(New page: a) <br> <math>H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}} = \frac{(1 - \frac{1}{\sqrt{2}}z^{-1})(1 + \frac{1}{\sqrt{2}}z^{-1})}{(1-\frac{1...) |
(No difference)
|
Revision as of 12:00, 1 March 2009
a)
$ H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}} = \frac{(1 - \frac{1}{\sqrt{2}}z^{-1})(1 + \frac{1}{\sqrt{2}}z^{-1})}{(1-\frac{1}{2\sqrt{2}}+\frac{j}{2\sqrt{2}}z^{-1})(1-\frac{1}{2\sqrt{2}}-\frac{j}{2\sqrt{2}}z^{-1})} $
$ zero = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} $
$ pole = \frac{1}{2\sqrt{2}}+\frac{j}{2\sqrt{2}},\frac{1}{2\sqrt{2}}-\frac{j}{2\sqrt{2}} $
b) check out the the Prof. Allebach's useful lecture note for this problem.
c) unstable, because it is recursive equation and some output will not be bounded from the certain input value which is bounded.
d)
