(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...) |
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| − | a) <br> | + | a) <br /> |
<math>H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}} | <math>H(z) = \frac{1 - \frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}}z^{-1}+\frac{1}{4}z^{-2}} | ||
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<math> zero = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}</math> | <math> zero = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}</math> | ||
| − | <math> pole = \frac{1}{2\sqrt{2}}+\frac{j}{2\sqrt{2}},\frac{1}{2\sqrt{2}}-\frac{j}{2\sqrt{2}}</math> | + | <math> pole = \frac{1}{2\sqrt{2}}+\frac{j}{2\sqrt{2}},\frac{1}{2\sqrt{2}}-\frac{j}{2\sqrt{2}}</math>--[[User:Kim415|Kim415]] 16:04, 1 March 2009 (UTC) |
| − | b) check out the the Prof. [http://cobweb.ecn.purdue.edu/~allebach/ece438/lecture/module_1/1.5_z_transform/1.5.4_zt_and_ccf_diff_eq.pdf Allebach's useful lecture note] for this problem. | + | b) check out the the Prof. [http://cobweb.ecn.purdue.edu/~allebach/ece438/lecture/module_1/1.5_z_transform/1.5.4_zt_and_ccf_diff_eq.pdf --[[User:Kim415|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. | + | c) unstable, because it is recursive equation and some output will not be bounded from the certain input value which is bounded.--[[User:Kim415|Kim415]] 16:04, 1 March 2009 (UTC) |
| − | d) <br> | + | d) <br /> |
| − | <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[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] = x[n] - \frac{1}{2}x[n-2] + \frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2]</math> | ||
Revision as of 11:04, 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] $
