Aidan's question via email about Fourier Transforms

[Prove It]

Find the Inverse Fourier Transform of $\displaystyle \frac{1}{6 + 5\,\mathrm{i}\,\omega - \omega ^2} $.

In order to factorise this quadratic, we will need to recognise that $\displaystyle \mathrm{i}^2 = -1 $, so we can rewrite this as

$\displaystyle \begin{align*} \frac{1}{6 + 5\,\mathrm{i}\,\omega - \omega ^2 } &= \frac{1}{\mathrm{i}^2\,\omega ^2 + 5\,\mathrm{i}\,\omega + 6} \\ &= \frac{1}{ \left( \mathrm{i}\,\omega \right) ^2 + 5\,\mathrm{i}\,\omega + 6 } \\
&= \frac{1}{\left( \mathrm{i}\,\omega + 2 \right) \left( \mathrm{i}\,\omega + 3 \right) } \end{align*} $

Now apply Partial Fractions:

$\displaystyle \begin{align*} \frac{A}{\mathrm{i}\,\omega + 2 } + \frac{B}{\mathrm{i}\,\omega + 3 } &\equiv \frac{1}{\left( \mathrm{i}\,\omega + 2 \right) \left( \mathrm{i}\,\omega + 3 \right) } \\
A\left( \mathrm{i}\,\omega + 3 \right) + B \left( \mathrm{i}\,\omega + 2 \right) &\equiv 1 \end{align*}$

Let $\displaystyle \mathrm{i}\,\omega = -2 \implies A = 1 $.

Let $\displaystyle \mathrm{i}\,\omega = -3 \implies -B = 1 \implies B = -1 $. Thus

$\displaystyle \begin{align*} \mathcal{F}^{-1}\,\left\{ \frac{1}{6 + 5\,\mathrm{i}\,\omega - \omega ^2 } \right\} &= \mathcal{F}^{-1}\,\left\{ \frac{1}{2 + \mathrm{i}\,\omega} - \frac{1}{3 + \mathrm{i}\,\omega} \right\} \\ &= \mathrm{e}^{-2\,t}\,H\left( t \right) - \mathrm{e}^{-3\,t} \,H\left( t \right) \end{align*} $

 

[jvicens]

where $\displaystyle H\left( t \right) = \begin{cases} 1 & t \geq 0 \\ 0 & t < 0 \end{cases} $ is the Heaviside step function.