Question 1 of 8: Digital (Sampled-Data) Proportional Control of a Heated Tank
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2017 — 04-Bio-A2 Process Dynamics and Control. Three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that only the first five questions in the answer book are marked, yet all eight problems are printed on the paper — all eight are solved below for completeness, since the full set is a study resource. Content spans sampled-data (digital) control, state-space-to-transfer-function modelling, Internal Model Control (IMC) of a non-minimum-phase dead-time process, Nyquist and Routh stability, frequency response (Bode/gain-margin) design and nonlinear-reactor linearisation.
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace/z-domain modelling, transfer functions, Routh and Jury stability, Nyquist/Bode frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist criterion, dead-time systems and sampled-data control. Standard control conventions (deviation variables; unity sensor/valve gains unless stated) are used throughout.
Problem 1: Digital (Sampled-Data) Proportional Control of a Heated Tank (20%)
Given. $G_p(s)=\dfrac{2.31}{(s+1)(5s+1)}$, proportional digital controller gain $k_c$, zero-order hold (ZOH), sampling period $T=0.5\ \text{s}$.
Find. (1) the digital control-loop block diagram; (2) the range of $k_c$ that keeps the sampled-data closed loop stable, via the Jury test.
Problem 1(1): the continuous error $R(s)-Y(s)$ is formed first, then sampled every $T=0.5\ \text{s}$; the digital controller applies gain $k_c$ to the sampled error, a ZOH holds the output constant between samples, and the analog process $G_p(s)$ sees the resulting staircase signal $m(t)$.
Approach. Reduce $G_p(s)/s$ to partial fractions, z-transform each first-order term with the standard table entry $\mathcal Z[1/(s+a)]=z/(z-e^{-aT})$, form the ZOH-plus-process pulse transfer function $G(z)=(1-z^{-1})\mathcal Z[G_p(s)/s]$, write the sampled-data characteristic equation $1+k_cG(z)=0$, and apply the Jury stability test to the resulting quadratic in $z$.
(1) Block diagram. See the figure: a continuous comparator forms $e(t)=r(t)-y(t)$; a sampler with period $T=0.5\ \text{s}$ produces $e^{*}(t)$; the digital controller multiplies by $k_c$; a zero-order hold converts the pulse train back to a piecewise-constant analog signal $m(t)$ that drives $G_p(s)$; the continuous output $y(t)$ is fed back to the comparator.
(2) Partial fractions of $G_p(s)/s$. Writing $G_p(s)=\dfrac{2.31}{(s+1)(5s+1)}=\dfrac{0.462}{(s+1)(s+0.2)}$ and expanding $G_p(s)/s$, $$\frac{G_p(s)}{s}=\frac{2.31}{s}+\frac{0.5775}{s+1}-\frac{2.8875}{s+0.2}.$$
z-transform and ZOH. Using $\mathcal Z[1/s]=z/(z-1)$ and $\mathcal Z[1/(s+a)]=z/(z-e^{-aT})$ with $T=0.5$ ($e^{-1(0.5)}=0.6065$, $e^{-0.2(0.5)}=0.9048$), and multiplying by $(1-z^{-1})$ to add the hold, the $s$-domain pole at the origin cancels and $$G(z)=(1-z^{-1})\mathcal Z\!\left[\frac{G_p(s)}{s}\right]=\frac{0.04755\,z+0.03894}{z^{2}-1.51137\,z+0.54881}.$$
Characteristic equation. With a pure-gain digital controller, $1+k_cG(z)=0$ gives $$\boxed{z^{2}+\big(0.04755k_c-1.51137\big)z+\big(0.03894k_c+0.54881\big)=0.}$$
Jury test (2nd order). For $z^{2}+a_1z+a_0=0$ all roots lie inside the unit circle iff $|a_0|<1$, $P(1)=1+a_1+a_0>0$ and $P(-1)=1-a_1+a_0>0$. Substituting $a_1(k_c)$, $a_0(k_c)$ above: $|a_0|<1\Rightarrow-39.77<k_c<11.59$; $P(1)>0\Rightarrow k_c>-0.433$; $P(-1)>0\Rightarrow k_c<355.3$. The tightest pair of bounds gives $$\boxed{-0.433<k_c<11.59.}$$
Cross-check. Direct root evaluation confirms the boundary: at $k_c=11$ the closed-loop pole magnitude is $0.989<1$ (stable), at $k_c=11.6$ it is $1.0003>1$ (unstable); at $k_c=0$ (open loop) the magnitude is $0.905<1$, consistent with the stable open-loop process.
