Discussion
An amplifier has poles at \( 1 \ \mathrm{MHz} \) and \( 4 \ \mathrm{MHz} \) and a unity-gain frequency of \( 8 \ \mathrm{MHz} \). Its phase margin is roughly
Answer: A
Each pole contributes the arctangent of the crossover frequency over its own:
\[ \begin{aligned} \phi &= \arctan 8 + \arctan 2 \\ &= 83^\circ + 63^\circ = 146^\circ \end{aligned} \]
leaving a margin of about thirty-four degrees - below the forty-five usually specified, so the step response would overshoot noticeably. Compensation is needed, either to bring the crossover down or to add a zero near it. Summing each pole's phase contribution at the crossover frequency is the whole calculation.
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