Test: Continuous Time Signals - Electrical Engineering (EE) MCQ

Concept:

• A continuous-time signal x(t) is said to be periodic if it repeats itself after its time period 'T₀'.
• A continuous-time signal x(t) satisfies:
   

​x(t) = x(t + T0) for all t; where T₀

The smallest value of T₀ that satisfies the periodicity condition is called the fundamental time period of x(t).

The signal denoted by x(t) is known as a continuous-time signal.

Continuous-Time Signal:

• A continuous-time (CT) signal is a function that is continuous, meaning there are no breaks in the signal.
• For all real values of 't' we will get a value f(t), t ⊂ R.
• CT signals are usually represented by using x(t), having parentheses and the variable 't'.

Discrete-time signal:

• A discrete-time signal is a signal whose value is taken at discrete measurements.
• With a discrete-time signal, there will be time periods of 'n' where we do not have a value.
• DT signals are represented using the form x[n].
• Discrete signals are approximations of CT signals

A system is continuous-time when its Input/Output signals are continuous-time. A system is discrete-time when its Input/Output signals are discrete-time.

Important Points:

Concept:

The Fourier transform of a signal in the time domain is given as:

The Laplace transform F(s) of a function f(t) is defined by:

The definition of z-transform is given by,

Calculations:

Unit impulse signal:

It is defined as, 

For f(t) = δ(t)

The spectrum is represented as:

The Laplace transform of unit impulse is 1 i.e. unity.

Discrete-time Laplace transform is Z- transform.

Given the Fourier transform pair,

The Inverse Fourier transform of F (ω) is given as

Replacing t by ‘ –t ‘ and ω by u in the above expression, we have,

If f (t) is even, then f (-t) = f (t)

∴ g (t) = 2π f (t)

⇒ g (t) will be proportional to f (t), if f (t) is an even function.

Concept:

Continious time signal x(t) is given as,

x(t) = x₁(t) + x₂(t) 

Then, Time period of x(t) is (N) = LCM (T₁, T₂,)

Time period of x₁₍t) is Given as:

T₁ = 2π / ω₀₁    

Time period of x₂(t) is Given as:

T₂ = 2π / ω₀₂   

Calculation:

Given:

T₁ = 2π / 2/3 = 3π

T₂ = 2π / .5 = 4π

T₃ = 2π / 13 = 6π

Time period = L.C.M. (T₁, T₂, T₃)

= L.C.M. (3π, 4π, 6π)

= 12π

Concept:

An LTI system is stable if and only if its impulse response is absolutely integrable.

Calculation:

Given,

h(t) = e^at u(t) + e^bt u(-t)

h(t) = h₁(t) + h₂(t)

∵ τ > 0 therefore, for absolute integrability a < 0

∵ τ < 0 therefore, for absolute integrability b > 0

∴ h(t) should be in the form shown in the figure below:

2: 4

The square law device produces an output signal that is equal to the square of the input signal. When a sinusoidal signal is passed through a square law device, the output signal will contain both the fundamental frequency and harmonics at higher frequencies. The DC component of the output signal is the average value of the output signal over time, and it is equal to the sum of the DC component of the input signal and the DC component of the harmonics.

In this case, the input signal is x(t) = 2sin(100t + π/3). The DC component of this signal is 0, since the average value of a sinusoidal signal over one period is 0. The output signal is y(t) = x^2(t) = (2sin(100t + π/3))^2 = 4cos^2(100t + π/3) = 4(1 - sin^2(100t + π/3)) = 4(1 - (2sin^2(100t + π/3)/2)) = 4(1 - sin^2(100t + π/3)/2). The DC component of this signal is 4/2 = 2.

Therefore, the correct answer is 2: 4.

Here the input is bounded and the output is also bounded.

Hence the system is stable.

The system is non-linear as it follows y(n) = x³(n).

The system is time-invariant as no operations are made on n.

Given, the signal

V (t) = 30 sin 100t + 10 cos 300 t + 6 sin (500t+π/4)

So, we have

ω₁ = 100 rads

ω₂ = 300 rads

ω₃ = 500 rads

∴ The respective time periods are

So, the fundamental time period of the signal is

as, 

∴ The fundamental frequency,

Concept:

Definite Integral:

If

Calculation: