# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 7 Examples  
Classical examples of linear mechanical control systems are mass-spring systems. We present the equations of motions of \(n\)-coupled mass-spring system, which consists of \(n\) bodies, where the position of \(i\)-th body is denoted \(x^{i}\), and \(m_{i}\) is the mass of \(i\)-th body. The bodies are connected by \(n+1\) springs with \(k_{i}\) being the spring constant of \(i\)-th spring, as depicted in Figure 1. The external forces (controls) \(u_{i}\) may, \(\dot{a}\) priori, be applied to each body. Figure 1: The \(n\)-coupled mass-spring system  
The dynamics of \(i\)-th body is given by the balance of forces acting on the body  
\[m_{i}\ddot{x}^{i}=-k_{i}\Big{(}x^{i}-x^{i-1}\Big{)}+k_{i+1}\Big{(}x^{i+1}-x^{i} \Big{)}+u_{i}, \tag{21}\]  
where \(x^{0}\equiv x^{n+1}\equiv 0\). The equations can be formulated in the form of (5), where  
\[M=\left( \begin{array}{ccccc}m_{1}&0&0&\dots&0\\ 0&m_{2}&0&\dots&0\\ 0&0&m_{3}&\dots&0\\ 0&0&0&\ddots&0\\ 0&0&0&\dots&m_{n}\end{array} \right)P=\left( \begin{array}{ccccc}k_{1}+k_{2}&- k_{2}&0&\dots&0\\ -k_{2}&k_{2}+k_{3}&-k_{3}&\dots&0\\ 0&-k_{3}&k_{3}+k_{4}&\dots&0\\ 0&0&0&\ddots&-k_{n}\\ 0&0&\dots&-k_{n}&k_{n}+k_{n+1}\end{array} \right)\]  
or as a Lagrangian system of the form (6), where  
\[E=E^{L}=\left( \begin{array}{ccccc}\frac{-k_{1}-k_{2}}{m_{1}}&\frac{k_{2}}{m _{1}}&0&\dots&0\\ \frac{k_{2}}{m_{2}}&\frac{-k_{2}-k_{3}}{m_{2}}&\frac{k_{3}}{m_{2}}&\dots&0\\ 0&\frac{k_{3}}{m_{3}}&\frac{-k_{3}-k_{4}}{m_{3}}&\dots&0\\ 0&0&0&\ddots&\frac{k_{n}}{m_{n-1}}\\ 0&0&\dots&\frac{k_{n}}{m_{n}}&\frac{-k_{n}-k_{n+1}}{m_{n}}\end{array} \right)B= \left( \begin{array}{ccccc}\frac{1}{m_{1}}&0&0&\dots&0\\ 0&\frac{1}{m_{2}}&0&\dots&0\\ 0&0&\frac{1}{m_{3}}&\dots&0\\ 0&0&0&\ddots&0\\ 0&0&0&\dots&\frac{1}{m_{n}}\end{array} \right). \tag{22}\]  
If all \(n\) controls are present, then the system is fully actuated. However, it is enough to apply only the control \(u_{n}\) in order for the system to be controllable. **Example 2** (The mass-spring system with one control). : _Consider the \(n\)-mass-spring system (21) with only one control \(u:=u_{n}\), i.e \(u_{i}=0\) for \(1\leq i\leq n-1\). The system is (2), where \(E\) is given by (22) and \(b=\left(0,\dots,\frac{1}{m_{n}}\right)^{T}\). It is straightforward to show that \(\operatorname{rank}\left(b,Eb,\dots,E^{n-1}b\right)=n\). What is more, introduce \(c=\left(\frac{n}{K},0,0,\dots,0\right)\), where \(\bar{m}=\prod_{i=1}^{n}m_{i}\) and \(\bar{k}=\prod_{i=2}^{n}k_{i}\), and apply the transformation \(T=(c,cE,\dots,cE^{n-1})^{T}\). The transformed system is a Lagrangian system of the form (19), given by:_  
\[E_{\mathrm{F}}^{L}=TET^{-1}=\left( \begin{array}{ccccc}0&1&\dots&0\\ &&\ddots&\\ 0&0&\dots&1\\ a_{1}&a_{2}&\dots&a_{n}\end{array} \right)\quad\text{and}\quad b_{\mathrm{F}}= Tb=\left( \begin{array}{c}0\\ \vdots\\ 0\\ 1\end{array} \right)\]  
_and \(E_{d}^{L}\) is nonexistent since the system is controllable. Note that, by applying the feedback \(u=-\sum_{i=1}^{n}a_{i}\dot{x}^{i}+\bar{u}\), we obtain the (non Lagrangian) mechanical canonical form (15) or we substitute \(a_{i}\dot{s}\) with any other \(\dot{a}_{i}\dot{s}\), and those \(\dot{a}_{i}\dot{s}\), for which the corresponding eigenvalues \(\tilde{\lambda}_{j}\) satisfy \(\tilde{\lambda}_{j}\in\mathbb{R}_{\leq 0}\) and are mutually distinct, give stable Lagrangian systems._  
**Example 3** (The uncontrollable mass-spring system). : _Consider the mass-spring system with 3 equal masses m and 4 equal springs with the spring constant \(k\). The external force is applied to the second mass only. The equations of motion are of the form (6) with:_  
\[E=E^{L}=\left( \begin{array}{ccccc}-\frac{2k}{m}&\frac{\dot{k}}{m}&0\\ \frac{\dot{k}}{m}&-\frac{2k}{m}&\frac{k}{m}\\ 0&\frac{\dot{k}}{m}&-\frac{2k}{m}\end{array} \right)\qquad\text{and}\qquad b= \left( \begin{array}{c}0\\ \frac{1}{m}\\ 0\end{array} \right).\]A direct calculation shows that \(\mathrm{rank}\ (b,Eb,E^{2}b)=2<n=3\).