Nyquist Stability of an Open-Loop-Unstable Process: 23-Chem-A6 Process Dynamics and Control, December 2013, Q7

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. $G_p(s)=\dfrac{20}{s-3}$ — a first-order process that is open-loop unstable (one pole at $s=+3$, so $P=1$ RHP pole). Open loop $L(s)=k_cG_p=\dfrac{20k_c}{s-3}$.

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

Approach. Apply the Nyquist criterion $Z=N+P$, where $Z$ is the number of closed-loop RHP poles, $P=1$ the open-loop RHP poles, and $N$ the number of clockwise encirclements of $-1$. Stability ($Z=0$) here demands exactly one counter-clockwise encirclement of $-1$. Cross-check with the closed-loop pole directly.

Key points of $L(j\omega)$ at $k_c=1$. $L(j\omega)=\dfrac{20}{j\omega-3}=\dfrac{-60-20j\omega}{\omega^{2}+9}$: at $\omega=0$, $L=-6.667$ (real axis); as $\omega\to\infty$, $L\to0$. The real part is always negative and the imaginary part negative for $\omega>0$, so the locus is a semicircle in the third quadrant. Over $-\infty<\omega<\infty$ it is a full circle through the origin and $-6.667$, centred at $-3.33$.
(a) Encirclement and verdict. The critical point $-1$ lies between the origin and $-6.667$, i.e. inside the circle. Tracing $\omega:-\infty\to+\infty$ the locus encircles $-1$ once counter-clockwise, so $N=-1$. Then $Z=N+P=-1+1=0$: the closed loop is stable at $k_c=1$. Direct check: the closed-loop pole is $s=3-20k_c=3-20=-17<0$. ✓
(b) Stabilising range. The circle crosses the real axis at $0$ and at $L(0)=-20k_c/3$. The point $-1$ is enclosed (giving the needed CCW encirclement) only when $|L(0)|>1$, i.e. $\dfrac{20k_c}{3}>1\Rightarrow k_c>0.15$. Equivalently the closed-loop pole $s=3-20k_c<0$ requires the same. There is no upper limit, so $$\boxed{k_c>\dfrac{3}{20}=0.15.}$$ For $k_c<0.15$ the $-1$ point falls outside the circle ($N=0$), leaving $Z=1$ — the open-loop instability is not removed.

Problem 7(a): Nyquist locus of $L=20/(s-3)$ at $k_c=1$ — a circle from $L(0)=-6.67$ through the origin ($\omega\to\infty$); the mirrored $\omega<0$ branch is dashed. The critical point $-1$ sits inside, giving one counter-clockwise encirclement; with $P=1$ open-loop RHP pole, $Z=N+P=-1+1=0$, so the loop is stable. Shrinking $k_c$ toward $0.15$ pulls $L(0)$ onto $-1$ (marginal).

Result	Value
$L(0)$ at $k_c=1$	$-6.667$
Open-loop RHP poles $P$	$1$
Encirclements of $-1$ at $k_c=1$	$N=-1$ (one CCW) ⇒ $Z=0$ (stable)
Closed-loop pole	$s=3-20k_c$
Stabilising range	$k_c>0.15$

Concept & Theory

Topic. Nyquist stability for an open-loop-unstable plant; the role of $P$ in $Z=N+P$.

Key relations. Nyquist: $Z=N+P$ with $N$ = clockwise encirclements of $-1$; closed-loop stable iff $Z=0$; here $P=1$ so one CCW encirclement ($N=-1$) is required.

Why this works. The Nyquist map counts, via the argument principle, the difference between closed-loop and open-loop RHP poles. When the plant is already unstable ($P=1$), feedback must actively supply an encirclement to cancel it — too little gain ($k_c<0.15$) cannot, and the instability persists.

Common pitfall. Assuming a plant that is stable at low gain and unstable at high gain (the usual case); here it is the reverse — low gain is unstable. Also miscounting the encirclement direction, which flips the sign of $N$ and the conclusion.

Source. Stephanopoulos, Ch. 17 (Nyquist criterion, unstable open loops); Seborg et al., Ch. 14 (Nyquist stability).

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

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