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b) One pole and one zero at z=1? Or is there multiplicity for the zeros? | 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( | + | 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>So <math>h(n)=\frac{1}{8}u(n)-\frac{1}{8}u(n-8)</math> | ||
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)
