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Homework #9

Problem Statement:
Show that, for a bandwidth limited signal ( x(t) with  f_{max} < {1\over {2T}} )

\sum_{k=-\infty}^{\infty} \left | x(kT) \right | ^2
=c\int_{-\infty}^{\infty} \left | x(t) \right | ^2\,dt
And find c.

Equations:
 
\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle=\int_{-\infty}^{\infty} \phi_k(t)^{*} \phi_l(t)\,dt

x(t)=\sum_{k=-\infty}^{\infty} x(kT)\phi_k(t)
Solution:

\begin{matrix}
\left \langle x(t) \vert x(t) \right \rangle & = & \int_{-\infty}^{\infty} x(t)^{*} x(t)\,dt
\\ \ & = & \int_{-\infty}^{\infty} \left | x(t) \right |^2\,dt
\end{matrix}

x(t)=\sum_{k=-\infty}^{\infty} x(kT)\phi_k(t)


\begin{matrix}
\Rightarrow \left \langle x(t) \vert x(t) \right \rangle & = &
\left \langle \sum_{k=-\infty}^{\infty} x(kT)\phi_k(t) \vert \sum_{l=-\infty}^{\infty} x(lT)\phi_l(t) \right \rangle
\\ \ & = & \sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty} x(kT)x(lT)
\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle
\end{matrix}
By earlier work: 
\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle
=T\delta_{l,k}

\Rightarrow
\sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty} x(kT)x(lT)
\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle
=T\sum_{k=-\infty}^{\infty} \left | x(kT) \right |^2

\Rightarrow
\sum_{k=-\infty}^{\infty} \left | x(kT) \right | ^2
={1\over T}\int_{-\infty}^{\infty} \left | x(t) \right | ^2\,dt

\Rightarrow
c={1\over T}

Homework #13

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