# For unit step $g(t) = u(t)$, why does $G(z) = \frac{z}{1-z}$, whereas $G(s) = \frac{1}{s}$?

Edit: The source of my confusion is over the existence of multiple s-z mappings. After researching these mappings and where they came from, I couldn't find why $$z=e^{sT}$$ can be ignored and replaced by the other mappings. Isn't $$z=e^{sT}$$ intrinsic to the very definition z-transform? $$G(s) = \int_{-\infty}^\infty {g(t) e^{-st}dt} \space\space \xrightarrow{\text{t=kT}} \space \space G(z)=\sum_{k=-\infty}^{\infty} {g(kT)e^{-skT}}= \sum_{k=-\infty}^{\infty} {g[k]z^{-k}}$$

One source explained $$z=e^{sT} \approx \frac{1+sT/2}{1-sT/2}$$ I didn't find this explanation satisfactory though, since $$z=e^{sT}$$ came from the Laplace Transform of a sampled function $$g(kT)$$, so how does an approximation of z result at all in the same transform. Additionally, the forward difference mapping ($$s=z-1$$) is in no way $$\approx z=e^{sT}$$.

After much pondering, my hypothesis is:

• The z-transform, similar to the Laplace transform, is simply a way to transform a discrete function from its original function space to another.
• The new space for ZT is based on the basis function $$z^{-k}$$, whereas in LT it is $$e^{-st}$$. $$z$$ can be anything. What the z-space represents, can be linked to the more familiar s-space representation based on the relationship between $$z$$ and $$s$$.
• The most intuitive map is $$z=e^{sT}$$, which comes out from the FT of a sampled function. However, there can exist other mappings.
• These mappings do not change the rules for stability and causality in z-plane, since these rules are derived based on analysing ROC properties of $$G(z)= \sum_{k=-\infty}^{\infty} {g[k]z^{-k}}$$ (see: https://www.vikramuniv.ac.in/files/academic/e-Resources2020-21/engg2020-21/BE_6_SEM_EL_DSP-ROC_Z_TRANSFORM-AMIT_THAKUR.pdf). They simply change the mapping of poles and zeroes from s to z plane.

Is this correct? If so, my final question is, why is there no one optimal mapping? What is wrong with any of the ones mentioned by Matt?

Original post:

I understand mathematically why $$G(s) = \frac{1}{s}$$ and $$G(z) = \frac{z}{1-z}$$ in this case (simply a case of working out the integrals/summations for $$g(t) = u(t)$$).

However, I tried to think about why $$G(s)$$ maps to $$G(z)$$ here, by mapping the poles and zeroes in $$s$$-domain to $$z$$-domain under the mapping function $$z = e^{sT}.$$ I see why we have a pole at $$z=1$$, but I do not see where the zero at $$z=0$$ comes from. This would imply we have a zero at $$s=-\infty$$ .

• You must be talking about the unit step, not the unit impulse, right? Nov 21, 2023 at 13:42
• Does this help? Nov 21, 2023 at 13:45
• @Matt correct. Thanks! Sorry for the delay - have been doing tutorial sheets. Nov 23, 2023 at 0:56
• BTW, even though we (disciples of O&S) like $g[n]$, we don't extend that to $G(z)$. The square-corner brackets are for discrete argument. Like an index. $g[n]$ is the DSP way of indicating a sequence, same as $g_n$ where $n$ is only an integer. If the argument is continuous, then round-corner parenths are used. So it's not "$g[kT]$", but it's $g(kT)=g[k]$ Nov 23, 2023 at 6:34

You wonder why there are several mappings from the continuous domain to the discrete domain, and why we don't just use the optimal mapping $$z=e^{sT}$$.

Let's see how we could use that mapping. From $$z=e^{sT}$$ we obtain

$$s=\frac{1}{T}\log z\tag{1}$$

If we have a rational transfer function $$H(s)$$ and we replace $$s$$ by $$(1)$$ we end up with a function of $$z$$ that is transcendental. However, using delays, additions, and multiplications we can only implement rational functions of $$z$$. That's why we are looking for transformations that map a rational function $$H(s)$$ to another rational function $$H_d(z)$$. Now it shouldn't be surprising that there are several methods that achieve that, each of them with specific advantages and drawbacks. But all of them result in an approximation of the original transfer function by a realizable rational function $$H_d(z)$$.

As a final remark, you mention that you don't see how the forward difference comes even close to the ideal mapping. However, it's just a linear approximation of that mapping:

$$z=e^{sT}\approx 1+sT,\qquad T\ll 1$$

from which we obtain the mapping

$$s=\frac{z-1}{T}$$

which is just the forward difference.

Note that in the original answer below I assumed a normalized $$T=1$$.

Answer to the original version of the question:

The transfer function $$G(s)=1/s$$ is the Laplace transform of the unit step function $$u(t)$$. $$G(s)=1/s$$ describes an ideal integrator. There is no single optimal mapping from continuous time to discrete time. If you want to discretize the continuous step function $$u(t)$$, there are several reasonable methods. Obviously, we can sample $$u(t)$$ at $$t=nT$$. This will result in a sequence with values $$1$$ for $$n>0$$ and values zero for $$n<0$$. The question is what we do at $$t=0$$ where $$u(t)$$ has a discontinuity. The three most obvious choices are

1. set that sample to $$1$$
2. set that sample to $$0$$
3. set that sample to $$1/2$$

The first choice results in the standard unit step sequence $$u[n]$$, which equals $$1$$ for $$n=0$$. The second choice gives a delayed unit step sequence $$u[n-1]$$, and the last choice results in the average of the previous two: $$\frac12u[n]+\frac12u[n-1]$$.

What does that mean in the $$z$$-domain? The first choice corresponds to the discrete-time transfer function

$$G_1(z)=\frac{z}{z-1}\tag{1}$$

A delay corresponds to a multiplication with $$z^{-1}$$, hence

$$G_2(z)=\frac{1}{z-1}\tag{2}$$

And finally, the average of $$(1)$$ and $$(2)$$ is

$$G_3(z)=\frac12\frac{z+1}{z-1}\tag{3}$$

The three discrete-time transfer function $$(1)-(3)$$ can be obtained from $$G(s)$$ via the following mappings:

1. $$s\rightarrow 1-z^{-1}$$
2. $$s\rightarrow z-1$$
3. $$s\rightarrow 2\displaystyle\frac{z-1}{z+1}$$

Method $$1$$ is the backward Euler method, method $$2$$ is the forward Euler method, and method $$3$$ is the bilinear transform (Tustin method). All three methods give valid approximations of an ideal continuous-time integrator. For discretizing other, more general continuous-time systems, the bilinear transform is often the most useful of the three. Note that there are also other ways for discretizing continuous-time systems. These other methods cannot generally be described by a mapping from the $$s$$-plane to the $$z$$-plane. Two examples of such methods are the impulse-invariant method and the step-invariant method, which sample the system's impulse response and step response, respectively.

Concerning stability, the backward Euler method and the bilinear transform always map a stable continuous-time system to a stable discrete-time system. This is not the case with the forward Euler method. A stable continuous-time system might be mapped to an unstable discrete-time system, depending on the pole locations in the $$s$$-plane.

• Thanks so much @Matt. Sorry it's taken a while to respond, I've been studying a lot. Please do see the updates made to my question! I believe the edits should show you where I am getting confused, and whether or not I have figured out the explanation. Nov 23, 2023 at 3:24
• thank you that answers my question brilliantly. For certainty, could you confirm whether my original hypothesis (excl. the parts which you have corrected/altered in your answer) was accurate? Nov 23, 2023 at 12:20
• @ThePhysicsOverthinker: I think your understanding is correct. If you have any doubts you could formulate a new question. Nov 23, 2023 at 12:53