Question 7 of 8: Nyquist Stability of an Open-Loop-Unstable Process

Problem 7: Nyquist Stability of an Open-Loop-Unstable Process (20%)

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. $G_p=\dfrac{1}{s^2-s-2}=\dfrac{1}{(s-2)(s+1)}$ — one RHP pole at $s=+2$ (so $P=1$) and one LHP pole at $s=-1$; open loop $L(s)=k_cG_p$.

Find. (a) the Nyquist verdict at $k_c=1$; (b) the stabilising range of $k_c$.

Approach. Evaluate $L(j\omega)$ at low, mid and high frequency to fix the locus shape, count encirclements of $-1$, apply $Z=N+P$, and cross-check against the characteristic polynomial by the Routh test.

1. Key points of $L(j\omega)$. $L(j\omega)=\dfrac{k_c}{(j\omega)^2-j\omega-2}=\dfrac{k_c}{-(\omega^2+2)-j\omega}$. At $\omega=0$: $L=\dfrac{k_c}{-2}=-\dfrac{k_c}{2}$ (for $k_c=1$, the point $-0.5$). As $\omega\to\infty$: $L\to0$ (approaching from the upper half-plane, since $\mathrm{Im}\,L=\dfrac{k_c\omega}{(\omega^2+2)^2+\omega^2}>0$). The real part is negative for all $\omega$, so the locus is a small arc confined to the left half-plane, running from $-0.5$ up to the origin.
2. (a) Encirclement count for $k_c=1$. The locus reaches only to $-0.5$ on the real axis — it stays to the right of the critical point $-1$ and therefore does not encircle it: $N=0$. With $P=1$ open-loop RHP pole, $$Z=N+P=0+1=1\neq0,$$ so there is one closed-loop RHP pole — the loop is unstable.
3. Direct check ($k_c=1$). Characteristic equation $s^2-s-2+k_c=s^2-s-1=0$ has roots $s=\dfrac{1\pm\sqrt5}{2}=+1.618,\,-0.618$: one root in the RHP — unstable, confirming $Z=1$. ✓
4. (b) Range of $k_c$ — Routh test. General characteristic equation: $s^2-s+(k_c-2)=0$. The Routh array is $$\begin{array}{c|cc}s^2&1&k_c-2\\ s^1&-1&0\\ s^0&k_c-2&\end{array}$$ The first column contains $-1<0$ for every $k_c$, so there is always a sign change — at least one RHP root regardless of gain.
5. (b) Conclusion. $$\boxed{\text{No value of }k_c\text{ stabilises the loop.}}$$ Equivalently, the sum of the closed-loop roots equals $+1>0$ (from the $-s$ term, independent of $k_c$), so the roots can never both lie in the LHP. A single proportional gain cannot move both poles left; stabilising this plant requires a controller that adds phase/zeros (e.g. PD or a lead compensator).