23-Chem-A6 Process Dynamics and Control · May 2015
Question 2 of 8: Second-Order ODE — Standard Form and Stability vs. $k$
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2015 — 04-Chem-A6 Process Dynamics & Control. Three-hour open-book examination; any non-communicating calculator is permitted. Eight problems are printed and any five constitute a complete paper (each worth 20%); all eight are solved below for completeness. Most parts are quantitative (dynamic modelling, transfer functions, step/pulse/ramp responses, Routh and Nyquist stability, and IMC design); qualitative sketches are drawn as real figures where the paper asks for them.
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout.
Problem 2: Second-Order ODE — Standard Form and Stability vs. $k$ (20%)
Given. A linear second-order system $\ddot y+k\dot y+2y=x$ with adjustable damping parameter $k$.
Find. (a) $Y/X$ as $K/(\tau^2 s^2+2\zeta\tau s+1)$; (b) convergence classification and response form over $-20<k<20$.
Approach. Transform the ODE (zero initial conditions), normalise the constant term to unity to expose $K$, $\tau$ and $\zeta$, then locate the two characteristic roots as functions of $k$ and classify by their real parts and by the discriminant.
Transfer function. Laplace-transforming, $(s^2+ks+2)Y=X$, so $\dfrac{Y}{X}=\dfrac{1}{s^2+ks+2}$. Divide numerator and denominator by $2$ to force a unity constant term: $$\boxed{\frac{Y(s)}{X(s)}=\frac{\tfrac12}{\tfrac12 s^2+\tfrac{k}{2}s+1}=\frac{K}{\tau^2s^2+2\zeta\tau s+1}}.$$
Read off the standard parameters. Matching coefficients: $$K=\tfrac12,\qquad \tau^2=\tfrac12\Rightarrow \tau=\tfrac{1}{\sqrt2}=0.707,\qquad 2\zeta\tau=\tfrac{k}{2}\Rightarrow \zeta=\frac{k}{2\sqrt2}.$$ The damping ratio is proportional to $k$; the natural period $\tau$ and steady gain $K$ do not depend on $k$.
Characteristic roots. $s^2+ks+2=0\Rightarrow s=\dfrac{-k\pm\sqrt{k^2-8}}{2}$. The product of roots is $+2$ and the sum is $-k$, so convergence (both real parts negative) requires the sum $-k<0$, i.e. $k>0$. The discriminant changes sign at $|k|=2\sqrt2\approx2.83$.
(b) Convergent range $0<k<20$. Real parts are negative, so the response decays to the steady value $Kx$. Sub-cases: for $0<k<2\sqrt2$ the roots are complex ($\zeta<1$) — a decaying oscillation $$y(t)=C_1+e^{-(k/2)t}\big(C_2\cos\omega_d t+C_3\sin\omega_d t\big),\quad \omega_d=\tfrac12\sqrt{8-k^2};$$ for $k>2\sqrt2$ the roots are real and negative ($\zeta>1$) — an overdamped $y(t)=C_1+C_2e^{r_1t}+C_3e^{r_2t}$ with $r_{1,2}<0$.
(b) Boundary $k=0$ and divergent range $-20<k<0$. At $k=0$ the roots are purely imaginary $\pm j\sqrt2$: a sustained (undamped) oscillation $y(t)=C_1+C_2\cos\sqrt2\,t+C_3\sin\sqrt2\,t$ — it does not converge. For $-2\sqrt2<k<0$ the roots are complex with positive real part $-k/2>0$: a growing oscillation (same form as step 4 but with $e^{+|k|/2\,t}$); for $-20<k<-2\sqrt2$ the roots are real and positive: a monotonically diverging $C_2e^{r_1t}+C_3e^{r_2t}$, $r_i>0$. All of these diverge.
