State-Space Model → Transfer Function & Step Response: 23-Chem-A6 Process Dynamics and Control, December 2013, Q3

Problem 3: State-Space Model → Transfer Function & 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 model (already in deviation form, zero initial conditions):

Equation	Coefficients
$\dot x_1$	$-2.4048\,x_1+7u$
$\dot x_2$	$0.8333\,x_1-2.2381\,x_2-1.117\,u$
output	$y=x_2$
input	$u=$ unit step (part b)

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

Approach. Laplace-transform each state equation (zero ICs), eliminate $X_1$, and assemble $Y=X_2$ over $U$. For part (b) multiply by $1/s$ and invert by partial fractions into a constant plus two decaying exponentials.

Transform and solve for $X_1$. $(s+2.4048)X_1=7U\Rightarrow X_1=\dfrac{7U}{s+2.4048}$.
Substitute into the $X_2$ equation. $(s+2.2381)X_2=0.8333X_1-1.117U=\left[\dfrac{0.8333\cdot7}{s+2.4048}-1.117\right]U=\dfrac{5.8331-1.117(s+2.4048)}{s+2.4048}\,U.$ The numerator collapses to $-1.117s+3.147$.
(a) Transfer function. $$\boxed{\dfrac{Y(s)}{U(s)}=\dfrac{-1.117s+3.147}{(s+2.4048)(s+2.2381)}=\dfrac{-1.117s+3.147}{s^{2}+4.643s+5.382}.}$$ The DC gain is $Y/U|_{s=0}=3.147/5.382=0.585$; note the negative-$s$ numerator term (a left-half-plane zero at $s=+2.82$, i.e. an RHP zero — this system shows a mild inverse-response tendency).
(b) Unit-step response. With $U=1/s$, expand $Y(s)=\dfrac{-1.117s+3.147}{s(s+2.4048)(s+2.2381)}=\dfrac{A}{s}+\dfrac{B}{s+2.4048}+\dfrac{C}{s+2.2381}$. The residues are $A=0.585$ (the DC gain), $B=+14.55$ and $C=-15.14$, so $$\boxed{y(t)=0.585+14.55\,e^{-2.405t}-15.14\,e^{-2.238t}.}$$
Sanity checks. $y(0)=0.585+14.55-15.14\approx0$ (correct, $x_2$ starts at rest) and $y(\infty)=0.585$ (the DC gain). The two closely-spaced poles give large, nearly-cancelling residues; the response rises, overshoots slightly toward $\approx0.60$ near $t\approx0.6\,$s, then settles to $0.585$ (see figure).

Problem 3(b): unit-step response $y(t)=0.585+14.55e^{-2.405t}-15.14e^{-2.238t}$. Starting from zero it rises quickly, peaks just above the final value near $t\approx0.6\,$s, and settles at the DC gain $0.585$.

Result	Expression
Transfer function	$\dfrac{-1.117s+3.147}{s^{2}+4.643s+5.382}$
DC gain	$0.585$
Poles	$s=-2.405,\,-2.238$
Step response	$y(t)=0.585+14.55e^{-2.405t}-15.14e^{-2.238t}$

Concept & Theory

Topic. Converting a state-space model to a transfer function; partial-fraction inversion of a step response.

Key relations. $G(s)=C(sI-A)^{-1}B+D$; for a step, $Y(s)=G(s)/s$; residue at a simple pole $p_i$ is $\lim_{s\to p_i}(s-p_i)Y(s)$; DC gain $=G(0)$.

Why this works. Eliminating $X_1$ is exactly the matrix inverse $(sI-A)^{-1}$ for this lower-triangular $A$; the shared denominator is the characteristic polynomial $\det(sI-A)$. The step response is the DC gain plus one decaying mode per pole.

Common pitfall. Sign errors on the $-1.117u$ feed-through when clearing fractions (it produces the numerator zero), and mislabelling which residue belongs to which pole — here the poles are close (2.24 vs 2.40), so the residues are large and opposite; a small algebra slip destroys the $y(0)=0$ check.

Source. Seborg et al., Ch. 3–6 (Laplace transforms, transfer functions, state-space); Coughanowr & LeBlanc, Ch. 4–7.

Question 3 of 8: State-Space Model → Transfer Function & Step Response

Problem 3: State-Space Model → Transfer Function & Step Response (20%)