# Why do we use this particular approximation for the bilinear transform?

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. 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. • 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? – 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. from http://scilab.ninja/study-modules/scilab-control-engineering-basics/module-7-continuous-to-discrete-conversion-methods/