Exercise 9.4 A 3-stage RC-phase shift oscillator

a)

An answer:

$$\begin{array}{llll}\hfill H_{\beta }(\mathit{j\omega })=\frac{v_{\mathit{in}}}{v_{\mathit{out}}}& ={\left(\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\right)}^3& \hfill & \end{array}$$

b)

An answer:

For the Bode plot, we defined ${\omega }_0=\frac{1}{\mathit{RC}}$.

c)

An answer:
Using symmetry is the easiest and leads to:

$$\begin{array}{rcll}& {\omega }_{\mathit{osc}}& =\frac{1}{\mathit{RC}}\mathit{tan}(30^{\circ })& \\ & & =\frac{1}{\sqrt{3}\mathit{RC}}& \end{array}$$

d)

An answer:
A = -8

e)

An answer:
The polar plot of $\mathit{A\beta }$ is shown in blue for positive $\omega$; the light blue curve is for negative frequencies and is required to make it a Nyquist plot.