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

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

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

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.

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.

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

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.

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

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

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

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.

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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).

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.

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.

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.

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.

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.

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.

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.

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: