Exercise 9.8 An harmonic oscillator?

(a)

(b)

An answer:
For this circuit, the opamp is ideal and then the signal transfer per stage is $H(\mathit{j\omega })=-Z_2∕Z_1$ where $Z_1$ is between the inverting input and the output node. This yields e.g.: $$\begin{array}{rcll}& H_1(\mathit{j\omega })& =-\frac{R}{R+\mathit{j\omega L}}=-\frac{1}{1+\mathit{j\omega L}∕R}=...& \\ & H_2\mathit{j\omega })& =-\frac{R}{R+1∕\mathit{j\omega C}}=-\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}=...& \\ & H_3(\mathit{j\omega })& =-\frac{R∕∕1∕(\mathit{j\omega C})}{R}=-\frac{1}{1+\mathit{j\omega RC}}=...& \end{array}$$

The loop gain is then:

$$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot \frac{1}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{1}{1+\mathit{j\omega RC}}& \end{array}$$

(c)

An answer:
The loop gain was derived earlier. If this can be equation to 1 for a finite non-zero $\omega$ then this can omplement a harmonic oscillator. Recycling the derived equation: $$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot \frac{1}{1+\mathit{j\omega L}∕R}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{1}{1+\mathit{j\omega RC}}& \\ & & -A_{\mathit{buf}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega }(L∕R+2\mathit{RC})+{(\mathit{j\omega })}^2(2\mathit{LC}+R^2{C}^2)+{(\mathit{j\omega })}^3\mathit{RL}{C}^2}& \end{array}$$

This has an imaginary numerator and a complex polynomial denominator. Setting the latter to a purely imaginary (complex) number:

$$\begin{array}{rcll}& & 1+{(j{\omega }_{\mathit{osc}})}^2(2\mathit{LC}+R^2{C}^2)\equiv 0\to & \\ & & {\omega }_{\mathit{osc}}=\frac{1}{\sqrt{2\mathit{LC}+R^2{C}^2}}& \end{array}$$

Hence it is possible to get harmonic oscillation (out of the box) for this circuit for a specific scalar $A_{\mathit{buf}}$.

(d)

An answer:
Recycling the results at (c), we only need to derive $A_{\mathit{buf}}$: $$\begin{array}{rcll}& A_{\mathit{loop}}(j{\omega }_{\mathit{osc}})& =-A_{\mathit{buf}}\cdot \frac{j{\omega }_{\mathit{osc}}\mathit{RC}}{j{\omega }_{\mathit{osc}}(L∕R+2\mathit{RC})+{(j{\omega }_{\mathit{osc}})}^3\mathit{RL}{C}^2}& \\ & & =-A_{\mathit{buf}}\cdot \frac{j{\omega }_{\mathit{osc}}\mathit{RC}}{j{\omega }_{\mathit{osc}}(L∕R+2\mathit{RC})+{(j{\omega }_{\mathit{osc}})}^3\mathit{RL}{C}^2}& \\ & & =-A_{\mathit{buf}}\cdot \frac{\mathit{RC}}{(L∕R+2\mathit{RC})-{\omega }_{\mathit{osc}}^2\mathit{RL}{C}^2}\equiv 1& \\ & & & \end{array}$$

yielding a nasty equation. Full derivation:

$$\begin{array}{rcll}& A_{\mathit{buf}}& =-\frac{(L∕R+2\mathit{RC})-{\omega }_{\mathit{osc}}^2\mathit{RL}{C}^2}{\mathit{RC}}& \\ & & =-\frac{(L∕R+2\mathit{RC})-\frac{\mathit{RL}{C}^2}{2\mathit{LC}+R^2{C}^2}}{\mathit{RC}}& \\ & & & \end{array}$$

Further cleanup/simplication is NOT required.

(e)

An answer:
Not applicable.