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* Nope, ROC is correct. Look at the second example I did on Friday and use the same trick. --[[User:Mboutin|Mboutin]] 18:12, 7 September 2009 (UTC) | * Nope, ROC is correct. Look at the second example I did on Friday and use the same trick. --[[User:Mboutin|Mboutin]] 18:12, 7 September 2009 (UTC) | ||
**I tried to do this, but I get stuck with this, which doesn't compare well with the formula of the Z-Transform: <math>(1/4) * \sum_{n=0}^\infty (-z/2 + 3/4)^n </math> --[[User:Rodrigaa|Rodrigaa]] 19:47, 7 September 2009 (UTC) | **I tried to do this, but I get stuck with this, which doesn't compare well with the formula of the Z-Transform: <math>(1/4) * \sum_{n=0}^\infty (-z/2 + 3/4)^n </math> --[[User:Rodrigaa|Rodrigaa]] 19:47, 7 September 2009 (UTC) | ||
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| + | * For 7a, the singularity occurs at z = -1/2. I also applied the trick and got the same ROC. I really think the ROC is wrong :( | ||
For problem 7d, the differentiation property might help. However, the ROC defined in the problem seems wierd. Negative magnitude? :( | For problem 7d, the differentiation property might help. However, the ROC defined in the problem seems wierd. Negative magnitude? :( | ||
Revision as of 18:44, 8 September 2009
I think the ROC for question 7a is incorrect. Should be |z| < 1/2.
- Nope, ROC is correct. Look at the second example I did on Friday and use the same trick. --Mboutin 18:12, 7 September 2009 (UTC)
- I tried to do this, but I get stuck with this, which doesn't compare well with the formula of the Z-Transform: $ (1/4) * \sum_{n=0}^\infty (-z/2 + 3/4)^n $ --Rodrigaa 19:47, 7 September 2009 (UTC)
- For 7a, the singularity occurs at z = -1/2. I also applied the trick and got the same ROC. I really think the ROC is wrong :(
For problem 7d, the differentiation property might help. However, the ROC defined in the problem seems wierd. Negative magnitude? :(
- Oops! Yeah the ROC was supposed to be $ |Z|<1 $. But, technically, the statement is correct, just a bit weird. --Mboutin 18:12, 7 September 2009 (UTC)
- I figured, out the residue method involving complex integration is not as bad as it looks. It also helps when you get stuck with a fraction you can't expand because the norm of 1/z or z is not < 1. At least it helped me. But I was curious whether we are allowed to use it on the exam and homework problems in general. The above method also helps with 7a --Dlamba 23:43, 8 September 2009 (UTC)
