Revision as of 14:50, 19 September 2010 by Han83 (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


Solution to Q4 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} \,\! $


Back to Lab Week 5 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang