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While reading about the root locus method, I thought of a question and couldn't seem to answer it.

Suppose for a negative feedback system, all the open-loop poles and zeroes and the centroid $\big(\frac{\sum{p_i}-\sum{z_i}}{n-m}\big)$ lie in the LHP (where $n$ is the number of poles ($p_i$) of the system , $m$ is the number of zeroes ($z_i$) of the system) and $n-m = 2$ so that the angle of asymptotes are $90^{\circ}$ and $-90^{\circ}$.

Question: Is it possible for the root locus of a system with the above conditions to cross the $j\omega$ axis multiple times ($\geq4$)?

I thought of constructing an example with loop transfer function $\frac{K(s+z_1)(s+z_2)(s+z_3)(s+z_4)}{(s+p_1)(s+p_2)(s+p_3)(s+p_4)(s+p_5)(s+p_6)}$ (with $n=6$ and $m=4$) but can't find the appropriate values for the poles and zeroes.

So, I need help in constructing an explicit example. Any help would be much appreciated.

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  • $\begingroup$ It may be, but I'm not sure. Is this homework? Are you just asking the plot can have two poles crossing at some gain, then two more later (definitely, although I'd have to work at getting an example). In general, root loci tend to be attracted to zeros even when they don't go to the zero, and are repelled by poles. I'm not sure if I could make a single pair of loci that wiggled enough to do what you wanted -- but it might be possible, either intuiting it with multiple tries, or by doing the math. $\endgroup$
    – TimWescott
    Apr 6, 2022 at 21:46
  • $\begingroup$ @TimWescott I was discussing this with a friend and we both believe it might be possible but couldn't find a concrete example satisfying the above conditions (I think it is never possible for a 4-th order system) and it gets a little complicated with higher order systems. (The equations are messy and playing around with a root locus plotter is going nowhere.) And yeah, the question is crossing the imaginary axis at least 4 times (2 pairs) with the centroid, poles and zeroes all in OLHP. $\endgroup$
    – LM2357
    Apr 7, 2022 at 4:31
  • $\begingroup$ This question popped up while studying for high gain feedback and varying $n-m$.For $n-m>2$ obviously, larger values of $K$ don't work and for $n-m=2$, large values of $K$ are possible and so, we thought if all values of $K$ would work or just large enough values naturally resulting in the above question. $\endgroup$
    – LM2357
    Apr 7, 2022 at 4:37
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    $\begingroup$ @TimWescott What I found interesting is the poles actually ALL do go to a zero, so there is no case of "even when they don't go to a zero". (When we include the zeros at infinity: all systems have the same number of poles and zeros) $\endgroup$ Apr 8, 2022 at 11:06
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    $\begingroup$ I meant when they don't go to that zero. In a complicated-enough root-locus, you can see a locus bending toward a zero, even if it is some other locus that terminates at that particular zero. $\endgroup$
    – TimWescott
    Apr 8, 2022 at 15:19

1 Answer 1

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Since the Asymptotes of the Root Locus tends to infinity through angles $(2r-1)\pi/n, r=1..n$, with $n+z$ poles and $z$ zeros, independently if they lie or not at the LHP, if you place the zeros enough far from the poles, and if you use an enough small gain $K$, you can keep the asymptotic shape of the poles only.

Hence, if you have p-multiple poles, for some small gain, they should be located through angles $(2p-1)\pi/4, p=1..4$. and for some large gain, they should be located through angles $(2(p-z)-1)\pi/4, p-z=1..2$.

A simple example would be:

$$ H(s)={1+0.2s+(0.32s)^2 \over (s+1)^4} $$

This system has $p_{1,2,3,4}=-1$, $z_{1,2}=-1\pm 3i$ and $k_0=1$.

It is Critically Stable for $K\approx 5.8$, at $p_{1,8} \approx \pm i$ for a gain keeping the asymptotic trend of the 4-multiple poles, and for $K\approx 275$, at $p_{1,8}\approx\pm 4i$ for a gain losting the asymptotic trend of that 4-multiple poles.

For an enough large gain, we recover the final asymptotic shape, which in our case, happens just after the closed loop poles reencounter.

As all the zeros and poles were chosen at a fixed imaginary line, the centroid lies in the same line, and the asymptotes angles of the whole feedback system at $\pi/2$ and $3\pi/2$, start as straight lines.

enter image description here

Many more examples can be found following this principle.

If we require to pass 4 times through the positive imaginary axis, then a similar system can be build, for example: $$ H(s)={(1+0.02s+(0.1s)^2)^3 \over (s+1)^8} $$

This time we use a 8-multiple poles $p_{1,2,3,4,5,6,7,8}=-1$ and a corresponding 3-multiple complex zeros, this time a bit farther $z_{1,2,3,4,5,6}=-1\pm10i$ than the previous example, for the 8-multiple pole asymptotes shape to appear freely for low gains, and then to recover the overall asymptotes for higher gains.

As the latter case, we have two gains $K\approx 1.9$ and $K\approx 2.6e^{9}$ for a Critically Stable System.

enter image description here

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  • $\begingroup$ Nice! Can you show the example of it crossing 4 times in the positive jw axis? $\endgroup$ Apr 11, 2022 at 13:35
  • $\begingroup$ @DanBoschen Sure, just check the edit. This works as far as the zeros are "enough" far to have the initial asymptotics to appear. $\endgroup$
    – Brethlosze
    Apr 11, 2022 at 16:44
  • $\begingroup$ Well done, wish I could upvote twice. $\endgroup$ Apr 11, 2022 at 16:45

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