角度條件

假設。若將控制方程改為極坐標表示

${\displaystyle e^{j2\pi }+{\textbf {G}}(s){\textbf {H}}(s)=0}$

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

其中${\displaystyle k=0,1,2,\ldots }$為方程式中的解。將${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)}$改寫為各因式相乘的形式

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

再將因式${\displaystyle (s-a_{p})}$及${\displaystyle (s-b_{q})}$用向量的形式${\displaystyle A_{p}e^{j\theta _{p}}}$及${\displaystyle B_{q}e^{j\varphi _{q}}}$表示，因此可以改寫${\displaystyle {\textbf {G}}(s){\textbf {H}}(s)}$如下。

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

簡化特徵方程式，

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

因此可以得到角度條件：

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

其中${\displaystyle k=0,1,2,\ldots }$,

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

為極點1至n的角度，而

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

為零點1至m的角度。

也可以用類似的方式推導量值條件。