## 4 Cases Study  
The results provided in the preceding section are illustrated through some interesting applications. **Application 1**. : _Consider the electronic circuit in Figure 4. In the hypothesis that the considered transistor is a BJT described by the equations:_  
\[ \begin{array}{l}i_{1}=\frac{I_{s}}{P_{F}}\Big{(}e^{v_{1}/v_{T}}-1\Big{)}+\frac {I_{s}}{P_{R}}\Big{(}e^{(v_{1}-v_{2})/v_{T}}-1\Big{)}=g_{1}(v_{1},\;v_{2})\\ i_{2}=I_{s}\Big{(}e^{v_{1}/v_{T}}-1\Big{)}-\frac{I_{s}}{\frac{P_{R}}{1+P_{R}}} \Big{(}e^{(v_{1}-v_{2})/v_{T}}-1\Big{)}=g_{2}(v_{1},\;v_{2})\end{array}  \tag{52}\]  
applying the [PERSON]'s principles and the [PERSON]'s theorem the model becomes:  
\[ \begin{array}{l}R_{c}g_{2}(v_{1},v_{2})+v_{2}+R_{e}(g_{1}(v_{1},v_{2})+g_{2 }(v_{1},v_{2}))-V_{a}=c_{1}(v_{1},v_{2})=0\\ \frac{R_{b1}R_{b2}}{R_{b1}+R_{b2}}g_{1}(v_{1},v_{2})+v_{1}+R_{e}(g_{1}(v_{1},v _{2})+g_{2}(v_{1},v_{2}))-\frac{V_{a}R_{b1}}{R_{b1}+R_{b2}}=c_{2}(v_{1},v_{2}) =0\end{array}  \tag{53}\]  
A goal is to determine the working point \(\hat{v}=\left[ \begin{array}{cc}\hat{v}_{1}&\hat{v}_{2}\end{array} \right]^{T}\) i.e., the solution of the nonlinear equations system:  
\[\left( \begin{array}{c}c_{1}(v_{1},v_{2})\\ c_{1}(v_{1},v_{2})\end{array} \right)=c(v_{1},v_{2})=0 \tag{54}\]  
Based on (17) and Theorems 1 and 2 the value of \(\hat{v}\) can be obtained by integrating the control scheme in Figure 5.  
Figure 4: The considered electronic circuit. Figure 5: Control scheme to determine \(\hat{v}\). Under the following hypotheses:  
\[ \begin{array}{l}I_{s}=1e^{-13}\,\mathrm{A},v_{T}=\frac{300}{11\cdot 60}\, \mathrm{V},\beta_{F}=100,\beta_{R}=0.5\\ R_{c}=3\,\mathrm{k}\Omega,R_{e}=900\,\Omega,R_{b1}=1.70\,\mathrm{k}\Omega,R_{b2 }=8.30\,\mathrm{k}\Omega,V_{a}=12\end{array}  \tag{55}\]  
by numerically computing the Jacobian matrix as follows:  
\[J=\left( \begin{array}{cc}\frac{c_{1}(v_{1}+0.01,v_{2})-c_{1}(v_{1}-0.01,v_{2 })}{0.02}&\frac{c_{1}(v_{1},v_{2}+0.1)-c_{1}(v_{1},v_{2}-0.1)}{0.2}\\ \frac{c_{2}(v_{1}+0.01,v_{2})-c_{2}(v_{1}-0.01,v_{2})}{0.02}&\frac{c_{2}(v_{1},v_{2}+0.1)-c_{2}(v_{1},v_{2}-0.1)}{0.2}\end{array} \right) \tag{56}\]  
with \(a=30\), \(\dot{v}_{0}=0\) and by using the Matlab solver ode45 with variable-step, after 1.2 s the results reported in Table 1 are obtained. **Application 2**. : _Consider the planar robot in Figure 6._  
The goal is to realize the object in Figure 7 with the following time histories in the workspace:  
\[y(t)=\left( \begin{array}{c}y_{1}(t)\\ y_{2}(t)\\ y_{2}(t)\end{array} \right)=\left( \begin{array}{c}0.3\sin(2\pi t/10)+0.15\sin( 6\pi t/10)+0.50\\ 0.50\sin(2\pi t/10)+0.30\\ -\frac{\pi}{10}\cos(\pi t/10)-\frac{\pi}{2}\end{array} \right)=c(\beta(t)),\ \beta(t)=\left( \begin{array}{c}\beta_{1}(t)\\ \beta_{2}(t)\\ \beta_{2}(t)\end{array} \right),\ t\in[0,10] \tag{57}\] \begin{table}
\begin{tabular}{c c c} \hline \(v_{0}\) & \(c(v_{0})\) & \(\dot{v}\) & \(c(\dot{v})\) \\ \hline \(\left[\begin{array}{c}0\\ 0\end{array}\right]\) & \(\left[\begin{array}{c}-12.000000000000000\\ -2.040000000000000000\\ \end{array}\right]\) & \(\left[\begin{array}{c}0.60687199846390\\ 5.931278656205906\end{array}\right]\) & \(\left[\begin{array}{c}0.127767840751858\\ 0.030172309095633\end{array}\right]\) \\ \hline \(\left[\begin{array}{c}0.5\\ 5\end{array}\right]\) & \(\left[\begin{array}{c}-6.902629890911657\\ -1.517006057883000\end{array}\right]\) & \(\left[\begin{array}{c}0.606871998469258\\ 5.931278656228555\end{array}\right]\) & \(\left[\begin{array}{c}-0.355271367880050\\ -0.088817841970013\end{array}\right]\) & \(10^{-14}\) \\ \hline \(\left[\begin{array}{c}1\\ 10\end{array}\right]\) & \(\left[\begin{array}{c}2.42541244463312\\ 0.572760874825921\end{array}\right]\) & \(\left[\begin{array}{c}0.606871998507988\\ 5.931278656392560\end{array}\right]\) & \(\left[\begin{array}{c}-0.925230914106123\\ -0.218493267922781\end{array}\right]\) & \(10^{-8}\) \\ \hline \end{tabular}
\end{table} 
Table 1: Values of \(\dot{v}\) and \(c(\dot{v})\) for several \(v_{0}\). Figure 6: The considered planar robot.