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.
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