(New page: Category:2010 Fall ECE 438 Boutin ---- From the first question, we knew that <math> -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\!</...) |
|||
| (4 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
[[Category:2010 Fall ECE 438 Boutin]] | [[Category:2010 Fall ECE 438 Boutin]] | ||
| − | + | ---- | |
| + | == Solution to Q2 of Week 5 Quiz Pool == | ||
---- | ---- | ||
| Line 24: | Line 25: | ||
\,\!</math> | \,\!</math> | ||
---- | ---- | ||
| − | Back to | + | <span style="color:green"> I saw many students made a mistake about this kind of question, such as </span> |
| + | |||
| + | <math> -a^{n}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \;\;\; \text{This is not correct!!} \,\!</math> | ||
| + | |||
| + | <span style="color:green"> which the time-shifting is applied only to the unit step and not to the power of <math>a</math>. </span> | ||
| + | <span style="color:green"> Thus, you should be very careful when you use the time-shifting property. You have to apply time-shifting to all <math>n</math>, when you find the inverse Z-transform. </span> -[[User:han83|Jaemin]] | ||
| + | ---- | ||
| + | Back to [[ECE438_Week5_Quiz|Lab Week 5 Quiz Pool]] | ||
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]] | Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]] | ||
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]] | Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]] | ||
Latest revision as of 16:39, 19 September 2010
Solution to Q2 of Week 5 Quiz Pool
From the first question, we knew that
$ -a^{n}u[-n-1] = \mathcal{Z}^{-1}\bigg\{\frac{1}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\! $
And the time-shifting property of Z-transform is defined as
$ x[n-k] = \mathcal{Z}^{-1}\bigg\{z^{-k}X(z)\bigg\} \text{ when } x[n] = \mathcal{Z}^{-1}\bigg\{X(z)\bigg\}\,\! $
Therefore, if we use the time-shifting property of Z-transform, then
$ -a^{n-3}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \,\! $
Combined with the result from the linearity of Z-transform, then
$ \begin{align} \mathcal{Z}^{-1}\bigg\{\frac{2z^{-3}}{1-az^{-1}}\bigg\} \text{ for } |z|<|a| &= -2a^{n-3}u[-(n-3)-1], \\ &= -2a^{n-3}u[-n+2] \end{align} \,\! $
I saw many students made a mistake about this kind of question, such as
$ -a^{n}u[-(n-3)-1] = \mathcal{Z}^{-1}\bigg\{\frac{z^{-3}}{1-az^{-1}}\bigg\} \text{ where } |z|<|a|. \;\;\; \text{This is not correct!!} \,\! $
which the time-shifting is applied only to the unit step and not to the power of $ a $. Thus, you should be very careful when you use the time-shifting property. You have to apply time-shifting to all $ n $, when you find the inverse Z-transform. -Jaemin
Back to Lab Week 5 Quiz Pool
Back to ECE 438 Fall 2010 Lab Wiki Page
Back to ECE 438 Fall 2010
