Definition of encirclement of -1+j*0 point re: Nyquist Stabiltiy Criterion

Hi All,
I am using a software package for designing RF amplifiers. To test the stabilitiy of the amplifier there is a tool in the package that plots
the Nyquist Stability contour for the positive imaginary points in the complex plane. Some notes that I have found on Nyquist stability state that
1. the contour only needs to cross the x-axis to the left of (-1, 0) in a clockwise sense for the system to be unstable.
Other notes simply state
2. that the contour must encircle the -1 + j*0 point for the system to be unstable.
I have found this confusing since there are many curves that can statisfy 1. but do not seem to satisfy 2. For example, when i plot the Nyquist contour for the positive imaginary points of my amplifier design I get a contour that first crosses the x-axis to the left of (-1,j*0) in an anticlockwise direction. Once in the lower-left-hand plane, the contour then turns around and proceeds to cross the x-axis to the left of (-1,j*0) in the clockwise direction. To me, this curve does satisfy 1. but it doesn't seem to satisfy 2.
Can anyone shed some light on whether this is an unstable system as per the Nyquist Stabiltiy Criterion? Cheers
Peter Vun.
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Hello Peter
I have yet to come across software that adequately deals with the limits implicit in the application of the Nyquist Criterion.
That aside - to answer your question, the actual Nyquist Criterion for stability for the open loop system is required. (I like to state the criterion in terms of number and sense of circulations around the -1+0j point required for stability).
The actual criterion (i.e. how many circulations, in which direction, for stability) depends of course on (a) the s-plane contour assumption, i.e. are poles on the jw-axis included or excluded (usually excluded - who knows how relevant this is, to amplifier design) (b) direction assumed for the construction of the s-plane contour, and (c) number of poles in the s-plane contour (RH s-plane) in the open loop transfer function.
I should imagine that for particular software, (a) and (b) will be assumed and documented somewhere, so that (c) is used to determine the Nyquist Criterion for Stability for any open loop transfer function, so that the L(s) can be plotted for all points s=sigma+jw on the s-plane contour, to see if the Nyquist Criterion established as above, is satisfied.
In my view, this is incredibly easily done with pen and paper and some understanding of limits, and quite a challenge to write analysis and visualization software to do the same.
In summary: It is impossible to inspect a curve of L(jw) around the point -1+0j in the L(s) plane, to decide whether the "Nyquist Criterion is satisfied", without knowing the Nyquist Criterion for Stability, dependent in the first place on the s-contour assumptions and in the second place on the poles of L(s) within that contour.
cheers,
John
Peter Vun wrote:

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Hi John,
Thanks for your reply. Here are some answers to the questions you posed.

As far as I understand the assumption made by the software is that there will be no poles on the jw-axis.

The direction of the s-plane contour is clockwise.

Again, I believe the software assumes that the number of RH poles of the open loop transfer function is zero. I believe that for most cases of RF amplifier design if the "open loop" part of the circuit is carefully defined by the user then this is assumption is probably true.
Unfortunately the open loop transfer function is not explicitly calculated and presented to the designer (at this point the design is way to complex for the designer to derive it by hand). Instead the open loop transfer function is calculated by the simulator and the Nyquist contour is then generated for the user.
Cheers
Peter Vun
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Hello again Peter - see inside:
Peter Vun wrote:

At this point, we can establish the Nyquist Criterion for stability to be "Number of circulations of the contour in the L(s) plane about the -1+0j point is equal to zero (which implies no poles of L(s)/(1+L(s)) in RH plane, necessary and sufficient for stability) minus number of poles of L(s) inside the contour in the s-plane" where clockwise circulations are positive, i.e. N=-P, and N is N(CW)-N(CCW)

For such systems, the Nyquist criterion (stability) is N=0.

Ah... I see that with knowledge of the number of poles at the origin, i.e. how many pure integrators 1/s in L(s), we can interpret your plot.
For no integrators, L(jw) starts from +ve real axis for w=0 and winds in to the origin as w-->inf. If the plot crosses the -ve real axis on the inside of -1+0j then there are no circulations about -1+0j therefore closed loop system is stable. In this case, gain and phase margins are expected to increase with decreasing magnitude at phase -180, i.e. where the plot crosses -ve real axis.
For one integrator, L(jw) "arrives into view" from the lag side of the -jw axis, i.e. bottom of the plot on LHS of -jw. In this case too, if the plot of L(jw) crosses the -ve real axis to the left of -1+0j then we have instability. Same expectations of increase in stability with gain decrease apply.
For two integrators, the same applies! Here, the only way for L(jw) to cross the -ve real axis at magnitude smaller than 1 (of course, I am describing absolute stability in yet another way) is for the L(jw) plot to "arrive into view" near the -1+0j point from the lead side of the -ve real axis, i.e. below it (implying a requirement for lower frequency lead action in L(s) for stability)
For three integrators, the plot arrives into view from near the +ve jw axis, and must bend to the left (implying more than 90 of lead) and cross the -ve real axis _outside_ of -1+0j for stability; because for N=N(CW)-N(CCW) to be 0, we need two anticlockwise circulations from L(jw) through to the L(-jw) parts of the contour to cancel out the two clockwise circulations caused by 1/s^3. In this case, system will be unstable for low gains, stable again for some range of higher gain (then generally unstable again for very high gains)
Hopefully this completes the picture - so for the special cases mentioned, a plot of L(jw) near -1+0j would be fine! (I actually find the Nyquist plot very useful also for identifying the kind of compensation I need, for stability requirements for a given system)
go well,
John
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Thanks heaps John. It will take me a day or so to digest your latest reply. In a few days time if I would like to ask you further questions re: your reply, would you mind if I emailed you directly or should I just post a reply to your latest posting on this thread.
Cheers
Peter Vun

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Hello Peter
Either reply here (...there are a few others monitoring this frequency who also appreciate Nyquist, and will no doubt pull me up if I make an error...) or else remove the 0-looking thing in my return address and reply directly.
cheers now,
John
Peter Vun wrote:

<all the rest snipped>
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The Nyquist stability criterion can be very confusing. Personally I prefer the Root Locus method which provides a clearer picture. If you have the open loop transfer function available, either as poles and zeros or as a ratio of two polynomials, it is not too difficult to plot the root locus diagram using well known rules. Alternativley you can plot it with my free root locus plotting program (RootLocs) available at http://www.biemke.freeserve.co.uk .
Good luck,
AAR
On Mon, 3 May 2004 1:41:44 +0100, Peter Vun wrote

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A.Robinson wrote:

The Nyquist criterion is useful if you're stabilizing a plant with experimental data -- but if you're doing that then you have to start with a stable realization and you can just use the Bode plot.
If this is an RF circuit program, why is it messing with the Nyquist criterion instead of drawing stability circles on a Smith chart?
--

Tim Wescott
Wescott Design Services
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Hi Tim,
Nyquist stability is used to check the internal stability of multi-stage amplifiers feedback where stability circles may fail. Stability circles require the assumption that any internal instability will propogate all the way to the input or output of the amplifier. This is not always the case as matching networks on the input or output of the amplifier could isolate the offending part of the amplifier from the input or output.
Cheers
Peter Vun

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Using just Bode plots can be dangerous, as it gives no information about encirclements, leading to incorrect conclusions about stability!
Fred.
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