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Consider the signal <math>x[n]=a^{n}u[n]</math> | Consider the signal <math>x[n]=a^{n}u[n]</math> | ||
| − | <math>X(z)=\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n}=</math> | + | <math>X(z)=\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n}=\sum_{n=0}^{\infty}(az^{-1})^{n}</math> |
| + | |||
| + | for convergence, | ||
| + | |||
| + | <math>\left| az^{-1}\right|<1</math>, or equivalently <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> | ||
Revision as of 19:49, 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} $
