(New page: Category:ECE438Spring2009mboutin) |
|||
| (7 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
[[Category:ECE438Spring2009mboutin]] | [[Category:ECE438Spring2009mboutin]] | ||
| + | |||
| + | a) <math>h[n] = \frac{1}{8}(\delta[n] + \delta[n-8]) + h[n-1]</math>--[[User:Kim415|Kim415]] 16:23, 1 March 2009 (UTC) | ||
| + | |||
| + | b) <math>H(z) = \frac{1}{8} \frac{\prod_{1}^{8}(z - z_{k})}{z - 1}</math> | ||
| + | |||
| + | This is the only thing that I can figure out. I'm still working on b) --[[User:Kim415|Kim415]] 16:23, 1 March 2009 (UTC) | ||
| + | |||
| + | c) | ||
| + | |||
| + | I have no idea.--[[User:Kim415|Kim415]] 16:23, 1 March 2009 (UTC) | ||
| + | |||
| + | |||
| + | ---- | ||
| + | Here's what I got: | ||
| + | |||
| + | a) <math>Y(z) = \frac{1}{8}[X(z)-X(z)z^{-8}]+Y(z)z^{-1}</math> | ||
| + | <math>H(z) = \frac{Y(z)}{X(z)} = \frac{1-z^{-8}}{8(1-z^{-1})}</math> | ||
| + | |||
| + | b) One pole and one zero at z=1? Or is there multiplicity for the zeros? | ||
| + | |||
| + | c) <br><math>H(z)=\frac{1}{8}(\frac{1-z^{-8}}{1-z^{-1}})=\frac{1-z^{-8}}{8}\sum_{n=0}^{\infty}(\frac{1}{z})^{n}</math><br><br><math>=\frac{1}{8}\sum_{n=0}^{\infty}z^{-n}-\frac{1}{8}\sum_{n=0}^{\infty}z^{-(n+8)}=\frac{1}{8}\sum_{n=-\infty}^{\infty}z^{-n}u(n)-\frac{1}{8}\sum_{m=-\infty}^{\infty}z^{-m}u(m-8)</math> | ||
| + | |||
| + | <br>So <math>h(n)=\frac{1}{8}u(n)-\frac{1}{8}u(n-8)</math> | ||
| + | |||
| + | <br>Assuming the ROC is <math>|z|>1\!</math>, I guess. Not sure if that is right. | ||
| + | <br>--[[User:Pjcannon|Pjcannon]] 19:29, 3 March 2009 (UTC) | ||
Latest revision as of 14:35, 3 March 2009
a) $ h[n] = \frac{1}{8}(\delta[n] + \delta[n-8]) + h[n-1] $--Kim415 16:23, 1 March 2009 (UTC)
b) $ H(z) = \frac{1}{8} \frac{\prod_{1}^{8}(z - z_{k})}{z - 1} $
This is the only thing that I can figure out. I'm still working on b) --Kim415 16:23, 1 March 2009 (UTC)
c)
I have no idea.--Kim415 16:23, 1 March 2009 (UTC)
Here's what I got:
a) $ Y(z) = \frac{1}{8}[X(z)-X(z)z^{-8}]+Y(z)z^{-1} $
$ H(z) = \frac{Y(z)}{X(z)} = \frac{1-z^{-8}}{8(1-z^{-1})} $
b) One pole and one zero at z=1? Or is there multiplicity for the zeros?
c)
$ H(z)=\frac{1}{8}(\frac{1-z^{-8}}{1-z^{-1}})=\frac{1-z^{-8}}{8}\sum_{n=0}^{\infty}(\frac{1}{z})^{n} $
$ =\frac{1}{8}\sum_{n=0}^{\infty}z^{-n}-\frac{1}{8}\sum_{n=0}^{\infty}z^{-(n+8)}=\frac{1}{8}\sum_{n=-\infty}^{\infty}z^{-n}u(n)-\frac{1}{8}\sum_{m=-\infty}^{\infty}z^{-m}u(m-8) $
So $ h(n)=\frac{1}{8}u(n)-\frac{1}{8}u(n-8) $
Assuming the ROC is $ |z|>1\! $, I guess. Not sure if that is right.
--Pjcannon 19:29, 3 March 2009 (UTC)
