23-Chem-A6 Process Dynamics and Control · December 2016
Question 7 of 8: State-Space Model to Transfer Function and Step Response
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
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. This December 2016 sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 7: State-Space Model to Transfer Function and Step Response (20%)
Given. A two-state linear model with $a_{11}=-2.4048$, $a_{21}=0.8333$, $a_{22}=-2.2381$, $b_1=7$, $b_2=-1.117$, output $y=x_2$.
Find. (a) $Y(s)/U(s)$; (b) $y(t)$ for a unit step.
Unit-step response. The output first dips below zero (initial slope negative) before climbing to its final value $0.585$ — the signature of the right-half-plane zero at $s=+2.82$ that the second state introduces through the negative $b_2$.
Approach. Laplace-transform each state equation (zero initial conditions), eliminate $X_1$, and read $Y/U$; then invert the step response by partial fractions over the two real poles and the pole at the origin.
Transform state 1. $(s-a_{11})X_1=b_1U\Rightarrow X_1=\dfrac{7U}{s+2.4048}$.
(a) Eliminate and assemble $Y/U$. $(s-a_{22})X_2=a_{21}X_1+b_2U$, so $$\frac{Y}{U}=\frac{a_{21}b_1+b_2(s-a_{11})}{(s-a_{11})(s-a_{22})}=\boxed{\frac{-1.117\,s+3.1469}{(s+2.4048)(s+2.2381)}.}$$ The numerator constant is $a_{21}b_1-b_2a_{11}=0.8333(7)-(-1.117)(-2.4048)=3.1469$.
Right-half-plane zero. $-1.117s+3.1469=0$ at $s=+3.1469/1.117=\boxed{2.817}$ — a positive zero, so expect inverse response.
(b) Set up the step. For $U=1/s$, $Y=\dfrac{-1.117s+3.1469}{s(s+2.4048)(s+2.2381)}=\dfrac{A}{s}+\dfrac{B}{s+2.4048}+\dfrac{C}{s+2.2381}$.
Residues. $A=\dfrac{3.1469}{(2.4048)(2.2381)}=0.5847$ (final value); $B=14.551$ at $s=-2.4048$; $C=-15.135$ at $s=-2.2381$. Check $A+B+C=0$ (so $y(0)=0$).
(b) Invert. $$\boxed{y(t)=0.5847+14.551\,e^{-2.4048t}-15.135\,e^{-2.2381t}.}$$ The two large, nearly cancelling exponentials produce the early negative excursion; $y\to0.5847$ as $t\to\infty$.
