# Z-Transform of a^|n|

I am wanting to compute the Z-transform of $f(n) = a^{|n|}$ . 'a' is a positive constant.

Looking at the transform table, I found that Z-transform for $a^n u(n)$ is available from the tables and is $\frac{Z}{Z-a}$. Where $u(n)$ is the unit step function.

I am trying to decompose $f(n)$ as -- $f(n) = a^n u(n) + a^{-n} u(-n) + \delta(n)$

Then using the table to find $Z( f(n) )$ as -- $\frac{Z}{Z-a} + \frac{Z^{-1}}{Z^{-1} - a} + 1$

Can anyone tell me if this approach is correct and/or suggest an alternate way. Thanks!

• For comparison: Wolfram Alpha: Z transform of a^|n| – rickhg12hs Nov 26 '13 at 18:11
• Don't think you need $\delta(n)$ in your decomposition. – rickhg12hs Nov 26 '13 at 20:55
• As per wolfram the z-transform is $\frac{z}{z-a}$. Is this same as ignoring the negative part ie. when n is negative. Please help about how this comes up. – mkuse Nov 27 '13 at 6:34
• Wolfram's Z-Transform is one-sided/unilateral. Here's the two-sided/bilateral Z-Transform sum calculated directly: Sum[a^|n| * z^(-n), n from -infinity to +infinity] – rickhg12hs Nov 29 '13 at 8:56
• I think you are decomposing properly but replace delta[n] by -delta[n] in f[n] equation. Just check for n=0 if you are unsure . – Phani Sep 5 '15 at 18:50

$f(n) = a^n u(n) + a^{-n} u(-n)$
$Z( f(n) ) = \frac{Z}{Z-a} + \frac{Z^{-1}}{Z^{-1} - a}$ EDIT: Looks like it is wrong answer. :(
Given sequence is $f(n) = a^{|n|}$ . 'a' is a positive constant.$1) Let's decompose it into sum of 2 sequences:$f(n) = a^n u(n) + a^{-n} u(-n-1)$2) If we define$f_1(n) = a^n u(n)$and$f_2(n) = a^{-n} u(-n-1)$Then$Z( f(n) ) = Z( f_1(n) )+Z( f_2(n) )$3)$Z( f_1(n) ) = \sum_{n=-\infty}^{\infty} a^n u(n) z^{-n} = \sum_{n=0}^{\infty} (az^{-1})^n = \frac{1}{1-az^{-1}} = \frac{z}{z-a} $4)$Z( f_2(n) ) = \sum_{n=-\infty}^{\infty} a^{-n} u(-n-1) z^{-n} = \sum_{n=-\infty}^{-1} (az)^{-n} = \sum_{n=1}^{\infty} (az)^{n} = -1+1+\sum_{n=1}^{\infty} (az)^{n} = -1+\sum_{n=0}^{\infty} (az)^{n} = -1 + \frac{1}{1-az} = \frac{-1+az+1}{1-az} = \frac{az}{1-az} $5)$Z( f(n) ) = Z( f_1(n) )+Z( f_2(n) ) = \frac{z}{z-a} + \frac{az}{1-az} = \frac{z}{z-a} - \frac{z}{z-\frac{1}{a}} $If there is anything unclear or wrong in this derivation, please comment. As for wolfram answer - I don't know why it is so. I never used wolfram before. But it perfoms symbolic computations and you must be sure that your assumptions about range of n (does$n\in[-\infty,\infty]$or$ n\in[0,\infty]$) for example should coinside with wolfram assumptions. • I believe this answer is not consistent with what we get with wolfram wolframalpha.com/input/?i=Z+transform+of+a%5E%7Cn%7C Can someone also explain why this discrepancy? – mkuse Nov 28 '13 at 3:28 • Ok, I edited original answer and put detailed derivation in it. – casador Nov 28 '13 at 14:25 • @casador, I'm wondering whether the$f_2(n)$in the above answer should be$a^{−n-1}u[−n−1]$? then can we use property of folding and time-shift to find the z-transform?$f_2(n)=a^{-n-1}u[-n-1]=a^{-(n+1)}u[-(n+1)] \xrightarrow{z-transform\ with \ fold\ and\ shift } z^{-(-1)}F(1/z)=zF(1/z)=z\frac{1}{1-az^{-(-1)}}=\frac{z}{1-az}\$ – Feng Shi Sep 5 '15 at 7:20