Transfer Function and Step Response from a State-Space Model: 20-Bio-A2 Process Dynamics and Control, Undated paper, Q4

Problem 4: 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 $\dot x_1=-2.6666x_1+u$, $\dot x_2=0.8333x_1-0.6250x_2$, $y=x_2$; 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); the first equation is decoupled from $x_2$, so solve it directly for $X_1(s)/U(s)$ and substitute into the second to eliminate $X_1$; for (b), multiply by $1/s$ and expand in partial fractions using the two real poles.

(a) State 1 (decoupled). $(s+2.6666)X_1(s)=U(s)\Rightarrow X_1(s)=\dfrac{1}{s+2.6666}U(s)$.
(a) State 2, eliminate $X_1$. $(s+0.6250)X_2(s)=0.8333X_1(s)=\dfrac{0.8333}{s+2.6666}U(s)$.
(a) Assemble the transfer function. $$\boxed{\frac{Y(s)}{U(s)}=\frac{0.8333}{(s+2.6666)(s+0.6250)}.}$$ Both poles are real and in the left half-plane ($s=-2.6666,-0.6250$) — a stable, non-oscillatory, overdamped second-order system with no zero (a pure two-pole cascade, unlike Problems 1 and 5 which carry a zero or an unstable pole).
(b) Partial fractions for the unit step. $Y(s)=\dfrac{0.8333}{s(s+2.6666)(s+0.6250)}=\dfrac As+\dfrac{B}{s+2.6666}+\dfrac{C}{s+0.6250}$, with residues evaluated at each pole: $$A=\frac{0.8333}{2.6666\times0.6250}=0.500,\qquad B=\frac{0.8333}{(-2.6666)(-2.6666+0.6250)}=0.1531,\qquad C=\frac{0.8333}{(-0.6250)(-0.6250+2.6666)}=-0.6531.$$ (Check: $A+B+C=0.500+0.153-0.653\approx0=y(0)$, consistent with zero initial conditions.)
(b) Result. $$\boxed{y(t)=0.500+0.1531\,e^{-2.6666t}-0.6531\,e^{-0.6250t}.}$$ At $t=1$: $y=0.161$; at $t=2$: $y=0.314$; at $t=5$: $y=0.471$, approaching the steady-state gain $y(\infty)=A=0.500$ (the slower pole, $s=-0.625$, dominates the settling time).

Result	Value
Transfer function	$Y(s)/U(s)=0.8333/[(s+2.6666)(s+0.6250)]$
Step response	$y(t)=0.500+0.1531e^{-2.6666t}-0.6531e^{-0.6250t}$
Steady-state gain	$y(\infty)=0.500$
Dominant time constant	$1/0.625=1.6\ \text{s}$ (slower pole)

Concept & Theory

Topic. Deriving an input–output transfer function from a linear cascade state-space model and inverting its step response.

Key relations. Laplace transform of $\dot x=Ax+bu$ with zero ICs; sequential elimination of states for a cascade (companion) structure where $x_1$ does not depend on $x_2$; standard partial-fraction residues $B_i=\lim_{s\to p_i}(s-p_i)Y(s)$.

Why this works. Because state 1's dynamics do not depend on state 2, its Laplace transform can be solved in isolation and substituted directly into state 2's equation, avoiding a full matrix inversion; with no direct feedthrough of $u$ into the second equation (unlike the closely related sibling problem where $u$ enters both states), the numerator is a pure constant and the transfer function has no zero.

Common pitfall. Forgetting that $x_2$'s equation has no direct $u$ term here (so $X_2$ depends on $u$ only through $X_1$), and mis-ordering the two poles when assigning $B$ and $C$.

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

Question 4 of 8: Transfer Function and Step Response from a State-Space Model

Problem 4: Transfer Function and Step Response from a State-Space Model (20%)