# Why is this system stable?

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.

• What is the gain, when phaseshift is -180 degrees? – analogsystemsrf Feb 13 '18 at 2:31
• The closed loop denominator is 2nd order and has no negative coefficients, so it must be stable. What's happening is the open loop zeros are dragging the open loop poles into the left half s-plane. – Chu Feb 13 '18 at 7:48
• I'm puzzled. So far all I can say is that your multiplicity intuition is disproven by some literature. From Astrom Murray: "Poles and zeros of of m multiplicity are counted m times". – raggot Feb 13 '18 at 13:15

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.

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.

• The Nyquist criterion relies on Cauchy’s argument principle. As far as I know, the only requirement is the function needs to be holomorphic (complex differentiable) everywhere except at the poles. There is no requirement for the function to be bounded. – user110971 Feb 13 '18 at 16:32
• I can't find any specific on English Websites. When I search for it on German websites (which is my native tongue) I find several sites were it is noted that the Nyquist criterion can only be applyed if the amplitude tends to zero for high frequencies. – Sorkfa Feb 14 '18 at 6:38
• I have been looking for sources for your point but could not find them. Can you share a good one, even if in German? Beware of forums and technical websites though. Sometimes they spread urban legends. – raggot Feb 14 '18 at 8:15
• In the german Wikipedia articel link at the chapter Grundlagen; last sentence. Also in this document from the EHT Zurich link at page seven. – Sorkfa Feb 14 '18 at 9:53
• Interesting. I just looked through Ogata and Astrom Murray but found no trace of that prerequisite. I did not see any examples using non-proper transfer functions though. It is assumed that L(s) is the open-loop function of a feedback system. In general, it could be silently assumed that it is proper, but I would find it weird that it's not stated in the assumptions for the theorem. So I'm puzzled, once again. – raggot Feb 15 '18 at 9:37