State-Space Model → Transfer Function and Step Response: 23-Chem-A6 Process Dynamics and Control, May 2015, Q3

Problem 3: 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. A coupled two-state linear model with a direct feed-through of $u$ into the $x_2$ equation:

Equation	Expression
State 1	$\dot x_1=-2.4048\,x_1+7u$
State 2	$\dot x_2=0.8333\,x_1-2.2381\,x_2-1.117u$
Output	$y=x_2$

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

Approach. Transform each state equation, eliminate $X_1$ to get $Y/U$, note the sign of the numerator zero, then invert the step-forced output by partial fractions.

Transform the states. With zero initial conditions, $X_1=\dfrac{7U}{s+2.4048}$ and $(s+2.2381)X_2=0.8333X_1-1.117U$. Substituting $X_1$: $$(s+2.2381)X_2=\Big[\tfrac{0.8333\times7}{s+2.4048}-1.117\Big]U.$$
(a) Transfer function. Clearing the denominator, $$\boxed{\frac{Y(s)}{U(s)}=\frac{-1.117\,s+3.1469}{(s+2.4048)(s+2.2381)}}.$$ (Numerator: $0.8333\times7-1.117\times2.4048=5.8331-2.6862=3.1469$.) The numerator has a right-half-plane zero at $s=+3.1469/1.117=+2.817$, which produces an inverse response.
(b) Step-forced output. For $U=1/s$, $$Y(s)=\frac{-1.117s+3.1469}{s\,(s+2.4048)(s+2.2381)}=\frac{A}{s}+\frac{B}{s+2.4048}+\frac{C}{s+2.2381}.$$
Residues. Evaluating by cover-up: $A=\dfrac{3.1469}{(2.4048)(2.2381)}=0.5847$; $B=\dfrac{-1.117(-2.4048)+3.1469}{(-2.4048)(-2.4048+2.2381)}=+14.55$; $C=\dfrac{-1.117(-2.2381)+3.1469}{(-2.2381)(-2.2381+2.4048)}=-15.14$. (Check: $A+B+C=0$, so $y(0)=0$ — consistent with a strictly proper transfer function.)
Response. Inverting, $$\boxed{y(t)=0.5847+14.55\,e^{-2.4048t}-15.14\,e^{-2.2381t}}.$$ Because the two exponentials nearly cancel at $t=0$ but the faster ($e^{-2.4048t}$) term with the large positive coefficient fades first, $y$ first dips negative (to $\approx-0.094$ near $t\approx0.3\,\mathrm s$) before climbing to its steady value $0.585$ — the classic inverse response of the RHP zero.

Problem 3(b): unit-step response. The right-half-plane zero at $s=+2.82$ makes the output first dip to $\approx-0.09$ before rising monotonically to its steady value $0.585$ — an inverse response.

Quantity	Value
$Y/U$	$(-1.117s+3.1469)/[(s+2.4048)(s+2.2381)]$
RHP zero	$s=+2.817$ (inverse response)
Step response	$y=0.585+14.55e^{-2.4048t}-15.14e^{-2.2381t}$
Undershoot / final value	$\approx-0.094$ / $0.585$

Concept & Theory

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

Key relations. $Y/U=C(sI-A)^{-1}B+D$; a numerator zero $s=+z>0$ gives $y(0^+)$ moving opposite to $y(\infty)$; step residues by cover-up $A_i=\lim_{s\to p_i}(s-p_i)Y(s)$.

Why this works. Eliminating $X_1$ is exactly the SISO reduction of the $2\times2$ state model; the feed-through $-1.117u$ into $\dot x_2$ supplies the numerator $s$-term that creates the RHP zero. Two competing decaying modes with opposite-sign residues and a near-common pole are the algebraic signature of inverse response.

Common pitfall. Dropping the $-1.117u$ term (losing the zero and the undershoot), or a sign slip in the numerator constant. Always cross-check $\sum A_i=y(0)=0$ for a strictly proper system.

Source. Seborg et al., Ch. 4 & 6 (state models, poles/zeros, inverse response); Stephanopoulos, Ch. 11 (partial-fraction inversion of step responses).

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

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