Question

Given a real valued function x (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, x(t) satisfies the equation ____________

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

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

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

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

The question was asked during an interview.

This intriguing question comes from Trigonometric Fourier Series in chapter Fourier Series of Signals and Systems

Answer 1

Right option is (d) x (t) = x (t-T) = x (t – \(\frac{T}{2}\))

To elaborate: For an even symmetry, x (t) = x (t-T)

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

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

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