角度條件

角度條件（angle condition）是自動控制的根軌跡圖中，有關角度的限制條件，根軌跡圖中的點和閉迴路極點、零點組成向量的角度會滿足角度條件。角度條件和量值條件可以完全確定根軌跡圖。 令系統的特徵方程為 1 + G ( s ) H ( s ) = 0 {\displaystyle 1+{\textbf {G}}(s){\textbf {H}}(s)=0} ，而 G ( s ) H ( s ) = P ( s ) Q ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)={\frac {{\textbf {P}}(s)}{{\textbf {Q}}(s)}}} ，可改寫為以下各因式相乘的形式 G ( s ) H ( s ) = P ( s ) Q ( s ) = K ( s − a 1 ) ( s − a 2 ) ⋯ ( s − a n ) ( s − b 1 ) ( s − b 2 ) ⋯ ( s − b m ) , {\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})}},} 則角度條件是指 ∠ ( G ( s ) H ( s ) ) = π + 2 k π {\displaystyle \angle ({\textbf {G}}(s){\textbf {H}}(s))=\pi +2k\pi } 其中 k {\displaystyle k} 為整數。 也就是說 ∑ i = 1 m ∠ ( s − b i ) − ∑ i = 1 n ∠ ( s − a i ) = π + 2 k π {\displaystyle \sum _{i=1}^{m}\angle (s-b_{i})-\sum _{i=1}^{n}\angle (s-a_{i})=\pi +2k\pi } 開迴路零點到 s {\displaystyle s} 點角度的和，減去開迴路極點到 s {\displaystyle s} 點角度的和，除 2 π {\displaystyle 2\pi } 後的餘數需等於 π {\displaystyle \pi } 。 假設。若將控制方程改為極坐標表示 e j 2 π + G ( s ) H ( s ) = 0 {\displaystyle e^{j2\pi }+{\textbf {G}}(s){\textbf {H}}(s)=0} G ( s ) H ( s ) = − 1 = e j ( π + 2 k π ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)=-1=e^{j(\pi +2k\pi )}} 其中 k = 0 , 1 , 2 , … {\displaystyle k=0,1,2,\ldots } 為方程式中的解。將 G ( s ) H ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)} 改寫為各因式相乘的形式 G ( s ) H ( s ) = P ( s ) Q ( s ) = K ( s − a 1 ) ( s − a 2 ) ⋯ ( s − a n ) ( s − b 1 ) ( s − b 2 ) ⋯ ( s − b m ) , {\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 ( s − a p ) {\displaystyle (s-a_{p})} 及 ( s − b q ) {\displaystyle (s-b_{q})} 用向量的形式 A p e j θ p {\displaystyle A_{p}e^{j\theta _{p}}} 及 B q e j φ q {\displaystyle B_{q}e^{j\varphi _{q}}} 表示，因此可以改寫 G ( s ) H ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)} 如下。 G ( s ) H ( s ) = K A 1 A 2 ⋯ A n e j ( θ 1 + θ 2 + ⋯ + θ n ) B 1 B 2 ⋯ B m e j ( φ 1 + φ 2 + ⋯ + φ 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})}}}} 簡化特徵方程式， e j ( π + 2 k π ) = K A 1 A 2 ⋯ A n e j ( θ 1 + θ 2 + ⋯ + θ n ) B 1 B 2 ⋯ B m e j ( φ 1 + φ 2 + ⋯ + φ m ) = K A 1 A 2 ⋯ A n B 1 B 2 ⋯ B m e j ( θ 1 + θ 2 + ⋯ + θ n − ( φ 1 + φ 2 + ⋯ + φ 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}}} 因此可以得到角度條件： π + 2 k π = θ 1 + θ 2 + ⋯ + θ n − ( φ 1 + φ 2 + ⋯ + φ m ) {\displaystyle \pi +2k\pi =\theta _{1}+\theta _{2}+\cdots +\theta _{n}-(\varphi _{1}+\varphi _{2}+\cdots +\varphi _{m})} 其中 k = 0 , 1 , 2 , … {\displaystyle k=0,1,2,\ldots } , θ 1 , θ 2 , … , θ n {\displaystyle \theta _{1},\theta _{2},\ldots ,\theta _{n}} 為極點1至n的角度，而 φ 1 , φ 2 , … , φ m {\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。