# Energy Spectal Density of rectangular pulse and solved problems

As we all know the Rectangular pulse is defined as $x(t)&space;=&space;rect(\frac{t}{T})$, exists for a duration of T sec symmetrical with respect to y-axis as shown

Fourier Transform is $X(f)$

$X(f)&space;=&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi&space;ft}dt$

$X(f)&space;=&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}&space;1.e^{-j2\pi&space;ft}dt$

$X(f)&space;=&space;\left&space;[&space;\frac{e^{-j2\pi&space;ft}}{j2\pi&space;f}&space;\right&space;]&space;^{\frac{T}{2}}_{\frac{-T}{2}}$

$X(f)&space;=&space;\frac{1}{\pi&space;f}\left&space;[&space;\frac{e^{j\pi&space;ft}-e^{-j\pi&space;ft}}{2j}&space;\right&space;]$

$X(f)&space;=&space;\frac{T&space;\sin&space;\pi&space;fT}{\pi&space;fT}$

$X(f)&space;=&space;X(f)&space;=&space;sinc&space;(\pi&space;fT)$

$\therefore$ The Energy Spectral Density of the given signal $x(t)$ will be $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

$s(f)=&space;T^{2}sinc^{2}(\pi&space;fT)$.

Pb2. Let $x(t)&space;=&space;sinc(T\pi&space;t)$ and  $y(t)&space;=&space;\frac{dx(t)}{dt}$ find the Energy Spectral Density $S(f)$ of $y(t)$.

Ans:-  The Energy Spectral Density of $y(t)$is $S(f)$

i.e,  $S(f)&space;=&space;\left&space;|&space;Y(f)&space;\right&space;|^{2}$

$y(t)&space;\leftrightarrow&space;\frac{dx(t)}{dt}$, then Fourier Tansform of  $y(t)$is

$Y(f)&space;\leftrightarrow&space;j2\pi&space;f&space;X(f)$

as $x(t)&space;=&space;sinc(\pi&space;Tt)\leftrightarrow&space;X(f)=&space;\frac{1}{T}rect(\frac{f}{T})$

then $Y(f)&space;\leftrightarrow&space;j2\pi&space;f&space;X(f)$

$\left&space;|&space;Y(f)&space;\right&space;|^{2}&space;\leftrightarrow&space;\left&space;|&space;j2\pi&space;f&space;\frac{1}{T}rect(\frac{f}{T})&space;\right&space;|&space;^{2}$

$\left&space;|&space;Y(f)&space;\right&space;|^{2}\leftrightarrow&space;\frac{4\pi&space;^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})$

$\therefore&space;S(f)&space;=&space;\frac{4\pi&space;^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})$

Pb3. The Energy contained with in the band $[0,&space;f]$, $f>0$ $E(f)&space;=&space;\frac{1}{\sigma&space;^{3}}(&space;{2-(2+2\sigma&space;f+\sigma&space;^{2}f^{2})e^{-\sigma&space;f}})$ then find the ESD $S(f)$, for any $f>0$ Energy of a signal can be defined $\int_{-\infty&space;}^{\infty&space;}\left&space;|&space;S(f)&space;\right&space;|&space;df$

Ans:- Given $E(f)&space;=&space;\int_{0}^{f}\left&space;|&space;S(f)&space;\right&space;|&space;df$

$S(f&space;)&space;=&space;\frac{dE(f)}{df}$

$E(f)&space;=&space;\frac{1}{\sigma&space;^{3}}(&space;{2-(2+2\sigma&space;f+\sigma&space;^{2}f^{2})e^{-\sigma&space;f}})$

$\frac{dE(f)}{df}&space;=&space;0-&space;\frac{1}{\sigma&space;^{3}}[2(-\sigma&space;)e^{-\sigma&space;f}+2\sigma&space;f(-\sigma&space;)e^{-\sigma&space;f}+2\sigma&space;e^{-\sigma&space;f}+\sigma&space;^{2}f^{2}(-\sigma&space;)e^{-\sigma&space;f}+\sigma&space;^{2}(2f&space;)e^{-\sigma&space;f}]$

$\frac{dE(f)}{df}&space;=&space;-&space;\frac{1}{\sigma&space;^{3}}[-2\sigma&space;e^{-\sigma&space;f}-2\sigma^{2}&space;fe^{-\sigma&space;f}+2\sigma&space;e^{-\sigma&space;f}+\sigma&space;^{3}f^{2}e^{-\sigma&space;f}+2\sigma&space;^{2}fe^{-\sigma&space;f}]$

$\frac{dE(f)}{df}&space;=&space;-&space;\frac{1}{\sigma&space;^{3}}\sigma&space;^{3}f^{2}e^{-\sigma&space;f}$

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

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

Ans:-  Auto Correlation Function $\leftrightarrow$Energy spectral Density

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

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

By using the above rule

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

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

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

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

Now the Energy Spectral Density $S(f)&space;=&space;\left&space;|&space;X(f)&space;\right&space;|^{2}$

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

by finding the inverse Fourier Transform of $S(f)$

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

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

Auto Correlation Function = $\frac{1}{2\sqrt{\pi&space;}}e^{\frac{-t^{2}}{4}}$

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