| (3 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
Should question 3 be | Should question 3 be | ||
| − | 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| > \ | + | 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 |
<math class="inline"> p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} k) ?</math> | <math class="inline"> p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} k) ?</math> | ||
| Line 10: | Line 10: | ||
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 | ||
| + | |||
| + | <math class="inline"> p(t)= \sum_{k=-\infty}^\infty \delta (t- \frac{2\pi}{\omega_s} n) ?</math> | ||
| + | |||
| + | :TA's comment: Yes. It should be k and not n. | ||
| + | |||
| + | Can you post a question similar to number 1 a)? | ||
| + | -mm | ||
Latest revision as of 16:06, 2 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_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} 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) ? $
- TA's comment: Yes. It should be k and not n.
Can you post a question similar to number 1 a)? -mm
