PI Control of a First-Order Process — Stability and Step Response: 23-Chem-A6 Process Dynamics and Control, December 2018, Q5

Problem 5: PI Control of a First-Order Process — Stability and Step Response (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. A single loop with proportional-integral control ($\tau_I=1$):

Element	Transfer function
Process	$G_p=1/(s+5)$
Controller	$G_c=K_c(1+1/s)=K_c(s+1)/s$
Loop	$L=K_c(s+1)/[s(s+5)]$
Set point	unit step $R=1/s$

Find. (1) stabilising range of $K_c$; (2) $c(t)$ for a unit set-point step at $K_c=1$.

Problem 5: the two closed-loop poles are real ($-0.172,-5.828$); the response is overdamped and, thanks to integral action, reaches the set point with zero offset. The slow pole ($\tau\approx5.8$ s) sets the settling time.

Approach. Form the characteristic polynomial $s(s+5)+K_c(s+1)$, apply the second-order positivity test for the range, then invert the closed-loop transfer function for a unit step at $K_c=1$.

(1) Characteristic equation. $1+L=0\Rightarrow s(s+5)+K_c(s+1)=0$, i.e. $$s^{2}+(5+K_c)s+K_c=0.$$
(1) Stability range. A second-order polynomial has both roots in the left half-plane iff all coefficients are positive: $5+K_c>0$ and $K_c>0$. The binding condition is $$\boxed{K_c>0.}$$ Any positive gain stabilises the loop — adding an integrator to a stable first-order plant cannot destabilise it for $K_c>0$ (the discriminant $(5+K_c)^2-4K_c=K_c^2+6K_c+25$ is always positive, so the two closed-loop poles are always real and negative — one on each side of the controller zero at $-1$).
(2) Closed-loop transfer function ($K_c=1$). $$\frac{C}{R}=\frac{L}{1+L}=\frac{K_c(s+1)}{s^{2}+(5+K_c)s+K_c}\Big|_{K_c=1}=\frac{s+1}{s^{2}+6s+1}.$$ Roots: $s=-3\pm2\sqrt2=-0.172,\,-5.828$ (overdamped).
Set-point step. With $R=1/s$, $$C(s)=\frac{s+1}{s\,(s+0.172)(s+5.828)}=\frac{A}{s}+\frac{r_1}{s+0.172}+\frac{r_2}{s+5.828}.$$ $A=\dfrac{0+1}{(0.172)(5.828)}=1.0$ (product of the poles $=1$), so the final value is $1$ — integral action removes offset.
Residues and response. $r_1=\dfrac{-0.172+1}{(-0.172)(-0.172+5.828)}=-0.853$; $r_2=\dfrac{-5.828+1}{(-5.828)(-5.828+0.172)}=-0.147$. Inverting, $$\boxed{c(t)=1-0.853\,e^{-0.172\,t}-0.147\,e^{-5.828\,t}.}$$ The fast mode decays within $\sim1$ s; the slow mode ($\tau\approx5.8$ s) governs the approach to the set point, which is reached with no steady-state error.

Quantity	Value
(1) Characteristic eqn	$s^2+(5+K_c)s+K_c=0$
(1) Stable range	$K_c>0$
(2) Closed-loop poles	$-0.172,\ -5.828$ (overdamped)
(2) $c(t)$	$1-0.853e^{-0.172t}-0.147e^{-5.828t}$
Steady-state offset	$0$ (integral action)

Concept & Theory

Topic. PI control of a first-order process; second-order stability test; offset elimination by integral action.

Key relations. Characteristic polynomial from $1+G_cG_p=0$; second-order rule — both roots LHP $\Leftrightarrow$ all coefficients positive; final-value theorem $\Rightarrow$ zero offset when the loop contains an integrator.

Why this works. The integrator adds a pole at the origin and a zero at $-1/\tau_I$; the resulting second-order closed loop keeps both poles in the LHP for every $K_c>0$, and the open-loop pole at the origin forces the steady-state error of a step to zero.

Common pitfall. Expecting an upper gain limit as for higher-order loops — a second-order characteristic polynomial can never be destabilised by increasing a positive gain. Also, do not forget the controller zero $(s+1)$ in the numerator of $C/R$; it speeds the response but does not affect the poles.

Source. Coughanowr & LeBlanc, Ch. 13–16 (PI control, offset); Seborg et al., Ch. 11.

Question 5 of 8: PI Control of a First-Order Process — Stability and Step Response

Problem 5: PI Control of a First-Order Process — Stability and Step Response (20%)