Question

In equation \(x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]\), if \(F^{-1} [2V(F)]=δ(t)+j/πt\) and \(F^{-1} [X(F)]\) = x(t). Then the value of ẋ(t) is?

(a) \(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t+τ} dτ\)

(b) \(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\)

(c) \(\frac{1}{π} \int_{-\infty}^\infty \frac{2x(t)}{t-τ} dτ\)

(d) \(\frac{1}{π} \int_{-\infty}^\infty \frac{4x(t)}{t-τ} dτ\)

This question was addressed to me in an interview for internship.

My query is from The Representation of Bandpass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing

Answer 1

Correct answer is (b) \(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\)

Easy explanation: \(x_+ (t)=[δ(t)+j/πt]*x(t)\)

\(x_+ (t)=x(t)+[j/πt]*x(t)\)

\(ẋ(t)=[j/πt]*x(t)\)

=\(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\)

Hence proved.