David's Fourier Transform Problem from YAnswers

[CaptainBlack]

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:
\[\mathfrak{F}\left[ e^{-i a t}p(t) \right]=P(\omega+a)\]

CB

 

[jvicens]

Using this theorem, we can find the inverse Fourier transform of \(\frac{6}{(\omega+2)^2 + 9^2}\) as follows:
\[\begin{aligned} \mathfrak{F}^{-1}\left[\frac{6}{(\omega+2)^2 + 9^2}\right]&=\mathfrak{F}^{-1}\left[\frac{6}{(\omega+3i)^2}\right] \\
&=\frac{6}{2\pi}\int_{-\infty}^{\infty}\frac{e^{i(\omega+3i)t}}{(\omega+3i)^2}\;d\omega \\
&=\frac{3}{\pi}\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2}\;d\omega \\
&=\frac{3}{\pi}\left(-\frac{1}{\omega}+i\pi\delta(\omega) \right) \\
&=-\frac{3}{\pi\omega}+i\frac{3}{\pi}\delta(\omega) \end{aligned}\]

Therefore, the inverse Fourier transform of \(\frac{6}{(\omega+2)^2 + 9^2}\) is \(-\frac{3}{\pi\omega}+i\frac{3}{\pi}\delta(\omega)\). I hope this helps with your problem. Let me know if you have any further questions. Good luck!