Exercise 6.1 Opamp configurations with reactive components

a)

An answer:
$$\begin{array}{rcll}& H(\mathit{j\omega })& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=Z_{C1}\cdot i_{C_1}& \\ & & i_{C_1}=\frac{0-v_{\mathit{in}}}{R_2}& \\ & & v_{\mathit{out}}=\frac{1}{\mathit{j\omega }{C}_1}\cdot -\frac{v_{\mathit{in}}}{R_2}& \\ & H(\mathit{j\omega })& =-\frac{1}{\mathit{j\omega }{C}_1{R}_2}& \end{array}$$

b)

An answer:
$$\begin{array}{rcll}& v_{\mathit{out}}& =v_{-}-v_{C1}(t)& \\ & & v_{-}=0& \\ & & v_{C1}(t)=v_{C1}(0)+\frac{1}{C_1}{\int \; <!--\; nolimits\; -->}_0^t{i}_C\mathit{dt}& \\ & & i_C=\frac{v_{\mathit{in}}(t)}{R_2}& \\ & v_{\mathit{out}}& =v_{\mathit{out}}(0)-\frac{1}{R_2{C}_1}{\int \; <!--\; nolimits\; -->}_0^t{v}_{\mathit{in}}(t)\mathit{dt}& \end{array}$$

c)

An answer:
$$\begin{array}{rcll}& H(\mathit{j\omega })& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=R_1\cdot i_{R_1}& \\ & & i_{R_1}=\frac{0-v_{\mathit{in}}}{\mathit{j\omega }{L}_2}& \\ & & v_{\mathit{out}}=R_1\cdot -\frac{v_{\mathit{in}}}{\mathit{j\omega }{L}_2}& \\ & H(\mathit{j\omega })& =-\frac{1}{\mathit{j\omega }\frac{L_2}{R_1}}& \end{array}$$

d)

An answer:
$$\begin{array}{rcll}& v_{\mathit{out}}& =v_{-}-v_{R1}(t)& \\ & & v_{-}=0& \\ & & v_{R1}(t)=R_1\cdot i_{R1}& \\ & & i_{R1}=i_{L2}(0)+\frac{1}{L_1}{\int \; <!--\; nolimits\; -->}_0^t{v}_{\mathit{in}}\mathit{dt}& \\ & v_{\mathit{out}}& =v_{\mathit{out}}(0)-\frac{R_1}{L_2}{\int \; <!--\; nolimits\; -->}_0^t{v}_{\mathit{in}}(t)\mathit{dt}& \end{array}$$