Update: I am not saying a dead band is always better than a filter; I am only saying it is another tool in the toolbox, and its benefits or detriments in any situation will come from knowledge of the entire system (process + control).
TL;DR
Do you know how to generate transfer functions using Laplace transforms? There is a significant difference between having an input filter and a output filter.
It's been awhile (decades), I am working on refreshing that. But I do roughly follow what you have done.
Yes, but an input low pass filter doesn't change the characteristic equation. An output filter just adds another pole and doesn't change the tuning.
Yah I could be wrong here, but I don't see how either one does not delay, or at least attenuate, the response.
Close but no. The dead band will ignore drifting from the set point within the dead band. If the noise is Gaussian the noise will filter out to the true value.
Sorry, I have to disagree, but a dead band that ignores the drift
has been set incorrectly. If the dead band is set correctly i.e. based on common mode variation i.e. actual
noise, not signal, then when the actual
signal drifts in one direction, the noise will "break through" that side of the dead band and the PID will respond. Noise that is Gaussian makes both approaches work.
An interesting aspect of this is that there is a delay, random to be sure, but a delay nonetheless, in how long it takes for the (first) breakthrough, so in that sense it has the problem as the filter. But since we are saying things like "if the sample time is fast then the noise averages itself out." then it does not matter.
The filter will not increase the noise.
I didn't say the
filter increases the noise. What I said was that, in a system with input noise, a
PID that responds to that noise, i.e. "chases the noise," will increase (amplify) the noise. Since a filter attenuates
but does not eliminate noise, the filter-attenuated noise that gets through will be amplified in the system (system = process + control)
by the PID.
Yes, but the system will not be a steady state. It will drift within the dead band. What if the noise is greater than the dead band? I made my simulation difficult on purpose. The dead band is 3 degrees but that is also the standard deviation of the noise so about 35% of the readings are outside the dead band.
That dead band was set incorrectly, which is why you got that result. But an incorrectly-designed experiment proves nothing. That is why I try to always remember to qualify that the dead band has to be set appropriately/correctly. It should be set to 2-σ (95%) or 3-σ (99.7%), not 1-σ (68%). Obviously there is a tradeoff here between how immune it will be to noise vs. how long it should take to respond to drift, as noted above.
No! How can you say that? Where is the dead band? There is no dead band. The goal is sample fast enough to get a true reading of the average pressure. Do you remember Nyquist? Actually one needs to sample faster than that.
If we sample at a high rate, and take the extrema over some period, in order declare their midpoint as the actual signal, then that is roughly akin to empirically determining the dead band limits. Yes, that is not an
actual dead band, because an actual dead band is fixed around the setpoint (only), but it is saying "ignore all variation between these two limits as noise," which is the same
concept,
roughly, as a dead band.