We all know that $a^nu(n)$ has unilateral $\mathcal Z$-transform. But what is the $\mathcal Z$-transform of $a^n$? (bilateral) When i tried to solve, i got answer as 'zero'.

But bilateral Laplace transform of $e^t$ doesn't exist. Both are exponentials in discrete and continuous domain respectively. Considering the similarity between Laplace and $\mathcal Z$-transform, how to explain the above problem?

Below, this is how I got 'zero'

$$a^n=a^nu(n) + a^nu(-n-1),$$

Now taking $\mathcal Z$-transform on both sides we get $$\frac{z}{z-a}\quad \text{and}\quad \frac{-z}{z-a}$$ respectively which add to 'zero'

  • $\begingroup$ thanks for the edit, can you please let me know how to use proper notations?where can i learn it? $\endgroup$
    – spectre
    Commented Aug 29, 2015 at 8:32
  • $\begingroup$ Click the 'edit' button to see how I changed your formulas. You basically enclose formulas like \$formula\$ and use Latex for math formatting. $\endgroup$
    – Matt L.
    Commented Aug 29, 2015 at 9:12

1 Answer 1


In complete analogy with the bilateral Laplace transform of $x(t)=e^{-at}$ (which doesn't exist), the bilateral $\mathcal{Z}$-transform of $a^n$ doesn't exist either. The series


converges nowhere, simply because $a^n$ grows without bounds for $n\rightarrow -\infty$ if $|a|<1$, or for $n\rightarrow\infty$ if $|a|>1$. Of course, for $|a|=1$ there series doesn't converge either.


As for your computation of the $\mathcal{Z}$-transform of $a^n$, the mistake lies in the fact that in addition to the algebraic expression of the transform you also need to consider the region of convergence. If you split $a^n$ (as you did) as


you can compute the $\mathcal{Z}$-transform of both right-hand side expressions separately:

$$\mathcal{Z}\{a^nu[n]\}=\frac{z}{z-a},\quad |z|>|a|\\ \mathcal{Z}\{a^nu[-n-1]\}=-\frac{z}{z-a},\quad |z|<|a|\tag{2}$$

Note that the region of convergence (ROC) for the first part is outside the circle with radius $|a|$, whereas the ROC of the second part is inside the circle with radius $|a|$. The ROC of the total expression would be the overlap of the two ROCs, which is zero. Consequently, the sum doesn't converge anywhere and the $\mathcal{Z}$-transform of the total expression doesn't exist.

  • $\begingroup$ if we break the problem into right and left sided transforms, we will get Z-transform as '0', where both of them cancel each other.Then how can you say Z-transform doesn't exist? $\endgroup$
    – spectre
    Commented Aug 29, 2015 at 8:34
  • $\begingroup$ @spectre: That's not the case. Show us how you did it by editing your question, and we'll tell you where the mistake is. $\endgroup$
    – Matt L.
    Commented Aug 29, 2015 at 9:10
  • $\begingroup$ @spectre: I've edited my answer to include an explanation why the $\mathcal{Z}$-transform of $a^n$ doesn't exist. $\endgroup$
    – Matt L.
    Commented Aug 29, 2015 at 10:26
  • $\begingroup$ L:So, when can we add 2 Z-transform polynomials? $\endgroup$
    – spectre
    Commented Aug 29, 2015 at 12:10
  • $\begingroup$ @spectre: Adding only makes sense if the ROCs of the separate terms overlap. The ROC of the total sum is then the overlap of the different ROCs. If the overlap is zero, as in your example, there is no single value of $z$ for the which the sum of the terms exists. $\endgroup$
    – Matt L.
    Commented Aug 29, 2015 at 14:46

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