Energy Spectal Density of rectangular pulse and solved problems

As we all know the Rectangular pulse is defined as x(t) = 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) = \int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi ft}dt

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

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

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

X(f) = \frac{T \sin \pi fT}{\pi fT}

X(f) = X(f) = sinc (\pi fT) 

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

s(f)= T^{2}sinc^{2}(\pi fT).

Pb2. Let x(t) = sinc(T\pi t) and  y(t) = \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) = \left | Y(f) \right |^{2}

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

Y(f) \leftrightarrow j2\pi f X(f)

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

then Y(f) \leftrightarrow j2\pi f X(f)

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

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

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

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

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

S(f ) = \frac{dE(f)}{df}

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

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

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

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

\therefore S(f)=f^{2}e^{-\sigma f}

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

Ans:-  Auto Correlation Function \leftrightarrowEnergy spectral Density

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

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

By using the above rule

\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=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}}

\therefore X(f) =e^{-2\pi ^{2}f^{2}}

Now 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)

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

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

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

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Author: Lakshmi Prasanna Ponnala

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