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

#### Prove It

##### Well-known member
MHB Math Helper
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*}