Second-Order ODE — Standard Form and Damping Regimes: 23-Chem-A6 Process Dynamics and Control, December 2018, Q6

Problem 6: Second-Order ODE — Standard Form and Damping Regimes (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. A linear second-order ODE with a variable damping term:

Quantity	Value
ODE	$\ddot y+k\dot y+10y=2x$
Static gain factor	output/input $=2/10$
Natural coefficient	$10$ ($=\omega_n^2$)
Damping parameter	$k$ (free)

Find. (a) $Y/X$ in standard form; (b) $k$-ranges for stable / underdamped / overdamped; (c) $\tau(k)$ and $\zeta(k)$ when underdamped.

Problem 6: the character of the response is set by where $k$ falls. $\omega_n=\sqrt{10}$ is fixed; increasing $k$ moves the system from unstable, through underdamped, to overdamped, crossing critical damping at $k=2\sqrt{10}$.

Approach. Laplace-transform to get $Y/X$, normalise to $K/(\tau^2s^2+2\zeta\tau s+1)$ to read $K,\tau,\zeta$, then classify by the sign of the discriminant and the coefficient positivity.

(a) Transfer function. With zero initial conditions, $(s^2+ks+10)Y=2X$, so $Y/X=\dfrac{2}{s^2+ks+10}$. Dividing numerator and denominator by $10$: $$\boxed{\frac{Y}{X}=\frac{0.2}{\tfrac{1}{10}s^{2}+\tfrac{k}{10}s+1},\qquad K=0.2,\ \tau=\frac{1}{\sqrt{10}}=0.316,\ \zeta=\frac{k}{2\sqrt{10}}.}$$ Matching $\tau^2=1/10$ gives $\tau=1/\sqrt{10}$; matching $2\zeta\tau=k/10$ gives $\zeta=k/(2\sqrt{10})$.
(b-i) Stable. Roots $s=\dfrac{-k\pm\sqrt{k^2-40}}{2}$ have negative real part iff the coefficients are positive: $k>0$ (with $10>0$ automatic). So stable for $k>0$ ($k=0$ is marginally stable, $k<0$ unstable).
(b-ii) Underdamped. Complex roots require $k^2-40<0\Rightarrow|k|<2\sqrt{10}$; combined with stability, underdamped for $0 ($\zeta<1$).

(b-iii) Overdamped. Real distinct roots require $k^2-40>0$; taking the stable branch, overdamped for $k>2\sqrt{10}\approx6.32$ ($\zeta>1$). Critical damping ($\zeta=1$) occurs exactly at $k=2\sqrt{10}$.

(c) Underdamped expressions. The standard-form parameters are independent of the input, so directly from step 1: $$\boxed{\tau=\frac{1}{\sqrt{10}}=0.316\ (\text{constant}),\qquad \zeta=\frac{k}{2\sqrt{10}}=0.158\,k.}$$ Equivalently $\omega_n=1/\tau=\sqrt{10}$ is fixed and only the damping ratio $\zeta$ scales with $k$; for $0

Quantity Value
(a) $K,\ \tau,\ \zeta$ $0.2,\ 1/\sqrt{10}=0.316,\ k/(2\sqrt{10})$
(b-i) Stable $k>0$
(b-ii) Underdamped $0
(b-iii) Overdamped $k>2\sqrt{10}\approx6.32$
(c) $\tau,\ \zeta$ (underdamped) $\tau=0.316,\ \zeta=0.158\,k$

Concept & Theory
Topic. Standard second-order form; natural frequency and damping ratio; classification of the step response by $\zeta$.
Key relations. $G=K/(\tau^2s^2+2\zeta\tau s+1)$; $\omega_n=1/\tau$; discriminant $k^2-4\omega_n^2$ sets real vs complex roots; $\zeta<1$ underdamped, $=1$ critical, $>1$ overdamped.
Why this works. Normalising the constant term to $1$ exposes $\tau$ (from the $s^2$ coefficient) and $\zeta$ (from the $s^1$ coefficient); since only $k$ multiplies $s$, $\tau$ is fixed and $\zeta$ is linear in $k$, so the whole qualitative behaviour is a one-parameter sweep in $\zeta$.
Common pitfall. Reading $\tau$ or $\zeta$ before normalising the constant term to unity, or forgetting the stability requirement $k>0$ when listing the underdamped range (a negative $k$ also gives complex roots but they grow, not decay).
Source. Seborg et al., Ch. 5 (second-order systems); Coughanowr & LeBlanc, Ch. 8.

Quantity	Value
(a) $K,\ \tau,\ \zeta$	$0.2,\ 1/\sqrt{10}=0.316,\ k/(2\sqrt{10})$
(b-i) Stable	$k>0$
(b-ii) Underdamped	$0
(b-iii) Overdamped	$k>2\sqrt{10}\approx6.32$
(c) $\tau,\ \zeta$ (underdamped)	$\tau=0.316,\ \zeta=0.158\,k$

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

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