State-Space Model — Transfer Function and Step Response: 20-Bio-A2 Process Dynamics and Control, May 2017, Q2

Problem 2: State-Space Model — Transfer Function and 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. Linear state-space model $\dot x_1=-2.4048x_1+7u$, $\dot x_2=0.8333x_1-2.2381x_2-1.117u$, $y=x_2$, zero initial conditions (deviation variables).

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

Approach. Laplace-transform each state equation from rest, eliminate $X_1(s)$ to get $Y/U$, then invert $Y(s)=[Y(s)/U(s)]\cdot(1/s)$ by partial fractions.

(a) Eliminate $X_1$. $sX_1=-2.4048X_1+7U\Rightarrow X_1=\dfrac{7U}{s+2.4048}$. Substituting into $sX_2=0.8333X_1-2.2381X_2-1.117U$ and solving for $X_2=Y$: $$\boxed{\frac{Y(s)}{U(s)}=\frac{5.8331/(s+2.4048)-1.117}{s+2.2381}=\frac{3.1469-1.117\,s}{(s+2.2381)(s+2.4048)}.}$$ The numerator has a zero at $s=+2.818$ — a right-half-plane (RHP) zero, since $3.1469/1.117=2.818>0$.
Recognise the inverse-response signature. An RHP zero makes $y(t)$ initially move opposite to its eventual direction before turning around — a classic "wrong-way" response seen in boiler drum level and other biomedical/process loops with competing fast and slow paths.
(b) Step response by partial fractions. With $U(s)=1/s$, $Y(s)=\dfrac{3.1469-1.117s}{s(s+2.2381)(s+2.4048)}=\dfrac{A}{s}+\dfrac{B}{s+2.2381}+\dfrac{C}{s+2.4048}$. Evaluating each residue at its own pole gives $A=0.5847$, $B=-15.135$, $C=14.551$, so $$\boxed{y(t)=0.5847-15.135\,e^{-2.2381t}+14.551\,e^{-2.4048t}.}$$
Check limits. $y(0)=0.5847-15.135+14.551\approx0$ (correct start), and $y(\infty)=A=0.5847$ (matches the final-value $Y(s)/U(s)$ evaluated at $s=0$). At $t=0.1\ \text{s}$, $y\approx-0.075$ — the response briefly dips negative before rising to $0.5847$, confirming the inverse response predicted by the RHP zero.

Result	Value
Transfer function	$Y/U=(3.1469-1.117s)/[(s+2.2381)(s+2.4048)]$
RHP zero	$s=+2.818$ (inverse response)
Step response	$y(t)=0.5847-15.135e^{-2.2381t}+14.551e^{-2.4048t}$
Steady-state gain	$y(\infty)=0.5847$

Concept & Theory

Topic. State-space to transfer-function reduction; partial-fraction step-response inversion; right-half-plane zeros and inverse response.

Key relations. Elimination of intermediate states via Laplace substitution; $Y(s)=[Y/U]\cdot U(s)$; residue formula $A_i=\lim_{s\to p_i}(s-p_i)Y(s)$.

Why this works. Because the numerator polynomial $3.1469-1.117s$ changes sign at a positive value of $s$, the initial slope $\dot y(0^+)=\lim_{s\to\infty}s^2Y(s)$ carries the opposite sign to the final steady-state gain, which is exactly what produces the observed dip before the rise.

Common pitfall. Mis-signing the residues (a common slip when three poles are close together), or missing the RHP zero and expecting a monotonic first-order-like rise.

Source. Seborg et al., Process Dynamics and Control, Ch. 3–4 (Laplace transforms, transfer functions from state equations, inverse response).

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

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