Multivariable Model — Transfer Functions and Response: 23-Chem-A6 Process Dynamics and Control, May 2013, Q7

Problem 7: Multivariable Model — Transfer Functions and Response (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. The coupled linear state model $\dot y_1=-y_1-y_2+x_1$, $\dot y_2=y_1-2y_2+x_1+x_2$, zero initial conditions (deviation variables).

Find. (a) $Y_2/X_1$ and $Y_2/X_2$; (b) $y_2(t)$ for $x_1=0$, unit step $x_2$.

Approach. Laplace-transform both ODEs, giving two algebraic equations in $Y_1,Y_2$; eliminate $Y_1$ (Cramer/substitution) to express $Y_2$ in terms of $X_1,X_2$, read off the two transfer functions, then invert the specified input by partial fractions.

Transform the model. With zero initial conditions, $sY_1=-Y_1-Y_2+X_1$ and $sY_2=Y_1-2Y_2+X_1+X_2$, i.e. $$(s+1)Y_1+Y_2=X_1,\qquad -Y_1+(s+2)Y_2=X_1+X_2.$$
Eliminate $Y_1$. From the first equation $Y_1=(X_1-Y_2)/(s+1)$; substituting into the second and clearing $(s+1)$: $-(X_1-Y_2)+(s+1)(s+2)Y_2=(s+1)(X_1+X_2)$. Collecting $Y_2$: $Y_2\big[1+(s+1)(s+2)\big]=(s+2)X_1+(s+1)X_2$, and $1+(s+1)(s+2)=s^2+3s+3$.
(a) Transfer functions. $$\boxed{\dfrac{Y_2}{X_1}=\dfrac{s+2}{s^2+3s+3}},\qquad \boxed{\dfrac{Y_2}{X_2}=\dfrac{s+1}{s^2+3s+3}}.$$ The common denominator $s^2+3s+3$ has roots $s=-1.5\pm j\,\tfrac{\sqrt3}{2}$ (damping ratio $\zeta=1.5/\sqrt3=0.866$, underdamped).
(b) Response to a unit step in $x_2$ ($x_1=0$). $Y_2=\dfrac{s+1}{s(s^2+3s+3)}$. Partial fractions give $\dfrac{1/3}{s}-\dfrac{1}{3}\cdot\dfrac{s}{s^2+3s+3}$. Completing the square, $s^2+3s+3=(s+1.5)^2+(\tfrac{\sqrt3}{2})^2$, and inverting: $$y_2(t)=\dfrac13-\dfrac13\,e^{-1.5t}\cos\!\big(\tfrac{\sqrt3}{2}t\big)+\dfrac{\sqrt3}{3}\,e^{-1.5t}\sin\!\big(\tfrac{\sqrt3}{2}t\big).$$
Interpret. The response starts at $y_2(0)=0$, overshoots slightly (a damped oscillation, $\omega_d=\sqrt3/2\approx0.866$), and settles at $\boxed{y_2(\infty)=\tfrac13}$ — the DC gain $Y_2/X_2\big|_{s=0}=1/3$. The combined transient amplitude is $\sqrt{(1/3)^2+(\sqrt3/3)^2}=\tfrac23$.

Problem 7(b): $y_2(t)$ for $x_1=0$ and a unit step in $x_2$. Complex poles $-1.5\pm j0.866$ produce a lightly damped oscillation that settles at the DC gain $y_2(\infty)=1/3$.

Result	Expression
$Y_2/X_1$	$(s+2)/(s^2+3s+3)$
$Y_2/X_2$	$(s+1)/(s^2+3s+3)$
Response $y_2(t)$ (step $x_2$)	$\tfrac13-\tfrac13e^{-1.5t}\cos\omega_d t+\tfrac{\sqrt3}{3}e^{-1.5t}\sin\omega_d t$
Damped frequency / final value	$\omega_d=\sqrt3/2\approx0.866$; $y_2(\infty)=1/3$

Concept & Theory

Topic. Deriving transfer functions from a coupled linear state model; underdamped second-order step response.

Key relations. Laplace of a linear ODE system $\Rightarrow$ $(sI-A)Y=BX$; a transfer function $Y_2/X_j=[\,\text{adj}(sI-A)B\,]_{2j}/\det(sI-A)$; second-order response with complex poles $-\sigma\pm j\omega_d$ gives $e^{-\sigma t}(\cos+\sin)$ terms.

Why this works. The characteristic polynomial $\det(sI-A)=s^2+3s+3$ is shared by every input-output pair; only the numerator changes with which input is excited. Completing the square converts the quadratic denominator into the standard damped-sinusoid Laplace pair, and the final-value theorem gives the offset directly as the DC gain.

Common pitfall. Dropping the coupling term when eliminating $Y_1$, or mis-identifying the numerator (each input carries a different zero: $s+2$ for $x_1$, $s+1$ for $x_2$).

Source. Seborg et al., Ch. 6 (transfer functions of dynamic systems), Ch. 5 (second-order response); Coughanowr & LeBlanc, Ch. 7–8.

Question 7 of 8: Multivariable Model — Transfer Functions and Response

Problem 7: Multivariable Model — Transfer Functions and Response (20%)