Question 5 of 6: Root locus of an unstable plant and the gain range for stability

Question 5: Root locus of an unstable plant and the gain range for stability

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. Loop gain function $G(s)=K(s^2+4s+8)/[s^2(s-1)]$ in a unity-feedback loop. The open-loop plant has a double pole at the origin and a real pole in the right half plane at $s=+1$, so it is open-loop unstable.

Find. The root-locus sketch as $K$ varies from 0 to $\infty$, and the range of $K$ giving a stable closed loop.

Approach. Construct the locus from the standard rules (real-axis segments, asymptote count and centroid, breakaway point from $dK/ds=0$, imaginary-axis crossing from the Routh array), then read the stability range directly off the crossing.

1. Catalogue poles, zeros and branches. The open-loop poles are $s=0$ (double) and $s=+1$; the zeros are the roots of $s^2+4s+8$, namely $s=-2\pm j2$. With $n=3$ poles and $m=2$ zeros the locus has three branches: two terminate on the finite zeros and $n-m=1$ runs off to infinity.
2. Locate the real-axis segments and asymptote. A real point lies on the locus if the number of real poles and zeros to its right is odd. Counting the double pole twice, every real $s \lt 1$ satisfies this, so the entire real axis to the left of $+1$ belongs to the locus. The single asymptote lies at $180^\circ$, with centroid $$\sigma_a=\frac{\sum p_i-\sum z_i}{n-m}=\frac{(0+0+1)-(-2-2)}{1}=+5$$
3. Find the breakaway point. Solving $1+G=0$ for the gain, $K=-s^2(s-1)/(s^2+4s+8)$, and setting $dK/ds=0$ yields $$-s\left(s^3+8s^2+20s-16\right)=0$$ Apart from the trivial root at the double pole, the only real root of the cubic is $s=+0.6292$, with $K=0.01345$ there. Two branches (one leaving the origin to the right, one leaving $s=+1$ to the left) meet at this point and break away into the complex plane at $\pm 90^\circ$.
4. Form the characteristic equation. From $1+G(s)=0$, $$s^2(s-1)+K(s^2+4s+8)=0 \quad\Longrightarrow\quad s^3+(K-1)s^2+4Ks+8K=0$$
5. Apply the Routh test. The array is $$\begin{array}{c|cc} s^3 & 1 & 4K\\ s^2 & K-1 & 8K\\ s^1 & \dfrac{4K(K-1)-8K}{K-1}=\dfrac{4K(K-3)}{K-1} & 0\\ s^0 & 8K &\end{array}$$ The $s^0$ row needs $K \gt 0$; the $s^2$ row needs $K \gt 1$; and with $K \gt 1$ the $s^1$ row needs $K \gt 3$. The binding condition is the last, so $$\boxed{\;K \gt 3\;\text{ for closed-loop stability}\;}$$ There is no upper limit: unusually, this plant is stabilised only by large gain.
6. Locate the imaginary-axis crossing. At $K=3$ the $s^1$ row vanishes and the auxiliary polynomial from the row above is $$(K-1)s^2+8K=2s^2+24=0 \quad\Longrightarrow\quad s=\pm j\sqrt{12}=\pm j3.464$$ Factoring the cubic at $K=3$ confirms it: $s^3+2s^2+12s+24=(s^2+12)(s+2)$, so the third pole is at $s=-2$. Below $K=3$ the pair sits in the right half plane; above it, in the left.
7. Read the sketch. As $K$ rises from zero the two right-hand branches leave $s=0^+$ and $s=1$, meet at $+0.629$, break away vertically, curve left, cross the imaginary axis at $\pm j3.464$ when $K=3$, and finally converge on the zeros $-2\pm j2$ as $K\to\infty$. The third branch leaves the origin travelling left and follows the $180^\circ$ asymptote to $-\infty$. Because the whole locus is in the right half plane for small $K$, the closed loop is unstable at low gain — the mirror image of the more usual situation.