Exercise 7.1 Stability for a trans impedance amplifier configuration

a)

We now define an (angular) frequency ${\omega }_1=\frac{1}{R_2\cdot C_1}$

b)

An answer:
For a phase margin of $45°$, the phase shift of each low pass transfer is $-67.5°$. With $A_o\gg 1$ this cannot occur for $|A_{\mathit{loop}}=1$. Hence: NO.

c)

An answer:
For a phase margin of $45°$, the summed phase shift of the two low pass transfers is $-135°$. With ${\omega }_0\gg {\omega }_1$ this can occur for for an $\omega \gg \mathit{omeg}{a}_1$, for a loop gain magnitude 1. Hence: YES.

d)

An answer:
For a phase margin of $45°$, the summed phase shift of the two low pass transfers is $-135°$. With ${\omega }_1\gg {\omega }_0$ this can occur for for an $\omega \gg \mathit{omeg}{a}_0$, for a loop gain magnitude 1. Hence: YES.

e)

An answer:

For the case that ${\omega }_1\gg {\omega }_0$:

$$\begin{array}{rcll}& \omega & \gg {\omega }_0& \\ & & \to \angle \frac{A_o}{1+\mathit{j\omega }∕{\omega }_0}\approx -90°& \\ & & \to & \\ & & \angle \frac{1}{1+\mathit{j\omega }∕{\omega }_1}\approx -45°& \\ & & |\frac{1}{1+\mathit{j\omega }∕{\omega }_1}|\approx \frac{1}{\sqrt{2}}& \\ & \omega ={\omega }_1& =\frac{A_0}{\sqrt{2}}\cdot {\omega }_0\approx A_0\cdot {\omega }_0& \end{array}$$

Using an asymptotic Bode plot, the $\sqrt{2}$ is easily lost. A graphical “derivation” would be similar to the plot below.