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* Oops! Yeah the ROC was supposed to be <math>|Z|<1</math>. But, technically, the statement is correct, just a bit weird. --[[User:Mboutin|Mboutin]] 18:12, 7 September 2009 (UTC) | * Oops! Yeah the ROC was supposed to be <math>|Z|<1</math>. But, technically, the statement is correct, just a bit weird. --[[User:Mboutin|Mboutin]] 18:12, 7 September 2009 (UTC) | ||
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[[ ECE438 (BoutinFall2009)|Back to ECE438 (BoutinFall2009)]] | [[ ECE438 (BoutinFall2009)|Back to ECE438 (BoutinFall2009)]] | ||
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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. | 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. | ||
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Revision as of 17:35, 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 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)
Back to ECE438 (BoutinFall2009)
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.
--Dlamba
