Exercise 9.1 AN RC-phase shift oscillator

a)

An answer:
This 3 stages consist of a unity gain voltage buffer that each have their own C-R high-pass section:

The signal transfer of each voltage buffer preceded by a CR-high pass is:

$$\begin{array}{llll}\hfill H(\mathit{j\omega })=\frac{R}{R+\frac{1}{\mathit{j\omega C}}}& =\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}& \hfill & \end{array}$$

b)

An answer:

With the previous answer, the signal transfer function for the 3 cascaded sections is:

$$\begin{array}{llll}\hfill \frac{v_z}{v_{\mathit{out}}}& ={\left(\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\right)}^3& \hfill & \\ \hfill & =\frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}\cdot \frac{\mathit{j\omega RC}}{1+\mathit{j\omega RC}}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{(1+\mathit{j\omega RC})\cdot (1+2\mathit{j\omega RC}-{\omega }^2{(\mathit{RC})}^2)}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{1+3\mathit{j\omega RC}-3{\omega }^3{(\mathit{RC})}^2-j{\omega }^3{(\mathit{RC})}^3}& \hfill & \\ \hfill & =\frac{-j{\omega }^3{(\mathit{RC})}^3}{\{1-3{\omega }^2{(\mathit{RC})}^2\}+j(3\mathit{\omega RC}-{\omega }^3{(\mathit{RC})}^3)}& \hfill & \end{array}$$

c)

An answer:

d)

An answer:
Node $x$. From the Bode pot for the 3 cascaded stages it follows that for oscillation at a finite non-zero frequency, the gain of the leftmost (amplifying) stage must be negative. There is just one (radian) frequency at which $\mathit{\angle H}(\mathit{j\omega })=\pm 180^{\circ }$; hence there is no need to check whether other frequencies would be more favourable for oscillation.

e)

An answer:

$$\begin{array}{llll}\hfill 1-3{\omega }_{\mathit{osc}}^2{(\mathit{RC})}^2& =0& \hfill & \\ \hfill \to {\omega }_{\mathit{osc}}^2& =\frac{1}{3{(\mathit{RC})}^2}& \hfill & \\ \hfill \to {\omega }_{\mathit{osc}}& =\frac{1}{\sqrt{3}\mathit{RC}}& \hfill & \end{array}$$

f)

An answer:

$$\begin{array}{llll}\hfill \frac{v_z}{v_{\mathit{out}}}\left({\omega }_{\mathit{osc}}=\frac{1}{\sqrt{3}\mathit{RC}}\right)& =\frac{-j{\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)}^3{(\mathit{RC})}^3}{j\left(3\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)\mathit{RC}-{\left(\frac{1}{\sqrt{3}\mathit{RC}}\right)}^3{(\mathit{RC})}^3)\right)}& \hfill & \\ \hfill & =\frac{-{(\frac{1}{\sqrt{3}})}^3}{\{\frac{3}{\sqrt{3}}-{(\frac{1}{\sqrt{3}})}^3\}}& \hfill & \\ \hfill & =-\frac{\frac{1}{3\sqrt{3}}}{\frac{3}{\sqrt{3}}-\frac{1}{3\sqrt{3}}}& \hfill & \\ \hfill & =-\frac{1}{8}& \hfill & \end{array}$$

The voltage gain of the leftmost gain stage must equal -8. For an opamp in negative feedback configuration then $\beta =8$.