Difference between revisions of "HW 5 Question 1" - Rhea

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)