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Given f(t) = 3e-4tu(t). Its Fourier transform F(ω) at ω = 4 is


1. \(\frac{1}{{1 + j}}\)
2. \(\frac{{\frac{3}{4}}}{{1 + j}}\)
3. \(\frac{1}{{1 + \frac{4}{3}j}}\)
4. \(\frac{{\frac{4}{3}}}{{1 + \frac{4}{3}j}}\)

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Correct Answer - Option 2 : \(\frac{{\frac{3}{4}}}{{1 + j}}\)

f(t) = 3e-4t u(t)

\(F\left( \omega \right) = \mathop \smallint \nolimits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\)

\({e^{ - at}}\mathop \leftrightarrow \limits^{F.T} \frac{1}{{a + j\omega }}\)

∴ \(3{e^{ - 4t}}\;\mathop \leftrightarrow \limits^{FT} \frac{3}{{4 + j\omega }}\)

At ω = 4,

\(F\left( \omega \right){\left. \right|_{\omega = 4}} = \frac{3}{{4 + j4}} = \left( {\frac{{\frac{3}{4}}}{{1 + j}}} \right)\)

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