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If x(t) = sin 2πt e– t u(t), then its Fourier transform is—
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1 
1 – 1 
2 1 + j(ω – 2π) 1 + j(ω + 2π) -
1 1 – 1 2j 1 + j(ω – 2π) 1 + j(ω + 2π) -
1 1 – 1 2j 1 + j(ω + 2π) 1 + j(ω – 2π) -
1 1 – 1 2j 1 + j(ω + 2π) 1 + j(ω – 2π)
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Correct Option: C
x(t) = sin 2πte–t u(t)
| x1(t) = e–t u(t) ←F.T.→ | = x1(jω) | |
| 1 + jω |
| ∴ sin 2πt = | |
| j2 |
So,
| x(t) = | | | e–t u(t) | |
| 2j |
By using frequency shifting property
| x(t) = | x1(t) – | x1(t) ←F.T.→ | ||
| j2 | j2 |
| − | | = x(jω) | |||
| 2j | 1 + j(ω – 2π) | 1 + j(ω + 2π) |
Hence, alternative (C) is the correct choice.
