As I understand it, for a signal \$f(t)\$ in time, its Laplace transform \$\mathfrak{L}\left\{f(t)\right\}=F_1(s)\$ and Z transform \$\mathfrak{Z}\left\{f(t)\right\}=F_2(z)\$ are related by a transformation \$z=e^{sT}\leftrightarrow s=1/T\,\log(z)\$ where \$T\$ is the sampling period (since the Z transform is discrete in time).

In practice, this is approximated to the first degree as follows $$\begin{align*}z&=e^{sT}\\&=\frac{e^{sT/2}}{e^{-sT/2}}\\&\approx\frac{1+sT/2}{1-sT/2}\end{align*}$$and thus \$(1-sT/2)z\approx1+sT/2\$ so \$sT/2\approx(z-1)/(z+1)\$ and ultimately \$s\approx\frac2T\frac{z-1}{z+1}=\frac2T\frac{1-z^{-1}}{1+z^{-1}}\$.

Now, I understand up to here, but I fail to understand why we use this particular first-order approximation over, say, \$z=e^{sT}\approx1+sT\leftrightarrow s\approx(z-1)/T=\frac{1-z^{-1}}{Tz^{-1}}\$.

Does this approximation 'behave' in some significantly poorer way for most purposes?

Sorry about the tags -- I tried various things like 'bilinear-transform' but they did not exist and I lack the points to create them.


The Forward Euler Transform $$z=e^{sT}\approx1+sT\leftrightarrow s\approx(z-1)/T=\frac{1-z^{-1}}{Tz^{-1}}$$ is easy to understand in that it is a direct translation and scaling from the \$s\$-domain to the z-domain. But the translation can transform stable \$s\$-domain poles into unstable \$z\$-domain poles.

To see this consider the diagram below.

The Left Half Plane in the \$s\$-domain (shaded) is scaled by \$T\$ and translated by \$1\$ to the \$z\$-domain. It should be clear to see that a pole X that is stable in \$s\$ can be unstable in \$z\$ by the Forward Euler transform.

enter image description here

In contrast, the bilinear transform $$s\approx\frac2T\frac{z-1}{z+1}=\frac2T\frac{1-z^{-1}}{1+z^{-1}}$$ translates the entire LHP of the s-domain in the unit circle of the \$z\$-domain by frequency warping via an intermediate w-plane.

Referring to the diagram below (ref: Ogata.K, Discrete Time Systems, 1995, Prentice-Hall), you can see that the entire LHP of the \$s\$-domain (a) is transformed to the unit-circle (b) via the w-plane scaling (c).

So that's why Bilinear is preferred in practice to the Forward Euler. However, there are other choices such as zero-pole matching (which I prefer), that may be employed over Bilinear due the frequency warping involved in Bilinear.

enter image description here

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    \$\begingroup\$ awesome! I understand now -- the bilinear transform as an approximation preserves system 'stability'. Both forward Euler and the bilinear result in frequency warping as they are simple approximations (as I understand) but pole-zero matching gives no such warping, correct? \$\endgroup\$ – oldrinb Dec 19 '14 at 21:37

I can answer your question only in a context with the switched capacitor (S/C) technique which uses the mathematical tools of digital signal processing. Here four different approximations are in use:

(1) Euler forward (EF), (2) Euler backward (EB), (3) bilinear (BI) and (4) LDI (Lossless discrete integrator).

For S/C circuits, it is common practice to use S/C circuits based on integrators. Here are the important differences:

(1) EF-integrator: For rising frequencies the approximation causes POSITIVE phase errors

(2) EB-integrator: For rising frequencies the approximation causes NEGATIVE phase errors

(3) BI-integrator: No phase and amplitude errors, however, for rising frequencies there is a kind of "shrinking" of the frequency axis based on an arctan function. For all lowpass and bandpass functions this effect causes a real zero for a finite frequency w=0.5*wcl (wcl: clock frequency). This effect is appreciated because all periodic spectral repetitions do not overlap and, thus, do not disturb each other.

(4) LDI-integrator: Combination of two integrators with EF and EB approximation, respectively.

I hope this helps to answer a part of your question.

EDIT: The approximation (z-1)/T as mentioned by you is equivalent to the the EF transformation.


In addition to the excellent point about stability made by akellyirl, another advantage of the bilinear transform (or Tustin method) compared to forward and backward differences is that it leads to a better approximation to the integral.

See the following image:

image of approximation for bilinear method compared to forward and backward differences from http://scilab.ninja/study-modules/scilab-control-engineering-basics/module-7-continuous-to-discrete-conversion-methods/


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