Homework

Homework #9

Problem Statement:
Show that, for a bandwidth limited signal ($x(t)$ with $f_{max}<\frac{1}{2T}$)
$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|x(kT)|}^2=c\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|x(t)|}^{2}dt$
And find c.

Equations:
$⟨{\varphi }_k(t)|{\varphi }_l(t)⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\varphi }_k(t{)}^{*}{\varphi }_l(t)dt$
$x(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}x(kT){\varphi }_k(t)$
Solution:
$$\begin{gathered} \begin{array}{ccc}⟨x(t)|x(t)⟩& =& \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t{)}^{*}x(t)dt\\ \\ & =& \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|x(t)|}^{2}dt\end{array} \end{gathered}$$
$x(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}x(kT){\varphi }_k(t)$
$$\begin{gathered} \begin{array}{ccc}\Rightarrow ⟨x(t)|x(t)⟩& =& ⟨\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}x(kT){\varphi }_k(t)|\sum _{l=-\mathrm{\infty }}^{\mathrm{\infty }}x(lT){\varphi }_l(t)⟩\\ \\ & =& \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{l=-\mathrm{\infty }}^{\mathrm{\infty }}x(kT)x(lT)⟨{\varphi }_k(t)|{\varphi }_l(t)⟩\end{array} \end{gathered}$$
By earlier work: $⟨{\varphi }_k(t)|{\varphi }_l(t)⟩=T{\delta }_{l,k}$
$\Rightarrow \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{l=-\mathrm{\infty }}^{\mathrm{\infty }}x(kT)x(lT)⟨{\varphi }_k(t)|{\varphi }_l(t)⟩=T\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|x(kT)|}^2$
$\Rightarrow \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|x(kT)|}^2=\frac{1}{T}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{|x(t)|}^{2}dt$
$\Rightarrow c=\frac{1}{T}$

Homework #13

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