Question 1 of 8: Overshoot and Inverse-Response Conditions for a Lead/Lag Process

Problem 1: Overshoot and Inverse-Response Conditions for a Lead/Lag 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. $G(s)=\dfrac{K(\tau_a s+1)}{(\tau_1 s+1)(\tau_2 s+1)}$ with $\tau_1>\tau_2>0$, $K>0$, subject to a step input of magnitude $M$.

Find. Prove (1) a genuine overshoot (response exceeding its final value) occurs iff $\tau_a/\tau_1>1$; (2) an inverse response (initial motion opposite the eventual direction of change) occurs iff $\tau_a<0$.

Approach. Expand $Y(s)=G(s)M/s$ by partial fractions in pole form, invert to $y(t)$, differentiate to locate the single interior critical point, and classify it (real? positive $t$? maximum above $y(\infty)$, or minimum below $y(0)$?) as a function of $\tau_a$.

1. Partial-fraction inversion. With poles at $s=-1/\tau_1,-1/\tau_2$, standard partial fractions give $$\boxed{\frac{y(t)}{KM}=1+\frac{\tau_a-\tau_1}{\tau_1-\tau_2}\,e^{-t/\tau_1}-\frac{\tau_a-\tau_2}{\tau_1-\tau_2}\,e^{-t/\tau_2}.}$$ Check: at $t=0$ the bracket is $1+\tfrac{\tau_a-\tau_1}{\tau_1-\tau_2}-\tfrac{\tau_a-\tau_2}{\tau_1-\tau_2}=1+\tfrac{-\tau_1+\tau_2}{\tau_1-\tau_2}=1-1=0=y(0)$, and as $t\to\infty$, $y\to KM$, matching $G(0)M$.
2. Initial slope (sign test for inverse response). Differentiating and evaluating at $t=0$: $$\left.\frac{dy}{dt}\right|_{t=0}=\frac{KM}{\tau_1-\tau_2}\left(-\frac{\tau_a-\tau_1}{\tau_1}+\frac{\tau_a-\tau_2}{\tau_2}\right)=\frac{KM\,\tau_a}{\tau_1\tau_2}.$$ Since $K,M,\tau_1,\tau_2>0$, the sign of the initial slope is exactly the sign of $\tau_a$: the response starts by moving in the same direction as its eventual increase ($\tau_a>0$) or in the opposite direction ($\tau_a<0$) — the definition of inverse response.
3. Interior critical point. Setting $dy/dt=0$ for $t>0$ and solving: $$e^{t\left(\frac1{\tau_1}-\frac1{\tau_2}\right)}=\frac{\tau_2(\tau_a-\tau_1)}{\tau_1(\tau_a-\tau_2)}\equiv R.$$ Because $\tau_1>\tau_2$, the exponent coefficient $\left(\tfrac1{\tau_1}-\tfrac1{\tau_2}\right)<0$, so the left side is a strictly decreasing function of $t$ running from $1$ (at $t=0$) down to $0$ (as $t\to\infty$). A finite root $t^{*}>0$ therefore exists iff $0<R<1$.
4. Sign of $R$. $R=\dfrac{\tau_2(\tau_a-\tau_1)}{\tau_1(\tau_a-\tau_2)}$ has the sign of $(\tau_a-\tau_1)(\tau_a-\tau_2)$ (a ratio of reals has the sign of the product). Since $\tau_2<\tau_1$, this product is positive exactly when $\tau_a<\tau_2$ or $\tau_a>\tau_1$, and non-positive (no real, finite $t^{*}>0$ — monotonic response) for $\tau_2\le\tau_a\le\tau_1$.
5. Branch $\tau_a>\tau_1$ (overshoot test). Both $(\tau_a-\tau_1)$ and $(\tau_a-\tau_2)$ are positive, so multiplying $R<1$ through by the positive quantity $\tau_1(\tau_a-\tau_2)$ preserves the inequality: $\tau_2(\tau_a-\tau_1)<\tau_1(\tau_a-\tau_2)\iff \tau_2\tau_a<\tau_1\tau_a\iff \tau_2<\tau_1$, which is always true. So whenever $\tau_a>\tau_1$ a critical point $t^{*}>0$ always exists; since $dy/dt|_0=KM\tau_a/(\tau_1\tau_2)>0$ here and there is only one interior critical point, it must be a maximum, and $y$ can only return to $y(\infty)=KM$ from above it — i.e. $$\boxed{y(t^{*})>KM=y(\infty):\ \text{genuine overshoot occurs iff } \tau_a>\tau_1\iff \tau_a/\tau_1>1.}$$
6. Branch $\tau_a<\tau_2$ (inverse-response test). Here $(\tau_a-\tau_1)<0$ and $(\tau_a-\tau_2)<0$; multiplying $R<1$ through by the negative quantity $\tau_1(\tau_a-\tau_2)$ flips the inequality: $\tau_2(\tau_a-\tau_1)>\tau_1(\tau_a-\tau_2)\iff \tau_a(\tau_2-\tau_1)>0\iff \tau_a<0$ (since $\tau_2-\tau_1<0$). So a finite $t^{*}>0$ exists on this branch only when $\tau_a<0$; with $dy/dt|_0<0$ there, the single critical point is a minimum below $y(0)=0$, and the response dips negative before rising to $KM>0$ — the defining signature of inverse response. For $0\le\tau_a<\tau_2$ no finite $t^{*}>0$ exists (response monotonic, no dip), so $$\boxed{\text{inverse response occurs iff } \tau_a<0.}$$
7. Numerical confirmation. With $\tau_1=2$, $\tau_2=1$, $K=M=1$ (arbitrary illustrative values with $\tau_1>\tau_2$): $\tau_a=3$ ($\tau_a/\tau_1=1.5>1$) simulates to a peak $y_{\max}=1.125>1$; $\tau_a=1$ ($\tau_a/\tau_1=0.5<1$) is monotonic ($y_{\max}=y(\infty)=1.000$); $\tau_a=-0.5<0$ dips to $y_{\min}=-0.042<0$ before rising; $\tau_a=+0.5>0$ never goes negative — all four cases match the boxed conditions exactly.