State-Space Model — Transfer Function and Unit-Step Response: 23-Chem-A6 Process Dynamics and Control, December 2018, Q2

Problem 2: State-Space Model — Transfer Function and Unit-Step 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. A two-state linear system in standard form $\dot{\mathbf x}=\mathbf A\mathbf x+\mathbf B u,\ y=\mathbf C\mathbf x$:

Element	Value
$\mathbf A$	$\begin{bmatrix}-2.4048 & 0\\ 0.8333 & -2.2381\end{bmatrix}$
$\mathbf B$	$[\,7,\ -1.117\,]^\mathsf T$
$\mathbf C$	$[\,0,\ 1\,]$
Input	unit step $u=1$, $U(s)=1/s$

Find. (a) $Y(s)/U(s)$; (b) $y(t)$ for a unit step.

Problem 2: the response starts with negative slope ($\dot y(0^+)=-1.117$), dips below zero, then recovers to the positive steady value $0.585$ — the signature of the right-half-plane zero at $s=+2.817$.

Approach. Transform each state equation, solve the triangular system for $X_2=Y$, then invert the resulting third-order (with the $1/s$ input) rational function by partial fractions.

Transform the states. With zero initial conditions, $sX_1=-2.4048X_1+7U\Rightarrow X_1=\dfrac{7U}{s+2.4048}$, and $(s+2.2381)X_2=0.8333X_1-1.117U$.
(a) Transfer function. Substituting $X_1$: $(s+2.2381)X_2=\Big[\dfrac{0.8333\cdot7}{s+2.4048}-1.117\Big]U$. Since $0.8333\cdot7=5.8331$ and $5.8331-1.117\,(s+2.4048)=3.1469-1.117\,s$, $$\boxed{\frac{Y(s)}{U(s)}=\frac{3.1469-1.117\,s}{(s+2.4048)(s+2.2381)}.}$$ The steady-state gain is $3.1469/(2.4048\cdot2.2381)=0.585$; the numerator zero is at $s=+2.817$ (right half-plane), so an inverse response is expected.
Set up the step. For $U=1/s$, $$Y(s)=\frac{3.1469-1.117\,s}{s\,(s+2.4048)(s+2.2381)}=\frac{A}{s}+\frac{B}{s+2.4048}+\frac{C}{s+2.2381}.$$
Residues. $A=\dfrac{3.1469}{(2.4048)(2.2381)}=0.5847$; $B=\dfrac{3.1469-1.117(-2.4048)}{(-2.4048)(-2.4048+2.2381)}=\dfrac{5.8331}{0.4009}=14.55$; $C=\dfrac{3.1469-1.117(-2.2381)}{(-2.2381)(-2.2381+2.4048)}=\dfrac{5.6469}{-0.3731}=-15.14$. (Check: $A+B+C\approx0$, consistent with $y(0^+)=0$.)
(b) Time response. Inverting term by term, $$\boxed{y(t)=0.585+14.55\,e^{-2.4048\,t}-15.14\,e^{-2.2381\,t}.}$$ The two exponentials are large and nearly cancel; because the faster decays quicker, their difference is negative at first (initial slope $\dot y(0^+)=-1.117<0$), producing an inverse response before $y$ climbs to its final value $0.585$.

Quantity	Value
(a) $Y/U$	$(3.147-1.117s)/[(s+2.4048)(s+2.2381)]$
Steady-state gain	$0.585$
RHP zero	$s=+2.817$ (inverse response)
(b) $y(t)$	$0.585+14.55e^{-2.4048t}-15.14e^{-2.2381t}$

Concept & Theory

Topic. State-space to transfer-function conversion; partial-fraction inversion; inverse response from a right-half-plane zero.

Key relations. $Y/U=\mathbf C(s\mathbf I-\mathbf A)^{-1}\mathbf B$; step response of a two-pole system $=A+Be^{-p_1t}+Ce^{-p_2t}$; a positive (RHP) zero $\Rightarrow$ $\dot y(0^+)$ opposite in sign to the final value.

Why this works. The direct feed-through of $u$ into $\dot x_2$ (the $-1.117u$ term) competes with the indirect path through $x_1$; because the fast direct path acts first and pulls $y$ down while the slower path later dominates, the trajectory reverses — exactly what the RHP zero encodes.

Common pitfall. Reporting only the final value or missing the sign of the initial slope; the near-cancelling residues ($+14.55,-15.14$) are physically meaningful, not arithmetic error. Always sanity-check $A+B+C=y(0^+)=0$.

Source. Seborg et al., Ch. 3–4 (Laplace, transfer functions), Ch. 6 (inverse response, RHP zeros).

Question 2 of 8: State-Space Model — Transfer Function and Unit-Step Response

Problem 2: State-Space Model — Transfer Function and Unit-Step Response (20%)