1
$\begingroup$

I encountered a question when using Bode stability criterion to analyze the closed-loop stability of a system. In a word, the Bode stability criterion says the system is stable but it turns out to be not. Here's the description. Thank you in advance for reading!

Given a feedback control system with a proportional controller $k$ and the open-loop transfer function $L(s) = \frac{1}{s(s+1)^2}$. I would like to analyze the system's stability using the Bode stability criterion.

The criterion reads: Consider an open-loop transfer function $L(s)$ controlled by a proportional controller $k$ and negative feedback. Assume that the open-loop system is stable and the amplitude and phase of the Bode plot are monotonically decreasing. Then, the closed-loop system is asymptotically stable if and only if $|kL(iω)| < 1$ where $∠L(iω) = −180◦$.

If I consider $k<0$, the bode plot from MATLAB shows (Bode plot of $k<0$ can be plotted using Bode($-L(s)$), namely, inverting $k$ is the same as inverting $L(s)$): enter image description here

The bode plots are monotonely decreasing and the phase plot never hit $-180◦$, MATLAB also says its gain margin is infinite. However, the system is not stable using the proportional gain $k<0$.

It can be shown using the Nyquist plot and the Nyquist stability criterion that this system can not be stable when $k<0$. Because when $k<0$, $(-\frac{1}{k},0)$ lays on the RHP, which is always encircled by the Nyquist plot, which makes it unstable. enter image description here

I wonder what makes this discrepency, namely, why the Bode criterion says it's stable but it's unstable however? There is one catch I found is that: Bode criterion applies when the feedback is negative. If $k<0$, it becomes a positive feedback so the Bode criterion no longer works. But I am not sure if this is correct. If so, does Nyquist criterion have this caveat? Are there any other caveats when using these criterion for stability, such as Bode, Nyquist?

Thank you so much for your time and help!

$\endgroup$

1 Answer 1

2
$\begingroup$

Considering $k<0$ is the same as considering $L = \frac{-1}{s(s+1)^2}$ with $k>0$. Multiplying a transfer function by $-1$ is the same as introducing a phase shift of $-180^\circ$. Thus, the initial phase shift of $L$ with the negative gain should be $-90^\circ-180^\circ = -270^\circ$. However, MATLAB does not help you here: the phase $-270^\circ$ is the same as $+90^\circ$. Subtracting $360^\circ$ from the phase plot you got from Matlab shows that your system is unstable for any negative $k$.

$\endgroup$
7
  • $\begingroup$ I have several questions: 1.So the Bode criterion using the Bode plot only works for $k>0$, because when $k<0$ we have to change the Bode plot and do a different analysis? 2. Why can we not introduce a phase shift of $+180\circ$ instead of a $-180\circ$? If we assume a negative sign introduce $+180\circ$, the gain margin is indeed infinite because it never reach $-180\circ$. Are there any convention not stated here? 3. If, as you said, the phase shift is $-180\circ$, how do we use the Bode criterion to conclude its stability? It still never reach $-180\circ$. $\endgroup$ Commented Nov 15, 2022 at 13:21
  • 2
    $\begingroup$ 1. Bode criterion is formulated for negative feedback with a positive gain. If you want to use a negative gain, you can multiply the plant by $-1$. $\endgroup$
    – Arastas
    Commented Nov 15, 2022 at 13:23
  • $\begingroup$ 2. You may consider the Bode criterion as a mag-phase interpretation of the Nyquist criterion that makes the phase and gain margins evident. Thus, for all doubts, we should look back to the Nyquist plot. The +90 or -270 phase is the same, but the Nyquist contour differs. $\endgroup$
    – Arastas
    Commented Nov 15, 2022 at 13:28
  • $\begingroup$ I could understand that the Bode plot can be interpreted as a part of the Nyquist plot. But why when we are using the Nyquist plot, we can analyze for both $k>0$ and $k<0$, but the bode plot needs to change when $k<0$? Why do we not also need to change the Nyquist plot? If we plot the Nyquist plot with -1 before the transfer function (when analyzing $k<0$), the Nyquist plot is also mirrored. $\endgroup$ Commented Nov 15, 2022 at 13:34
  • 1
    $\begingroup$ Bode's plot is exactly the Nyquist plot. However, the criterion of stability holds only for "simple" cases. The Nyquist criterion considers the encirclement of the critical point. For some cases, it corresponds to Bode's criterion. $\endgroup$
    – Arastas
    Commented Nov 16, 2022 at 10:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .