| Line 4: | Line 4: | ||
I guess that the ROCs are in 3 cases. | I guess that the ROCs are in 3 cases. | ||
| − | 1) <math> |z| < \frac{1}{2}, |z| < 1 </math> | + | 1) |
| + | <math> |z| < \frac{1}{2}, |z| < 1 </math> | ||
<br> | <br> | ||
| − | 2) <math> |z| > \frac{1}{2}, |z| < 1 </math> | + | 2) |
| + | <math> |z| > \frac{1}{2}, |z| < 1 </math> | ||
<br> | <br> | ||
| − | 3) <math> |z| > \frac{1}{2}, |z| > 1 </math> | + | 3) |
| + | <math> |z| > \frac{1}{2}, |z| > 1 </math> | ||
If not, please reply.--[[User:Kim415|Kim415]] 16:33, 24 February 2009 (UTC) | If not, please reply.--[[User:Kim415|Kim415]] 16:33, 24 February 2009 (UTC) | ||
Revision as of 11:33, 24 February 2009
At first I was confused by this question. I think it is fairly straight forward to see that you need to use partial fraction expansion and then take the inverse z-transform to obtain x[n]. However, I did not understand how it was possible to have three separate solutions for x[n]. Looking at the transform pairs in ECE 438 Essential Definitions for the z-tranform, you notice that 2. and 3. have the same z-transform but different x[n]'s and different ROC's. Looking at the ROC's, you can figure out three different possible solutions and derive the corresponding x[n]'s from there. --Babaumga 16:27, 24 February 2009 (UTC)
I guess that the ROCs are in 3 cases.
1)
$ |z| < \frac{1}{2}, |z| < 1 $
2)
$ |z| > \frac{1}{2}, |z| < 1 $
3)
$ |z| > \frac{1}{2}, |z| > 1 $
If not, please reply.--Kim415 16:33, 24 February 2009 (UTC)
