State-space model of a translational mechanical system
Let’s try an example of a higher-order translational mechanical system.
For the translational mechanical system shown, find a state-space model with outputs the spring forces and mass momenta.
0. Linear graph model. Assigning the coordinate arrow in the direction of Vs, we obtain the linear graph shown.
1a. The normal tree is shown on the linear graph.
1b.
| Primary variables: | Vs, vm1, vm2, fk1, fk2, fB1, fB2. |
| Secondary variables: | Fs, fm1, fm2, vk1, vk2, vB1, vB2. |
1c. State variables: vm1, vm2, fk1, fk2 (system order n = 4).
1d. State, input, and output vectors: $$\begin{aligned} \bm{x} = \begin{bmatrix} v_{m_1}\\v_{m_2}\\f_{k_1}\\f_{k_2} \end{bmatrix} \qquad \bm{u} = \begin{bmatrix} V_s \end{bmatrix} \qquad \bm{y} = \begin{bmatrix} f_{k_1}\\f_{k_2}\\p_{m_1}\\p_{m_2} \end{bmatrix}. \end{aligned}$$
1e. Elemental equations.
| m1 | $\dot{v}_{m_1} = \frac{1} {m_1} f_{m_1}$ |
| m2 | $\dot{v}_{m_2} = \frac{1} {m_2} f_{m_2}$ |
| k1 | ḟk1 = k1vk1 |
| k2 | ḟk2 = k2vk2 |
| B1 | fB1 = B1vB1 |
| B2 | fB2 = B2vB2 |
1f. Continuity equations.
| Branch | Equation |
| m1 (con1) | fm1 = fk1 + fB1 − fk2 − fB2 |
| m2 (con2) | fm2 = fk2 + fB2 |
1f. Compatibility equations.
| Link | Equation |
| k1 | vk1 = Vs − vm1 |
| B1 | vB1 = Vs − vm1 |
| k2 | vk2 = vm1 − vm2 |
| B2 | vB2 = vm1 − vm2 |
2a. Eliminate secondary variables from the elemental equations.
| m1 | $\dot{v}_{m_1} = \frac{1} {m_1} (f_{k_1} + f_{B_1} - f_{k_2} - f_{B_2})$ |
| m2 | $\dot{v}_{m_2} = \frac{1} {m_2} (f_{k_2} + f_{B_2})$ |
| k1 | ḟk1 = k1(Vs−vm1) |
| k2 | ḟk2 = k2(vm1−vm2) |
| B1 | fB1 = B1(Vs−vm1) |
| B2 | fB2 = B2(vm1−vm2) |
2b. Eliminate non-state and non-input variables from the elemental equations. In this case, fB1 and fB2 are the only variables remaining to eliminate, and they are already expressed as state- and input-variables!
2c. Write the state equation in standard form $$\begin{aligned} \bm{\dot{x}} &= \begin{bmatrix} -(B_1+B_2)/m_1 & B_2/m_1 & 1/m_1 & -1/m_1 \\ B_2/m_2 & -B_2/m_2 & 0 & 1/m_2 \\ -k_1 & 0 & 0 & 0 \\ k_2 & -k_2 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{m_1}\\v_{m_2}\\f_{k_1}\\f_{k_2} \end{bmatrix} + \begin{bmatrix} B_1/m_1 \\ 0 \\ k_1 \\ 0 \end{bmatrix} \begin{bmatrix} V_s \end{bmatrix}. \end{aligned}$$
2d. Outputs in terms of states and inputs. $$\begin{aligned} f_{k_1} = f_{k_1} && f_{k_2} = f_{k_2} && p_{m_1} = m_1 v_{m_1} && p_{m_2} = m_2 v_{m_2}. \end{aligned}$$
2e. Output in standard form. $$\begin{aligned} \bm{y} &= \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ m_1 & 0 & 0 & 0 \\ 0 & m_2 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{m_1}\\v_{m_2}\\f_{k_1}\\f_{k_2} \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} V_s \end{bmatrix}. \end{aligned}$$
Online Resources for Section 3.5
No online resources.