Discrepancy in stability conditions when calculating via RH criterion and Nyquist criteria

Question

I have the following open loop transfer function for a unity feedback system.

$$G(s)=\frac{K(s+20)^2}{s^3}$$

1.When using RH criterion it can be easily proved that the closed loop transfer function is valid for $K>10$.

2.Trying with Nyquist criteria for stability : $$|G(jw)_{w=w_{pc}}| \leq 1$$

Here $w_{pc}$ is the phase cross over frequency ($\angle G(jw)_{w=w_{pc}}=-180^\circ$). First solving for $w_pc$, I get the value as $20$ rad/sec and $|G(jw)_{w=20}|=\frac {K}{10}$ thus, using it with the nyquist criteria gives me $K<10$.

I don't understand why there is a difference. What am I missing?

Answer 1

As pointed out in fibonatic's answer, you didn't actually use the Nyquist stability criterion by just computing gain or phase margins. The latter cannot generally be used to check the stability of a system. Instead, gain and phase margins are used to describe the robustness of a stable system with respect to changes in gain or phase. This means that the use of gain and phase margins assumes stability a priori.

The Nyquist stability criterion uses the trace of the open loop transfer function $G(s)$ in the complex plane to check the stability of the corresponding closed-loop transfer function.

The figures below show the traces of $G(s)$ for two different values of $K$ ($K=5$ and $K=15$). The traces show the behavior of $G(s)$ for $s=j\omega$ with $\omega$ moving from $-\infty$ to $\infty$. What is not shown in the plots is the path connecting the branches labeled $\omega=0^-$ and $\omega=0^+$. That path is a circle with an infinite radius moving clockwise $1.5$ times around the origin of the complex plane. That means that in the left plot, which corresponds to an unstable system, the point $-1+j0$ (the red cross) is encircled twice in a clockwise fashion. Consequently, according to the Nyquist stability criterion, the closed-loop transfer function has two poles in the right half-plane. The plot on the right shows the trace of a stable system. There is no clockwise encirclement of the point $-1+j0$, i.e., that point is always to the left of the trace.

Note that for the stable system (the plot on the right), the point where the phase of $G(j\omega)$ equals $\pm 180$ degrees, i.e., the point where the trace crosses the real axis lies to the left of $-1+j0$, i.e., it corresponds to a magnitude $|G(j\omega)|>1$. This means that the gain margin is negative even though the system is stable. Consequently, a negative gain margin does not generally imply that the system is unstable, and, vice-versa, form the left plot we see that a positive gain margin does not generally imply stability.

Answer 2

The full Nyquist criterion can be summarized to the equation $N=Z-P$, where $N$ is equal to the number of encirclements in the clockwise direction of the minus one point, $Z$ is the number of right half plane closed loop poles and $P$ is the number of open loop unstable poles. So for closed loop stability you want $Z=0$, so $P$ counterclockwise encirclements of the minus one point. If the open loop is stable and you want the closed loop the be stable as well, then this simplifies to that you always keep the minus one point on your left, or in other words rotate around the minus one point in counterclockwise direction.

It can be noted that scaling the open loop transfer function by a constant $K$ then the Nyquist plot also gets scaled by this constant. When determining the closed loop stability this is equivalent to scaling the minus one point by $1/K$.