Overshoot Condition for a Lead-Zero Second-Order Process: 20-Bio-A2 Process Dynamics and Control, Undated paper, Q1

Problem 1: Overshoot Condition for a Lead-Zero Second-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. $G(s)=\dfrac{2(1+cs)}{(s+1)(s+2)}$, subject to a unit step input ($M=1$); $c$ is a free (real) parameter.

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

Approach. Rewrite $G(s)$ in standard time-constant (zero/pole) form, invert the unit-step response by partial fractions, then locate and classify the single interior critical point of $y(t)$ as a function of $c$ to determine when it is a genuine overshoot.

(a) Standard form. Since $(s+2)=2(0.5s+1)$, $$G(s)=\frac{2(1+cs)}{2(s+1)(0.5s+1)}=\frac{1+cs}{(s+1)(0.5s+1)}=\frac{\tau_a s+1}{(\tau_1 s+1)(\tau_2 s+1)},\qquad K=1,\ \tau_1=1,\ \tau_2=0.5,\ \tau_a=c.$$
(a) Partial-fraction inversion. With poles at $s=-1/\tau_1=-1$ and $s=-1/\tau_2=-2$, the standard lead/lag step response gives $$\boxed{y(t)=1+\frac{\tau_a-\tau_1}{\tau_1-\tau_2}e^{-t/\tau_1}-\frac{\tau_a-\tau_2}{\tau_1-\tau_2}e^{-t/\tau_2}=1+(2c-2)e^{-t}-(2c-1)e^{-2t}.}$$ Check: $y(0)=1+(2c-2)-(2c-1)=0$ and $y(\infty)=1=G(0)\times1$, both independent of $c$ as required.
(b) Locate the interior critical point. $\dfrac{dy}{dt}=-(2c-2)e^{-t}+2(2c-1)e^{-2t}=0\ \Rightarrow\ e^{-t^{*}}=\dfrac{c-1}{2c-1}.$ A finite $t^{*}>0$ (i.e. $0 < e^{-t^{*}} < 1$) exists only when $c>1$ (for $0\le c\le1$ the right side is $\le0$ or the response is monotonic to $y(\infty)=1$ from below); this identifies the overshoot condition as $$\boxed{c>1.}$$
(b) Peak time and overshoot magnitude. For $c>1$: $t^{*}=\ln\!\left(\dfrac{2c-1}{c-1}\right)$. Substituting back into $y(t)$ and simplifying (using $e^{-t^{*}}=\tfrac{c-1}{2c-1}$, $e^{-2t^{*}}=\left(\tfrac{c-1}{2c-1}\right)^2$): $$y(t^{*})=1+(2c-2)\frac{c-1}{2c-1}-(2c-1)\left(\frac{c-1}{2c-1}\right)^2=1+\frac{(c-1)^2}{2c-1}.$$ So the overshoot (peak above the final value $y(\infty)=1$) is $$\boxed{\text{OS}(c)=y(t^{*})-1=\frac{(c-1)^2}{2c-1}\ ,\qquad c>1.}$$
Numerical check. $c=2$: $t^{*}=\ln3=1.099$, $\text{OS}=1^2/3=0.333$ (peak $y=1.333$) — confirmed by direct evaluation of $y(t)=1+2e^{-t}-3e^{-2t}$ at $t=1.099$. $c=1.5$: $\text{OS}=0.25/2=0.125$. $c=5$: $\text{OS}=16/9=1.778$ (a large zero relative to both poles produces a large overshoot, as expected for $\tau_a\gg\tau_1$).

Result	Value
Step response	$y(t)=1+(2c-2)e^{-t}-(2c-1)e^{-2t}$
Overshoot condition	$c>1$
Peak time	$t^{*}=\ln\!\left[(2c-1)/(c-1)\right]$
Overshoot magnitude	$\text{OS}(c)=(c-1)^2/(2c-1)$

Concept & Theory

Topic. Effect of a real lead zero's location on the overshoot of a second-order-plus-zero step response.

Key relations. Standard lead/lag partial-fraction step response of $K(\tau_as+1)/[(\tau_1s+1)(\tau_2s+1)]$; single-interior-critical-point analysis via $dy/dt=0$.

Why this works. The zero at $s=-1/c$ differentiates and re-adds a component of the response; when the zero's time constant exceeds the slower pole's ($c=\tau_a>\tau_1=1$), that extra component pushes the peak above the final value before the slower pole's decay wins out, producing a genuine overshoot whose size grows with how far $c$ exceeds $1$.

Common pitfall. Testing $c>\tau_2$ instead of $c>\tau_1$ (the larger time constant is the relevant threshold, since $\tau_1>\tau_2$ here), or mistaking any non-monotonic response for an overshoot when $c<0$ would instead give an inverse response (a dip below $y(0)=0$, not a peak above $y(\infty)$).

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

Question 1 of 8: Overshoot Condition for a Lead-Zero Second-Order Process

Problem 1: Overshoot Condition for a Lead-Zero Second-Order Process (20%)