Question

The Fourier transform of the signal te^-3|t-1| is _____________

(a) \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\)

(b) \(\frac{6e^{-jω}}{9+ω^2}\)

(c) –\(\frac{12jωe^{-jω}}{9+ω^2}\)

(d) \(\frac{6e^{-jω}}{9+ω^2} + \frac{12jωe^{-jω}}{9+ω^2}\)

I had been asked this question during an interview.

Origin of the question is Exponential Fourier Series and Fourier Transforms topic in division Fourier Series of Signals and Systems

Answer 1

Correct answer is (a) \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\)

The explanation is: e^-3|t| = e^-3t, when t>0 and

 e^3t, when t<0

Now, Fourier transform of e^-3|t| = \(\frac{6}{9+ω^2}\)

Again, x (t-1) ↔ e^-jω  X(jω)

And t x (t) ↔ \(j\frac{d}{dω}\) X(jω)

∴ X (jω)= \(j\frac{d}{dω} [e^{-jω}  \frac{6}{9+ω^2}]\)

= \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\).