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I am trying to calculate the fourier transform of the following function: $$ f(t) =\begin{cases} e^{-kt},& t \geq 0 \\ 0,& \text{otherwise}\end{cases} $$

where $k > 0$ is a real number.

I can plot the equation fine but am unsure of where the limits should lie, I think I should be doing something like this:

$$ f(\omega) = \int_0^T e^{-t(k+j\omega)}dt\ $$ Would this be correct or am I going the wrong way about completing this?

stuart194
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  • Start with $2\pi\hat{f}(s):=\int_{-\infty}^\infty f(t)e^{-ist}dt=\int_{0}^\infty f(t)e^{-ist}dt=\int_{0}^\infty e^{-kt}e^{-ist}dt=...$, by definition of $f$. – Avitus Nov 10 '14 at 12:11
  • this would give $\frac{1}{j\omega + K}$ I simply presumed this was wrong as none of the example questions I looked at resemble it. – stuart194 Nov 10 '14 at 12:23
  • The above gives $\int_0^\infty e^{-(k+is)t}dt$: can you integrate it? – Avitus Nov 10 '14 at 12:29
  • Yes it gives $\frac{1}{j\omega + k}$ – stuart194 Nov 10 '14 at 12:35

1 Answers1

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Rather than the limits being T and 0 they had to be infinity and zero as the equation continued onto infinity asymptotic to the x axis.

stuart194
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