The Fourier transform of x(t) = te-|t| (where ‘t’ indicates continuous time domain)

Question

TuteeHUB · Accepted Answer

$X\left( \omega \right) = \frac{{ - j4}}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}$

TuteeHUB · Answer

$X\left( \omega \right) = \frac{4}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}$

TuteeHUB · Answer

$X\left( \omega \right) = \frac{{ - j4\omega }}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}$

TuteeHUB · Answer

$X\left( \omega \right) = \frac{{ - 4}}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}$

1.	The Fourier transform of x(t) = te-\|t\| (where ‘t’ indicates continuous time domain)
A.	\(X\left( \omega \right) = \frac{4}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}\)
B.	\(X\left( \omega \right) = \frac{{ - j4\omega }}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}\)
C.	\(X\left( \omega \right) = \frac{{ - j4}}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}\)
D.	\(X\left( \omega \right) = \frac{{ - 4}}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}\)
Answer» C. \(X\left( \omega \right) = \frac{{ - j4}}{{{{\left( {1 + {\omega ^2}} \right)}^2}}}\)

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