Nyquist Analysis of an Open-Loop-Unstable Process: 23-Chem-A6 Process Dynamics and Control, December 2018, Q4

Problem 4: Nyquist Analysis 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. An open-loop-unstable first-order process under proportional control:

Quantity	Value
Process	$G_p(s)=20/(s-3)$
Open-loop pole	$s=+3$ (RHP $\Rightarrow$ $P=1$)
Controller	$G_c=K_c$ (proportional)
Loop TF	$L(s)=20K_c/(s-3)$

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

Problem 4: $L(j\omega)$ traces a circle through the origin ($\omega\to\infty$) and $-6.67$ ($\omega=0$), encircling $-1$ once counter-clockwise ($N=-1$). With $P=1$ open-loop RHP pole, $Z=N+P=0$: the closed loop is stable.

Approach. Map $L(j\omega)$, count encirclements $N$ of the $-1$ point (CCW negative), and apply $Z=N+P$ with $P=1$; require $Z=0$. Cross-check with the closed-loop characteristic equation.

Key points ($K_c=1$). $L(j\omega)=\dfrac{20}{j\omega-3}=\dfrac{20}{-3+j\omega}$. At $\omega=0$: $L=-20/3=-6.67$ (negative real axis). At $\omega\to\infty$: $L\to0$. At $\omega=3$: $L=\dfrac{20}{-3+3j}=4.71\angle{-135^\circ}=-3.33-3.33j$.
Shape. The reciprocal of the vertical line $\mathrm{Re}=-3$ is a circle through the origin; scaled by $20$ it is a circle on the real axis from $0$ to $-6.67$ (centre $-3.33$, radius $3.33$). As $\omega$ runs $-\infty\to+\infty$ the point sweeps this circle counter-clockwise.
Encirclements. The critical point $-1$ lies inside the circle (between $0$ and $-6.67$ on the real axis), and is encircled once counter-clockwise, so $N=-1$.
(a) Stability at $K_c=1$. The open loop has $P=1$ pole in the RHP ($s=+3$). The Nyquist count gives $Z=N+P=-1+1=0$ closed-loop RHP poles — the system is stable. Direct check: characteristic equation $1+\dfrac{20}{s-3}=0\Rightarrow s=3-20=-17<0$. Confirmed stable.
(b) Range of $K_c$. From $1+\dfrac{20K_c}{s-3}=0$, the single closed-loop pole is $s=3-20K_c$. Stability requires $$3-20K_c<0\ \Rightarrow\ \boxed{K_c>\tfrac{3}{20}=0.15.}$$ Equivalently, the $\omega=0$ point $-20K_c/3$ must lie left of $-1$ so that $-1$ is encircled, i.e. $20K_c/3>1$ — the same condition. Below $0.15$ the pole stays in the RHP and the loop cannot be stabilised by proportional action alone.

Quantity	Value
$L(0)$	$-20K_c/3$ ($=-6.67$ at $K_c=1$)
Encirclements at $K_c=1$	$N=-1$ (CCW), $P=1$, $Z=0$
(a) Stable at $K_c=1$?	Yes (pole at $-17$)
(b) Stable range	$K_c>0.15$

Concept & Theory

Topic. Nyquist stability criterion for an open-loop-unstable plant; the role of $P$ (RHP open-loop poles); proportional stabilisation of an unstable process.

Key relations. $Z=N+P$ ($N$ = CW encirclements of $-1$, negative if CCW); closed-loop stability $\Leftrightarrow Z=0$; here characteristic pole $s=3-20K_c$.

Why this works. An unstable plant ($P=1$) requires exactly one counter-clockwise encirclement of $-1$ to be stabilised. High enough gain enlarges the Nyquist circle until $-1$ is enclosed — the opposite of the usual “lower the gain” intuition for stable plants.

Common pitfall. Forgetting $P\ne0$ and concluding instability because $-1$ is encircled; for an unstable plant that encirclement is precisely what is required. Also, the encirclement must be counter-clockwise — a clockwise one would give $Z=2$.

Source. Stephanopoulos, Ch. 17–18 (Nyquist criterion); Seborg et al., Ch. 14.

Question 4 of 8: Nyquist Analysis of an Open-Loop-Unstable Process

Problem 4: Nyquist Analysis of an Open-Loop-Unstable Process (20%)