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

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The discussion focuses on finding the Fourier Transform of the function 3H(t - 1)e^(-2t) using the Second Shift Theorem. To apply this theorem, the function is rewritten in terms of v = t - 1, leading to the expression e^(-2(t - 1))e^(-2). The Fourier Transform is computed as 3e^(-2) multiplied by the Fourier Transform of H(t - 1)e^(-2(t - 1)), resulting in 3e^(-2 - iω)(1/(2 + iω)). The Heaviside unit step function H(t) is crucial in this calculation, and the Fourier Transform of H(t)e^(-at) is referenced for clarity. The discussion effectively demonstrates the application of the Second Shift Theorem in Fourier analysis.
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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*} $
 
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