PI Control of a First-Order Process: 23-Chem-A6 Process Dynamics and Control, May 2016, Q8

Problem 8: PI Control 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. $G_p=\dfrac{1}{s+5}$; PI controller $G_c=K_c\left(1+\dfrac{1}{s}\right)=K_c\,\dfrac{s+1}{s}$ (integral time $\tau_I=1\,$s).

Find. (a) the stabilising range of $K_c$; (b) the unit-step set-point response at $K_c=1$.

Approach. Form the loop transfer function, write the closed-loop characteristic polynomial, and apply the (trivial for second order) Routh condition that all coefficients be positive. Then form the closed-loop set-point transfer function and invert the unit-step response by partial fractions.

(a) Characteristic equation and stability range. $L=K_c\dfrac{s+1}{s}\cdot\dfrac{1}{s+5}=\dfrac{K_c(s+1)}{s(s+5)}$, so $1+L=0$ gives $$s^{2}+(5+K_c)s+K_c=0.$$ For a second-order polynomial, stability requires every coefficient positive: $5+K_c>0$ and $K_c>0$. Hence $$\boxed{K_c>0}$$ — the loop is stable for all positive gains (PI control of a single first-order lag cannot be destabilised).
(b) Closed-loop set-point transfer function ($K_c=1$). $\dfrac{C}{R}=\dfrac{L}{1+L}=\dfrac{K_c(s+1)}{s^{2}+(5+K_c)s+K_c}=\dfrac{s+1}{s^{2}+6s+1}.$ The poles are the roots of $s^{2}+6s+1=0$: $s=-3\pm2\sqrt2=-0.1716,\,-5.828$ (real, overdamped).
Invert the unit-step response. With $R=1/s$, $C(s)=\dfrac{s+1}{s(s+0.1716)(s+5.828)}=\dfrac{A}{s}+\dfrac{B}{s+0.1716}+\dfrac{D}{s+5.828}$, giving $A=1$ (set-point value — PI removes offset), $B=-0.8536$, $D=-0.1464$, so $$\boxed{c(t)=1-0.8536\,e^{-0.1716t}-0.1464\,e^{-5.828t}.}$$
Interpret. $c(0)=1-0.8536-0.1464=0$ and $c(\infty)=1$ (zero offset, as integral action guarantees). The response is overdamped, dominated by the slow pole $s=-0.1716$ (time constant $\approx5.8\,$s); the fast mode $e^{-5.828t}$ decays almost immediately. No overshoot (see figure).

Problem 8(b): closed-loop unit-step response at $K_c=1$, $c(t)=1-0.8536e^{-0.1716t}-0.1464e^{-5.828t}$. A smooth overdamped rise to the set point with no offset and no overshoot; the dominant time constant is $\approx5.8\,$s.

Result	Value
Characteristic equation	$s^{2}+(5+K_c)s+K_c=0$
Stability range	$K_c>0$ (all positive gains)
Closed-loop poles ($K_c=1$)	$s=-0.1716,\,-5.828$
Step response	$c(t)=1-0.8536e^{-0.1716t}-0.1464e^{-5.828t}$
Steady-state offset	$0$ (integral action)

Concept & Theory

Topic. PI control of a first-order process; Routh stability; offset-free step response.

Key relations. PI controller $G_c=K_c(1+1/(\tau_I s))$; closed loop $C/R=L/(1+L)$; second-order Routh: all coefficients $>0$; integral action forces $C/R|_{s=0}=1$ (zero offset).

Why this works. Adding integral action raises the loop order to two but keeps both closed-loop poles in the LHP for any $K_c>0$, because the constant term ($K_c$) and the $s$-coefficient ($5+K_c$) are always positive. The integrator drives the steady-state error to zero regardless of gain.

Common pitfall. Expecting an upper gain limit (there is none here — only for higher-order or dead-time plants does PI go unstable), or forgetting the controller zero at $s=-1$, which appears in the numerator of $C/R$ and slightly speeds the response.

Source. Seborg et al., Ch. 11 (PI/PID control, closed-loop stability); Coughanowr & LeBlanc, Ch. 13–15 (controller modes, offset).

Question 8 of 8: PI Control of a First-Order Process

Problem 8: PI Control of a First-Order Process (20%)