Question

The Fourier series coefficient for the periodic signal x(t) = sin^2t is _____________

(a) –\(\frac{1}{4}\)  δ[k-1] +  \(\frac{1}{2}\)  δ[k] –  \(\frac{1}{4}\)  δ[k+1]

(b) –\(\frac{1}{4}\)  δ[k-2] +  \(\frac{1}{2}\)  δ[k] –  \(\frac{1}{4}\)  δ[k+2]

(c) –\(\frac{1}{2}\)  δ[k-1] + δ[k] –  \(\frac{1}{2}\)  δ[k+1]

(d) –\(\frac{1}{2}\)  δ[k-2] + δ[k] –  \(\frac{1}{2}\)  δ[k+2]

This question was addressed to me during an interview.

My doubt stems from Fourier Series Coefficients in division Fourier Series of Signals and Systems

Answer 1

Right option is (a) –\(\frac{1}{4}\)  δ[k-1] +  \(\frac{1}{2}\)  δ[k] –  \(\frac{1}{4}\)  δ[k+1]

To explain: sin^2t = \((\frac{e^{jt} – e^{-jt}}{2j})^2\)

= –\(\frac{1}{4}\) (e^2jt – 2 + e^-2jt)

The fundamental period of sin2t is π and ω = \(\frac{2π}{π}\) = 2

∴ X[k] = –\(\frac{1}{4}\)  δ[k-1] +  \(\frac{1}{2}\)  δ[k] –  \(\frac{1}{4}\)  δ[k+1].