Second-Order ODE — Standard Form, Stability and Damping: 23-Chem-A6 Process Dynamics and Control, May 2017, Q6

Problem 6: Second-Order ODE — Standard Form, Stability 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. $\ddot y+k\dot y+10y=2x$.

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

Approach. Transform and normalise the constant term to read $K,\tau,\zeta$ by inspection, then classify the poles of $s^2+ks+10$.

(a) Standard form. $(s^2+ks+10)Y=2X$; divide by 10: $$\boxed{\dfrac{Y}{X}=\dfrac{0.2}{0.1s^2+0.1k\,s+1}=\dfrac{K}{\tau^2s^2+2\zeta\tau s+1}}$$ with $K=0.2$, $\tau^2=0.1\Rightarrow\tau=1/\sqrt{10}=0.316$, and $2\zeta\tau=0.1k$.
(b) Regimes (poles of $s^2+ks+10=0$):
- (i) Stable: both roots in LHP $\Rightarrow k>0$ (constant term $10>0$).
- (ii) Underdamped: $\zeta<1\Rightarrow 0<k<2\sqrt{10}=6.32$.
- (iii) Overdamped: $\zeta>1\Rightarrow k>2\sqrt{10}=6.32$ (critically damped exactly at $k=6.32$).
(c) Underdamped $\tau$ and $\zeta$. $$\boxed{\tau=\frac{1}{\sqrt{10}}=0.316\ \text{(independent of }k),\qquad \zeta=\frac{k}{2\sqrt{10}}=0.158\,k.}$$ Only the damping ratio depends on $k$; the natural time constant is fixed by the $10y$ term. As $k$ rises from 0 to 6.32 the response moves from sustained oscillation toward critical damping.

Problem 6: unit-step response family ($K=0.2$, $\tau=0.316$). Underdamped ($k=2$) overshoots and rings; critical ($k=6.32$) is the fastest non-oscillatory approach; overdamped ($k=12$) is sluggish. All share the same final value $0.2$.

Quantity	Value
(a) $Y/X$	$0.2/(0.1s^2+0.1ks+1)$; $K=0.2,\ \tau=0.316$
(b) Stable / underdamped / overdamped	$k>0$ / $0<k<6.32$ / $k>6.32$
(c) $\tau,\ \zeta$	$0.316$ (const), $0.158k$

Concept & Theory

Topic. Reading $K,\tau,\zeta$ off a second-order transfer function; stability and damping classification.

Key relations. Canonical form $K/(\tau^2s^2+2\zeta\tau s+1)$; $\omega_n=1/\tau$; $\zeta<1$ underdamped, $=1$ critical, $>1$ overdamped; poles $-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}$.

Why this works. Normalising the constant term to 1 lets the constant set $\tau$, the first-order coefficient set $\zeta$, and the numerator set $K$ — all by inspection. Here the "spring" term $10y$ fixes $\tau$ regardless of $k$, while $k$ (the "damper") tunes only $\zeta$, sweeping the system through the three regimes.

Common pitfall. Forgetting to divide through so $\tau$ and $\zeta$ are misread; confusing stability ($k>0$) with the underdamped boundary ($k=6.32$) — they are different thresholds.

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

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

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