Second-Order ODE — Standard Form, Stability and Damping: 20-Bio-A2 Process Dynamics and Control, May 2017, Q6

Problem 6: Second-Order ODE — Standard Form, Stability and Damping (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. $\ddot y+k\dot y+10y=2x$ (deviation variables, zero initial conditions).

Find. (a) $Y(s)/X(s)$ in standard second-order form; (b) the ranges of $k$ giving a stable / underdamped / overdamped open-loop step response; (c) $\tau(k)$ and $\zeta(k)$ when underdamped.

Approach. Laplace-transform, normalise the constant term to $1$ to read off the standard-form gain, time constant and damping ratio by inspection, then classify the roots of the characteristic quadratic against $k$.

(a) Standard form. $(s^2+ks+10)Y=2X$; dividing by $10$, $$\boxed{\frac{Y}{X}=\frac{0.2}{0.1s^2+0.1k\,s+1}=\frac{K}{\tau^2s^2+2\zeta\tau s+1}}$$ with $K=0.2$, $\tau=\dfrac{1}{\sqrt{10}}=0.316$, and $2\zeta\tau=0.1k$.
(b) Classify the roots of $s^2+ks+10=0$. (i) Stable iff both coefficients are positive (the constant term $10$ already is), so $k>0$. (ii) Underdamped ($\zeta<1$, complex roots) iff the discriminant $k^2-40<0$, i.e. $0<k<2\sqrt{10}=6.325$. (iii) Overdamped ($\zeta>1$, real distinct roots) iff $k>2\sqrt{10}=6.325$ (critically damped exactly at $k=6.325$).
(c) Underdamped $\tau$ and $\zeta$. From $2\zeta\tau=0.1k$ with $\tau=1/\sqrt{10}$ fixed, $$\boxed{\tau=\frac{1}{\sqrt{10}}=0.316\ \text{(independent of }k),\qquad \zeta=\frac{k}{2\sqrt{10}}=0.1581\,k.}$$ Only the damping ratio depends on $k$; the natural period is fixed entirely by the $10y$ term. As $k$ rises from $0$ toward $6.325$ the response moves from sustained oscillation toward critical damping.

Result	Value
Standard form	$Y/X=0.2/(0.1s^2+0.1ks+1)$
(i) Stable	$k>0$
(ii) Underdamped	$0<k<6.325$
(iii) Overdamped	$k>6.325$
(c) $\tau$, $\zeta$	$\tau=0.316$ (const.); $\zeta=0.1581k$

Concept & Theory

Topic. Standard second-order gain–time-constant form; classification of a quadratic's roots by discriminant.

Key relations. Standard form $K/(\tau^2s^2+2\zeta\tau s+1)$; discriminant sign decides real vs. complex roots; $\zeta<1\Leftrightarrow$ underdamped.

Why this works. Normalising the constant term to $1$ is what makes the coefficients of $s$ and $s^2$ read directly as $2\zeta\tau$ and $\tau^2$; because the $10y$ term is fixed, $\tau$ never changes with $k$ — only the damping (the coefficient that actually contains $k$) does.

Common pitfall. Reading $\zeta$ off the un-normalised equation (forgetting to divide by $10$ first), or claiming the system is unstable for large $k$ — large $k$ only makes it more overdamped, never unstable.

Source. Seborg et al., Process Dynamics and Control, Ch. 5 (second-order systems, standard form, damping regimes).

Question 6 of 8: Second-Order ODE — Standard Form, Stability and Damping

Problem 6: Second-Order ODE — Standard Form, Stability and Damping (20% total)