Cuestion 6



El segundo operacional es un no inversor.


\begin{displaymath}v_{o_{2}}=(1+\frac{R}{R})\,V_{o_{1}}=2\,v_{o_{1}}\end{displaymath}




\begin{displaymath}i_{in}=\frac{v_{in}}{R}=-\frac{v_{o_{1}}}{3\,R}-\frac{v_{o_{2}}}{3\,R}\end{displaymath}


\begin{displaymath}v_{in}=-\frac{v_{o_{1}}}{3}-\frac{v_{o_{2}}}{3}=-\frac{v_{o_{1}}}{3}-\frac{2\,v_{o_{1}}}{3}=-v_{o_{1}}\end{displaymath}


\begin{displaymath}A_{v_{1}}=\frac{v_{o_{1}}}{v_{in}}=-1\end{displaymath}