Revision as of 12:10, 19 September 2010 by Han83 (Talk | contribs)



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} \,\! $


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