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== Solution to Q2 of Week 5 Quiz Pool == | == Solution to Q2 of Week 5 Quiz Pool == | ||
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Revision as of 15:42, 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 in 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
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