(New page: =Homework 9 Solutions, ECE301 Spring 2011 Prof. Boutin= Students should feel free to make comments/corrections or ask questions directly on this page. ==Qu...) |
|||
| Line 14: | Line 14: | ||
Using frequency shift property of FT, we get: | Using frequency shift property of FT, we get: | ||
| + | |||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\mathcal{X}(\omega)&=\left\{\begin{array}{ll} | \mathcal{X}(\omega)&=\left\{\begin{array}{ll} | ||
| − | \pi, & \ | + | \pi , & \mbox{ for } |\omega - \pi|<\pi \\ |
| − | 0, & \ | + | 0, & \mbox{ elsewhere} |
| − | \end{array} | + | \end{array}\right. \\ |
&=\left\{\begin{array}{ll} | &=\left\{\begin{array}{ll} | ||
| − | \pi, & \ | + | \pi , & \mbox{ for } -\pi <\omega-\pi<\pi \\ |
| − | 0, & \ | + | 0, & \mbox{ elsewhere} |
| − | \end{array} | + | \end{array}\right. \\ |
&=\left\{\begin{array}{ll} | &=\left\{\begin{array}{ll} | ||
| − | \pi, & \ | + | \pi , & \mbox{ for } 0<\omega<2\pi \\ |
| − | 0, & \ | + | 0, & \mbox{ elsewhere} |
| − | \end{array} | + | \end{array}\right. |
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | |||
| + | After sampling with a sampling period T, we get: | ||
| + | |||
| + | <math>\mathcal{Y}(\omega)=\frac{1}{2\pi}\mathcal{X}(\omega)*\mathcal{P}(\omega)=\frac{1}{T}\sum_{k=-\infty}^{k=\infty}\mathcal{X}(\omega-\frac{2\pi k}{T})</math> | ||
| + | |||
| + | (a) The FT of the signal after sampling with sampling period <math class="inline">T=\frac{1}{4}</math> is shown in the figure below. | ||
| + | |||
| + | [[Image:301_sp2011_boutin_hw9_figure1.jpg]] | ||
| + | |||
| + | From the figure, we can see that there is no aliasing and the signal can be recovered by filtering the sampled signal using the following filter: | ||
| + | |||
| + | <math> | ||
| + | \mathcal{H}(\omega)=\left\{\begin{array}{ll} | ||
| + | \frac{1}{4} , & \mbox{ for } |\omega - \pi|<\pi \\ | ||
| + | 0, & \mbox{ elsewhere} | ||
| + | \end{array}\right. | ||
| + | </math> | ||
| + | |||
| + | (b) The FT of the signal after sampling with sampling period <math class="inline">T=\frac{2}{3}</math> is shown in the figure below. | ||
| + | |||
| + | [[Image:301_sp2011_boutin_hw9_figure2.jpg]] | ||
| + | |||
| + | From the figure, we can see that there is no aliasing and the signal can be recovered by filtering the sampled signal using the following filter: | ||
| + | |||
| + | <math> | ||
| + | \mathcal{H}(\omega)=\left\{\begin{array}{ll} | ||
| + | \frac{2}{3} , & \mbox{ for } |\omega - \pi|<\pi \\ | ||
| + | 0, & \mbox{ elsewhere} | ||
| + | \end{array}\right. | ||
| + | </math> | ||
| + | |||
| + | (c) The FT of the signal after sampling with sampling period <math class="inline">T=\frac{2}{3}</math> is shown in the figure below. | ||
| + | |||
| + | [[Image:301_sp2011_boutin_hw9_figure3.jpg]] | ||
| + | |||
| + | From the figure, we can see that there is aliasing and thus the signal cannot be recovered. | ||
| + | |||
---- | ---- | ||
[[HW9 ECE301 Spring2011 Prof Boutin| HW9]] | [[HW9 ECE301 Spring2011 Prof Boutin| HW9]] | ||
Revision as of 19:48, 7 April 2011
Homework 9 Solutions, ECE301 Spring 2011 Prof. Boutin
Students should feel free to make comments/corrections or ask questions directly on this page.
Question 1
(a) The given signal has finite duration and hence it is not band-limited. (b) The given signal has finite duration and hence it is not band-limited. (c) The given signal has infinite duration and hence it is band-limited. (d) The given signal has infinite duration and hence it is band-limited.
Question 2
$ x(t)=e^{j\pi t}\frac{\sin(\pi t)}{t}=\pi e^{j\pi t}\frac{\sin(\pi t)}{\pi t} $
Using frequency shift property of FT, we get:
$ \begin{align} \mathcal{X}(\omega)&=\left\{\begin{array}{ll} \pi , & \mbox{ for } |\omega - \pi|<\pi \\ 0, & \mbox{ elsewhere} \end{array}\right. \\ &=\left\{\begin{array}{ll} \pi , & \mbox{ for } -\pi <\omega-\pi<\pi \\ 0, & \mbox{ elsewhere} \end{array}\right. \\ &=\left\{\begin{array}{ll} \pi , & \mbox{ for } 0<\omega<2\pi \\ 0, & \mbox{ elsewhere} \end{array}\right. \end{align} $
After sampling with a sampling period T, we get:
$ \mathcal{Y}(\omega)=\frac{1}{2\pi}\mathcal{X}(\omega)*\mathcal{P}(\omega)=\frac{1}{T}\sum_{k=-\infty}^{k=\infty}\mathcal{X}(\omega-\frac{2\pi k}{T}) $
(a) The FT of the signal after sampling with sampling period $ T=\frac{1}{4} $ is shown in the figure below.
From the figure, we can see that there is no aliasing and the signal can be recovered by filtering the sampled signal using the following filter:
$ \mathcal{H}(\omega)=\left\{\begin{array}{ll} \frac{1}{4} , & \mbox{ for } |\omega - \pi|<\pi \\ 0, & \mbox{ elsewhere} \end{array}\right. $
(b) The FT of the signal after sampling with sampling period $ T=\frac{2}{3} $ is shown in the figure below.
From the figure, we can see that there is no aliasing and the signal can be recovered by filtering the sampled signal using the following filter:
$ \mathcal{H}(\omega)=\left\{\begin{array}{ll} \frac{2}{3} , & \mbox{ for } |\omega - \pi|<\pi \\ 0, & \mbox{ elsewhere} \end{array}\right. $
(c) The FT of the signal after sampling with sampling period $ T=\frac{2}{3} $ is shown in the figure below.
From the figure, we can see that there is aliasing and thus the signal cannot be recovered.
