Fourier Transform of a Gaussian Signal

Signals and SystemsElectronics & ElectricalDigital Electronics

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

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

Fourier Transform of Gaussian Signal

Gaussian Function - The Gaussian function is defined as,

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

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

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

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

Let,

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

Then,

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

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

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

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

Therefore, the Fourier transform of the Gaussian function is,

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

Or, it can also be written as,

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

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

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

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

Therefore, the Fourier transform of the Gaussian modulated signal is

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

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

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

And

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

Also, the Fourier transform of Gaussian function is,

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

Therefore, the Fourier transform of Gaussian modulated function is,

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

Or, it can also be represented as,

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

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

raja
Updated on 17-Dec-2021 07:09:18

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