Question 7 of 8: Analytical Step Response from a State-Space Model

Question 7: Analytical Step Response from a State-Space Model (20 marks)

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. $A = \begin{bmatrix} -1 & 1 \\ -3 & -5 \end{bmatrix}$, $B = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$, $C = \begin{bmatrix} 0 & 1 \end{bmatrix}$, $D = 0$, unit step input and zero initial state.

Find. The complete output response $y(t)$ for $t \ge 0$ in closed form.

Approach. With zero initial conditions the response is entirely the forced term, so the quickest exact route is to transform to the s-domain: obtain $G(s) = C(sI-A)^{-1}B$, multiply by the step transform $1/s$, expand in partial fractions and invert term by term using the supplied Laplace table.

1. Form $sI - A$ and its determinant. $$sI - A = \begin{bmatrix} s+1 & -1 \\ 3 & s+5 \end{bmatrix}$$ $$\det(sI-A) = (s+1)(s+5) - (-1)(3) = s^2+6s+5+3 = s^2+6s+8 = (s+2)(s+4)$$ The system eigenvalues are therefore $-2$ and $-4$: both in the left half plane, so the response will be a stable, non-oscillatory decay to a constant.
2. Invert the matrix. For a $2\times2$ matrix the adjugate is immediate: $$(sI-A)^{-1} = \frac{1}{(s+2)(s+4)}\begin{bmatrix} s+5 & 1 \\ -3 & s+1 \end{bmatrix}$$
3. Compute the transfer function. First $$(sI-A)^{-1}B = \frac{1}{(s+2)(s+4)} \begin{bmatrix} s+5 & 1 \\ -3 & s+1 \end{bmatrix}\begin{bmatrix} -1 \\ 0 \end{bmatrix} = \frac{1}{(s+2)(s+4)}\begin{bmatrix} -(s+5) \\ 3 \end{bmatrix}$$ and premultiplying by $C = \begin{bmatrix} 0 & 1 \end{bmatrix}$ selects the second entry: $$\boxed{\,G(s) = \frac{Y(s)}{U(s)} = \frac{3}{(s+2)(s+4)}\,}$$ Note the numerator is a constant — the system has no finite zeros, and the $-(s+5)$ term is discarded because the output measures $x_2$ only.
4. Apply the step input. With $U(s) = 1/s$ and zero initial conditions, $$Y(s) = \frac{3}{s(s+2)(s+4)}$$
5. Expand in partial fractions. Writing $Y(s) = \dfrac{A_0}{s} + \dfrac{A_1}{s+2} + \dfrac{A_2}{s+4}$ and using the cover-up rule, $$A_0 = \frac{3}{(2)(4)} = 0.375, \qquad A_1 = \frac{3}{(-2)(-2+4)} = \frac{3}{-4} = -0.75, \qquad A_2 = \frac{3}{(-4)(-4+2)} = \frac{3}{8} = 0.375$$
6. Invert term by term. Using $1/s \leftrightarrow 1(t)$ and $1/(s+a) \leftrightarrow e^{-at}1(t)$ from the supplied table, $$\boxed{\,y(t) = 0.375 - 0.75\,e^{-2t} + 0.375\,e^{-4t}, \qquad t \ge 0\,}$$
7. Check the endpoints. At $t = 0$, $y(0) = 0.375 - 0.75 + 0.375 = 0$, which is correct because the state starts at rest and there is no direct feedthrough ($D = 0$). Differentiating, $\dot{y}(0) = 1.5 - 1.5 = 0$ as well — consistent with a relative degree of two, since the step must propagate through both integrators before the output can move. As $t \to \infty$ both exponentials vanish and $$y(\infty) = 0.375 = \frac{3}{8} = G(0)$$ matching the DC gain, exactly as the final-value theorem requires. The response is overdamped, rising monotonically to 0.375 with the slower $e^{-2t}$ mode dominating the approach.