Transfer Function and Step Response From a State-Space Model: 20-Bio-A2 Process Dynamics and Control, December 2019, Q2

Problem 2: Transfer Function and Step Response From a State-Space Model (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. Two-state linear ODE system with $\dot x_1=-2.4048x_1+7u$, $\dot x_2=0.8333x_1-2.2381x_2-1.117u$, $y=x_2$; 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 (zero ICs), solve the first (uncoupled) equation for $X_1(s)/U(s)$, substitute into the second to eliminate $X_1$, and combine over a common denominator; for (b) multiply by $1/s$ and expand in partial fractions using the two real poles.

(a) State 1. $(s+2.4048)X_1(s)=7U(s)\Rightarrow X_1(s)=\dfrac{7}{s+2.4048}U(s)$.
(a) State 2, eliminate $X_1$. $(s+2.2381)X_2(s)=0.8333X_1(s)-1.117U(s)=\left[\dfrac{0.8333\times7}{s+2.4048}-1.117\right]U(s)$. Combining over $(s+2.4048)$: $$\frac{0.8333\times7-1.117(s+2.4048)}{s+2.4048}=\frac{5.8331-1.117s-2.6858}{s+2.4048}=\frac{3.1473-1.117s}{s+2.4048}.$$
(a) Assemble the transfer function. $$\boxed{\frac{Y(s)}{U(s)}=\frac{3.1473-1.117s}{(s+2.4048)(s+2.2381)}=\frac{-1.117(s-2.818)}{(s+2.4048)(s+2.2381)}.}$$ Both poles are in the left half-plane ($s=-2.4048,-2.2381$, stable), but the numerator has a right-half-plane zero at $s=+2.818$ — by Problem 1's criterion this predicts an inverse response to a step in $u$.
(b) Partial fractions for the unit step. $Y(s)=\dfrac{3.1473-1.117s}{s(s+2.4048)(s+2.2381)}=\dfrac As+\dfrac{B}{s+2.4048}+\dfrac{C}{s+2.2381}$, with residues evaluated at each pole: $$A=\frac{3.1473}{2.4048\times2.2381}=0.5848,\quad B=\frac{3.1473-1.117(-2.4048)}{-2.4048(-2.4048+2.2381)}=14.552,$$ $$C=\frac{3.1473-1.117(-2.2381)}{-2.2381(-2.2381+2.4048)}=-15.137.$$ (Check: $A+B+C=0.5848+14.552-15.137\approx0=y(0)$ — consistent with zero initial conditions.)
(b) Result. $$\boxed{y(t)=0.5848+14.552\,e^{-2.4048t}-15.137\,e^{-2.2381t}.}$$ At $t=0.5$: $y=0.0137$ (still near zero — the two nearly-equal-time-constant exponentials almost cancel at first, the RHP-zero footprint); by $t=5$: $y=0.5846\approx y(\infty)=0.5848$ (settled to the DC gain).

Result	Value
Transfer function	$Y(s)/U(s)=(3.1473-1.117s)/[(s+2.4048)(s+2.2381)]$
RHP zero	$s=+2.818$
Step response	$y(t)=0.5848+14.552e^{-2.4048t}-15.137e^{-2.2381t}$
Steady-state gain	$y(\infty)=0.5848$

Concept & Theory

Topic. Deriving an input–output transfer function from a linear state-space model and inverting a step response with real poles and a right-half-plane zero.

Key relations. Laplace transform of $\dot x=Ax+bu$ with zero ICs; sequential elimination of states for a companion/cascade structure; standard partial-fraction residues $B_i=\lim_{s\to p_i}(s-p_i)Y(s)$.

Why this works. Because state 1 does not depend on state 2, its Laplace transform can be solved in isolation first and substituted into state 2's equation, avoiding a full matrix inversion; the resulting numerator sign (negative $s$-coefficient, positive constant) places a zero in the right half-plane, which is exactly the $\tau_a<0$ condition from Problem 1.

Common pitfall. Sign errors when combining the two fractions over $(s+2.4048)$, and forgetting that the near-cancellation of the two close poles ($-2.4048$ vs. $-2.2381$) makes the response deceptively slow to visibly depart from zero.

Source. Seborg et al., Process Dynamics and Control, Ch. 3–4 (state-space to transfer-function conversion, partial-fraction inversion).

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

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