Question 9.1: z transform of a noncausal signal Find the z transform of x[......
z transform of a noncausal signal
Find the z transform of x[n]=K\alpha^{|n|},\alpha\in\mathbb{R}
Its variation with n depends on the value of α (Figure 9.9). It can be written as
\mathbf{x}[n]=K({\alpha}^{n}\mathbf{u}[n]+\alpha^{-n}\mathbf{u}[-n]-1)
If |\alpha|\geq1 then |\alpha|\geq|\alpha^{-1}|, no ROC can be found and it does not have a z transform. If |\alpha|\lt 1 then |\alpha|\lt |\alpha^{-1}|, the ROC is |\alpha|\lt z\lt \left|\alpha^{-1}\right| and the z transform is
K\alpha^{|n|}\overset{\mathcal Z}{\longleftrightarrow } K\sum_{n=-\infty}^{\infty}\alpha^{|n|}z^{-n}=K\left[\sum_{n=0}^{\infty}(\alpha z^{-1})^{n}+\sum_{n=-\infty}^{0}(\alpha^{-1}z^{-1})^{n}-1\right],\;|\alpha|\lt z\lt \left|\alpha^{-1}\right|
K\alpha^{\left|n\right|}\overset{\mathcal Z}{\longleftrightarrow } K\left[\sum_{n=0}^{\infty}\left(\alpha z^{-1}\right)^{n}+\sum_{n=0}^{\infty}\left(\alpha z\right)^{n}-1\right],\;\left|\alpha\right|\lt z\lt \left|\alpha^{-1}\right|
This consists of two summations plus a constant. Each summation is a geometric series of the form \;_{}\sum_{n=0}^{\infty}r^{n} and the series converges to 1/(1-r)\ {\mathrm{if~}}|r|\lt 1.
K\alpha^{|n|}\overset{\mathcal Z}{\longleftrightarrow } \left\lgroup\frac{z}{1-\alpha z^{-1}}+\frac{1}{1-\alpha z}-1\right\rgroup =K\left\lgroup\frac{z}{z-\alpha}-\frac{z}{z-\alpha^{-1}}\right\rgroup ,\;\;|\alpha|\lt z\lt \left|\alpha^{-1}\right|