Nyquist Stability of an Open-Loop-Unstable Process: 23-Chem-A6 Process Dynamics and Control, May 2013, Q4

Problem 4: 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. $G_p(s)=\dfrac{100}{s-10}$ — an open-loop-unstable first-order process (one pole at $s=+10$, so $P=1$ RHP pole). Proportional controller $G_c=k_c$; open loop $L(s)=k_c G_p=\dfrac{100k_c}{s-10}$.

Find. (a) Nyquist verdict at $k_c=1,\ 0.1$; (b) the stability limit on $k_c$.

Problem 4: Nyquist plot of $L=100/(s-10)$ at $k_c=1$ — a circle from $L(0)=-10$ to the origin. The critical point $-1$ lies inside, giving one counter-clockwise encirclement; with $P=1$ open-loop RHP pole, $Z=N+P=-1+1=0$, so the loop is stable. At $k_c=0.1$ the circle shrinks to pass through $-1$ (marginal).

Approach. Apply the Nyquist criterion $Z=N+P$, where $Z$ is the number of closed-loop RHP poles, $P$ the open-loop RHP poles, and $N$ the number of clockwise encirclements of the $-1$ point by $L(j\omega)$; because $P=1$, stability ($Z=0$) requires exactly one counter-clockwise encirclement of $-1$. Cross-check against the closed-loop pole directly.

Shape of the Nyquist plot. $L(j\omega)=\dfrac{100k_c}{j\omega-10}=\dfrac{100k_c(-10-j\omega)}{100+\omega^2}$. The real part is always negative and the locus is a circle through the origin ($\omega\to\infty$) and through $L(0)=-10k_c$ ($\omega=0$), i.e. a circle on the real axis from $-10k_c$ to $0$, traversed counter-clockwise as $\omega:-\infty\to\infty$.
(a) $k_c=1$. The circle runs from $-10$ to $0$; the critical point $-1$ lies inside it, so $L$ makes one counter-clockwise encirclement: $N=-1$. Then $Z=N+P=-1+1=0$ — stable. Direct check: closed-loop pole from $s-10+100k_c=0\Rightarrow s=10-100(1)=\boxed{-90}$ (LHP). ✓
(a) $k_c=0.1$. The circle now runs from $-1$ to $0$, so it passes through $-1$: the locus is exactly on the critical point — marginally stable (a closed-loop pole at the origin). Direct check: $s=10-100(0.1)=\boxed{0}$. ✓
(b) Limiting gain. Encirclement of $-1$ (hence $Z=0$) requires $10k_c>1$, i.e. the circle must extend past $-1$. The boundary is $L(0)=-10k_c=-1$: $$\boxed{k_{c,\lim}=0.1}$$ The system is stable for $k_c>0.1$ and unstable for $k_c<0.1$ — note the reverse sense: an open-loop-unstable process needs enough gain to be stabilised. (Direct: $s=10-100k_c<0\Leftrightarrow k_c>0.1$.)

Gain $k_c$	Closed-loop pole $s=10-100k_c$	Verdict
$1$	$-90$	stable ($-1$ encircled once CCW)
$0.1$	$0$	marginal ($L$ passes through $-1$)
$>0.1$	$<0$	stable
$<0.1$	$>0$	unstable

Concept & Theory

Topic. The Nyquist criterion for an open-loop-unstable plant; encirclement counting with $P\ne0$.

Key relations. $Z=N+P$; closed-loop stability $\Leftrightarrow Z=0$. For $P=1$, stability demands one counter-clockwise encirclement of $-1$ ($N=-1$).

Why this works. The Nyquist map turns the closed-loop characteristic equation $1+L=0$ into a graphical count of how the open-loop locus wraps the $-1$ point; the argument principle links that winding number to the difference between closed- and open-loop RHP poles. For a plant with a RHP pole, a proportional controller must supply enough loop gain to pull the locus around $-1$ — too little gain leaves the instability uncorrected.

Common pitfall. Assuming "more gain = less stable" (true for many stable plants, but here it is inverted), or ignoring the open-loop pole $P=1$ and concluding the encircling locus at $k_c=1$ is unstable.

Source. Stephanopoulos, Chemical Process Control, Ch. 17–18 (Nyquist criterion, stability of unstable systems); Seborg et al., Ch. 14 (frequency-response stability).

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

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