# 量值條件

${\displaystyle {\textbf {G}}(s)}$改寫為各因式相乘的形式

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

${\displaystyle {\frac {|{\textbf {P}}(s)|}{|{\textbf {Q}}(s)|}}=1}$

${\displaystyle K{\frac {|s-a_{1}||s-a_{2}|\cdots |s-a_{n}|}{|s-b_{1}||s-b_{2}|\cdots |s-b_{m}|}}=1,}$

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

${\displaystyle {\textbf {G}}(s)=-1=e^{j(\pi +2k\pi )}}$其中 ${\displaystyle (k=0,1,2,...)}$，這些是方程式所有的解。

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

{\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(\phi _{1}+\phi _{2}+\cdots +\phi _{m})}}}\\&=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}-(\phi _{1}+\phi _{2}+\cdots +\phi _{m}))},\end{aligned}}}

${\displaystyle 1=K{\frac {A_{1}A_{2}\cdots A_{n}}{B_{1}B_{2}\cdots B_{m}}}.}$