t}}^2}}}{2}{\rm{u}}\left( {\rm{t}} \right)\)

Calculation:

x(t) = e^-t and y(t) = e^-2t

Now convolution

z(t) = x(t) * y(t)

z(t) = e^-t – e^-2t

Concept:

An LTI system is stable if and only if the ROC of the impulse function H(s) includes the jω axis.

For Causal System → ROC is to the right side of the rightmost pole.

For Anti Causal System → ROC is to the left side of the left-most pole.

Observations:

• For a causal system to be stable, the poles must lie on the left half of the complex plane (to include the jω axis)
• A causal system with a pole on the right side cannot be BIBO stable because it's ROC can never include the jω axis. (Statement (II) is therefore correct)
• A BIBO system with a pole in the right half of the complex plane is stable if the system is anti-causal, as this will include the jω axis. (Statement (I) is therefore incorrect)

Concept:

A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)

The time period for a signal in the form of A cos (ωt + ϕ) is \(T = \frac{{2\pi }}{\omega}\)

For the sum of the two continuous-time signals with period T1 and T2 to be periodic, there must exist non-zero integers a, b such that:

a × T1 = b × T2

The fundamental period of a signal = LCM (T1, T2)

Calculation:

​​\(x\left( t \right) = \cos \frac{\pi }{3}t + \sin \frac{\pi }{4}t\)

​​​\({T_1} = \frac{{2\pi }}{{\frac{{\pi }}{3}}} = 6\)​

\({T_1} = \frac{{2\pi }}{{\frac{{\pi }}{4}}} = 8\)

Fundamental time period = LCM (T1, T2)

= LCM (6, 8) = 24 sec

A signal is said to be Eigen function if the output is scalar multiple of input signal and that scalar is referred as Eigen value.

Let input signal is x(t) and it is an Eigen function then