Why define the z-transform differently from the Laplace transform?

The mapping between the $$z$$-plane and the $$s$$-plane is defined by $$z=e^{sT}$$ where $$T$$ is the sampling period. A result of this mapping is that the shapes of the planes are quite different, e.g., the stable region of the $$z$$-plane is inside the unit circle, whereas the stable region of the $$s$$-plane is the left half-plane (LHP).

Why have people chosen to define the $$z$$-plane like this? Why not something like this instead: $$z=sT$$ so that the $$z$$-transform becomes $$X(z) = \sum_{k=0}^{\infty} x[k] e^{-zk}$$ Defining $$z$$ in this way makes the $$z$$-transform much more similar to the Laplace transform, and makes the structure of each plane pretty much the same, so you don't have to 'relearn' as much when moving to the $$z$$-transform.

Furthermore, if it is more beneficial to use $$z=e^{sT}$$ in the discrete domain, why wouldn't it also be beneficial to use a similar form in the time domain? If you did that, the Laplace transform would look like: $$X(s) = \int_0^{\infty} x(t) s^{-t}\,dt$$

So I guess my question is really: why define the Laplace and $$z$$-transforms differently? What is it about discrete vs. continuous that makes one form preferable over another in a specific domain?

• Maybe useful: electronics.stackexchange.com/questions/86489/…. Commented Mar 31, 2019 at 10:44
• Thanks but that doesn't really help. I understand the difference between the transforms fine. My question is just why people have chosen to define the $z$-transform differently from the Laplace transform (i.e.: why not just change the integral to a sum and set $z=sT$? - that seems much more natural and has the benefit of preserving the structure of the $s$-plane in the discrete domain). Commented Apr 1, 2019 at 4:05
• The main reason why they are defined like that is because many operations and concepts have different meaning on continuous and discrete time domains. Applying a time shift to a continuous-time signal is not the same as applying a time shift to a discrete-time signal. Differentiating a discrete-time signal is not the same as differentiating a continuous-time signal. Periodicity is not quite the same for continuous-time and discrete-time. Therefore, it does not make much sense to define identically two transformations that deal with essentially two different things Commented Apr 3, 2019 at 9:25
• @bertozzijr how is periodicity different and how does the standard definition of the $z$-transform address that? Similarly with differentiation. Thanks. It might be better to write it out as an answer if it takes up a lot of space. Commented Apr 7, 2019 at 10:34

I think I figured out a satisfactory answer to this myself.

Why define the Laplace transform with $$e^{-st}$$?

You might think that you could define the Laplace transform as follows: $$F(s) = \int_0^{\infty}f(t)s^{-t}\,dt$$ to be consistent with the $$z$$-transform. This has issues however. Using the above definition, $$s = e^{\ln{|s|}}e^{i\,\mathrm{Arg}\,s} = e^{\sigma} e^{i\omega}$$ where $$\sigma=\ln{|s|}$$ and $$\omega = \mathrm{Arg}\,s$$.

The problem here is that $$\omega=\mathrm{Arg}\,s$$ will only give values in the range $$(-\pi,\pi]$$ (the principal range), so using $$s^{-t}$$ to define the Laplace transform prevents you from analyzing most frequencies. This highlights an important point: $$e^{-(\sigma+i\omega) t}$$ is not equal to $$\left(e^{\sigma+i\omega}\right)^{-t}$$ for all $$t\in\mathbb{R}$$. In fact, these two expressions are only equal when $$t$$ is some multiple of the period of $$\exp(i\omega)$$ (i.e., $$\omega/2\pi$$). If we use the definition of complex exponentiation, we find that $$\left(e^{\sigma+i\omega}\right)^{-t} = e^{-\sigma t}e^{-i\,\mathrm{Arg}(e^{i\omega})t}$$ which is almost the same as $$e^{-(\sigma+i\omega)t}$$ except for that fact that $$\mathrm{Arg}\,{e^{i\omega}}$$ is not equal to $$\omega$$, but instead equal to $$\omega$$ when wrapped around into the principal range of $$(-\pi,\pi]$$.

Why define the $$z$$-transform with $$z^{-n}$$?

It's possible to define the $$z$$-transform like the Laplace transform: $$F(z) = \sum_{n=0}^{\infty}f[n]e^{-zn}$$ however there are a few benefits to using $$z=e^{sT}$$. First of all, when dealing with discrete systems, differential equations turn into difference equations. A result of this is that instead of getting derivatives in the equation, we get delays.

We use the Laplace transform on differential equations since it turns differentiation into multiplication: $$\mathcal{L}\left\{\frac{dx}{dt}\right\} = sX(s)$$

On the other hand, we use the $$z$$-transform on difference equations, since it has a similar property for delays: $$\mathcal{Z}\left\{x[n-1]\right\} = z^{-1}X(z)$$

If we defined the $$z$$-transform like the Laplace transform, then we'd end up with a multiplication by an exponential instead of the above, and $$z$$-domain transfer functions would be much uglier as a result.

Another convenience of defining the $$z$$-transform the way it is, is that it "enforces" the notion that sampling causes aliasing of frequencies. Due to the definition $$z=e^{sT}$$, the $$z$$ transform cannot distinguish between frequencies separated by a multiple of the Nyquist limit (half the sampling rate).

Summary

• The Laplace transform cannot be defined in the same manner as the $$z$$-transform, since it would prevent you from analyzing frequencies outside the range $$(-\pi,\pi]$$.
• The above point is not a problem for $$z$$-transforms, since $$z$$-transforms deal with discrete systems, which cannot distinguish between frequencies separated by a multiple of $$2\pi/T$$ (where $$T$$ is the sampling period) due to aliasing.
• The definition of $$z=e^{sT}$$ is more suitable for discrete systems since they involve delays instead of derivatives, so using $$z=e^{sT}$$ makes the resulting transfer functions simpler.
• The point is the inverse transform : the Laplace transform of a sequence will be $2\pi$-periodic so the inverse Laplace transform integral diverges. That's the same idea as for the Fourier series vs Fourier transform. Letting $z= e^{-s}$ means we are looking only at one period. Commented Apr 11, 2019 at 11:32
• @reuns Why would periodicity make it not invertible? Isn’t the DTFT invertible? (which would equal the Laplace transform of a sequence along the imaginary axis if in the ROC.) (In the sense that you can recover the sampled impulse train from it, not the original signal). Commented Apr 11, 2019 at 11:41
• Similarly, isn’t the Fourier transform of a sampled signal invertible (again, in the sense that you can recover the samples from it, not some original continuous signal). Commented Apr 11, 2019 at 11:44
• Just because it’s periodic, doesn’t mean you can’t recover the original discrete signal. Commented Apr 11, 2019 at 11:45
• The inverse Laplace transform integral en.wikipedia.org/wiki/… Commented Apr 11, 2019 at 11:46