# Homework

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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: $\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle =T\delta_{l,k}$
$\Rightarrow \sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty} x(kT)x(lT) \left \langle \phi_k(t) \vert \phi_l(t) \right \rangle =T\sum_{k=-\infty}^{\infty} \left | x(kT) \right |^2$
$\Rightarrow \sum_{k=-\infty}^{\infty} \left | x(kT) \right | ^2 =1/T\int_{-\infty}^{\infty} \left | x(t) \right | ^2\,dt$

## Homework #13

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