Fourier Transform of a Gaussian Signal

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For a continuous-time function $\mathrm{x(t)}$, the Fourier transform of $\mathrm{x(t)}$ can be defined as,

$$\mathrm{X(\omega)\:=\:\int_{-\infty }^{\infty} x\left(t\right)\:e^{-j\omega t}\:dt}$$

Fourier Transform of Gaussian Signal

Gaussian Function - The Gaussian function is defined as,

$$\mathrm{g_{a}\left(t\right) \:=\: e^{-at^{2}} ;\:\:for\:all \:t}$$

Therefore, from the definition of Fourier transform, we have,

$$\mathrm{X(\omega) \:=\: F\left [e^{-at^2} \right ]\:=\:\int_{-\infty }^{\infty}e^{-at^2} \:e^{-j\omega t} \:dt}$$

$$\mathrm{\Rightarrow\: X\left(\omega\right) \:=\:\int_{-\infty}^{\infty} \:e^{-\left(at^2\:+\:j\omega t\right) }\:dt \:=\: e^{-\left(\omega^2/4a\right)}\int_{-\infty}^{\infty}\:e^{\left [{-t\sqrt{a}\:+\:(j\omega/2\sqrt{a})}\right]^{2}}dt }$$

Let,

$$\mathrm{\left [t\sqrt{a}\:+\:(j\omega 2\sqrt{a})\right ]\:=\: u}$$

Then,

$$\mathrm{du \:=\: \sqrt{a} \:dt\: and \: \:dt \:=\: \frac{du}{\sqrt{a}}}$$

$$\mathrm{\therefore\: X\left(\omega\right)\:=\:e^{-\left(\omega^2/4a\right)}\int_{-\infty }^{\infty}\: \frac{e^{-u^{2}}}{\sqrt{a}}\:du\: =\: \frac{e^{-\left(\omega^2/4a\right)}}{\sqrt{a}}\int_{-\infty }^{\infty}\:e^{-u^{2}} \:du}$$

$$\mathrm{\because\:\int_{-\infty }^{\infty}\:e^{-u^{2}} \:du \:=\: \sqrt{\pi}}$$

$$\mathrm{\therefore\: X\left(\omega\right) \:=\:\frac{e^{-\left(\omega^2/4a\right)}}{\sqrt{a}}\:\cdot\: \sqrt{\pi} \:=\: \sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left(\omega^2/4a\right)}}$$

Therefore, the Fourier transform of the Gaussian function is,

$$\mathrm{F\left [e^{-at^{2}}\right ] \:=\:\sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left ( \omega^2/4a\right )}}$$

Or, it can also be written as,

$$\mathrm{e^{-at^2}\overset{FT}{\leftrightarrow} \:\sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left (\omega^2/4a\right )}}$$

The graphical representation of Gaussian function and its frequency spectrum is shown in Figure-1.

Fourier Transform of Gaussian Modulated Function

The Gaussian modulated signal is defined as

$$\mathrm{x\left(t \right) \:=\: e^{-at^{2}}\: cos \:\omega_{0}t}$$

$$\mathrm{\Rightarrow\: x\left(t \right)\ \:=\: e^{-at^{2}} \left (\frac{e^{j\omega_{0}t} \:+\: e^{-j\omega _{0}t}}{2}\right); \left\{\because\: cos \:\omega _{0}t \:=\:\left (\frac{e^{j\omega _{0}t} \:+\: e^{-j\omega_{0}t}}{2} \right) \right \}}$$

Therefore, the Fourier transform of the Gaussian modulated signal is

$$\mathrm{X\left( \omega\right) \:=\: \frac{1}{2} F\left [ e^{-at^{2}}e^{j\omega_{0}t} \right ] \:+\: \frac{1}{2}F\left [ e^{-at^{2}}e^{-j\omega _{0}t} \right ]}$$

By using frequency shifting property [i.e.$\mathrm{e^{-j\omega _{0}t}x\left (t\right)\overset{FT}{\leftrightarrow}X \left(\omega \:+\: \omega_{0}\right)}$] of Fourier transform, we get,

$$\mathrm{F\left[e^{-at^{2}}e^{j\omega _{0}t} \right] \:=\: F\left [e^{-at^{2}} \right]|_{\omega \:=\: \left ( \omega\:-\:\omega _{0}\right )}}$$

And

$$\mathrm{F\left[e^{-at^{2}}e^{-j\omega _{0}t} \right] \:=\:F\left [e^{-at^{2}} \right]|_{\omega \:=\: \left ( \omega \:+\: \omega _{0}\right )}}$$

Also, the Fourier transform of Gaussian function is,

$$\mathrm{F\left [e^{-at^{2}}\right ] \:=\: \sqrt{\frac{\pi}{a}} \cdot e^{-\left ( \omega^2/4a\right )}}$$

Therefore, the Fourier transform of Gaussian modulated function is,

$$\mathrm{X\left( \omega\right) \:=\: \frac{1}{2}\left[\sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left [\left(\omega\:-\:\omega _{0}\right)^{2}/4a\right]} \:+\: \sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left [\left(\omega \:+\: \omega _{0}\right)^{2}/4a\right ] } \right ]}$$

Or, it can also be represented as,

$$\mathrm{e^{-at^{2}}\: cos \:\omega _{0}t\overset{FT}{\leftrightarrow}\frac{1}{2}\left[\sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left [\left(\omega \:-\:\omega _{0}\right)^{2}/4a\right]} \:+\: \sqrt{\frac{\pi}{a}} \:\cdot\: e^{-\left [\left(\omega \:+\: \omega _{0}\right)^{2}/4a\right]}\right ]}$$

The graphical representation of the Gaussian modulated signal and its frequency spectrum is shown in Figure-2.