# 角度條件

${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)={\frac {{\textbf {P}}(s)}{{\textbf {Q}}(s)}}=K{\frac {(s-a_{1})(s-a_{2})\cdots (s-a_{n})}{(s-b_{1})(s-b_{2})\cdots (s-b_{m})}},}$

${\displaystyle \angle ({\textbf {G}}(s){\textbf {H}}(s))=\pi +2k\pi }$

${\displaystyle \sum _{i=1}^{m}\angle (s-b_{i})-\sum _{i=1}^{n}\angle (s-a_{i})=\pi +2k\pi }$

## 推導

${\displaystyle e^{j2\pi }+{\textbf {G}}(s){\textbf {H}}(s)=0}$
${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)=-1=e^{j(\pi +2k\pi )}}$

${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)={\frac {{\textbf {P}}(s)}{{\textbf {Q}}(s)}}=K{\frac {(s-a_{1})(s-a_{2})\cdots (s-a_{n})}{(s-b_{1})(s-b_{2})\cdots (s-b_{m})}},}$

${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)=K{\frac {A_{1}A_{2}\cdots A_{n}e^{j(\theta _{1}+\theta _{2}+\cdots +\theta _{n})}}{B_{1}B_{2}\cdots B_{m}e^{j(\varphi _{1}+\varphi _{2}+\cdots +\varphi _{m})}}}}$

{\displaystyle {\begin{aligned}e^{j(\pi +2k\pi )}&=K{\frac {A_{1}A_{2}\cdots A_{n}e^{j(\theta _{1}+\theta _{2}+\cdots +\theta _{n})}}{B_{1}B_{2}\cdots B_{m}e^{j(\varphi _{1}+\varphi _{2}+\cdots +\varphi _{m})}}}\\[6pt]&=K{\frac {A_{1}A_{2}\cdots A_{n}}{B_{1}B_{2}\cdots B_{m}}}e^{j(\theta _{1}+\theta _{2}+\cdots +\theta _{n}-(\varphi _{1}+\varphi _{2}+\cdots +\varphi _{m}))},\end{aligned}}}

${\displaystyle \pi +2k\pi =\theta _{1}+\theta _{2}+\cdots +\theta _{n}-(\varphi _{1}+\varphi _{2}+\cdots +\varphi _{m})}$

${\displaystyle \theta _{1},\theta _{2},\ldots ,\theta _{n}}$

${\displaystyle \varphi _{1},\varphi _{2},\ldots ,\varphi _{m}}$