When numerically simulating a system, usually some kind of discretization is necessary, obtained by some kind of z-transform, such as, for instance, the bilinear transform $s\mapsto \frac{2}{\triangle t}\frac{z-1}{z+1}$, which is kind of the same as using the Trapezoidal rule to approximate integration. It also has the very nice property that it preserves BIBO-(in)stability from the s into the z-domain. However, when it is implicitly used as a differentiator, there is a very real possibility of unwanted numerical oscillations.

Now take, for example, the Euler backwards method, corresponding to the substitution $s\mapsto \frac{z-1}{\triangle t}$. This method reduces numerical oscillations quite effectively, which would (ignoring for now the slower convergence) seem to be quite a big advantage. However, if we do look at the BIBO-stability areas before and after the z-transformation, this method leads to an effectively increased stability region, i.e. there are some transfer functions that are unstable in the s-, but stable in the z-domain.

1) That the latter method is able to dampen numerical oscillations seems intuitively connected to a greater "numerical stability". However, as I understand, not all systems where numerical oscillations occur after using the bilinear transform are inherently unstable. A simple RL-circuit with transfer function $\frac{U}{I}=\frac{1}{R+Ls}$ will a current source attached will, for instance, show numerical oscillations when the source is suddenly switched on or off. So are these phenomena - increased stability domain after the transformation, and damping of numerical oscillations - connected?

2) In literature, I do find a lot about choosing a suitable z-transformation such that the resulting discrete system is stable. However, there was no regard given to possible cases where the original system might have been unstable. To me, this seems like something of an oversight - when simulating an unstable system, I would expect the discrete system to also be unstable. Because, well, the output signals for stable and unstable systems would possibly differ a lot. Have I misunderstood something there? Or is there a reason why one can safely assume the systems in question (namely, electrical power systems) to be stable?

I do know that many algorithms (say, eg, EMTP and all related) usea mix of the above, but while this decreases the difference in stability regions, it is still there.

Edit: By numerical oscillations using the Trapezoidal rule I mean phenomena such as those outlined in http://www.ece.uidaho.edu/ee/power/ECE524/spring14/Lectures/L39/numerosc.pdf , or similar (found using google, there seem to be different texts about the same).

  • $\begingroup$ I have honestly no idea what you are talking about and what do you mean by "numerical oscillations". When you discretize you need to pick a sample rate, which also means that you need to pick a maximum bandwidth and some low pass filter, otherwise you get aliasing. The exact nature of the low pass may add some ringing that you wouldn't necessarily see in a time continuous system, but that's not numerical and a principle limitation of sampling analog systems. $\endgroup$ – Hilmar Feb 6 '19 at 14:35
  • $\begingroup$ @Hilmar I have added a link to a document outlining one example for such oscillations before. These oscillations occur in the discrete, but obviously not in the continuous model. $\endgroup$ – Some Math Student Feb 6 '19 at 15:03
  • $\begingroup$ @Hilmar, i wonder if this is about the transition one gets with a filter that has all states set to zero and $x[n] = 0$ for $n<0$. now suppose when the switch is flipped on that $x[0]$ is some reasonably large non-zero value. That edge may create ringing or a transient response (added to the steady-state response) that wouldn't be there if the edge was gone and the signal continued back in time before $n=0$ so that $x[n]$ was the signal instead of always zero. $\endgroup$ – robert bristow-johnson Oct 29 '20 at 2:59
  • $\begingroup$ So the different $s \mapsto z$ mappings (Euler backward, Euler forward, Predictor-Corrector, Tustin or bilinear) might have for their dominant pole in the $z$-plane different distances from the unit circle (where the stable region ends). $\endgroup$ – robert bristow-johnson Oct 29 '20 at 3:03

The Simscape Power systems Toolbox from Simulink uses the Tustin Backward-Euler method to fix this problem.

I hate to simply post links, but here it is https://www.mathworks.com/help/physmod/sps/powersys/ug/simulating-discretized-electrical-systems.html

"To perform a discrete simulation, open the powergui block and set Simulation type to Discrete, and specify the sample time. The electrical system is discretized using the Tustin/Backward Euler (TBE) method. This method combines the Tustin method and the Backward Euler method. It allows you to simulate snubberless diode and thyristor converters. It eliminates numerical oscillations seen with the Tustin method and provides better accuracy than the Backward Euler method.

The TBE discretization method combines the accuracy of the Tustin method and the numerical oscillation damping property of the Backward-Euler method. It allows you to simulate power electronic circuits with virtually no snubber, using purely resistive snubbers with a very large resistance value resulting in negligible leakage currents."

EDIT : From what I understand, they generally use the tustin method but switch to backward Euler when there is a quick variation such as a switch being open in series with an inductor for example.

  • $\begingroup$ Thank you for your answer. However, this does not actually address my question. Sorry if I was not clear: I am aware of different methods how these numerical oscillations can be handled. My question is about a possible connection of this kind of oscillation to stability, and why there seems to be (from what I've read) noone bothered by discretization transforming unstable systems into stable ones. $\endgroup$ – Some Math Student Feb 6 '19 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.