Why is this system stable?

Question

I was given the system defined by the open-loop transfer function:

\$L(s)=\frac{5(s^2+1.4s+1)}{(s-1)^2}\$

I was told to use the Nyquist criterion to determine stability. Examining the Nyquist plot shows one CCW encirclement of the (-1,0) point. The open-loop system has two RHP poles (although they are repeated poles). Using the equation:

\$Z=N+P\$

Where N=-1 and P=2, we see Z=1. This means there should be 1 RHP pole in the closed-loop transfer function and we would expect the system to be unstable. However, upon inspection of the impulse response, and step response the system does in fact appear stable. I'm struggling to understand why this system is not unstable. My initial thought is that the repeated pole at s=1 should only be counted once, but I haven't been able to find any literature to suggest this.

Answer 1

You are actually encircling the point -1 twice. Your Nyquist plot overlaps two encirclements. If you try to apply the Nyquist criterion to

$$L(s)=\frac{5(s^2+1.4+1)}{(s-1)(s-0.99)}$$

you will see that what looked like a single encirclement was actually the convergence of two trajectories. Therefore, N=-2, and Z=0 as it was meant to be.

Answer 2

Haven't done something like this in over a year, but here is my simple answer:

You are not allowed to use Nyquist in this situation. The numerator and the denominator share the same polynomial degree, therefore the Nyquist criterion can't be applied.

This must be true because the open-loop system should tend to an amplitude of zero for high frequencies.

If you have a look at the bode plot, the amplitude goes up and not down.

Hope this helps.