Test: Time Shifting & Scaling Property - Electrical Engineering (EE) MCQ

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.
For example:

y(t) = x(t) + e^jtx(t)
x(t) ↔ X(jω)
e^jtx(t) ↔ X(j(ω - 1))

y(jω) at ω = 0.5 rad/sec = X(jω) + X(j(ω - 1))
= 1 + 0.5 = 1.5

Fourier transform of e^−tu(t) 

Fourier transform of e^-(t-1)u(t-1)

y(t) = x(t)∗h(t)
Taking the Fourier Transform, we get:
Y(f) = X(f)H(f)
Y(f) = X(f)e^−j4πf
Taking the inverse Fourier Transform of Y(f), we get:
y(t) = x(x - 2)

Let x₁(t) = x(t − 1)

Now,

Taking Fourier transform
 X₂(jω)=(jω)²X₁(jω)
⇒X₂(jω) = −ω²X(jω)e^−jω
Thus,

From the given pictures of x(t) and y(t)
We get,
y(t) = - x(2t + 2)
y(t) is time scaled and time shifted version of x(t)
Step 1:
If x(t) ↔ X(ω)
Then, x(t + 2) ↔ e^j.2.ω X(ω)
Step 2:
Using time shifting property
x(2t + 2) ↔ ½ e^jω X(ω/2)
Step 3:
Using time scaling property:

Concept:
Time shifting property of Fourier series states that:

Since ω₀ = 2π/T₀, the above expression can be written as:

Application:
Observing the two figures, we can write:
y(t) = x(t - 1)

Where c_n = Fourier series coefficient of x(t)
Since, T₀ = 2

Using the above expression, we get:

Also for even values of n, e^-jπn = 1

We know that, expansion in frequency domain result in compression in the time domain and vice versa.
In the given question, compression is done frequency domain. So there will be expansion in time domain by same amount.
A x(t/2) ↔ 2A X(2f)
2A = 3 ⇒ A = 3/2

Power density has no effect of shifting. It is affected only by scaling
We know,
E[X(t)X(t + τ)] = R_x(τ)
and
E[X(2t + 1)X(2(t + τ) + 1)] = E[X(2t + 1)X(2t + 2τ + 1)] = R_x(2τ)

then using the scaling property of Fourier transforms we have,

Thus, the power spectral density of 

using the properties of fourier transform