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How should I go about trying to solve this particular inverse F.T.:

$$ \mathcal G(f) = e^{\frac{-(f-f_0)^2}{(2\sigma)^2}} $$

From the definition of the continuous inverse F.T.

$$ \mathcal{F^{-1}} [G(f)] = g(t) = \int_{-\infty}^{\infty} (e^{j2\pi ft})G(f) df $$

I'm somewhat grasping the forward direction of going from time to frequency domains but it's a little more difficult for me to go the other way. It looks like a frequency shift to me but the $(2\sigma)^2$ is throwing me off.

I mistakenly took it as a Gaussian in the time domain, any help or advice on tackling this would be great.

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$$ \mathcal{F}^{-1} \{G(f)\} = g(t) = \int_{-\infty}^{\infty} \mathrm e^{j2\pi ft}\,G(f) \,\mathrm df= \int_{-\infty}^{\infty} \mathrm e^{j2\pi ft-\frac{(f-f_0)^2}{(2\sigma)^2}} \,\mathrm df $$ Completing the square $$ \begin{align} j2\pi ft-\frac{(f-f_0)^2}{4\sigma^2}&=-\frac{f_0^2-(4 j \pi t \sigma^2 + f_0)^2}{4\sigma^2}-\frac{[f -(4 j \pi t \sigma^2 + f_0)]^2}{4 σ^2}\\ &=j (4 \pi^2 j \sigma^2 t^2 + 2 \pi f_0 t)-\frac{[f -(4 j \pi t \sigma^2 + f_0)]^2}{4 \sigma^2} \end{align} $$

then we have $$ \begin{align} g(t) &= \mathrm e^{j (4 \pi^2 j \sigma^2 t^2 + 2 \pi f_0 t)} \underbrace{\int_{-\infty}^{\infty} \exp\left(-\frac{[f -(4 j \pi t \sigma^2 + f_0)]^2}{4 \sigma^2}\right) \,\mathrm df}_{\sqrt{\pi}\,2\sigma}\\ &=\sqrt{\pi}\,2\sigma\,\mathrm e^{j (4 \pi^2 j \sigma^2 t^2 + 2 \pi f_0 t)} \end{align} $$

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  • $\begingroup$ You can't use a complex change of variable without discussing it $\endgroup$
    – reuns
    Oct 1 '17 at 20:31

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