Question

Given a real valued function y (t) with period T. Its trigonometric Fourier series expansion contains no term of frequency ω = 2π \(\frac{(2k)}{T}\); where, k = 1, 2….. Also no terms are present. Then, y(t) satisfies the equation ____________

(a) y (t) = y (t+T) = -y (t+\(\frac{T}{2}\))

(b) y (t) = y (t+T) = y (t+\(\frac{T}{2}\))

(c) y (t) = y (t-T) = -y (t-\(\frac{T}{2}\))

(d) y (t) = y (t-T) = y (t-\(\frac{T}{2}\))

I have been asked this question in semester exam.

My question comes from Fourier Analysis topic in portion Fourier Analysis on Complex Spaces of Signals and Systems

Answer 1

Right answer is (d) y (t) = y (t-T) = y (t-\(\frac{T}{2}\))

Easiest explanation: For an even symmetry, y (t) = y (t-T)

Thus no sine component will exist because bn=0 and by half wave symmetry condition odd harmonics will exist.

Now, y (t) = y (t-\(\frac{T}{2}\))

Combining the two conditions, we get, y (t) = y (t-T) = y (t-\(\frac{T}{2}\)).