Exercise 9.17 An harmonic oscillator with opampe?

(a)

An answer:
The closed-loop transfer is $H(\mathit{j\omega })=\frac{A}{1+A\cdot \beta (\mathit{j\omega })}$. For $A\cdot \beta (\mathit{j\omega })=1$, $\frac{A}{1+A\cdot \beta (\mathit{j\omega })}\to \infty$, hence a sine would be sustained indefinitely without any input signal. This is harmonic oscillation, with in equation $v_{\mathit{out}}=H(\mathit{j\omega })\cdot v_{\mathit{in}}=\infty \cdot 0$.

(b)

An answer:
The derivation of $H(\mathit{j\omega })$ can be done in many ways. One of these is shown below. $$\begin{array}{rcll}& v_{\mathit{out}}& =A(v_{+}-v_{-})& \\ & & v_{-}=v_{\mathit{out}}& \\ & & v_{+}=v_x\frac{1}{1+\mathit{j\omega RC}}& \\ & & v_{-}=v_{+}(A\to \infty )& \end{array}$$

after which $v_x$ can be calculated in a variety of ways. E.g. applying KCL on node $v_x$ gives:

$$\frac{v_{\mathit{in}}-v_x}{R}=\frac{v_x-v_{\mathit{out}}}{1∕\mathit{j\omega C}}+\frac{v_x-v_{+}}{R}$$

Rewriting $v_{+}=v_x\frac{1}{1+\mathit{j\omega RC}}$ into $v_x=(1+\mathit{j\omega RC})v_{\mathit{out}}$ and using that along with $v_{+}=v_{\mathit{out}}$ yields

$$\begin{array}{rcll}& & v_{\mathit{out}}=\frac{v_{\mathit{in}}+\mathit{j\omega RC}\cdot v_{\mathit{out}}}{1+3\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2}& \\ & & (1+3\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2)v_{\mathit{out}}=v_{\mathit{in}}+\mathit{j\omega RC}\cdot v_{\mathit{out}}& \\ & H(\mathit{j\omega })& =\frac{1}{1+2\mathit{j\omega RC}+j^2{\omega }^2{R}^2{C}^2}& \end{array}$$

(c)

An answer:
One of the many correct derivation of $H(\mathit{j\omega })$ is shown below.

$$\begin{array}{rcll}& v_{\mathit{out}}& =A(v_{+}-v_{-})& \\ & & v_{-}=v_{\mathit{in}}\cdot \frac{\mathit{j\omega L}}{1∕\mathit{j\omega C}+\mathit{j\omega L}}+v_{\mathit{out}}\cdot \frac{1∕\mathit{j\omega C}}{1∕\mathit{j\omega C}+\mathit{j\omega L}}& \\ & & v_{+}=v_{\mathit{in}}\frac{\mathit{j\omega L}}{R+\mathit{j\omega L}}& \\ & & v_{-}=v_{+}(A\to \infty )& \end{array}$$

Yielding

$$\begin{array}{rcll}& & v_{\mathit{in}}\cdot \frac{j^2{\omega }^2\mathit{LC}}{1+j^2{\omega }^2\mathit{LC}}+v_{\mathit{out}}\cdot \frac{1}{1+j^2{\omega }^2\mathit{LC}}=v_{\mathit{in}}\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \\ & & v_{\mathit{out}}\cdot \frac{1}{1+j^2{\omega }^2\mathit{LC}}=v_{\mathit{in}}\frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}-v_{\mathit{in}}\cdot \frac{j^2{\omega }^2\mathit{LC}}{1+j^2{\omega }^2\mathit{LC}}& \\ & H(\mathit{j\omega })& =\frac{\mathit{j\omega L}∕R-j^2{\omega }^2\mathit{LC}}{1+\mathit{j\omega L}∕R}& \end{array}$$

(d)

An answer:

$$\begin{array}{rcll}& & H_1=\frac{1}{{(1+\mathit{j\omega RC})}^2}(\mathit{see}\mathit{before})& \\ & & H_2=\frac{\mathit{j\omega }\frac{L}{R}(1-\mathit{j\omega RC})}{\mathit{j\omega }\frac{L}{R}+1}(\mathit{see}\mathit{before})& \\ & & H_3=-\frac{R}{R+\mathit{j\omega L}}=-\frac{1}{1+\mathit{j\omega }\frac{L}{R}}& \end{array}$$

(e)

An answer:

(f)

An answer:
The open-loop transfer function is:

$$A_{\mathit{loop}}(\omega )=A_{\mathit{buf}}\frac{\mathit{j\omega }\frac{L}{R}(1-\mathit{j\omega RC})}{{(1+\mathit{j\omega RC})}^2{(1+\mathit{j\omega }\frac{L}{R})}^2}$$

Assuming the same corner frequencies for the sections,

$$A_{\mathit{loop}}(\omega )=A_{\mathit{buf}}\frac{j\frac{\omega }{{\omega }_c}(1-j\frac{\omega }{{\omega }_c})}{{(1+\mathit{j\omega }\frac{\omega }{{\omega }_c})}^4}$$

(g)

An answer:
yes. $$\begin{array}{rcll}& & {\omega }_{\mathit{osc}}=\mathit{tan}\left(\frac{\pi }{10}\right){\omega }_c& \\ & & A_{\mathit{buf}}=\frac{1+\mathit{tan}\left(\frac{\pi }{10}\right)}{\mathit{tan}\left(\frac{\pi }{10}\right)}& \end{array}$$

(h)

An answer:
yes. $$\begin{array}{rcll}& & {\omega }_{\mathit{osc}}=\mathit{tan}\left(\frac{3\pi }{10}\right){\omega }_c& \\ & & A_{\mathit{buf}}=-\frac{1+\mathit{tan}\left(\frac{3\pi }{10}\right)}{\mathit{tan}\left(\frac{3\pi }{10}\right)}& \end{array}$$