Question

The Fourier transform of 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) \(\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)

I have been asked this question in quiz.

Enquiry is from Trigonometric Fourier Series in chapter Fourier Series of Signals and Systems

Answer 1

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

The explanation is: 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) = \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\).