Test: Fourier Series Representation of Continuous Periodic Signals - Electrical Engineering (EE) MCQ

The average value

CASE 1: Let us assume that x(t) is real, then x(t) = x*(t). This implies that for x(t) to be real, ak = a*_-k. Since this is not true. So, x(t) is not real.
CASE 2: Let us assume that x(t) is even, then x(t) = x(-t) and a_k = a_–k. Since this is true for this case, hence x(t) is even.
CASE 3: Let us assume

Therefore,

Since bk is not even, So, g(t) is not even.

We know that,

Given form is,

So, we can write it as,

Comparing both equations, we get,

Taking Inverse Fourier Transform,

Comparing with given signal, g(t) = Ay (Bt)
A = 1/3 and B = 3

We know that from Fourier Series coefficient of square periodic signal, power is proportional to (1/n²).



-  P₃   =  (1/3)²

- P₅  = (1/5)²

Ratio of Powers =  

Given x(t) = u(sint)
Drawing waveform

We know for waveform like y(t) shown

DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics