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<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) | <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 --[[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.--[[User:Kim415|Kim415]] 16:07, 1 March 2009 (UTC) |
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| + | c) unstable, because it is recursive equation and some output will go to the infinite from the certain input value which is bounded.--[[User:Kim415|Kim415]] 16:04, 1 March 2009 (UTC) | ||
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| + | I'm not sure of my answer of c). If you guys have any good answer, please post yours on this page. | ||
| − | c | + | For c you may also be able to say that for left-sided input signals, the ROC of the Z-transform is inside the unit circle, and thus contains poles. So, for certain bounded left-sided input signals, the output will be unbounded (I believe). --[[User:Ghadley|Ghadley]] 17:43, 2 March 2009 (UTC) |
d) <br /> | d) <br /> | ||
| − | <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><br /> |
<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(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><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><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><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><br /> --[[User:Kim415|Kim415]] 16:05, 1 March 2009 (UTC) |
Latest revision as of 13:44, 2 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. Allebach's useful lecture note for this problem.--Kim415 16:07, 1 March 2009 (UTC)
c) unstable, because it is recursive equation and some output will go to the infinite from the certain input value which is bounded.--Kim415 16:04, 1 March 2009 (UTC)
I'm not sure of my answer of c). If you guys have any good answer, please post yours on this page.
For c you may also be able to say that for left-sided input signals, the ROC of the Z-transform is inside the unit circle, and thus contains poles. So, for certain bounded left-sided input signals, the output will be unbounded (I believe). --Ghadley 17:43, 2 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] $
--Kim415 16:05, 1 March 2009 (UTC)
