Remove ads量值条件(magnitude condition)是自动控制的根轨迹图中,有关量值的限制条件,根轨迹图中的点和闭回路极点、零点组成向量的量值会满足量值条件。量值条件和角度条件可以完全确定根轨迹图。 此条目没有列出任何参考或来源。 (2017年9月) 根轨迹图上的每一点和极点、零点组成向量的量值会满足量值条件 令系统的特征方程为 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)}}} ,可改写为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})}},} , 则量值条件是指找到K值使下式成立: | G ( s ) H ( s ) | = 1 {\displaystyle |G(s)H(s)|=1} 也就是说 K | s − a 1 | | s − a 2 | ⋯ | s − a m | | s − b 1 | | s − b 2 | ⋯ | s − b n | = 1 {\displaystyle K{\frac {|s-a_{1}||s-a_{2}|\cdots |s-a_{m}|}{|s-b_{1}||s-b_{2}|\cdots |s-b_{n}|}}=1} . Remove ads推导 令 G ( s ) H ( s ) = P ( s ) Q ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)={\frac {{\textbf {P}}(s)}{{\textbf {Q}}(s)}}} 。 将 G ( s ) H ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)} 改写为各因式相乘的形式 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})}},} 量值条件即是 | P ( s ) | | Q ( s ) | = 1 {\displaystyle {\frac {|{\textbf {P}}(s)|}{|{\textbf {Q}}(s)|}}=1} 也就是 K | s − a 1 | | s − a 2 | ⋯ | s − a n | | s − b 1 | | s − b 2 | ⋯ | s − b m | = 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,} 若将控制方程改为极坐标表示: 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,...)} ,这些是方程式所有的解。 将每一个因子 ( s − a p ) {\displaystyle (s-a_{p})} 及 ( s − b q ) {\displaystyle (s-b_{q})} 都用等效的向量 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\phi _{q}}} 来表示,因此 G ( s ) H ( s ) {\displaystyle {\textbf {G}}(s){\textbf {H}}(s)} 可以再作整理。 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(\phi _{1}+\phi _{2}+\cdots +\phi _{m})}}}} 简化特征方程式。 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(\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}}} 因此可以推导出另一个形式的量值条件: 1 = K A 1 A 2 ⋯ A n B 1 B 2 ⋯ B m . {\displaystyle 1=K{\frac {A_{1}A_{2}\cdots A_{n}}{B_{1}B_{2}\cdots B_{m}}}.} 也可以用类似的方式推导角度条件。 Remove adsLoading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads