Exercise 9.9 An harmonic oscillator?

(a)

An answer:
This is an intro question - meant to get you going.

(b)

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

The loop gain is then:

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

(c)

An answer:
Recycling the (partial) results from (b) and simplifying: $$\begin{array}{rcll}& A_{\mathit{loop}}(\mathit{j\omega })& =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega L}∕R)\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega L}∕R}{1+\mathit{j\omega L}∕R}& \\ & & =-A_{\mathit{buf}}\cdot \frac{1+\mathit{j\omega RC}}{\mathit{j\omega RC}}\cdot \mathit{j\omega L}∕R& \\ & & =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega RC})\cdot L∕R^{2}C& \end{array}$$

for oscillation:

$$\begin{array}{rcll}& & =-A_{\mathit{buf}}\cdot (1+\mathit{j\omega RC})\cdot L∕R^{2}C=1& \end{array}$$

yields $\omega =0$ which is NOT a harmonic oscillator. $\to$ (e)

(d)

(e)

An answer:
We can (sensibly) swap in just 1 way per stage. And the swap should be done in only one stage. Swapping in one stage the transfer for that stage becomes: $$\begin{array}{rcll}& H_1(\mathit{j\omega })& =-\frac{2R}{\mathit{j\omega L}}=-\frac{2}{\mathit{j\omega L}∕R}=...& \\ & H_2(\mathit{j\omega })& =-\frac{2R}{1∕(\mathit{j\omega C})}=-2\mathit{j\omega RC}=...& \\ & H_3\mathit{j\omega })& =-\frac{R∕2}{\mathit{j\omega L}}=-\frac{1}{2\mathit{j\omega L}∕R}=...& \end{array}$$

These transfer functions should replace their original version in

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

And it can be derived that also after replacing, the circuit will still not oscillate. Any good proof is OK. This can be done by showing the equating the resulting $A_{\mathit{loop}}$ cannot be accomplished for any non-zero finite (radian) frequency.

IF you swap in multiple stages the circuit may get to oscillate!