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I know a closed-loop system $\frac{G}{1+G}$ is unstable when the open-loop transfer function $G$ equals -1 or $0dB\angle-180^{\circ}$. The denominator becomes 0 so the gain is infinite. This is why there is the concept of phase margin and gain margin.

But I don't understand, for this example $\frac{10}{s(s+1)(s+1)}$:

  1. Why does the negative phase margin mean the system is unstable? In the example, the phase margin is $-37^{\circ}$. So at some frequency, the open-loop gain $G = 1\angle-217^{\circ}$, but how does this make the closed-loop $\frac{G}{1+G}$ unstable?
  2. Does a negative phase margin always mean the closed-loop system is unstable, or only in some cases?
  3. If the closed-loop system has infinite gain only at a particular frequency (the frequency that causes the open-loop gain $G$ to be -1), why is it also unstable at all other frequencies?
  4. I plot the closed-loop Bode plot in Matlab, but I cannot see any infinite gain on the plot. Why?
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  • $\begingroup$ Besides Bode plots, also you also familiar with Nyquist plots? $\endgroup$
    – fibonatic
    Apr 17 at 7:13
  • $\begingroup$ Yes, I am familiar with Nyquist plots. I know number of encirclements around -1 for the open-loop transfer function can tell us about the closed-loop stability. $\endgroup$
    – teba
    Apr 18 at 4:41

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A closed loop linear system is unstable if some Laplace s with positive real part makes the denominator of the transfer function zero, but the nominator is non-zero. We say there's pole in the right half plane. The system starts to oscillate with exponentially growing amplitude. The imaginary part of the pole determines the frequency. The bode plot, when done accurately enough, can predict the existence of a right half plane pole although it doesn't show its place. Search for Nyquist stability criteria to learn more of it.

There's also possible the limit case - a pole of closed loop transfer function just on the imaginary axis i.e. the gain becomes infinite at a certain frequency. The system may oscillate sinusoidally in that frequency if some impulse input or state variable initial value starts the oscillation. In practice such thing never happens exactly, but you searched only for it and ignored the exponentially growing oscillations.

ADD due the comment:

You said your process G has phase margin -37 degrees. It has no phase margin at all. Phase margin tells how much there's room to insert more phase lag before the phase lag grows to 180 degrees assuming the absolute value of G is =1. This is how a control engineer thinks it. He empirically assumes that if there's no phase margin just like in your case, there's a good change to exist an exponentially growing sine which is amplified infinitely in your closed loop with transfer function G/(1+G).

That assumption can be wrong in certain cases. Nyquist showed what's the exact procedure to check the stability by inspecting the transfer function only at imaginary s-values. You may even find a ready to use Nyquist plot function from your programs. It's freely available online for ex. in this website: https://mathlets.org/mathlets/bode-and-nyquist-plots/

The system is stable or it's not stable. Thinking it's stable at some frequencies and it's unstable at some other frequencies is your own way to use the term stability. It's like you could say that a woman is pregnant every day at 13:05 o'clock but she's not pregnant at other times.

BTW. Nyquist developed his stability criteria when no computers were available. There was no practical way to simulate the closed loop behaviour of complex transfer functions nor to find the poles of the closed loop system.

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  • $\begingroup$ I think I'm familiar with most of this including Nyquist stability criteria. But I still don't understand the specific questions in my post. The theories make sense to me until I think about those questions. For example, as I mention, phase margin comes from the idea that closed-loop $H=\frac{G}{1+G}$ would be infinite if $G=-1$, but if $G=-1$ only at a particular frequency, why is $H$ unstable at all frequencies? And why doesn't the infinite gain appear in the $H$ Bode plot? $\endgroup$
    – teba
    Apr 16 at 23:35
  • $\begingroup$ @teba when the magnitude of the Bode plot of a transfer function goes to infinity then that system might just be on the border of instability. This does not mean the magnitude of the Bode plot to infinity is a requirement for instability. For example consider $1/(s-1)$. $\endgroup$
    – fibonatic
    Apr 18 at 17:52

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