Question 1 of 6: Root Locus and Range of Stabilising Gain

Question 1: Root Locus and Range of Stabilising Gain (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A unity negative-feedback loop (Figure 1) whose forward transfer function is

The open loop therefore has $n=2$ poles at $s=0$ and $s=-2$, and $m=2$ finite zeros at $s=\pm j2$ (the imaginary roots of $s^{2}+4=0$).

Find. An accurate root-locus sketch, and the set of positive gains $K$ for which every closed-loop pole lies strictly in the left half-plane.

[Figure not reproduced: Figure 1 — the unity-feedback configuration as printed on page 2 of the examination paper. See the official exam paper.]

Approach. Form the closed-loop characteristic polynomial, apply the Routh–Hurwitz coefficient test to settle stability for all $K$ at once, then construct the locus from the standard rules (real-axis segments, breakaway points from $dK/ds=0$, and the terminal behaviour as $K\to\infty$); because $n=m=2$ the locus turns out to be an exact circle, which is what makes an "accurate" sketch possible.

1. Write the characteristic equation. Closed-loop poles satisfy $1+G(s)H(s)=0$, i.e. $s(s+2)+K(s^{2}+4)=0$. Collecting powers of $s$, $$(1+K)\,s^{2} \;+\; 2\,s \;+\; 4K \;=\; 0 .$$ The loop is second order for every finite $K$, and the coefficient of $s^{2}$ grows with $K$ — the tell-tale signature of a plant with as many zeros as poles.
2. Apply the Routh–Hurwitz test. For a quadratic $a_{2}s^{2}+a_{1}s+a_{0}$ the necessary and sufficient condition for both roots to lie in the left half-plane is simply that $a_{2},a_{1},a_{0}$ all share the same sign. Here $a_{2}=1+K$, $a_{1}=2$ and $a_{0}=4K$. Since $a_{1}=2>0$ is fixed, the requirement is $1+K>0$ and $4K>0$, both of which hold for every $K>0$. Hence $$\boxed{\;0 < K < \infty \;\Longrightarrow\; \text{the closed loop is stable for every positive gain.}\;}$$ There is no finite gain margin to compute: the locus never crosses into the right half-plane.
3. Confirm this directly from the pole locations. Solving the quadratic, $$s \;=\; \frac{-2 \pm \sqrt{4-16K(1+K)}}{2(1+K)} .$$ Whenever the roots are complex their real part is $\sigma=-\dfrac{1}{1+K}$, which is negative for all $K>0$ and tends to $0^{-}$ as $K\to\infty$. Whenever the roots are real, both the sum $-2/(1+K)$ and the product $4K/(1+K)$ are respectively negative and positive, which forces both roots to be negative. Stability for all positive gain is therefore confirmed twice over.
4. Locate the real-axis segments. A point on the real axis belongs to the $K>0$ locus when the total number of real poles and real zeros to its right is odd. The only real critical points are the poles $s=0$ and $s=-2$; the zeros sit on the imaginary axis and do not count. The segment $-2 < \sigma < 0$ has exactly one critical point ($s=0$) to its right and so lies on the locus, while $\sigma>0$ and $\sigma<-2$ do not.
5. Find the breakaway and break-in points. Solving the characteristic equation for the gain gives $K=-\dfrac{s(s+2)}{s^{2}+4}$. Setting $dK/ds=0$, $$(2s+2)(s^{2}+4)-2s\,(s^{2}+2s)=0 \;\Longrightarrow\; -2s^{2}+8s+8=0 \;\Longrightarrow\; s^{2}-4s-4=0 ,$$ whose roots are $s=2\pm2\sqrt{2}$, that is $s=-0.8284$ and $s=+4.8284$. Only $s=-0.8284$ lies on the $K>0$ real-axis segment, so it is the breakaway point; the other root is a break-in belonging to the negative-gain (complementary) locus. Substituting back, $$\boxed{\;s_{b}=2-2\sqrt{2}=-0.8284,\qquad K_{b}=\frac{\sqrt{2}-1}{2}=0.2071 . \;}$$ The same gain follows from setting the discriminant $4-16K(1+K)$ to zero, which is the independent check that the two branches meet exactly there.
6. Show that the complex branches form a circle. Writing a complex root as $s=\sigma+j\omega$, we have $\sigma=-1/(1+K)$ and $\sigma^{2}+\omega^{2}=|s|^{2}=4K/(1+K)$. Eliminating $K$ between the two relations gives $\sigma^{2}+\omega^{2}=4+4\sigma$, i.e. $$\boxed{\;(\sigma-2)^{2}+\omega^{2}=8\;}$$ — a circle centred at $s=+2$ with radius $2\sqrt{2}=2.8284$. It cuts the real axis exactly at the breakaway and break-in points $2\mp2\sqrt{2}$ and passes through the zeros $\pm j2$, since $(0-2)^{2}+2^{2}=8$. Only the left-hand arc ($\sigma\le 0$) is traced by positive gain.
7. Assemble the sketch. For $K$ increasing from $0$ the two poles move along the real axis towards each other, meet at $-0.8284$ when $K=0.2071$, split into a complex-conjugate pair that runs along the circular arc, and finally converge on the finite zeros $\pm j2$ as $K\to\infty$. The loop is therefore stable for every positive gain but becomes marginally stable only in the limit, where the poles reach the imaginary axis and the response degenerates into an undamped 2 rad/s oscillation.