Z transform of $ 2^{-|n|} $

Question

Dears,

I'm trying to compute the Z-transform of $$ x(n) = 2^{-|n|} $$. My procedure is as follows:

Using definition of Z transform:

$$ X(z) = \sum_{n=-\infty}^{\infty}2^{-|n|} z^{-n} = \sum_{n=-\infty}^{0}2^{-|n|} z^{-n} + \sum_{n=1}^{\infty}2^{-|n|} z^{-n} $$

$$ X(z) = \sum_{n=-\infty}^{0}2^{-n} z^{-n} + \sum_{n=1}^{\infty}2^{-n} z^{-n} $$

$$ X(z) = \sum_{n=0}^{\infty}2^{n} z^{n} + \sum_{n=0}^{\infty}2^{-n} z^{-n} - 1 $$

$$ X(z) = \sum_{n=0}^{\infty}(2z)^{n} + \sum_{n=0}^{\infty}(2z)^{-n} - 1 = \sum_{n=0}^{\infty}(2z)^{n} + \sum_{n=0}^{\infty}(\frac{1}{2z})^{n} - 1 $$

Using Geometric Series Formula:

$$ X(z) = \frac{1}{1-2z} + \frac{1}{1-\frac{1}{2z}} - 1 $$

But according to the geometric series formula, there is a region of convergence (ROC):

For the left fraction i get ROC: $$|2z|<1$$ and for right fraction i get ROC: $$|\frac{1}{2z}|<1$$

Solving the inequalities:

Left fraction ROC:

$$ (- \frac{1}{2} , \frac{1}{2} )$$

Right fraction ROC:

$$ (\frac{1}{2},\infty) , (-\infty, -\frac{1}{2}) $$

Therefore, there is no intersection ROC for both fractions.

My question is: It means the Z-transform does not exist?

Thanks!!

Answer 1

Z-transform exists because for $n<0$, $|n|=-n$, the mistake is in in your second line (in the eq): $$\begin{align}X(z) &= \sum_{n=-\infty}^{0}2^{n} z^{-n} + \sum_{n=1}^{\infty}2^{-n} z^{-n}\\&=\sum_{n=0}^\infty (z/2)^n + \sum_{n=0}^\infty (1/2z)^n-1\\&=\frac{1}{1-z/2}+\frac{1}{1-1/2z}-1\end{align}$$

And the ROC is $|z/2|<1 \cap |1/2z|<1\rightarrow 1/2<|z|<2$.