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

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

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

ACF\leftrightarrow ESD

R_{xx}(t)\leftrightarrow S(f)

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

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

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }}.\sqrt{2\pi }e^{-2 \pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow e^{-2 \pi ^{2}f^{2}}

x(t)\leftrightarrow X(f)

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

S(f) = e^{-4\pi ^{2}f^{2}}

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

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

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

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

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


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Author: Lakshmi Prasanna

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.