Exercise 6.3 An opamp configuration with feedback to both inputs

a)

An answer:

$$\begin{array}{rcll}& z_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}& \\ & & i_{\mathit{in}}=\frac{v_{\mathit{in}}-v_{\mathit{out}}}{R_2}& \\ & & v_{\mathit{out}}=v_i\cdot (1+\frac{Z}{R_1})& \\ & & i_{\mathit{in}}=\frac{v_{\mathit{in}}(1-(1+\frac{Z}{R_1}))}{R_2}& \\ & & \Rightarrow & \\ & z_{\mathit{in}}& =\frac{v_{\mathit{in}}}{i_{\mathit{in}}}=-\frac{R_1{R}_2}{Z}& \end{array}$$

b)

An answer:
$$\begin{array}{rcll}{z}_{\mathit{in}}=-\frac{R_1{R}_2}{\frac{1}{\mathit{jwC}}}=-\mathit{jw}{R}_1{R}_{2}C& & & \end{array}$$

c)

An answer:

In general, the Nyquist plot should not right-handedly encircle the -1 point. For a frequency independent loop the Nyquist point should stay at the right hand side of the -1 point.

$$\begin{array}{rcll}& & A\cdot \beta >-1& \end{array}$$

which translates into $\beta >0$ for $A\to \infty$ and $\mathit{beta}$ is the net feedback as defined for negative feedback systems. In this system, both positive feedback ${\beta }_{\mathit{pos}}$ and negative feedback ${\beta }_{\mathit{neg}}$ is applied:

$$\begin{array}{rcll}& {\beta }_{\mathit{neg}}& >{\beta }_{\mathit{pos}}& \\ & & {\beta }_{\mathit{neg}}=\frac{R_1}{R_1+R_{Z_R}}& \\ & & {\beta }_{\mathit{pos}}=\frac{R_6}{R_G+R_2}& \\ & \Rightarrow & & \\ & R_G& <\frac{R_1{R}_2}{R_{Z_R}}& \end{array}$$

This is the same as $R_G∕∕z_{\mathit{in}}>0$