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Aidan's question via email about Fourier Transforms (2)

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Find the Fourier Transform of $\displaystyle 3\,H\left( t - 1 \right) \mathrm{e}^{-2\,t} $.
In order to use the Second Shift Theorem, the function needs to be entirely of the form $\displaystyle f\left( t - 1 \right) $. To do this let $\displaystyle v = t - 1 \implies t = v + 1 $, then

$\displaystyle \begin{align*}
\mathrm{e}^{-2\,t} &= \mathrm{e}^{-2 \, \left( v + 1 \right) } \\
&= \mathrm{e}^{-2\,v - 2 } \\
&= \mathrm{e}^{-2\,\left( t - 1 \right) - 2 } \\
&= \mathrm{e}^{-2\,\left( t - 1 \right) } \,\mathrm{e}^{-2}
\end{align*} $

And so

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