# few problems on Auto correlatioon Function(ACF) and Energy Spectral Density(ESD)

1. Find the Auto correlation function of $x(t)&space;=&space;\frac{1}{\sqrt{2\pi&space;}}\exp&space;^{\frac{-t^{2}}{2}}$.

Ans. We know that  Auto correlation function forms fourier transform pair with Energy Spectral Density function

$ACF\leftrightarrow&space;ESD$

$R_{xx}(t)\leftrightarrow&space;S(f)$

the Fourier Transform of  $e^{-ct^{2}}\leftrightarrow&space;\frac{\sqrt{\pi&space;}}{c}e^{-\pi&space;^{2}f^{2}}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;\frac{1}{\sqrt{2\pi&space;}}.\sqrt{\frac{\pi&space;}{(1/2)}}e^{\frac{-\pi&space;^{2}f^{2}}{(1/2)}}$ here $c&space;=&space;\frac{1}{2}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;\frac{1}{\sqrt{2\pi&space;}}.\sqrt{2\pi&space;}e^{-2&space;\pi&space;^{2}f^{2}}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;e^{-2&space;\pi&space;^{2}f^{2}}$

$x(t)\leftrightarrow&space;X(f)$

$\therefore$ the Fourier Transform of x(t) is X(f) and is $X(f)&space;=&space;e^{-2\pi&space;^{2}f^{2}}$ and the Energy Spectral Density $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

$S(f)&space;=&space;e^{-4\pi&space;^{2}f^{2}}$

By finding the inverse Fourier Transform of S(f) gives the Auto Correlation Function

$S(f)&space;=&space;e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}$

$e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}\leftrightarrow&space;\frac{e^{\frac{-t^{2}}{4}}}{\sqrt{4\pi&space;}}$

$e^{\frac{-\pi&space;^{2}f^{2}}{(1/4)}}\leftrightarrow&space;\frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi&space;}}$

$\therefore$ the ACF of the given signal is inverse Fourier Transform of S(f) which is $R_{xx}(t)&space;=&space;\frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi&space;}}$.

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