Homework

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:
${\begin{matrix}\left\langle x(t)\vert x(t)\right\rangle &=&\int _{-\infty }^{\infty }x(t)^{*}x(t)\,dt\\\ &=&\int _{-\infty }^{\infty }\left|x(t)\right|^{2}\,dt\end{matrix}}$
$x(t)=\sum _{k=-\infty }^{\infty }x(kT)\phi _{k}(t)$
${\begin{matrix}\Rightarrow \left\langle x(t)\vert x(t)\right\rangle &=&\left\langle \sum _{k=-\infty }^{\infty }x(kT)\phi _{k}(t)\vert \sum _{l=-\infty }^{\infty }x(lT)\phi _{l}(t)\right\rangle \\\ &=&\sum _{k=-\infty }^{\infty }\sum _{l=-\infty }^{\infty }x(kT)x(lT)\left\langle \phi _{k}(t)\vert \phi _{l}(t)\right\rangle \end{matrix}}$