Difference between revisions of "Homework"

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(Homework #9)
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<br>
 
<br>
 
<math>
 
<math>
\left \langle \phi_k(t) \| \phi_l(t) \right \rangle=\int_{-\infty}^{\infty} \phi_k(t)* \phi_l(t)\,dt
+
\left \langle \phi_k(t) \vert \phi_l(t) \right \rangle=\int_{-\infty}^{\infty} \phi_k(t)^{*} \phi_l(t)\,dt
  +
</math>
  +
<br>
  +
<math>
  +
x(t)=\sum_{k=-\infty}^{\infty} x(kT)\phi_k(t)
 
</math>
 
</math>

Revision as of 09:10, 10 December 2004

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)