Question

The Fourier transform of the signal sin(2πt) e^-t u (t) is ____________

(a) \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} + \frac{1}{1+j(ω+2π)}\right)\)

(b) \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\)

(c) \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)

(d) \(\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)

This question was posed to me in class test.

My question is taken from Common Fourier Transforms in division Fourier Transform of Signals and Systems

Answer 1

The correct choice is (b) \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\)

To explain I would say: The Fourier Transform of e^-tu (t) = \(\frac{1}{1+jω}\)

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

So, x (t-1) ↔ X {j (ω-2π)}

∴ X (jω) = \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\).