Routh Stability and Closed-Loop Step Response of a PI-Controlled First-Order Process: 20-Bio-A2 Process Dynamics and Control, December 2019, Q8

Problem 8: Routh Stability and Closed-Loop Step Response of a PI-Controlled First-Order 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. $G_p(s)=\dfrac{1}{s+2}$, 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 $s$, build the Routh array and require every first-column entry positive; for (b), substitute $k_c=1$, take the closed-loop transfer function times a unit step, and invert by partial fractions using the two real roots of the resulting quadratic.

(a) Characteristic equation. $1+\dfrac{k_c(s+1)}{s}\cdot\dfrac{1}{s+2}=0\Rightarrow s(s+2)+k_c(s+1)=0\Rightarrow$ $$\boxed{s^2+(2+k_c)s+k_c=0.}$$
(a) Routh array (2nd-order shortcut). For $as^2+bs+c=0$ with $a=1>0$, stability requires all coefficients positive: $b=2+k_c>0$ and $c=k_c>0$. The first condition holds automatically whenever the second does, so $$\boxed{k_c>0\ \text{for closed-loop stability}}$$ (the integral action guarantees the controller cannot itself destabilize this simple first-order process for any positive gain).
(b) Closed-loop transfer function at $k_c=1$. $$\frac{Y(s)}{Y_{sp}(s)}=\frac{G_cG_p}{1+G_cG_p}=\frac{k_c(s+1)}{s^2+(2+k_c)s+k_c}\Big|_{k_c=1}=\frac{s+1}{s^2+3s+1}.$$
(b) Roots and partial fractions. $s^2+3s+1=0\Rightarrow s=\dfrac{-3\pm\sqrt5}{2}=\{-0.38197,\,-2.61803\}$ (real, distinct, both stable — consistent with part (a) since $k_c=1>0$). For a unit step, $Y(s)=\dfrac{s+1}{s(s^2+3s+1)}=\dfrac As+\dfrac{B}{s-s_1}+\dfrac{C}{s-s_2}$ with $s_1=-0.38197$, $s_2=-2.61803$: $$A=\frac{1}{s_1s_2}=1\ (\text{zero offset, as expected with integral action}),$$ $$B=\frac{s_1+1}{s_1(s_1-s_2)}=-0.72361,\qquad C=\frac{s_2+1}{s_2(s_2-s_1)}=-0.27639.$$
(b) Result. $$\boxed{y(t)=1-0.72361\,e^{-0.38197t}-0.27639\,e^{-2.61803t}.}$$ Check: $y(0)=1-0.72361-0.27639=0$ (matches zero initial output); $y(\infty)=1$ (matches the setpoint — zero offset from the PI's integral action). Sample values: $y(1)=0.486$, $y(2)=0.661$, $y(3)=0.770$, $y(5)=0.893$.

Result	Value
Characteristic equation	$s^2+(2+k_c)s+k_c=0$
Stability range	$k_c>0$
Closed-loop poles ($k_c=1$)	$s=-0.38197,\,-2.61803$
Step response ($k_c=1$)	$y(t)=1-0.72361e^{-0.38197t}-0.27639e^{-2.61803t}$
Offset	zero (integral action)

Concept & Theory

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

Key relations. Characteristic equation $1+G_cG_p=0$; second-order Routh shortcut (all coefficients positive $\iff$ stable); zero offset guaranteed whenever $G_c$ contains a pure integrator (final-value theorem gives $y(\infty)=1$ for any stabilizing $k_c$).

Why this works. Adding integral action to a stable first-order process converts the closed loop from first- to second-order and introduces a pole at the origin in the open loop, which by the final-value theorem forces exactly zero steady-state offset for a step in set point — the price is that the closed loop can now (in general) be non-monotonic, though here both roots remain real for all $k_c>0$ up to the point where the discriminant $(2+k_c)^2-4k_c$ turns negative.

Common pitfall. Forgetting to verify $A=1$ (zero offset) as a consistency check, or mis-signing the Routh array coefficients when converting $G_c=k_c(1+1/s)$ to the polynomial form $k_c(s+1)/s$.

Source. Seborg et al., Process Dynamics and Control, Ch. 11 (Routh stability), Ch. 8 (PI controller, offset elimination).

Question 8 of 8: Routh Stability and Closed-Loop Step Response of a PI-Controlled First-Order Process

Problem 8: Routh Stability and Closed-Loop Step Response of a PI-Controlled First-Order Process (20%)