So I've been struggling with describing this Quarter Car Suspension with a differential equation:
The main point of difficulty for me is the fact that there are two x variables (x1 and x2) and I'm not sure which one I'm supposed to use for velocity and acceleration. I was told I should use Newton's 2nd law so I used it and ended up with these equations:
None of these look like differential equations. There are always some irrelevant x1 or x2 terms in each one. In a later part I am asked to find the space-state (the ABCD matrices) using x={x1, x'1, x2, x'2}T as a space-state vector and with w and u as inputs. I'm not sure how to go about finding the space-state with so many variables. If I'm not mistaken, the input is multiplied by D, but in this case the inputs are two (u and w).
Using Newton and assuming that $u$ is a force active suspension control, we have
$$ \cases{ M\ddot x_1 = k_1(x_2-x_1)+b_1(\dot x_2-\dot x_1) + u\\ m\ddot x_2 = -k_1(x_2-x_1)-b_1(\dot x_2-\dot x_1) - u +k_2(w-x_2)+b_2(\dot w-\dot x_2) } $$
or
$$ \left(\matrix{ M& 0 \\ 0&m}\right)\left(\matrix{ \ddot x_1 \\ \ddot x_2}\right)=\left(\matrix{ -b_1& b_1 \\ b_1&-(b_1+b_2)}\right)\left(\matrix{ \dot x_1 \\ \dot x_2}\right)+\left(\matrix{ -k_1& k_1 \\ k_1&-(k_1+k_2)}\right)\left(\matrix{ x_1 \\ x_2}\right)+\left(\matrix{ u \\ k_2w+b_2\dot w-u}\right) $$
now calling $y = \left(\matrix{ x_1 \\ x_2}\right)$ and $v = \left(\matrix{ u \\ k_2w+b_2\dot w - u}\right)$ we have
$$ \mathcal{M} \ddot y + \mathcal{B} \dot y + \mathcal{K} y = v $$
or
$$ \ddot y + \mathcal{M}^{-1}\mathcal{B}\dot y + \mathcal{M}^{-1}\mathcal{K} y = \mathcal{M}^{-1}v $$