Question

What are the Fourier series coefficients for the signal x(n)=cosπn/3?

(a) c1=c2=c3=c4=0,c1=c5=1/2

(b) c0=c1=c2=c3=c4=c5=0

(c) c0=c1=c2=c3=c4=c5=1/2

(d) none of the mentioned

The question was asked during an interview for a job.

My enquiry is from Frequency Analysis of Discrete Time Signal topic in division Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer 1

The correct option is (a) c1=c2=c3=c4=0,c1=c5=1/2

Easiest explanation: In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.

Given signal is x(n)=cosπn/3=cos2πn/6=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{-j2πn/6}\)

We know that -2π/6=2π-2π/6=10π/6=5(2π/6)

Therefore, x(n)=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{j2π(5)n/6}\)

Compare the above equation with x(n)=\(\sum_{k=0}^{N-1}c_k e^{j2πkn/N}\)

So, we get c1=c2=c3=c4=0 and c1=c5=1/2.