Fourier Transform of Complex Gaussian

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Theorem:

$\displaystyle \zbox {e^{-pt^2} \;\longleftrightarrow\;\sqrt{\frac{\pi}{p}} \, e^{-\frac{\omega^2}{4p}},\quad \forall p\in {\bf C}: \; \mbox{re}\left\{p\right\}>0}$

(D.16)

Proof: [202, p. 211] The Fourier transform of $e^{-pt^2}$ is defined as

$\displaystyle \int_{-\infty}^\infty e^{-pt^2} e^{-j\omega t}dt \eqsp \int_{-\infty}^\infty e^{-(pt^2+j\omega t)} dt.$

(D.17)

Completing the square of the exponent gives

$\begin{eqnarray*} pt^2 + j\omega t - \frac{\omega^2}{4p} + \frac{\omega^2}{4p} &=& p\left(t+j\frac{\omega}{2p}\right)^2 + \frac{\omega^2}{4p} \end{eqnarray*}$

Thus, the Fourier transform can be written as

$\displaystyle e^{-\frac{\omega^2}{4p}} \int_{-\infty}^\infty e^{-p\left(t+j\frac{\omega}{2p}\right)^2} dt \eqsp \sqrt{\frac{\pi}{p}}\, e^{-\frac{\omega^2}{4p}}$

(D.18)

using our previous result.

Alternate Proof

The Fourier transform of a complex Gaussian can also be derived using the differentiation theorem and its dual (§B.2).^D.1

Proof: Let

$\displaystyle g(t)\isdefs e^{-pt^2} \;\longleftrightarrow\;G(\omega).$

(D.19)

Then by the differentiation theorem (§B.2),

$\displaystyle g^\prime(t) \;\longleftrightarrow\;j\omega G(\omega).$

(D.20)

By the differentiation theorem dual (§B.3),

$\displaystyle -jtg(t) \;\longleftrightarrow\;G^\prime(\omega).$

(D.21)

Differentiating gives

$\displaystyle g^\prime(t) \eqsp -2ptg(t) \eqsp \frac{2p}{j}[-jtg(t)] \;\longleftrightarrow\;\frac{2p}{j}G^\prime(\omega).$

(D.22)

Therefore,

$\displaystyle j\omega G(\omega) \eqsp \frac{2p}{j}G^\prime(\omega)$

(D.23)

$\displaystyle \left[\ln G(\omega)\right]^\prime \eqsp \frac{G^\prime(\omega)}{G(\omega)} \eqsp -\frac{\omega}{2p} \eqsp \left(-\frac{\omega^2}{4p}\right)^\prime.$