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!

  • $\begingroup$ For comparison: Wolfram Alpha: Z transform of a^|n| $\endgroup$
    – rickhg12hs
    Nov 26, 2013 at 18:11
  • $\begingroup$ Don't think you need $\delta(n)$ in your decomposition. $\endgroup$
    – rickhg12hs
    Nov 26, 2013 at 20:55
  • $\begingroup$ 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. $\endgroup$
    – mkuse
    Nov 27, 2013 at 6:34
  • $\begingroup$ 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] $\endgroup$
    – rickhg12hs
    Nov 29, 2013 at 8:56
  • $\begingroup$ 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 . $\endgroup$
    – Phani
    Sep 5, 2015 at 18:50

1 Answer 1


I believe the right way to decompose this function is:

$f(n) = a^n u(n) + a^{-n} u(-n)$

Then the answer for your question is:

$Z( f(n) ) = \frac{Z}{Z-a} + \frac{Z^{-1}}{Z^{-1} - a}$ EDIT: Looks like it is wrong answer. :(

Ok, let's start from the beggining and try to find right answer step by step

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.

  • $\begingroup$ 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? $\endgroup$
    – mkuse
    Nov 28, 2013 at 3:28
  • $\begingroup$ Ok, I edited original answer and put detailed derivation in it. $\endgroup$
    – casador
    Nov 28, 2013 at 14:25
  • $\begingroup$ @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? $f_2(n)=a^{-n-1}u[-n-1]=a^{-(n+1)}u[-(n+1)] \xrightarrow{z-transform\ with \ fold\ and\ shift } z^{-(-1)}F(1/z)=zF(1/z)=z\frac{1}{1-az^{-(-1)}}=\frac{z}{1-az}$ $\endgroup$
    – Feng Shi
    Sep 5, 2015 at 7:20

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