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David's Fourier Transform Problem from Y!Answers


Well-known member
Jan 26, 2012
Here's my problem;

Find the Fourier transform \(P(\omega)\) of the function;

\[ p(t)=\left\lbrace \begin{array}{ll} e^{-9t} & \text{for } t \ge 0 \\ e^{9t} & \text{for } t \lt 0 \end{array} \right.\]

Hence (use one of the shift theorems) find the inverse Fourier transform of; \( \frac{6}{(\omega+2)^2 + 9^2} \)

many thanks!


Well we start by guessing which definition of the Fourier transform you are working with. I will assume you want:
\[P(\omega)=\int_{-\infty}^{\infty}p(t)e^{-i \omega t}\;dt\]

With our function this becomes:
\[\begin{aligned} P(\omega)&=\int_{0}^{\infty}e^{-9t}e^{-i \omega t}\;dt+\int_{-\infty}^{0}e^{9t}e^{-i \omega t}\;dt\\
&=\left. \frac{e^{-(9+i \omega)t}}{-9-i \omega} \right|_0^{\infty}+\left. \frac{e^{-(-9+i \omega)t}}{+9-i \omega} \right|^0_{-\infty} \\
&=\frac{1}{9+i\omega}+\frac{1}{9-i\omega} \\
&=\frac{18}{9^2+\omega^2} \end{aligned}\]

To complete this the shift theorem you need is:
\[\mathfrak{F}\left[ e^{-i a t}p(t) \right]=P(\omega+a)\]

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