| Line 17: | Line 17: | ||
<math>H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}}</math> | <math>H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}}</math> | ||
| − | <math>Y(z)(1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}) = X(z)(1 - \frac{1}{2}z^{-2})</math> | + | <math>Y(z)(1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}) = X(z)(1 - \frac{1}{2}z^{-2})</math><br /> |
| − | <math>y[n] - \frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] =x[n] - \frac{1}{2}x[n-2]</math> | + | <math>y[n] - \frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] =x[n] - \frac{1}{2}x[n-2]</math><br /> |
| − | <math>y[n] = x[n] - \frac{1}{2}x[n-2] + \frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2]</math> | + | <math>y[n] = x[n] - \frac{1}{2}x[n-2] + \frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2]</math><br /> |
Revision as of 12:04, 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}} $--Kim415 16:04, 1 March 2009 (UTC)
b) check out the the Prof. --Kim415 16:04, 1 March 2009 (UTC)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.--Kim415 16:04, 1 March 2009 (UTC)
d)
$ H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}} $
$ Y(z)(1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}) = X(z)(1 - \frac{1}{2}z^{-2}) $
$ y[n] - \frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] =x[n] - \frac{1}{2}x[n-2] $
$ y[n] = x[n] - \frac{1}{2}x[n-2] + \frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2] $
