Determine stability of feedback system from open loop transfer function and Nyquist stability criterion gives different results

Question

I'm confused due to the fact that the Nyquist stability criterion and looking at the transfer function doesn't give the same results whether a feedback system is stable or not. When I have the system of this block diagram the closed loop transfer function is:

$$ T_{CL}(s) = \frac{G}{1+GH} $$ Due to the Nyquist stability criterion this closed loop transfer function will be marginally stable when GH = -1 + 0j and unstable when $GH = < -1 + 0j$ (if the open loop transfer function is stable, in this case there aren't any poles in the right half plane thus the nyquist plot should not encircle the point -1, so this plot should stay right from this -1 point).

However when I look at the closed loop transfer function, I would say that this system is unstable for $GH = -1$. In this case the transfer function becomes infinity so a bounded input will result in a unboundend (=infinity) output.

In my train of thought the point $GH = -2$ would again be stable since $T_{CL}$ will be finite again, however conform the Nyquist stability criterion this point wil still be unstable?

I know that Nyquist is correct but what's the problem with my way of thinking

Answer 1

However when I look at the closed loop transfer function, I would say that this system is unstable for 𝐺𝐻=−1. In this case the transfer function becomes infinity so a bounded input will result in a unbounded (=infinity) output.

This depends on your definition of stability. $GH = -1$ is called marginally stable because depending on how you look at it, it could be stable, or it could be unstable.

In the Lyapunov sense that @Petrus1904 mentions, it is stable. But if you take bounded-input bounded-output (BIBO) stability to mean that the input can go on infinitely long but within bounds, then the output can, indeed, go to infinity. So in the BIBO sense it's unstable.

Hence, "marginally stable".

In my train of thought the point 𝐺𝐻=−2 would again be stable since 𝑇𝐶𝐿 will be finite again, however conform the Nyquist stability criterion this point will still be unstable?

An unstable system can have a transfer function that never goes infinite for $s \in j \omega$. All it requires is that there are no poles on the stability boundary.

I apologize for not taking the time to do so, but I know that I could dummy up a system that has $GH = -2$ and was still stable, because I've designed such systems. All you need to do is wrap a double integrator with a PID controller: let $G = 1/s^2$ and $H = k_i / s + k_p + \frac{s}{\tau_d s + 1} k_d$ and tune for stability. The resulting system will have a spot in its open-loop Bode plot which has a phase shift of 180$^\circ$ and a gain greater than one -- to get that gain equal to two you'll just need to jigger the numbers around.

Addendum: Use the PID transfer function I give above with $k_i = 0.002$, $k_p = 0.02$, $k_d = 0.2$ and $\tau_d = 0.01$ and you should get a system that's stable, with $GH \simeq -2$ at somewhere around $\omega = 0.1 \mathrm{\frac{rad}{sec}}$.

Assuming I got my math right...

Answer 2

There are a few things I can note about your question. As far as I have always learned, the nyquist stability criterion is taken over the openloop transfer function. if you take the closed loop transfer function, you should count the encirclements of 0 instead (if i recall correctly).

The formal definition of stability, as expressed by the Lyapunov's stability criterion is the following (in layman terms): a system is considered stable if the amount of energy in the system is less or equal to the amount of energy put into the system. Again, very this is the very basic and there are some formal mathematical rules bound to this that I will not bother you with. But it means that if you stop exciting the system, the amount of energy in the system will not increase.

For example take the following system: $H = 1, G =1/s^2$. $T_{cl}$ has an infinite magnitude if excited with a signal carrying a frequency of 1 rad/s. However, if you stop exciting the system, the magnitude of the signal will not increase anymore (due to the nature of this system, it will not decrease either). As such, the energy in the system remains bounded and does not increase. In fact, if you excite this system with any other frequency than this resonance frequency, its magnitude will not increase either. As such the system is (marginally) stable. Its the same example as with a single mass in space. if you push it, it will move forever forward. But the energy in the system does not increase without excitation.