# A Vibrational Energy Harvesting Sensor Based on Linear and Rotational Electromechanical Effects
## 3 Magnetically Coupled Double-Spring Design
The physical parameters are as follows:  
\[m_{n} \text{are the masses of the magnets for }n=1,2,\] \[x_{n} \text{are the vertical displacements of the magnets for }n=1,2,\] \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]0 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]1 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]2 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]3 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]4 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]5 Note: \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]6 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]7 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]8 \[\text{K.E.} = \frac{1}{2}m_{1}\dot{x}_{1}^{2}+\frac{1}{2}m_{2}\dot{x}_{2}^{2}+ \frac{1}{2}J_{1}\dot{\theta}_{1}^{2}+\frac{1}{2}J_{2}\dot{\theta}_{2}^{2}, \tag{2}\]9 \[\text{P.E.} = \frac{1}{2}k_{11}(x_{1}-l_{11}\sin\theta_{1})^{2}+\frac{1}{2}k_{ 12}(x_{1}+l_{12}\sin\theta_{1})^{2}+ \tag{3}\]0 \[\text{P.E.} = \frac{1}{2}k_{11}(x_{1}-l_{11}\sin\theta_{1})^{2}+\frac{1}{2}k_{ 12}(x_{1}+l_{12}\sin\theta_{1})^{2}+ \tag{3}\]1 \[\text{P.E.} = \frac{1}{2}k_{11}(x_{1}-l_{11}\sin\theta_{1})^{2}+\frac{1}{2}k_{ 12}(x_{1}+l_{12}\sin\theta_{1})^{2}+ \tag{3}\]2 \[\text{P.E.} = \frac{1}{2}k_{11}(x_{1}-l_{11}\sin\theta_{1})^{2}+\frac{1}{2}k_{ 12}(x_{1}+l_{12}\sin\theta_{1})^{2}+ \tag{3}\]3 \[\text{P.E.} = \frac{1}{2}k_{11}(x_{1}-l_{11}\sin\theta_{1})^{2}+\frac{1}{2}k_{ 12}(x_{1}+l_{12}\sin\theta_{1})^{2}+ \tag{3}\]4  
Refer to Appendix A for the physical units  
Define the state vector as,  
\[\vec{x}(t) = \left[ \begin{array}{c}x_{1}(t)\\ \dot{x}_{1}(t)\\ x_{2}(t)\\ \dot{x}_{2}(t)\\ \theta_{1}(t)\\ \theta_{2}(t)\\ \dot{\theta}_{2}(t)\\ \dot{i}(t)\end{array} \right], \tag{18}\]  
and express Equation (13) through (17) as a state-space system of ODEs,  
\[\frac{d}{dt}\vec{x}(t)=A_{\text{linear}}\vec{x}(t)+A_{\text{nonlinear}}(\vec{ x}(t))+\vec{u}(t), \tag{19}\]  
where  
\[A_{\text{linear}}=\left[ \begin{array}{ccccccccc}0&1&0&0&0&0&0&0\\ -\frac{1}{m_{1}}(k_{11}+k_{12})&-\frac{c_{1}}{m_{1}}&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&-\frac{1}{m_{2}}(k_{21}+k_{22})&-\frac{c_{2}}{m_{2}}&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&-\frac{R_{\text{J}}+R_{\text{J}}}{L_{\text{c}}}\end{array} \right], \tag{20}\]  
\[A_{\text{nonlinear}}=\left[