- #1

Prove It

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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*} $