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A necessary condition for absolute stability of a system is

Gain cross over frequency < Phase cross over frequency

Where Gain cross over frequency is the frequency at which Gain = 1 or 0 dB

Phase cross over frequency is the frequency is the frequency at which phase = -180 deg

Can anyone please tell me why?

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  • \$\begingroup\$ "Gain > Phase" does not make any sense. \$\endgroup\$
    – pipe
    Aug 12, 2016 at 11:42

5 Answers 5

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Any system (without negative feedback) will eventually roll-off the amplitude at the higher frequencies and, at some point, the gain will fall to unity and get smaller as frequency increases more. If the phase angle hasn't fallen to 180 degrees before the gain has become zero then, when applying non-phase-changing negative feedback, the system will not oscillate.

Lets call that frequency point F1.

If the gain (without negative feedback) is still greater than unity when the phase has degraded to 180 degrees, it is fairly certain that when applying non-phase-changing feedback, the system will oscillate.

Lets call this F2

Quite simply if F1 occurs then F2 cannot occur and the system with feedback is stable. Alternatively if F2 occurs it CANNOT be lower than F1.

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The other posters make references to the misconception that if the gain is larger than 1 when the phase shift has reached 180 degrees in the open loop, the system will oscillate when the loop is closed. Let me provide a counter example. Consider the open loop transfer function: $$L = \frac{10 (s + 1)^2}{s^3}$$ Here is the bode plot: bode Clearly the system should oscillate when the loop is closed, right? Well, not quite: $$H = \frac{L}{L + 1} = \frac{10(s+1)^2}{s^3 + 10(s+1)^2}$$ The poles are: poles Looks stable to me.

The thing is there is not an easy explanation. Here is something from mit. To quote the first sentence: "The Barkhausen Stability Criterion is simple, intuitive, and wrong."

The truth is you would want to look at Nyquist, and the theory behind it, i.e. Complex analysis, winding numbers etc., in order to understand what is going on.

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  • \$\begingroup\$ Yes, there is not an easy explanation in general. As I mentioned in my answer, the analysis only holds for minimum phase systems. The example you use is not minimum phase. \$\endgroup\$ Aug 12, 2016 at 19:09
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    \$\begingroup\$ At first, the quoted sentence ("The Barkhausen Stability Criterion is simple, intuitive, and wrong.") is pure garbage. The author did not fully understand this criterion (it is a necessary one only, not sufficient). Secondly, the closed loop system neither will be stable nor oscillate - it will go immediately into saturation (because the loop gain it has a rising phase in the critical frequency range). \$\endgroup\$
    – LvW
    Feb 7, 2018 at 11:55
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    \$\begingroup\$ @anhnha it is a counter example because the criterion states 180 degrees phase shift. The idea behind it is that negative feedback becomes positive feedback. I don’t understand what you mean by the negative sign of the loop gain? \$\endgroup\$
    – user110971
    Feb 7, 2018 at 11:57
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    \$\begingroup\$ @LvW you can see the step response at wolframalpha.com/input/…. It does not go into saturation. \$\endgroup\$
    – user110971
    Feb 7, 2018 at 12:33
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    \$\begingroup\$ @anhnha you draw a small semi-circle around zero in the contour. Hence the contour encircles no poles or zeros. In the Nyquist plot -1 is encircled once clockwise and once counter-clockwise. Thus the system is stable. Alternatively just look at the poles of the closed loop system, which is much easier. \$\endgroup\$
    – user110971
    Feb 7, 2018 at 13:53
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Trying to understand this using the Nyquist plot is very visual.

On the Nyquist plot, the gain crossover frequency is when the plot crosses the unit circle and the phase crossover frequency is when it crosses the imaginary axis.

The analysis only holds for minimum phase systems, because for such systems the Nyquist plot originates on the real axis, moves cw, and eventually winds up at the origin.

By the Nyquist stability criterion there should be no encirclements of the -1 point.

In the figure on the left, the plot first crosses the circle, then the imaginary axis, and finally winds up at the origin. Here the gain crossover frequency is less than the phase crossover frequency and there are no encirclements of -1. Thus the closed-loop system is stable.

In the figure on the right, the plot first crosses the imaginary axis, then the circle, and finally winds up at the origin. Here the phase crossover frequency is less than the gain crossover frequency and there are encirclements of -1. Thus the closed-loop system is unstable.

enter image description here

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  • \$\begingroup\$ You forgot to mention that the transfer function for the Nyquist plot is the LOOP GAIN (the transfer function of the open loop). \$\endgroup\$
    – LvW
    Aug 12, 2016 at 18:22
  • \$\begingroup\$ @LvW: could you answer my question to user110971 in the comment section above? \$\endgroup\$
    – emnha
    Feb 7, 2018 at 11:16
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    \$\begingroup\$ OK - I have tried to comment. \$\endgroup\$
    – LvW
    Feb 7, 2018 at 11:55
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At phase -180 deg there should be enough gain margin. Methods like \$\lambda_{max}\$ say a gain margin of 2.4dB for the closed loop.

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This gentleman does an excellent job explaining phase/gain margin and gets into nyquist plots and how they play an imporant role in stability. https://www.youtube.com/watch?v=ThoA4amCAX4

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