Dgdrxrt

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?

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:
$$
F(s) = int_0^inftyf(t)s^-t,dt
$$
to be consistent with the $z$-transform. This has issues however. Using the above definition,
$$
s = e^lne^iargs = e^sigma e^iomega
$$
where $sigma=ln$ and $omega = args$.

The problem here is that $omega=args$ 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:
$$
e^-(sigma+iomega) t
$$
is not equal to
$$
left(e^sigma+iomegaright)^-t
$$
for all $tinmathbbR$. In fact, these two expressions are only equal when $t$ is some multiple of the period of $exp(iomega)$ (i.e., $omega/2pi$). If we use the definition of complex exponentiation, we find that
$$
left(e^sigma+iomegaright)^-t = e^-sigma te^-iarg(e^iomega)t
$$
which is almost the same as $e^-(sigma+iomega)t$ except for that fact that $arge^iomega$ 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:
$$
F(z) = sum_n=0^inftyf[n]e^-zn
$$
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:
$$
mathcalLleftfracdxdtright = sX(s)
$$

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

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 $2pi/T$ (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.