Question 12.19: Objective: Determine the required feedback transfer function...

A phase margin of 45 degrees implies that the phase of the loop gain is −135 degrees at the frequency at which the magnitude of the loop gain is unity. The phase of the loop gain is
\phi = − \left[ \tan^{−1} \left( \frac{f}{10^{3}}\right)  +  \tan^{−1} \left( \frac{f}{5  ×  10^{4}}\right)  +  \tan^{−1} \left( \frac{f}{10^{6}}\right) \right]
Since the three poles are far apart, the frequency at which the phase is −135 degrees is approximately equal to the frequency of the second pole, as shown in Figure 12.53.
In this example, f_{135} \cong 5 × 10^{4}  Hz, so we have that
\phi = − \left[ \tan^{−1} \left( \frac{5  ×  10^{4}}{10^{3}}\right)  +  \tan^{−1} \left( \frac{5  ×  10^{4}}{5  ×  10^{4}}\right)  +  \tan^{−1} \left( \frac{5  ×  10^{4}}{10^{6}}\right) \right]

or
\phi = − [88.9° + 45° + 2.86°] \cong −135°
Since we want the loop gain magnitude to be unity at this frequency, we have
|T (f)| = 1 = \frac{β(1000)}{\sqrt{1  + \left(\frac{5  ×  10^{4}}{10^{3}}\right)^{2}} \sqrt{1  + \left(\frac{5  ×  10^{4}}{5  ×  10^{4}}\right)^{2}} \sqrt{1  + \left(\frac{5  ×  10^{4}}{10^{6}}\right)^{2}}}
or
1 \cong \frac{β(1000)}{(50)(1.41)(1)}
which yields β = 0.0707.
The closed-loop low-frequency gain for this case is
A_{f o} = \frac{A_{o}}{1  +  β A_{o}} = \frac{1000}{1  +  (0.0707)(1000)} = 13.9
Comment: For this value of β, if the frequency is greater than 5 × 10^{4}  Hz, the loop gain magnitude is less than unity. If the frequency is less than 5 × 10^{4}  Hz, the phase of the loop gain is |\phi| < 135° (phase margin of 45 degrees). These conditions imply that the system is stable