Phase Margin Puzzler

You are correct in believing that your thinking could be wrong.

Phase (and gain) margin make the most difference where the open loop gain is close to -1 (or +1 when you take the action of the summing junction with it's implied subtraction). Ringing etc. happens because the denominator of the system transfer function is equal to 1 + Hol, where Hol is the open loop transfer function. The closer that Hol gets to -1, the closer the system transfer function gets to 0, and the more the system rings.

It is not uncommon to have systems with three integrators. In such cases the phase shift approaches 270 degrees at DC, yet the system will be stable with the right compensation. In these cases one actually has _two_ gain margins -- one from the low-frequency point where the phase crosses 180 degrees with gain > 1, and the other at a higher frequency point where the phase crosses 180 degrees with gain < 1.

I give an example of such a system in Chapter 5 of my book, "Applied Control Theory for Embedded Systems", in example 5.3 on pages 110-111. For more information see

This quality of systems, that they can be stable with phase shifts over

180 degrees, points out a shortcoming of the Bode plot: you can't just look at the Bode plot of a system's open-loop gain and tell if the thing will be stable when the loop is closed! At best you can tell if the system is close to going from stable to unstable, if you know it to be stable from the start.

Here the Nyquist plot comes to your rescue. The same fellow who brought us the early glimmers of the Nyquist-Shannon sampling theory also gave us a plot that enables us (by counting how many times a plot of Hol circles -1) to determine if a system will be stable when the loop is closed -- but only if we know from the start how many unstable zeros there are in Hol. If we know that, then we can determine system stability.

Thanks Tim, a very swift and comprehensive reply. I shall study it in some depth, but essentially you have jogged my memory that feedback not only flattens gain variations but also irons out phase shift up to near where it hits -180 degrees. I actually wrote a formula in an Excel spreadsheet which plotted the effect of various amounts of feedback on phase shift and the effect was dramatic.

Like you say, where the gain has dropped right off, the feedback won't be helping reduce phase-shift any ( which isn't actually what you said, but is my finger-jabbing interpretation of it ). In fact I suspect I was looking at an open loop phase plot and thinking it was closed loop, duh!

I need to look into Nyquist a bit further, I got stuck on Chapter 9 of Dorff 'Modern Control Systems' when he introduced Nyquist with some stuff on conformal mapping!

cheers,

Andy.

Dnia 08-01-2007 o 22:02:32 Tim Wescott napisa³(a):

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Do you mean the transfer function tends to inf?

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Do you mean if we know from the start how many unstable poles there are in Hol?

"the closer the system transfer function _denominator_ gets to zero". I was thinking it really hard, but somehow my keyboard missed it.

No, I mean unstable zeros. Do check any good reference on the Nyquist plot. The cool thing about the Nyquist plot is that you _don't_ have to know ahead of time if the system is stable. The less-than-cool part is that you _do_ have to know how many unstable zeros there are. This isn't that big of a deal, because for most systems that you'd want to use the Nyquist plot the number of unstable zeros is known (the answer is almost always "none").

Dnia Mon, 15 Jan 2007 23:51:33 +0100, Tim Wescott napisa³:

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I must disagree. Nyquist criterion says that:

number of clockwise encirclements of open loop tf Hol over (-1,0j) (let's say N) (or 1+Hol Nyquist plot over origin )

equals number of unstable zeros of 1+Hol (call it Z) minus number of unstable poles of 1+Hol (and P)

Legend: Mind that zeros of 1+Hol are the poles of closed loop Hol/(1+Hol) (Z) poles of 1+Hol are also poles of open loop Hol. (P)

N is the value that we see on the plot P we should know from the start (and that is why I disagree with you) number of unstable pols of Hol Z we are looking for

N=Z-P Z=N-P

If number of clockwise encirclements of -1 N equals number of unstable poles of open loop TF then closed loop is stable (Z=0).

If open loop is stable (P=0) and there is no encirclements N=0 then Z=0 and close loop is stable.

[Remark. 1 counterclockwise encirclement : N=-1]

Did I mention that I don't actually use the Nyquist plot to determine initial stability? I think I'm rusty.

I'll have to go back and check my references, but I think you may indeed be right.

Dangit.