Second-Order ODE — Standard Form and Damping: 23-Chem-A6 Process Dynamics and Control, May 2014, Q6

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

Find. (a) $Y/X$ in standard form; (b) the $k$-ranges for stability / under- / over-damping; (c) $\tau$ and $\zeta$ when underdamped.

Approach. Laplace-transform the ODE, then match the denominator to the canonical $\tau^2s^2+2\zeta\tau s+1$ to read off $K,\tau,\zeta$; the sign/discriminant of the characteristic roots then classifies the response.

(a) Transfer function. Transforming (zero IC): $(s^2+ks+10)Y=2X$, so $Y/X=\dfrac{2}{s^2+ks+10}$. Dividing top and bottom by $10$ to force the constant term to unity: $$\boxed{\dfrac{Y}{X}=\dfrac{0.2}{\tfrac{1}{10}s^2+\tfrac{k}{10}s+1}=\dfrac{K}{\tau^2s^2+2\zeta\tau s+1}}$$ with gain $K=0.2$, $\tau^2=\tfrac1{10}$ and $2\zeta\tau=\tfrac{k}{10}$.
Read off the parameters. $\omega_n=\sqrt{10}$, so $$\tau=\dfrac{1}{\omega_n}=\dfrac{1}{\sqrt{10}}\approx0.316,\qquad \zeta=\dfrac{k}{2\sqrt{10}}\approx\dfrac{k}{6.32}.$$ The characteristic equation is $s^2+ks+10=0$ with roots $s=\dfrac{-k\pm\sqrt{k^2-40}}{2}$.
(b-i) Stability. For $s^2+ks+10$ both roots have negative real part iff every coefficient is positive; since $10>0$ always, the condition is $$\boxed{k>0}.$$ At $k=0$ the poles sit on the imaginary axis ($\pm j\sqrt{10}$, sustained oscillation); for $k<0$ the response grows unbounded.
(b-ii,iii) Under- vs over-damped. The roots are complex ($\zeta<1$) when the discriminant is negative: $k^2-40<0\Rightarrow|k|<2\sqrt{10}$. With stability ($k>0$): $$\underbrace{0<k<2\sqrt{10}}_{\text{underdamped }(\approx6.32)}\ ;\qquad \underbrace{k>2\sqrt{10}}_{\text{overdamped}}\ ;\qquad k=2\sqrt{10}\ (\text{critically damped}).$$
(c) Underdamped parameters. Directly from the standard form (valid for any $k$, and underdamped when $0<k<2\sqrt{10}$): $$\boxed{\tau=\dfrac{1}{\sqrt{10}}\approx0.316\ \text{(time units)}},\qquad \boxed{\zeta=\dfrac{k}{2\sqrt{10}}}.$$ The time constant is fixed by the $10y$ term (independent of $k$); only the damping coefficient scales with $k$.

Problem 6(b): unit-step response for representative $k$. $k=2$ is underdamped (overshoot and ringing), $k=2\sqrt{10}\approx6.32$ is critically damped, $k=12$ is overdamped (sluggish, no overshoot); all settle at the gain $K\cdot\Delta x=0.2$.

Quantity	Expression / value
Transfer function	$Y/X=0.2/(\tfrac1{10}s^2+\tfrac{k}{10}s+1)$
Gain $K$	$0.2$
Natural frequency / time constant	$\omega_n=\sqrt{10}$, $\tau=1/\sqrt{10}\approx0.316$
Damping ratio	$\zeta=k/(2\sqrt{10})\approx k/6.32$
Stable / underdamped / overdamped	$k>0$ / $0<k<2\sqrt{10}$ / $k>2\sqrt{10}$

Concept & Theory

Topic. Reducing a linear second-order ODE to standard form; classifying the step response by the damping ratio.

Key relations. Standard form $\tau^2\ddot y+2\zeta\tau\dot y+y=Kx$ with $\omega_n=1/\tau$; roots $s=-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}$; underdamped $\zeta<1$, critical $\zeta=1$, overdamped $\zeta>1$; stable $\Leftrightarrow$ all characteristic coefficients positive.

Why this works. Forcing the constant term to $1$ exposes $K$ (DC gain), $\tau$ (from the $s^2$ coefficient) and $\zeta$ (from the $s$ coefficient) by direct comparison. The discriminant $k^2-40$ decides whether the poles are real (over) or complex (under); its sign flips exactly at $\zeta=1$.

Common pitfall. Forgetting to normalise the constant term before reading $\tau$ and $\zeta$, or claiming $\tau$ depends on $k$ (it does not — only $\zeta$ does).

Source. Coughanowr & LeBlanc, Ch. 7–8 (second-order systems); Seborg et al., Ch. 5 (second-order dynamics, damping).

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

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