# Why not use a complex number in the exponent in the z-transform?

I am trying to comprehend how the z-transform has come to be similar but different from its continuous counterpart, namely the Laplace transform.

It seems to me that the most parallel and intuitive approach to the deduction of a generalized form of the DTFT would be to substitute $$j\omega$$ for $$z = \sigma + j\omega$$, i. e.: $$X(z) = \sum_{n = -\infty}^{+\infty} x[n]z^{-n} = \sum_{n = -\infty}^{+\infty} x[n]e^{-(\sigma + j\omega)n}$$

Instead, we find normally that $$z$$ is a polar-form complex number which is the product of its magnitude and $$e^{j\omega}$$. I can guess that probably such representation leads to a rectangular representation of the z-transform, eventually rendering the analysis of poles and zeros impractical due to the periodic nature of the DTFT, but I am not sure. Could anyone enlighten me?

• So just as you can think of the Fourier Transform as existing along the imaginary axis of the Laplace Transform's s-plane, you can think of the DTFT and DFT as existing on the unit circle of the z-plane. This also explains their cyclic nature quite nicely. DC (0 Hz) is at 1+0j; the most positive amd most negative frequencies are -1+0j. Jun 16, 2019 at 23:12

$$z^{-1}$$ is the unit delay operator.

$$z$$ is the unit advance operator.

$$s$$ is the derivative operator.

$$1/s$$ is the integration operator.

the utility of the Z and Laplace transforms is to solve (analyze, predict) constant coefficient differential and difference equations algebraically.

• just curious about notation, Stanley. why is the reciprocal of $z$ denoted $z^{-1}$ but the reciprocal of $s$ is $1/s$? just seems curious. Jun 18, 2019 at 3:59
• @robertbristow-johnson nothing special, just the way it's burnt in over the years
– user28715
Jun 18, 2019 at 4:04

Both the Laplace and Z transform are generalized versions of their discrete or continuous Fourier counter parts. To get the respective Fourier transform you would evaluate the Z-transform on the unit circle and the Laplace transform on the imaginary axis.

Hence it's most practical to represent Z-transform argument in polar coordinates since "frequency" is the phase of the complex number and to represent Laplace in rectangular coordinates since "frequency" is the imaginary part.

Unfortunately, it is the way Z-transform has been defined. Your definition would make it more intuitive.

However, since the Z-transform is used for discrete signals, you would deal with time delays more often than integrations and differentiations. For example, multiplying $$z^{-d}$$ for delay $$d$$ would be easier than multiplying $$e^{-zd}$$.

Hence, from a practical standpoint the current definition, $$X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n}$$ becomes easier to use.

You could also read the answer to a similar question explaining this.