# 稳定性与频率补偿作业

\begin{align*} \newcommand{\dif}{\mathop{}\!\mathrm{d}} \newcommand{\belowarrow}[1]{\mathop{#1}\limits_{\uparrow}} \newcommand{\bd}{\boldsymbol} \newcommand{\tx}{\text} \newcommand{\p}{\partial\,} \end{align*}
1. What is the requirement of PM for a stable frequency-domain system and time-domain system respectively? In these cases, what is the relationship between $f_2$ and $\tx{GBW}$ respectively?
• for a stable system, PM must be positive, therefore, $90^\circ - \arctan (\tx{GBW}/\omega_2)$ must greater than 0. Usually, $f_2$ is about three times of the $\tx{GBW}$

10.3 An amplifier has a forward gain of $A_0 = 1000$ and two poles at $ω_{p1}$ and $ω_{p2}$.For $ω_{p1} = 1 \tx{MHz}$, calculate the phase margin of a unity-gain feedback loop if (a) $ω_{p2} = 2ω_{p1}$ and (b) $ω_{p2} = 4ω_{p1}$

(a)

$\omega_{p2}=2\omega_{p1}=2\tx{MHz}\\ \tx{GBW} = \sqrt{A_0 \omega_{p1}\omega_{p2}} = 44.72 \tx{MHz}\\ \tx{PM}=180^\circ - \arctan (\frac{\tx{GBW}}{\omega_{p1}}) - \arctan (\frac{\tx{GBW}}{\omega_{p2}})=3.84^\circ$

(b)

$\omega_{p2}=4\omega_{p1}=4\tx{MHz}\\ \tx{GBW} = \sqrt{A_0 \omega_{p1} \omega_{p2}} = 63.25 \tx{MHz}\\ \tx{PM} = 180^\circ - \arctan (\frac{\tx{GBW}}{\omega_{p1}}) - \arctan (\frac{\tx{GBW}}{\omega_{p2}}) = 4.52^\circ$

(c) The calculation below is just for fun.

\tx{if } \omega_{p2} = k\omega_{p1}\\ \tx{GBW} = \sqrt{A_0 \omega_{p1}\omega_{p2}} = \omega_{p1} \sqrt{kA} \\ \begin{aligned} \tx{PM} &= 180^\circ - \arctan \sqrt{kA} - \arctan \sqrt{\frac{A}{k}} \\ &=\arctan\frac{\sqrt{kA}+\sqrt{\frac{A}{k}}}{A-1}\\ &= \arctan \left[\frac{\sqrt{A}}{A-1}(\sqrt{k}+\sqrt{\frac{1}{k}})\right] \end{aligned}\\ \Rightarrow (\sqrt{A}-\sqrt{\frac{1}{A}})\tan \tx{PM} = \sqrt{k}+\sqrt{\frac{1}{k}}\\

if $A,k\gg 1$, than $\sqrt{A} \tan \tx{PM} = \sqrt{k}$ or $\tx{PM} = \arctan \sqrt{\dfrac{k}{A}}$