Question 6 of 8: PI Control of a First-Order Process — Stability and Step Response

Problem 6: 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. $G_p=\dfrac{1}{s+5}$; PI controller $G_c=K_c\!\left(1+\dfrac1s\right)=K_c\dfrac{s+1}{s}$ (integral time $\tau_I=1$); unity valve/sensor.

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

Approach. Form the loop transfer function, write the characteristic polynomial, apply the second-order Routh conditions for the range, then for $K_c=1$ factor the closed loop and invert the step by partial fractions.

1. Loop and characteristic equation. $L=G_cG_p=\dfrac{K_c(s+1)}{s(s+5)}$. The closed-loop characteristic equation $1+L=0$ gives $$s(s+5)+K_c(s+1)=s^2+(5+K_c)s+K_c=0.$$
2. (a) Stability range. For a monic quadratic $s^2+a_1s+a_0$, both roots lie in the LHP iff $a_1>0$ and $a_0>0$: here $5+K_c>0$ and $K_c>0$. Hence $$\boxed{K_c>0}$$ (any positive gain is stable; the process pole at $-5$ gives generous margin).
3. (b) Closed loop at $K_c=1$. Characteristic equation $s^2+6s+1=0$, roots $s=-3\pm2\sqrt2=-0.1716,\,-5.828$ (both real, overdamped). The servo transfer function is $$\frac{C}{R}=\frac{L}{1+L}=\frac{K_c(s+1)}{s^2+(5+K_c)s+K_c}=\frac{s+1}{s^2+6s+1}.$$
4. Step-forced output. For $R=1/s$, $C(s)=\dfrac{s+1}{s\,(s+0.1716)(s+5.828)}=\dfrac{A}{s}+\dfrac{B}{s+0.1716}+\dfrac{C}{s+5.828}$. Cover-up residues: $A=\dfrac{1}{(0.1716)(5.828)}=1.000$; $B=\dfrac{-0.1716+1}{(-0.1716)(-0.1716+5.828)}=-0.854$; $C=\dfrac{-5.828+1}{(-5.828)(-5.828+0.1716)}=-0.146$. (Check $A+B+C=0\Rightarrow c(0)=0$.)
5. Response. Inverting, $$\boxed{c(t)=1-0.854\,e^{-0.1716t}-0.146\,e^{-5.828t}}.$$ The output starts at $0$, rises without overshoot, and reaches the set point with zero offset — the integral action guarantees $c(\infty)=1$. The slow mode ($e^{-0.1716t}$, time constant $\approx5.8\ \mathrm s$) dominates the approach; the fast mode disappears within $\sim1\ \mathrm s$.