For a (complex valued) sequence $(a_n)_{n\in\mathbb{N}}$ there is the associated generating function $$ f(z) = \sum_{n=0}^\infty a_nz^n$$ and the $z$-Transform $$ Z(a)(z) = \sum_{n=0}^\infty a_nz^{-n}$$ which only differ by the sign of the exponent of $z$, that is, both are essentially the same and carry the same information about the sequence, though encoded slightly differently. The basic idea is the same: associate a holomorphic function with the sequence and use complex calculus (or formal power series).

However, the engineering books I know which treat the $Z$-transform do not even mention the word "generating function" (well one does but means the generator of a multiscale analysis...) and the mathematics books on generating function do not mention the $Z$-transform (see for example "generatingfunctionology").

I am wondering: Why is that? Has one formulation some advantage over the other? Or is it just for historical reasons?

(BTW: There is not a tag for the $Z$-transform, and the closest thing I found was "integral-transforms"...)

Peter Mortensen
  • 627
  • 5
  • 11
  • 11,270
  • 34
  • 57
  • 1
    I believe the Z-transform sums over all integers $n$, not just positive $n$ ... sometimes? ... Engineering definitions are a bit slippery, and change meaning to whatever is useful at the time. – GEdgar Apr 26 '12 at 12:46
  • 1
    Maybe because the rules obeyed by the Z-transform look similar to what one is used to from working with the Laplace transform? – Hans Lundmark Apr 26 '12 at 13:36

2 Answers2


I see three questions here:

  • Shouldn't exist more awareness about the fact that Z-transform (ZT) and generating functions (GF) are almost the same thing?

I think so. I've always found this strange and unfortunate, and I'd like to see in every textbook about ZT or GF a footnote ("The 'generating functions' employed in combinatorial mathematics are basically the same thing as the Z-transform" and viceversa).

  • Are they (apart from the change of sign) really the same thing?

Formally, they are obviously the same thing, but the context is different:

In the Z-transform $x[n] \leftrightarrow X(z) $, the input is usually double-sided (the sum runs over all integers), the "right sided" transform is less used. Further, in signal processing, $x[n]$ is almost always one of these: 1) a signal, 2) the impulse response of a LTI filter (causal or not), 3) a (auto/cross) correlation function. Hence, $x[n]$ is typically either bounded and decreasing for $n\to \pm \infty$ (for the case of filters and correlations) or (for the case of stochastic signals) stationary zero-mean sequences.

The generating function, instead, is usually applied to right-sided sequences (i.e. any $f:\mathbb{N} \to \mathbb{R}$). Apart from that, they are arbitrary; they often grow without bounds.

Because the ZT is applied to double-sided input, then the mapping $x[n] \leftrightarrow X(z) $ is not one-to-one: to have a unique inverse, we need to specify a ROC (region of convergence) of $X(z)$, in the complex plane. For GF the problem of unicity does not arise, the ROC is implied. (However, as pointed out in a comment, the radius of convergence can be relevant to characterize some sequence properties).

The Z-transform $X(z)$ is not usually regarded as a formal series, but as a "true" complex function. And because of the AR/MA/ARMA models that are usually considered in classical signal processing, we almost always deal with rational functions, which can be characterized in terms of zeros and poles.

The ZT transform is naturally thought as a generalization of the Fourier transform, as typically $x[n]$ is square summable (with perhaps the addition of sinusoids - or countable Dirac deltas in the transform). This correspondence is given by the natural mapping $z \leftrightarrow e^{jw}$, i.e. the DTF is the ZT along the unit circle in the complex plane (same as the continuous Fourier transform is the Laplace transform along the $y$ axis). And the classic concepts (e.g. energy per frequency band) are normally pertinent and useful. In the GF scenario, we don't often think of Fourier transforms.

  • Why the different sign?

The different convention can be understood from the previous difference. Regarding the ZT as a generalization of the DFT, the negative sign is more natural (the input is expressed as a "synthesis" of sinusoids). BTW: this gives a ROC that for causal signals -or right handed transform- extends "to the exterior" of the largest pole; which in turns implies the common rule: a stable causal filter must have its poles inside the unit circle. For the GF, being just a formal series, it feels more natural to use positive exponents.

  • 56,395
  • 9
  • 64
  • 139
  • 9
    One small note: the radius of convergence of generating functions (and in particular, their poles) comes up regularly in analysis of algorithms, where the location of the first pole usually gives a pretty good asymptotic estimate for the coefficients. – Steven Stadnicki May 02 '12 at 18:28
  • http://math.stackexchange.com/users/785/steven-stadnicki : Could you give a reference for that? – kjetil b halvorsen Jun 13 '14 at 21:35
  • 3
    @kjetilbhalvorsen, check out e.g. Wilf's "generatingfunctionology" or (for a much more in depth discussion) Flajolet and Sedgewick's "Analytic combinatorics" (both are available for free as PDFs from their author's web pages). – vonbrand Jun 15 '14 at 02:01

Given a sequence of numbers $\{x[n] \colon n \in \mathbb Z\}$ the $z$-transform is defined as $$X(z) = \sum_n x[n]z^{-n}$$ which when evaluated at $z = \exp(j\omega)$ (where $j = \sqrt{-1}$ is what electrical engineers typically use for what mathematicians denote by $i$) gives $${X}(\exp(j \omega)) = \sum_n x[n] \exp(-j\omega n)$$ which is called the discrete-time Fourier Transform (DTFT) of the sequence. Engineers view this as slightly easier to use and remember than evaluating the generating function $$\hat{X}(D) = \sum_n x[n]D^{n}$$ (where $D$ denotes delay) at $D = \exp(-j\omega)$ to arrive at the same result. So, it is essentially a matter of convention.

Dilip Sarwate
  • 23,545
  • 2
  • 42
  • 109
  • 5
    This appears to restate the question more than it answers it, doesn't it? – hmakholm left over Monica Apr 26 '12 at 11:51
  • 8
    @HenningMakholm There are three questions asked by the OP. Why is that? Has one formulation some advantage over the other? Or is it just for historical reasons? My answer to the second question is that engineers think that their convention is slightly easier to use and remember. The _implicit_ answer to the third is "Yes": conventions tend to form and crystallize over time, and so the difference is essentially for historical reasons. So I don't think that I have merely restated the question. I have answered at least two of them. – Dilip Sarwate Apr 26 '12 at 12:07