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<math>X(z)=\sum_{n=0}^{\infty}(az^{-1})^{n}=\frac{1}{1-az^{-1}}=\frac{z}{z-a}</math>, <math>\left| z\right|>\left| a\right|</math> | <math>X(z)=\sum_{n=0}^{\infty}(az^{-1})^{n}=\frac{1}{1-az^{-1}}=\frac{z}{z-a}</math>, <math>\left| z\right|>\left| a\right|</math> | ||
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| + | Now let <math>x[n]=-a^{n}u[-n-1]</math> | ||
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| + | <math>X(z)=-\sum_{n=-\infty}^{\infty}a^{n}u[-n-1]z^{-n}=-\sum_{n=-\infty}^{-1}a^{n}z^{-n}</math> | ||
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| + | <math>=-\sum_{n=1}^{\infty}a^{-n}z^{n}=1-\sum_{n=0}^{\infty}(a^{-1}z)^{n}</math> | ||
| + | |||
| + | <math>\left| a^{-1}z\right|<1</math>, or equivalently math>\left| z\right|<\left| a\right|</math> | ||
Revision as of 19:58, 3 December 2008
Consider the signal $ x[n]=a^{n}u[n] $
$ X(z)=\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n}=\sum_{n=0}^{\infty}(az^{-1})^{n} $
for convergence,
$ \left| az^{-1}\right|<1 $, or equivalently $ \left| z\right|>\left| a\right| $
$ X(z)=\sum_{n=0}^{\infty}(az^{-1})^{n}=\frac{1}{1-az^{-1}}=\frac{z}{z-a} $, $ \left| z\right|>\left| a\right| $
Now let $ x[n]=-a^{n}u[-n-1] $
$ X(z)=-\sum_{n=-\infty}^{\infty}a^{n}u[-n-1]z^{-n}=-\sum_{n=-\infty}^{-1}a^{n}z^{-n} $
$ =-\sum_{n=1}^{\infty}a^{-n}z^{n}=1-\sum_{n=0}^{\infty}(a^{-1}z)^{n} $
$ \left| a^{-1}z\right|<1 $, or equivalently math>\left| z\right|<\left| a\right|</math>
