Monday, April 29, 2019

z transform - Derive the Forward Euler substitution for transfer function


In the book "The control handbook. Volume 1 " by Levine, the author shows that the transfer function:


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can be aproximated and discretized in the transfer function:


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using the forwar euler integral aproximation. In this way Levine shows that if i have a transfer function $G(s)=\frac{1}{s}$ i can simply substitute $s=\frac{z-1}{T}$ to apply the Forward Euler aproximation. At this point the author doesn't show how to generalize this particular case to a generic transfer function $H(s)$. Are you able to show me ?



Answer



To the extent you can factor the transfer function into individual integrator sections of the general form $\frac{1}{s}$ you can make this substitution, which is an approximation of the Matched-$z$ Transform where you substitute every $s$ for $s=\frac{\ln(z)}{T}$. (map from $s$ to $z$ using $z =e^{sT}$).


This results in first order forms given by $H_\mathrm{a}(s) = k_\mathrm{a}\frac{s-b}{s-a}$ being replaced with $H_\mathrm{d}(z) = k_\mathrm{d}\frac{z-e^{bT}}{z-e^{aT}}$ where $k_\mathrm{a}$ and $k_\mathrm{d}$ are fixed gain constants. $k_\mathrm{d}$ can be determined by using $H_\mathrm{a}(s=0) = H_\mathrm{d}(z=1)$ for low-pass forms or $H_\mathrm{a}(s=j \pi/T ) = H_\mathrm{d}(z=\pi)$ for high-pass forms.


For all pole forms this is also equivalent to what you would get using the mapping from $s$ to $z$ using the method of Impulse Invariance. The Matched-$z$ Transform is not recommended for higher order polynomials with zeros and poles since neither the time or frequency domain response is preserved (The method of Impulse Invarance or Bilinear Transform would more often be used, each with their own advantages).


This post summarizes various mapping techniques further: Are there alternatives to the bilinear transform?


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