Root locus, read from the equation

You will turn $1+KG(s)H(s)=0$ into one condition on angles and one on lengths. You will use them to find the real segments of the locus, where the branches head for large $K$, and where two branches leave the real axis. You will describe a locus as a table of exact closed-loop poles at chosen $K$, up to where it crosses the $j\omega$ axis.

Take a negative-feedback loop with loop gain $KG(s)H(s)$. All the gain sits in $K\ge 0$, and $GH=N(s)/D(s)$ with monic factors: $N={\prod }_{k=1}^m(s-z_k)$ and $D={\prod }_{i=1}^n(s-p_i)$, with no common factor and $m<n$. Here $s=\sigma +j\omega$, both parts measured per second ($\omega$ in rad/s). By Chapter 6 the closed-loop poles are the roots of the characteristic equation $1+KGH=0$. Multiplying it by $D$ gives the characteristic polynomial $D(s)+KN(s)$, with the same roots; at $K=0$ they are the open-loop poles.

For $K>0$, a closed-loop pole $s_0$ needs $KGH(s_0)=-1$. A complex number equals $-1$ exactly when its length is 1 and its angle is an odd multiple of $180^{\circ }$. A positive real $K$ adds length but no angle. So $K|GH(s_0)|=1$, which gives $K=1/|GH(s_0)|$, and the angle of $GH(s_0)$ must be an odd multiple of $180^{\circ }$.

Two questions need no plotting: which real points are on the locus, and where the branches go for large $K$. For the first, take a real point $s_0$. A real pole or zero to its left gives a vector along the positive real axis, angle $0^{\circ }$; one to its right gives $180^{\circ }$. A complex pole or zero and its conjugate give angles $\theta$ and $-\theta$, which cancel. So only the real poles and zeros to the right of $s_0$ matter.

For large $K$, write $D=s^n+\alpha s^{n-1}+\dots$ and $N=s^m+\beta s^{m-1}+\dots$; by Vieta, $\alpha =-\sum p_i$ and $\beta =-\sum z_k$. At a root of $D+KN$, $N(s)=-D(s)/K$. A root that stays within a fixed distance of the origin keeps $D(s)$ bounded, so $N(s)\to 0$: it closes in on a zero. The far-pole argument below finds $n-m$ poles that grow without bound; the other $m$ settle on the zeros. If $n-m\ge 2$, $KN$ has degree at most $n-2$, so the coefficient of $s^{n-1}$ in $D+KN$ stays $\alpha$ for every $K$. By Vieta, so does the sum of the roots.

For a far pole, write the locus equation as $D/N=-K$. Long division gives $D/N=s^{n-m}+(\alpha -\beta )s^{n-m-1}+\dots$, which agrees with $(s+c{)}^{n-m}=s^{n-m}+(n-m)c{s}^{n-m-1}+\dots$ in its first two terms when $(n-m)c=\alpha -\beta$. The rest is smaller than $|s{|}^{n-m}$ by a factor of order $|s{|}^2$. So at a far pole $(s+c{)}^{n-m}=-K(1+\epsilon )$, with $\epsilon$ shrinking like $1/|s{|}^2$. Taking roots, $s+c$ lies within about $|s||\epsilon |$, which shrinks like $1/|s|$, of a point whose $(n-m)$th power is $-K$. Those points lie on rays from $s=-c$.

When two branches on one real segment move towards each other as $K$ rises, they meet and then leave the axis as a complex pair. At the meeting point the characteristic polynomial has a double root. Each real point $s$ of the locus is a pole for exactly one gain, $K(s)$ below. On a locus segment between two real poles with no zero between, $K(s)$ rises from 0 at each end, so the branches meet where $K(s)$ stops rising and turns back.

Each row of the table comes from working backwards. Pick a real point $r$ of the locus and take $K=K(r)=-r(r+2)(r+4)$. Then $s-r$ divides $s^3+6s^2+8s+K$, and the quadratic left over gives the other two poles. For $r=-5$: $K=15$, the quotient is $s^2+s+3$, and its roots are $-0.5\pm j\sqrt{11}/2$.