Step Response and Overshoot Condition for a Process With an Adjustable Zero: 20-Bio-A2 Process Dynamics and Control, May 2018, Q4

Problem 4: Step Response and Overshoot Condition for a Process With an Adjustable Zero (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. $G(s)=\dfrac{2(1+cs)}{(s+1)(s+2)}$, poles fixed at $s=-1,-2$, adjustable zero at $s=-1/c$.

Find. (a) $y(t)$ for a unit step, as a function of $c$; (b) the values of $c$ giving overshoot and the overshoot magnitude as a function of $c$.

Approach. Expand $Y(s)=G(s)/s$ by partial fractions parametrized by $c$, invert to get $y(t;c)$, then locate the extremum by $dy/dt=0$ and test when it lies at $t>0$ and exceeds the final value $y(\infty)=1$ (true overshoot, as opposed to an inverse-response dip below zero from a right-half-plane zero).

(a) Partial fractions. $Y(s)=\dfrac{2(1+cs)}{s(s+1)(s+2)}=\dfrac{A}{s}+\dfrac{B}{s+1}+\dfrac{C}{s+2}$ with $A=1$, $B=2c-2$, $C=1-2c$, so $$\boxed{y(t)=1+(2c-2)e^{-t}+(1-2c)e^{-2t}.}$$ Check: $y(0)=1+(2c-2)+(1-2c)=0$ for every $c$, and $y(\infty)=1=G(0)$.
(b) Critical point. Setting $dy/dt=-(2c-2)e^{-t}-2(1-2c)e^{-2t}=0$ and solving gives $e^{t^*}=\dfrac{2c-1}{c-1}$, i.e. $$t^*=\ln\!\left(\frac{2c-1}{c-1}\right),$$ real and positive only when $\dfrac{2c-1}{c-1}\gt1\iff \dfrac{c}{c-1}\gt0\iff c\gt1\ \text{or}\ c\lt0.$
(b) Evaluate $y(t^*)$. Substituting $e^{-t^*}=(c-1)/(2c-1)$ into $y(t)$ and simplifying, $$\boxed{y(t^*)=1+\frac{(c-1)^2}{2c-1}}\quad\Rightarrow\quad \boxed{\Delta(c)=y(t^*)-1=\frac{(c-1)^2}{2c-1}}\ \ (\text{overshoot magnitude}).$$
(b) Sign analysis. For $c\gt1$: $(c-1)^2\gt0$ and $2c-1\gt0$, so $\Delta(c)\gt0$ — a genuine overshoot above the final value. For $c\lt0$: $2c-1\lt0$, so $\Delta(c)\lt0$ — this is instead the trough of an inverse response (the zero at $s=-1/c\gt0$ is in the right half-plane), not an overshoot. For $0\le c\le1$ no interior extremum exists ($t^*\le0$): the response is monotonic. At $c=1$ the zero cancels the pole at $s=-1$ (reduces to $2/(s+2)$, $\Delta=0$); at $c=0.5$ the zero cancels the pole at $s=-2$ (reduces to $1/(s+1)$), both first-order, no overshoot — consistent boundary checks.
Worked example. At $c=2$: $t^*=\ln3=1.099$, $\Delta(2)=(1)^2/3=0.333$, so $$\boxed{y_{\max}=1.333\ \text{at}\ t^*=1.099\ \text{s} \ (c=2).}$$

Result	Value
Step response	$y(t)=1+(2c-2)e^{-t}+(1-2c)e^{-2t}$
Overshoot condition	$c\gt1$
Overshoot magnitude	$\Delta(c)=(c-1)^2/(2c-1)$
Example ($c=2$)	$y_{\max}=1.333$ at $t=1.099\ \text{s}$

Concept & Theory

Topic. Effect of transfer-function zero location on the shape of a step response.

Key relations. Parametrized partial-fraction inversion; extremum condition $dy/dt=0$; zero location $s=-1/c$.

Why this works. A left-half-plane zero that is "slower" (closer to the origin) than the dominant poles ($c\gt1\Rightarrow$ zero between $-1$ and $0$) differentiates and re-adds a fast component of the response, producing overshoot; a right-half-plane zero ($c\lt0$) instead produces the classic inverse-response undershoot.

Common pitfall. Treating any nonzero critical point as "overshoot" without checking whether $y(t^*)$ exceeds or falls below the final value — the $c\lt0$ branch is a dip below the initial value, not an overshoot above the final one.

Source. Seborg et al., Process Dynamics and Control, Ch. 6 (effect of zeros on dynamic response, overshoot and inverse response).

Question 4 of 8: Step Response and Overshoot Condition for a Process With an Adjustable Zero

Problem 4: Step Response and Overshoot Condition for a Process With an Adjustable Zero (20%)