劳斯阵列的推导

给定系统

{\displaystyle {\begin{aligned}f(x)&{}=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}&{}\quad (1)\\&{}=(x-r_{1})(x-r_{2})\cdots (x-r_{n})&{}\quad (2)\\\end{aligned}}}  

假设${\displaystyle f(x)=0}$ 的根都不在虚轴上，并且令

${\displaystyle N}$  = 是${\displaystyle f(x)=0}$ 的根的实部为负数的个数，

${\displaystyle P}$  = 是${\displaystyle f(x)=0}$ 的根的实部为正数的个数，

因此可得

${\displaystyle N+P=n\quad (3)}$ 

将${\displaystyle f(x)}$ 以极座标型式表示，可得

${\displaystyle f(x)=\rho (x)e^{j\theta (x)}\quad (4)}$ 

其中

${\displaystyle \rho (x)={\sqrt {{\mathfrak {Re}}^{2}[f(x)]+{\mathfrak {Im}}^{2}[f(x)]}}\quad (5)}$ 

且

${\displaystyle \theta (x)=\tan ^{-1}{\big (}{\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]{\big )}\quad (6)}$ 

根据(2)会发现

${\displaystyle \theta (x)=\theta _{r_{1}}(x)+\theta _{r_{2}}(x)+\cdots +\theta _{r_{n}}(x)\quad (7)}$ 

其中

${\displaystyle \theta _{r_{i}}(x)=\angle (x-r_{i})\quad (8)}$ 

若${\displaystyle f(x)=0}$ 的第i个根的实部为正，则（用y=(RE[y],IM[y])的表示法）

${\displaystyle {\begin{aligned}\theta _{r_{i}}(x){\big |}_{x=-j\infty }&=\angle (x-r_{i}){\big |}_{x=-j\infty }\\&=\angle (0-{\mathfrak {Re}}[r_{i}],-\infty -{\mathfrak {Im}}[r_{i}])\\&=\angle (-|{\mathfrak {Re}}[r_{i}]|,-\infty )\\&=\pi +\lim _{\phi \to \infty }\tan ^{-1}\phi ={\frac {3\pi }{2}}\quad (9)\\\end{aligned}}}$ 

且

${\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (-|{\mathfrak {Re}}[r_{i}]|,0)=\pi -\tan ^{-1}0=\pi \quad (10)}$ 

且

${\displaystyle \theta _{r_{i}}(x){\big |}_{x=j\infty }=\angle (-|{\mathfrak {Re}}[r_{i}]|,\infty )=\pi -\lim _{\phi \to \infty }\tan ^{-1}\phi ={\frac {\pi }{2}}\quad (11)}$ 

同样地，若${\displaystyle f(x)=0}$ 的第i个根的实部为负，

${\displaystyle {\begin{aligned}\theta _{r_{i}}(x){\big |}_{x=-j\infty }&=\angle (x-r_{i}){\big |}_{x=-j\infty }\\&=\angle (0-{\mathfrak {Re}}[r_{i}],-\infty -{\mathfrak {Im}}[r_{i}])\\&=\angle (|{\mathfrak {Re}}[r_{i}]|,-\infty )\\&=0-\lim _{\phi \to \infty }\tan ^{1}\phi =-{\frac {\pi }{2}}\quad (2)\\\end{aligned}}}$ 

且

${\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (|{\mathfrak {Re}}[r_{i}]|,0)=\tan ^{-1}0=0\,\quad (13)}$ 

且

${\displaystyle \theta _{r_{i}}(x){\big |}_{x=j\infty }=\angle (|{\mathfrak {Re}}[r_{i}]|,\infty )=\lim _{\phi \to \infty }\tan ^{-1}\phi ={\frac {\pi }{2}}\,\quad (14)}$ 

由(9)至(11)式可知，若${\displaystyle f(x)}$ 的第i个根实部为正，则${\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=-\pi }$ ，由(12)至(14)式可知，若${\displaystyle f(x)}$ 的第i个根实部为负，则${\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\pi }$ 。因此

${\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\angle (x-r_{1}){\Big |}_{x=-j\infty }^{x=j\infty }+\angle (x-r_{2}){\Big |}_{x=-j\infty }^{x=j\infty }+\cdots +\angle (x-r_{n}){\Big |}_{x=-j\infty }^{x=j\infty }=\pi N-\pi P\quad (15)}$ 

若定义

${\displaystyle \Delta ={\frac {1}{\pi }}\theta (x){\Big |}_{-j\infty }^{j\infty }\quad (16)}$ 

则可以得到以下的关系

${\displaystyle N-P=\Delta \quad (17)}$ 

结合(3)式及(17)式可得

${\displaystyle N={\frac {n+\Delta }{2}}}$ 且${\displaystyle P={\frac {n-\Delta }{2}}\quad (18)}$ 

因此，给定${\displaystyle n}$ 次的方程${\displaystyle f(x)}$ ，只需要计算${\displaystyle \Delta }$ ，就可以得到根的实部为负的个数${\displaystyle N}$ ，以及根的实部为正的个数${\displaystyle P}$ 。

配合(6)式及图1，${\displaystyle \tan(\theta )}$ 相对${\displaystyle \theta }$ 的图，将${\displaystyle x}$ 在区间(a,b)之间变化，其中${\displaystyle \theta _{a}=\theta (x)|_{x=ja}}$ ，而${\displaystyle \theta _{b}=\theta (x)|_{x=jb}}$ ，都是${\displaystyle \pi }$ 的整数倍，若此变化会使函数${\displaystyle \theta (x)}$ 增加${\displaystyle \pi }$ ，表示在从点a到点b的过程中，${\displaystyle \tan \theta (x)={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$ 从${\displaystyle +\infty }$ “跳到”${\displaystyle -\infty }$ 的次数比从${\displaystyle -\infty }$ “跳到”${\displaystyle +\infty }$ 的次数多一次。相反的，此变化会使函数${\displaystyle \theta (x)}$ 减少${\displaystyle \pi }$ ，表示在从点a到点b的过程中，${\displaystyle \tan(\theta )}$ 从${\displaystyle +\infty }$ “跳到”${\displaystyle -\infty }$ 的次数比从${\displaystyle -\infty }$ “跳到”${\displaystyle +\infty }$ 的次数少一次。

因此，${\displaystyle \theta (x){\Big |}_{-j\infty }^{j\infty }}$ 是${\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$ 从${\displaystyle -\infty }$ 跳到${\displaystyle +\infty }$ 的次数，减掉同函数从${\displaystyle +\infty }$ 跳到${\displaystyle -\infty }$ 的次数，两者差的${\displaystyle \pi }$ 倍。假设在${\displaystyle x=\pm j\infty }$ 处，${\displaystyle \tan[\theta (x)]}$ 有定义

若起始点是在不连续点（${\displaystyle \theta _{a}=\pi /2\pm i\pi }$ , i = 0, 1, 2, ...），则因为公式(17)（${\displaystyle N}$ 和${\displaystyle P}$ 都是整数，因此${\displaystyle \Delta }$ 也是整数），其结束点也会在不连续点。此时可以调整指标函数（正跳跃和负跳跃的差值）的计算方式，将正切函数的X轴移动${\displaystyle \pi /2}$ ，也就是在${\displaystyle \theta }$ 上加${\displaystyle \pi /2}$ 。此时的指标函数在各种${\displaystyle f(x)}$ 的系数组合下都有定义，就是在起始点（及结束点）连续的区间(a,b) = ${\displaystyle (+j\infty ,-j\infty )}$ 内计算${\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$ ，再在起始点连续的区间，计算

${\displaystyle \tan[\theta '(x)]=\tan[\theta +\pi /2]=-\cot[\theta (x)]=-{\mathfrak {Re}}[f(x)]/{\mathfrak {Im}}[f(x)]\quad (19)}$ 

差值${\displaystyle \Delta }$ 是${\displaystyle x}$ 从正跳跃和负跳跃的差值，若计算从${\displaystyle -j\infty }$ 到${\displaystyle +j\infty }$ 所产生的差值，即为相角正切的柯西指标（英语：Cauchy Index），其相角为${\displaystyle \theta (x)}$ 或${\displaystyle \theta '(x)}$ ，视${\displaystyle \theta _{a}}$ 是否是${\displaystyle \pi }$ 的整数倍而定。

为了要推导劳斯准则，会将${\displaystyle f(x)}$ 的奇次方项和偶次方项分开来列：

${\displaystyle f(x)=a_{0}x^{n}+b_{0}x^{n-1}+a_{1}x^{n-2}+b_{1}x^{n-3}+\cdots \quad (20)}$ 

因此可得到

${\displaystyle {\begin{aligned}f(j\omega )&=a_{0}(j\omega )^{n}+b_{0}(j\omega )^{n-1}+a_{1}(j\omega )^{n-2}+b_{1}(j\omega )^{n-3}+\cdots &{}\quad (21)\\&=a_{0}(j\omega )^{n}+a_{1}(j\omega )^{n-2}+a_{2}(j\omega )^{n-4}+\cdots &{}\quad (22)\\&+b_{0}(j\omega )^{n-1}+b_{1}(j\omega )^{n-3}+b_{2}(j\omega )^{n-5}+\cdots \\\end{aligned}}}$ 

若${\displaystyle n}$ 为偶数：

${\displaystyle {\begin{aligned}f(j\omega )&=(-1)^{n/2}{\big [}a_{0}\omega ^{n}-a_{1}\omega ^{n-2}+a_{2}\omega ^{n-4}-\cdots {\big ]}&{}\quad (23)\\&+j(-1)^{(n/2)-1}{\big [}b_{0}\omega ^{n-1}-b_{1}\omega ^{n-3}+b_{2}\omega ^{n-5}-\cdots {\big ]}&{}\\\end{aligned}}}$ 

若${\displaystyle n}$ 为奇数：

${\displaystyle {\begin{aligned}f(j\omega )&=j(-1)^{(n-1)/2}{\big [}a_{0}\omega ^{n}-a_{1}\omega ^{n-2}+a_{2}\omega ^{n-4}-\cdots {\big ]}&{}\quad (24)\\&+(-1)^{(n-1)/2}{\big [}b_{0}\omega ^{n-1}-b_{1}\omega ^{n-3}+b_{2}\omega ^{n-5}-\cdots {\big ]}&{}\\\end{aligned}}}$ 

可以看出若${\displaystyle n}$ 为奇数，根据(3)式，${\displaystyle N+P}$ 为奇数。若${\displaystyle N+P}$ 为奇数，${\displaystyle N-P}$ 也是奇数。同样的，若 ${\displaystyle n}$ 是偶数，${\displaystyle N-P}$ 也是偶数。(15)式可以看出若${\displaystyle N-P}$ 是偶数，${\displaystyle \theta }$ 是${\displaystyle \pi }$ 的整数倍。因此在${\displaystyle n}$ 为偶数时，${\displaystyle \tan(\theta )}$ 有定义，是n为偶数时使用的正确指标，在而在${\displaystyle n}$ 为奇数时，${\displaystyle \tan(\theta ')=\tan(\theta +\pi )=-\cot(\theta )}$ 有定义，也是n为奇数时使用的正确指标。

因此，根据(6)式及(23)式，${\displaystyle n}$ 为偶数时：

${\displaystyle \Delta =I_{-\infty }^{+\infty }{\frac {-{\mathfrak {Im}}[f(x)]}{{\mathfrak {Re}}[f(x)]}}=I_{-\infty }^{+\infty }{\frac {b_{0}\omega ^{n-1}-b_{1}\omega ^{n-3}+\cdots }{a_{0}\omega ^{n}-a_{1}\omega ^{n-2}+\ldots }}\quad (25)}$ 

因此，根据(19)式及(24)式，${\displaystyle n}$ 为奇数时：

${\displaystyle \Delta =I_{-\infty }^{+\infty }{\frac {{\mathfrak {Re}}[f(x)]}{{\mathfrak {Im}}[f(x)]}}=I_{-\infty }^{+\infty }{\frac {b_{0}\omega ^{n-1}-b_{1}\omega ^{n-3}+\ldots }{a_{0}\omega ^{n}-a_{1}\omega ^{n-2}+\ldots }}\quad (26)}$ 

因此可以计算相同的柯西指标：

${\displaystyle \Delta =I_{-\infty }^{+\infty }{\frac {b_{0}\omega ^{n-1}-b_{1}\omega ^{n-3}+\ldots }{a_{0}\omega ^{n}-a_{1}\omega ^{n-2}+\ldots }}\quad (27)}$ 

• Hurwitz, A., "On the Conditions under which an Equation has only Roots with Negative Real Parts", Rpt. in Selected Papers on Mathematical Trends in Control Theory, Ed. R. T. Ballman et al. New York: Dover 1964
• Routh, E. J., A Treatise on the Stability of a Given State of Motion. London: Macmillan, 1877. Rpt. in Stability of Motion, Ed. A. T. Fuller. London: Taylor & Francis, 1975
• Felix Gantmacher (J.L. Brenner translator) (1959) Applications of the Theory of Matrices, pp 177–80, New York: Interscience.