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If I have an equation of a signal in the frequency domain given by

$L(x)=\frac{1}{\pi}\frac{\frac{1}{2}\Gamma}{((\omega-\omega_0)^2+(\frac{1}{2}\Gamma)^2)}$

and I want to convert this to the time domain, can I do this using the inverse Fourier transform? I think the result should be an exponential, but when I tried to calculate this on mathematica I got a strange result.

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  • $\begingroup$ Why don't you show us what Mathematica gave you? This looks like the square modulus of a typical signal's Fourier transform, not the transform itself. Did Mathematica give you a damped sinusoid? $\endgroup$ – garyp Apr 11 '18 at 11:14
  • $\begingroup$ I just used the mathematica expression $FourierTransform[2.5/(Pi*((x - 97.5)^2 + 2.5^2)), x, t]$ which returns $ 0.795775 exp((-2.5 + 97.5 I) t) (0.501326 exp(5. t) HeavisideTheta[-t] + 0.501326 HeavisideTheta[t])$ $\endgroup$ – JJH Apr 11 '18 at 11:59
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    $\begingroup$ The result looks reasonable as it reflects the distinction of case for $t>0$ and $t<0$ (the Heaviside function shows this) which has to be applied to make the complex path integral convergent. $\endgroup$ – Frederic Thomas Apr 11 '18 at 15:13
  • $\begingroup$ All ok. Except that $L(x)$ looks like a power spectral density (PSD) rather than a spectrum. If that's correct, the Fourier transform will give the autocorelation function of the original signal, not the original signal itself. I'd expect the autocorrelation function to be a damped sinusoid. And that's what Mathematica produced. $\endgroup$ – garyp Apr 11 '18 at 16:46
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You can do the Fourier integral $\int L \exp(i\omega t) \frac{d\omega}{2\pi}$ easily by the application of the theorem of residues. First you look for the poles in the complex plane, second you choose the integration path (either a half-circle in the upper or in the lower complex plane, actually a distinction of cases has to be done for $t>0$ and $t<0$) checking that the integrand converges to zero on the half-circular part at infinity), and finally can apply the formula of the residue theorem which is essentially

$\oint f(\omega) d\omega = 2\pi i \sum\limits_{k=1,\ldots n} Res(f,\omega_k)$ with $f(\omega) = \frac{1}{\pi}\frac{1}{2\pi} \frac{a \exp(i\omega t )}{(\omega-\omega_1)(\omega-\omega_2)}$ with $\omega_1 = \omega_0 +ia$ and $\omega_2= \omega_0 -ia$ and $a = \frac{1}{2}\Gamma$

where $\omega_k$ is the k-th pole which is surrounded by the integral path. As the chosen half-circle (the integration path) only encircles one pole the expression Res(f) gives: $Res(f,\omega_{1/2}) = f(\omega)\cdot (\omega-\omega_{1/2})|_{\omega=\omega_{1/2}}$ 1 or 2 has to be chosen depending on which pole is encircled by the integration path.

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