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The definition of the z-transform is defined as \$z = e^{sT}\$ where "s" is complex frequency for continuous-time systems and "T" is the sample period. Why are rules such as the forward rectangular rule, or Tustin's method used instead of the definition?

Forward rectangular rule: \$s \leftarrow \frac{z-1}{T}\$ Tustin's rule: \$s \leftarrow \frac{2}{T} \frac{z - 1}{z + 1}\$

Edit to clarify my question: I am asking about the characteristics of the transform methods. If I transform G(s) to G(z), then why use Tustin's rule or forward rectangular rule instead of the definition? I would think the integration rules are just an approximation to the definition.

Edit #2:
What makes any given "s-to-z" map better or worse than any other? What are the characteristics of these maps and how do they compare with each other? In other words, if I map G(s) to G(z) using Tustin's method, how/why will this compare with mapping with the definition \$z = e^{sT}\$? The same question applies to other maps (forward rectangular, backward rectangular, etc). If I get different pole/zero locations using different maps, then how/why would I select a given result (G(z))? To reiterate my previous question, why use an approximation (Tustin's method etc) when one can use the definition?

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  • \$\begingroup\$ Not to the author, but interested readers: look up Lagrange-Boole and the shift operator and perhaps operator calculus, to start. Euler's backward (described by the OP as forward rectangular) and Tustin's just fall out with ease. \$\endgroup\$ – jonk Dec 4 '18 at 18:07
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\$\small z^{-1}\$ is the delay operator in the z-domain, ie, multiplying the z transform of a signal by \$\small z^{-1}\$ delays the signal by one time increment. For example, if \$\small X(z)\$ is the z-transform of the unit step sequence:

$$\small X(z)= z^{0}+z^{-1}+z^{-2}+z^{-3} ...$$

then $$\small z^{-1}X(z)= \:z^{-1}+z^{-2}+z^{-3} ...$$ which is clearly the delayed unit step (since \$\small z^0=1\$)

Similarly, multiplying a the laplace transform of the unit step by \$\small e^{-sT}\$ delays the step by \$\small T\: sec\$ in the time domain. Hence the relationship: $$\small z^{-1} \leftrightarrow e^{-sT}\:\: or\:\:z\leftrightarrow e^{sT}$$

This relationship is useful for pole-zero mapping between the s-, and z- domains, hence can be used for obtaining a digital controller from a continuous, Laplace prototype.

The bilinear transform ('Tustin' or many other names) is an algebraic method of doing, essentially, the same thing, but with generally better perfomance compared with pole-zero mapping, impulse invariant, step invariant, etc

It can be shown quite simply that the 1st order z-transfer function of integration (i.e. trapezoidal integration) is $$\small \frac{T}{2}\frac{(z+1)}{(z-1)}$$ and this corresponds to the Laplace integration operator: \$\large \frac{1}{s} \$

This leads to the very convenient algebraic substitution (i.e. the bilinear transform):$$\small s\leftrightarrow \frac{2}{T}\frac{(z-1)}{(z+1)}$$

Note that this transform preserves DC gain.

For example,say we require digital equivalent of the low-pass filter: $$\small G(s)= \frac{1}{3+s} $$

given a sampling increment, \$\small T=0.1\:sec\$

$$\small G(s)= \frac{Y(s)}{X(s)}=\frac{1}{3+s}\rightarrow G^*(z)=\frac{Y(z)}{X(z)}= \frac{1}{3+\frac{20(z-1)}{(z+1)}}=\frac{z+1}{23z-17} =\frac{0.043+0.043z^{-1}}{1-0.739z^{-1}}$$

The digital filter is thus implemented by the difference equation: $$ \small y(k)=0.043\left(x(k)+x(k-1)\right)+0.739y(k-1)$$

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  • \$\begingroup\$ I think Tustin's still needs sufficiently small sampling intervals, \$T\$. The discrete time closed loop system performance is more likely to follow the continuous time approach when \$T\lt\frac{\pi}{5\,\omega_c}\$, with \$\omega_c\$ the cross-over frequency, memory serving. \$\endgroup\$ – jonk Dec 4 '18 at 18:16
  • \$\begingroup\$ @jonk Of course the time increment is important; the example was only intended to show the procedure. I've changed T to 0.1sec which is a bit more realistic. \$\endgroup\$ – Chu Dec 4 '18 at 18:51
  • \$\begingroup\$ Good answer +1 nice and clear. \$\endgroup\$ – Andy aka Dec 4 '18 at 19:29
  • \$\begingroup\$ @Andyaka Agreed. +1, as well. \$\endgroup\$ – jonk Dec 4 '18 at 19:42
  • \$\begingroup\$ @Chu Thank you for your thoughtful and detailed answer. You make a comment that is what I am really interested in "The bilinear transform ('Tustin' or many other names) is an algebraic method of doing, essentially, the same thing, but with generally better perfomance compared with pole-zero mapping, impulse invariant, step invariant, etc" \$\endgroup\$ – Help Appreciated Dec 4 '18 at 19:58
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Because if you use one of those rules then you can go from a rational ratio of polynomials in \$s\$ to a rational ratio of polynomials in \$z\$. This leaves you with a system model upon which you can use all of the existing analysis methods that work with ratios of polynomials, and it leads you to filter designs that can be realized with difference equations.

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Why are rules such as the forward rectangular rule, or Tustin's method used instead of the definition?

Such rules are used to design a digital (better: time-discrete) circuit that behaves nearly like an already known analogue (better: time-continous) circuit.

Example:

You have an existing circuit ("old circuit") where some analogue signal processing is done using some filter \$G_1(s)\$. This signal is then converted to a digital signal using an Analogue-to-Digital-Converter (ADC).

You want to design a new circuit where the processing is done using a digital filter (\$G_2(z)\$) instead of an analogue one:

schematic

simulate this circuit – Schematic created using CircuitLab

Of course the outputs of the "old circuit" and the "new" one should be identical (or at least as similar as possible).

Tustin's method (and similar) are not used to calculate \$G_2(z)\$ from \$G_\color{red}{2}(s)\$!

Such methods are used to calculate some \$G_2(z)\$ from \$G_\color{red}{1}(s)\$ in a way that both the "new circuit" and the "old circuit" have the same output.

And simply using \$G_2(s)=G_1(s)\$ will not lead to the correct result!

Think about \$G_1(s)=s\$:

The inputs of the ADCs will be analogue signals and the outputs of the ADCs will be step functions(*).

If you'll pass a step function to a filter \$G(s)=s\$ (a differentiator), you'll get a series of Dirac pulses as output of the "new circuit". However you want to have the same output as in the "old circuit" - which is a step function.

Therefore using \$G_1(s)=G_2(s)\$ will not lead to the correct result.

Using Tustin's method for calculating \$G_2(z)\$ from \$G_\color{red}{1}(s)\$ however will result in an output of the "new circuit" which is close to the output of the "old circuit".

(*) Some people define the output of an ADC as series of Dirac pulses instead of a step function. In this case you could argue with an integrator instead of a differentiator...

Yes, but I am asking about the characteristics of the transform methods. If I transform G(s) to G(z), then why use Tustin's rule or forward rectangular rule instead of the definition?

Note that you are typically never interested of \$G_2(s)\$ in the case of a filter that processes time-discrete signals. You are only interested of \$G_2(z)\$. Therefore there is no need for some rule how to write \$G_2(s)\$ when \$G_2(z)\$ is known.

However in most cases your digital signal will represent some analogue one. Example: An audio signal.

In this case you are not interested in how some filter \$G_2(z)\$ is influencing the digital signal. You are not interested in the digital signal at all. You are interested on how this filter will influence the analogue signal in the end (e.g. your audio signal coming out of the loudspeakers).

Therefore you calculate an analogue filter \$G_1(s)\$ which will have the same influence on the analogue signal that your digital filter \$G_2(z)\$ has.

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  • \$\begingroup\$ Yes, but I am asking about the characteristics of the transform methods. If I transform G(s) to G(z), then why use Tustin's rule or forward rectangular rule instead of the definition? I would think the integration rules are just an approximation to the definition. \$\endgroup\$ – Help Appreciated Dec 4 '18 at 16:19
  • \$\begingroup\$ Your assertions are not correct. You can use a formula such as G(z) = .... perfectly fine in analogue circuits, it is present in very common and extremely precise everyday circuits. Such formula only implies time discretization, there is nothing "digital" implied by it. I guess you have heard of switched-capacitor circuits? Most sigma-delta converters achieve their extremely-high-precision via time-discretized analogue feedback loops. \$\endgroup\$ – Edgar Brown Dec 4 '18 at 17:09
  • \$\begingroup\$ @EdgarBrown Counterexample: An "echo effect" can be done using a DSP using the following transfer function: \$G(z)=1+4400z\$. All you need are 17600 bytes of RAM - assuming CD quality sound. I know that the same effect was done using magnetic tapes in analogue circuits before digital sound effects were available commercially. A magnetic tape is not an electronic part but a mechanical one - and rather expensive. This indicates to me that it is not that easy - or even impossible - to produce an analogue signal delay using electronic parts that are available commercially. \$\endgroup\$ – Martin Rosenau Dec 4 '18 at 19:19
  • \$\begingroup\$ Actually analog delays are rather common. They are called “wires” or “transmission lines” in some cases, or electro acoustic devices in others (close relatives of the crystal oscillators). Take a look at SAW filters, a very common way of implementing analog FIR filters. \$\endgroup\$ – Edgar Brown Dec 4 '18 at 19:24
  • \$\begingroup\$ @MartinRosenau Mmmm... that was a fully continuous example, but there are discrete time examples as well. There are the ubiquitous charge-bucket-brigade devices that surround us all over the place. They are called "CCD sensors" which are also used in audio. I myself once implemented (all the way to silicon and electrical testing) an experimental charge-cascade ASIC as a class project for fully-analog ultrasound triangulation. \$\endgroup\$ – Edgar Brown Dec 4 '18 at 19:42

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