DSPRelated.com
Free Books

Fourier Transform of Complex Gaussian


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 $ g(t)$ 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)

or

$\displaystyle \left[\ln G(\omega)\right]^\prime \eqsp \frac{G^\prime(\omega)}{G(\omega)} \eqsp -\frac{\omega}{2p} \eqsp \left(-\frac{\omega^2}{4p}\right)^\prime.$ (D.24)

Integrating both sides with respect to $ \omega$ yields

$\displaystyle \ln G(\omega) \eqsp -\frac{\omega^2}{4p} + \ln G(0).$ (D.25)

In §D.7, we found that $ G(0)=\sqrt{\pi/p}$ , so that, finally, exponentiating gives

$\displaystyle G(\omega) \eqsp \sqrt{\frac{\pi}{p}}\,e^{-\frac{\omega^2}{4p}}$ (D.26)

as expected.

The Fourier transform of complex Gaussians (``chirplets'') is used in §10.6 to analyze Gaussian-windowed ``chirps'' in the frequency domain.


Next Section:
Why Gaussian?
Previous Section:
Gaussian Integral with Complex Offset