Question

The Fourier transform of signal e^-2t u(t-3) is ___________

(a) \(\frac{e^{-3(2-jω)}}{2-jω}\)

(b) \(\frac{e^{-3(2+jω)}}{2+jω}\)

(c) \(\frac{e^{3(2-jω)}}{2-jω}\)

(d) \(\frac{e^{3(2+jω)}}{2+jω}\)

I have been asked this question in my homework.

This key question is from Exponential Fourier Series and Fourier Transforms topic in section Fourier Series of Signals and Systems

Answer 1

The correct answer is (b) \(\frac{e^{-3(2+jω)}}{2+jω}\)

Easy explanation: X (jω) = \(\int_{-∞}^∞ x(t) e^{-jωt} \,dt\)

Given u (t-3), hence value of this will be equal to 1 when t>=3

∴ X (jω) = \(\int_3^∞ e^{-2t} e^{-jωt} \,dt \)

= \(\frac{e^{-3(2+jω)}}{2+jω}\).