Routh Stability and Step Response of a PI-Controlled Unstable Process: 20-Bio-A2 Process Dynamics and Control, Undated paper, Q5

Problem 5: Routh Stability and Step Response of a PI-Controlled Unstable Process (20% total)

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. $G_p(s)=\dfrac{1}{s-5}$ — an open-loop-unstable first-order process (pole at $s=+5$); PI controller $G_c(s)=k_c\!\left(1+\dfrac1s\right)=\dfrac{k_c(s+1)}{s}$; unity feedback assumed.

Find. (a) the range of $k_c$ for closed-loop stability via Routh; (b) $y(t)$ for a unit step in set point at $k_c=1$.

Approach. Form the closed-loop characteristic equation $1+G_cG_p=0$, clear fractions to a polynomial in $s$, apply the second-order Routh shortcut (all coefficients positive); for (b), substitute $k_c=1$ into the closed-loop transfer function, multiply by a unit step, and invert by partial fractions using the two real roots — then compare the result against part (a)'s stability boundary as a consistency check.

(a) Characteristic equation. $1+\dfrac{k_c(s+1)}{s}\cdot\dfrac{1}{s-5}=0\Rightarrow s(s-5)+k_c(s+1)=0\Rightarrow$ $$\boxed{s^2+(k_c-5)s+k_c=0.}$$
(a) Routh array (2nd-order shortcut). For $as^2+bs+c=0$ with $a=1>0$, stability requires every coefficient positive: $b=k_c-5>0$ and $c=k_c>0$. The first condition is the binding one (it implies $k_c>5$, which already satisfies the second), so $$\boxed{k_c>5\ \text{for closed-loop stability}}$$ (the PI controller must supply enough proportional gain to overcome the process's own instability at $s=+5$ before the integral action can help).
(b) Closed-loop transfer function at $k_c=1$ (below the stability threshold). Since $k_c=1<5$, part (a) already predicts an unstable closed loop at this gain; part (b) computes that response explicitly. $$\frac{Y(s)}{Y_{sp}(s)}=\frac{k_c(s+1)}{s^2+(k_c-5)s+k_c}\bigg|_{k_c=1}=\frac{s+1}{s^2-4s+1}.$$
(b) Roots and partial fractions. $s^2-4s+1=0\Rightarrow s=2\pm\sqrt3=\{3.732,\,0.268\}$ — both real and positive (both closed-loop poles in the RHP), confirming instability. For a unit step, $Y(s)=\dfrac{s+1}{s(s-s_1)(s-s_2)}$ with $s_1=3.732$, $s_2=0.268$: $$A=\frac{1}{s_1s_2}=1,\qquad B=\frac{s_1+1}{s_1(s_1-s_2)}=0.3660,\qquad C=\frac{s_2+1}{s_2(s_2-s_1)}=-1.3660.$$
(b) Result. $$\boxed{y(t)=1+0.3660\,e^{3.732t}-1.3660\,e^{0.268t}.}$$ Check: $y(0)=1+0.366-1.366=0$, matching zero initial output. Unlike a stabilizing gain, $y(t)$ does not settle: the $e^{3.732t}$ term grows without bound ($y(0.5)=1.80$, $y(1)=14.5$, $y(1.5)=97.8$, $y(2)=637$), so the "response to a unit step" at $k_c=1$ is a diverging (unstable) trajectory — a direct demonstration, via time response, of the $k_c>5$ boundary found in part (a).

Result	Value
Characteristic equation	$s^2+(k_c-5)s+k_c=0$
Stability range	$k_c>5$
Closed-loop poles ($k_c=1$)	$s=3.732,\,0.268$ (both RHP — unstable)
Step response ($k_c=1$)	$y(t)=1+0.3660e^{3.732t}-1.3660e^{0.268t}$ (diverges)

Concept & Theory

Topic. Routh–Hurwitz stability and closed-loop step response for PI control of an open-loop-unstable first-order process.

Key relations. Characteristic equation $1+G_cG_p=0$; second-order Routh shortcut (all coefficients positive $\iff$ stable); a chosen gain below the Routh-derived threshold produces real, positive (right-half-plane) closed-loop poles rather than complex ones.

Why this works. Unlike PI control of an already-stable process (where any $k_c>0$ stabilizes), here the controller must first supply enough proportional action to move the process's own unstable pole into the left half-plane; the Routh test's binding inequality ($k_c>5$) says exactly this, and evaluating the actual step response at a gain below that threshold ($k_c=1$) reproduces the instability directly as an unbounded exponential, cross-validating parts (a) and (b) against each other.

Common pitfall. Assuming (as with a stable process) that "$k_c>0$" is automatically sufficient because the controller has integral action — here the process's own instability raises the threshold to $k_c>5$; also, mis-signing the quadratic when substituting $G_p=1/(s-5)$ (a $-5$, not $+5$, appears as the coefficient of $s$).

Source. Seborg et al., Process Dynamics and Control, Ch. 11 (Routh stability), Ch. 8 (PI controller); Stephanopoulos, Ch. 16 (stabilizing an open-loop-unstable process).

Question 5 of 8: Routh Stability and Step Response of a PI-Controlled Unstable Process

Problem 5: Routh Stability and Step Response of a PI-Controlled Unstable Process (20% total)