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Let x(t) be a continuous-time signal with <math class="inline"> \left| {\mathcal X} (\omega)\right| =0</math> for  <math class="inline"> \left| \omega)\right| > \omega_m </math>. Can one recover the signal x(t) from the signal <math class="inline"> y(t)=x(t) p(t-3) </math>, where
 
Let x(t) be a continuous-time signal with <math class="inline"> \left| {\mathcal X} (\omega)\right| =0</math> for  <math class="inline"> \left| \omega)\right| > \omega_m </math>. Can one recover the signal x(t) from the signal <math class="inline"> y(t)=x(t) p(t-3) </math>, where
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 +
<math class="inline"> p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} n) ?</math>

Revision as of 04:36, 1 April 2011

Should question 3 be

Let x(t) be a continuous-time signal with $ \left| {\mathcal X} (\omega)\right| =0 $ for $ \left| \omega\right| > \omega_s $. Can one recover the signal x(t) from the signal $ y(t)=x(t) p(t-3) $, where

$ p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} k) ? $


instead of this?


Let x(t) be a continuous-time signal with $ \left| {\mathcal X} (\omega)\right| =0 $ for $ \left| \omega)\right| > \omega_m $. Can one recover the signal x(t) from the signal $ y(t)=x(t) p(t-3) $, where

$ p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} n) ? $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009