Routh Test and Closed-Loop Response of a First-Order Process: 23-Chem-A6 Process Dynamics and Control, May 2013, Q5

Problem 5: Routh Test and Closed-Loop Response of a 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. $G_p=\dfrac{1}{s+5}$; controller $G_c=k_c\!\left(1+\dfrac1s\right)$; $G_v=G_m=1$.

Find. (a) the range of $k_c$ for stability (Routh); (b) the set-point step response at $k_c=1$.

Check — controller form as printed

The paper labels $G_c=k_c(1+1/s)$ a "PD" controller, but the $1/s$ term is an integral action — this is the standard PI form (with $\tau_I=1$). The solution uses the controller exactly as printed ($k_c(1+1/s)$); the integral term is what delivers the zero steady-state offset seen in part (b). A true PD controller $k_c(1+\tau_D s)$ would leave a finite offset.

Approach. Form the closed-loop characteristic equation $1+G_c G_p=0$, apply the Routh test to the resulting polynomial for part (a), then invert the closed-loop transfer function by partial fractions for the $k_c=1$ step response.

Open loop and characteristic equation. $L=G_c G_p=\dfrac{k_c(1+1/s)}{s+5}=\dfrac{k_c(s+1)}{s(s+5)}$. Setting $1+L=0$: $$s(s+5)+k_c(s+1)=0\ \Rightarrow\ s^2+(5+k_c)s+k_c=0.$$
(a) Routh test. For a quadratic $s^2+a_1 s+a_0$ the necessary and sufficient condition is $a_1>0$ and $a_0>0$: $a_1=5+k_c>0\Rightarrow k_c>-5$ and $a_0=k_c>0\Rightarrow k_c>0$. The binding condition is $$\boxed{k_c>0}$$ i.e. the closed loop is stable for every positive controller gain (a first-order process with a PI controller can never be destabilised here).
(b) Closed-loop transfer function, $k_c=1$. $\dfrac{Y}{Y_{sp}}=\dfrac{L}{1+L}=\dfrac{k_c(s+1)}{s^2+(5+k_c)s+k_c}=\dfrac{s+1}{s^2+6s+1}$. The poles are $s=-3\pm2\sqrt2=\{-0.1716,\,-5.828\}$ — both real and negative (overdamped).
Step response by partial fractions. With $Y_{sp}=1/s$, $Y(s)=\dfrac{s+1}{s(s+0.1716)(s+5.828)}=\dfrac{1}{s}-\dfrac{0.854}{s+0.1716}-\dfrac{0.146}{s+5.828}$, giving $$y(t)=1-0.854\,e^{-0.172t}-0.146\,e^{-5.83t}.$$ The response starts at $y(0)=0$, rises monotonically (overdamped, dominated by the slow $0.172\ \mathrm{min^{-1}}$ mode) and settles at $\boxed{y(\infty)=1}$ — no offset, the hallmark of the integral action.

Problem 5(b): closed-loop unit-step set-point response at $k_c=1$. Two real poles ($-0.172,\,-5.83$) give an overdamped monotonic rise to $y(\infty)=1$ with no offset (integral action).

Quantity	Value
Characteristic equation	$s^2+(5+k_c)s+k_c=0$
Stability range (Routh)	$k_c>0$
Closed-loop poles ($k_c=1$)	$-0.172,\ -5.83$ (overdamped)
Step response	$y(t)=1-0.854e^{-0.172t}-0.146e^{-5.83t}$
Steady-state value	$1$ (no offset)

Concept & Theory

Topic. Routh–Hurwitz stability test and closed-loop step response of a first-order process with PI (integral) control.

Key relations. Characteristic equation $1+G_cG_pG_vG_m=0$; Routh array (for a quadratic, all coefficients same sign $\Leftrightarrow$ stable); zero offset for a step input requires an integrator in the loop ($\lim_{s\to0}L=\infty$).

Why this works. The Routh test reports the number of RHP roots without solving the polynomial; for a second-order system it collapses to "all coefficients positive." The integral term adds an open-loop pole at the origin, so the loop gain is infinite at DC and the steady-state error to a step vanishes — which is why $y(\infty)=1$ exactly.

Common pitfall. Reading $k_c(1+1/s)$ as a derivative (PD) action and expecting a finite offset, or forgetting that $G_v=G_m=1$ so the characteristic equation involves only $G_cG_p$.

Source. Stephanopoulos, Ch. 15 (Routh test); Seborg et al., Ch. 11 (dynamic behaviour of feedback loops, offset).

Question 5 of 8: Routh Test and Closed-Loop Response of a First-Order Process

Problem 5: Routh Test and Closed-Loop Response of a First-Order Process (20%)