The Z-transform for a discrete signal is defined as $X(z) = \sum_{n = -\infty}^{\infty} x[n]z^{-n}$. I was wondering why we invert the exponent of $z$, rather than define it as $X(z) = \sum_{n = -\infty}^{\infty} x[n]z^{n}$, which to my mind seems like a more natural definition.

I've not been able to find an answer for this anywhere. Both definitions seem to give pretty much the same theory (but different domains of convergence in applications).

  • $\begingroup$ If you try factoring the denominator of $$\dfrac{1}{R^2-5R+6}$$ and decompose it as partial fractions, you will see why we need $z^{-n}$ instead of $z^{n}$ $\endgroup$
    – AgentS
    Jul 7, 2018 at 14:29
  • 1
    $\begingroup$ @rsadhvika That shows nothing. The partial fraction decomposition doesn't force any choice of Laurent series expansion. Just like you can expand on $|R|>3$, you can also expand in $2<|R|<3$ and in $|R|<2$. $\endgroup$
    – user566930
    Jul 7, 2018 at 14:33
  • $\begingroup$ @minghan that doesn't show anything yet, that's just a start. If the OP responds, I'd love to guide him/her $\endgroup$
    – AgentS
    Jul 7, 2018 at 14:34
  • $\begingroup$ @rsadhvika There is no theoretical reason. So, nothing that you intend to deduce from there will be an explanation. $\endgroup$
    – user566930
    Jul 7, 2018 at 14:36
  • 3
    $\begingroup$ I would be surprised if it isn''t a convention chosen to align with exp(-j2*pi*ft) in the forward Fourier transform (where the opposite sign works equally well). $\endgroup$ Jul 7, 2018 at 14:41

2 Answers 2


The Wikipedia article Z-Transform explains that

In geophysics, the usual definition for the $Z$-transform is a power series in $z$ as opposed to $z^{−1}$.

and continues

The two definitions are equivalent; however, the difference results in a number of changes. For example, the location of zeros and poles move from inside the unit circle using one definition, to outside the unit circle using the other definition

Thus it is a matter of convention or preference. Something analogous applies to the Laplace transform and Fourier transform, especially for Fourier transform with differing factors.

{EDIT: 2023] The English Wikipedia article has been edited and now the quotes I used in 2018 are no longer present in the article, however the book reference to E. R. Kanasewich, Time Sequence Analysis in Geophysics, 3rd edition, 1981, page 186 still exists as confirmed by Google Books.

The $z$ transform was reintroduced by Hurewicz (1947) using $z^{-1}$ instead of $z$ as defined in 13.2-1. This sign convention is common in engineering texts and such authors as Jury (1964), Ford and Hearne (1966). Bath (1974), Kulhanek (1976) and Claerbout (1976), etc. The classical definition is used in publications by Robinson (1963), Robinson and Treitel (1965), Van Nostrand (1971) and in this book. The ambiguity in sign should not cause confusion as long as a consistent definition is used throughout.

I think that the reason for the convention is due to the focus on causal systems

In control theory, a causual system (also known as a physical or nonanticipative system) is a system where the output depends on past and current inputs but not future inputs -- i.e., the output $y(t_0)$ depends only on the input $x(t)$ for values of $t\le t_0.$

The causality of systems also plays an important role in digital signal processing, where filters are constructed so that they are causal, sometimes by altering a non-causal formulation to remove the lack of causality so that it is realizable.

Notice the prototypical case of the signal $x[n]$ given by the unit step function $u[n] = 1$ iff $n\ge0$. Its $Z$-transform is given as $1/(1-z^{-1})$ with ROC (region of convergence) given as $|z|>1$. This is an example of a causal system since $u[n] = 0$ if $n<0$. Notice here that the rational function $1/(1-z^{-1})$ is expanded as a power series in $z^{-1}$ as $1/(1-z^{-1}) = 1 + z^{-1} + z^{-2} + \dots.$ The non-causual signal $x[n] = -u[-n-1]$ is given the $Z$-transform $1/(1-z^{-1})$ with ROC (region of convergence) given as $|z|<1$. In this case, the expansion is in powers of $z$ as follows: $ 1/(1-z^{-1}) = z/(z-1) = -z - z^2 - z^3 - \dots. $


I'll try to explain why $z$ transform is easy to use in control systems and not power series. Consider a system with block diagram

enter image description here

It has the system functional $$\dfrac{Y}{X}=\mathcal{H(R)}=\dfrac{1}{1-\mathcal{R-R^2}}=\sum\limits_{n=-\infty}^{\infty}h[n]\mathcal{R}^{\color{red}{n}}$$

You could try long division or partial fractions and get the unit sample response $h[n]$. They all work pretty good if all you want is just the unit sample response. The only drawback is you can't see the poles directly in above form. So we engineers, being lazy, replace $\mathcal{R}$ by $\dfrac{1}{z}$ so that the roots of the quadratic in $z$ in the denominator give the poles. After this replacement, we call it the system function :

$$H(z) = \dfrac{Y(z)}{X(z)}=\dfrac{z^2}{z^2-z-1}=\sum\limits_{n=-\infty}^{\infty}h[n]z^{\color{red}{-n}}$$

  • $\begingroup$ @Somos with the system functional, you have to factor it as $(1-\alpha R)(1-\beta R)$ then $\alpha, \beta$ will be poles right ? $\endgroup$
    – AgentS
    Jul 7, 2018 at 16:57
  • $\begingroup$ With the system function, the factoring is $(z-\alpha)(z-\beta)$ and the poles are easy to see. My professor says this is the only reason for doing $\mathcal{R} \to \dfrac{1}{z}$ for DT signals $\endgroup$
    – AgentS
    Jul 7, 2018 at 17:01
  • $\begingroup$ $a,b$ are the roots, they are not the poles. If we do partial fractions and expand, they will not be the base of geometric series right ? Poles in that case will be $1/a$ and $1/b$ $\endgroup$
    – AgentS
    Jul 7, 2018 at 17:03
  • $\begingroup$ We do a similar thing for CT signals. We replace the integrator $\mathcal{A}\to \dfrac{1}{s}$ and use the laplace transform $\endgroup$
    – AgentS
    Jul 7, 2018 at 17:06
  • $\begingroup$ @Somos I think we're in same page. You want to solve it in terms of R and take the reciprocal of the roots to get the poles. I want to replace R by 1/z and solve to get the poles directly. $\endgroup$
    – AgentS
    Jul 7, 2018 at 17:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .