Question 7 of 8: Stability and Step Response of a PI-Controlled First-Order Process

Problem 7: Stability and Step Response of a PI-Controlled First-Order Process (20% total)

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. $G_p(s)=\dfrac1{s+5}$, $G_c(s)=k_c\big(1+\tfrac1s\big)$ (the exam text labels this "proportional-derivative (PI)" — the acronym and the given equation are both for a Proportional-Integral controller; the equation, not the mismatched label, is what is solved here).

Find. (a) the range of $k_c$ for closed-loop stability; (b) the unit-step closed-loop response with $k_c=1$.

Approach. Form the open-loop $L(s)=G_cG_p$, clear fractions to a quadratic characteristic equation, apply the (trivial, for a quadratic) Routh positive-coefficient test, then invert the closed-loop step response by partial fractions at $k_c=1$.

1. (a) Characteristic equation. $L(s)=\dfrac{k_c(s+1)}{s(s+5)}$, so $1+L=0\Rightarrow s(s+5)+k_c(s+1)=0$, i.e. $$\boxed{s^2+(5+k_c)s+k_c=0.}$$
2. (a) Stability. A quadratic is stable iff both coefficients are positive: $5+k_c\gt0$ and $k_c\gt0$; the second condition already implies the first, so $$\boxed{k_c\gt0.}$$
3. (b) Poles at $k_c=1$. $s^2+6s+1=0\Rightarrow s=-3\pm2\sqrt2=-0.1716,\,-5.8284$ (both real and negative — an overdamped, non-oscillatory closed loop).
4. (b) Closed-loop TF and step response. $Y/Y_{sp}=\dfrac{k_c(s+1)}{s^2+(5+k_c)s+k_c}\Big|_{k_c=1}=\dfrac{s+1}{s^2+6s+1}$. With $Y_{sp}(s)=1/s$, partial fractions give residues $A=1$, $B=-0.8536$, $C=-0.1464$, so $$\boxed{y(t)=1-0.8536\,e^{-0.1716t}-0.1464\,e^{-5.8284t}.}$$
5. Check. $y(0)=1-0.8536-0.1464=0$ (correct start) and $y(\infty)=1$ (zero steady-state offset), as expected from the controller's integral action acting on a type-0 process.