23-Chem-A6 Process Dynamics and Control · December 2014
Question 2 of 8: State-Space Model — Transfer Function and Step Response
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2014 — 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. The parts are quantitative throughout — two-capacitance thermal modelling, state-space transfer functions, IMC design with a right-half-plane zero, Nyquist stability of an open-loop-unstable plant, ultimate gain with sensor dead time, second-order damping regimes, a non-linear CSTR, and Bode gain-margin design — and every requested plot (step response, IMC servo response, Nyquist locus, damping family, Bode diagram) is drawn as a real figure.
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, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist and Bode stability, dead-time systems and controller design; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first- and second-order dynamics, linearisation of non-linear balances. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout.
Problem 2: State-Space Model — Transfer Function and Step Response (20%)
Given. A two-state linear model $\dot x_1=-2.4048x_1+7u$, $\dot x_2=0.8333x_1-2.2381x_2-1.117u$, output $y=x_2$, zero initial conditions.
Find. (a) the transfer function $Y/U$; (b) the unit-step response $y(t)$.
Approach. Laplace-transform each state equation, eliminate $x_1$ to get $Y/U$, then invert $Y=(Y/U)\cdot(1/s)$ by partial fractions.
(a) Transfer function. From the first state, $x_1=\dfrac{7u}{s+2.4048}$. Substituting into $(s+2.2381)x_2=0.8333x_1-1.117u$ gives $$\boxed{\dfrac{Y}{U}=\dfrac{-1.117\,s+3.1469}{(s+2.4048)(s+2.2381)}.}$$ The constant $3.1469=0.8333\times7-1.117\times2.4048=5.8331-2.6862$. The numerator vanishes at $s=+2.817$ — a right-half-plane zero, so the system shows inverse response.
(b) Partial-fraction expansion. With $U=1/s$, $Y=\dfrac{-1.117s+3.1469}{s(s+2.4048)(s+2.2381)}$. Residues at $s=0,-2.4048,-2.2381$ are $0.5847$, $+14.551$ and $-15.135$, so $$\boxed{y(t)=0.5847-15.135\,e^{-2.2381t}+14.551\,e^{-2.4048t}.}$$
Consistency checks. $y(0)=0.5847-15.135+14.551\approx0$ (correct for a strictly proper system) and $y(\infty)=0.5847=\dfrac{3.1469}{2.4048\times2.2381}$, the DC gain. Because the two exponentials nearly cancel at $t=0^{+}$ but carry opposite signs, the output first dips below zero (inverse response) before climbing to $0.585$ — the plot below and a direct numerical integration of the state equations reproduce this.
Problem 2(b): unit-step response. The RHP zero produces the characteristic early dip below zero before the output settles to the steady-state gain $y(\infty)=0.585$.
