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

Question

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	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.	\(\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)
Answer» C. \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)

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