勞斯陣列的推導

給定系統

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

假設${\displaystyle f(x)=0}$的根都不在虛軸上，並且令

${\displaystyle N}$ = 是${\displaystyle f(x)=0}$的根的實部為負數的個數，

${\displaystyle P}$ = 是${\displaystyle f(x)=0}$的根的實部為正數的個數，

因此可得

${\displaystyle N+P=n\quad (3)}$

將${\displaystyle f(x)}$以極座標型式表示，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

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

根據(2)會發現

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

其中

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

若${\displaystyle f(x)=0}$的第i個根的實部為正，則（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

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

同樣地，若${\displaystyle f(x)=0}$的第i個根的實部為負，

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

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個根實部為正，則${\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=-\pi }$，由(12)至(14)式可知，若${\displaystyle f(x)}$的第i個根實部為負，則${\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\pi }$。因此

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

若定義

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

則可以得到以下的關係

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

結合(3)式及(17)式可得

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

因此，給定${\displaystyle n}$次的方程${\displaystyle f(x)}$，只需要計算${\displaystyle \Delta }$，就可以得到根的實部為負的個數${\displaystyle N}$，以及根的實部為正的個數${\displaystyle P}$。

圖1

${\displaystyle \tan(\theta )}$相對${\displaystyle \theta }$的圖

配合(6)式及圖1，${\displaystyle \tan(\theta )}$相對${\displaystyle \theta }$的圖，將${\displaystyle x}$在區間(a,b)之間變化，其中${\displaystyle \theta _{a}=\theta (x)|_{x=ja}}$，而${\displaystyle \theta _{b}=\theta (x)|_{x=jb}}$，都是${\displaystyle \pi }$的整數倍，若此變化會使函數${\displaystyle \theta (x)}$增加${\displaystyle \pi }$，表示在從點a到點b的過程中，${\displaystyle \tan \theta (x)={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$從${\displaystyle +\infty }$「跳到」${\displaystyle -\infty }$的次數比從${\displaystyle -\infty }$「跳到」${\displaystyle +\infty }$的次數多一次。相反的，此變化會使函數${\displaystyle \theta (x)}$減少${\displaystyle \pi }$，表示在從點a到點b的過程中，${\displaystyle \tan(\theta )}$從${\displaystyle +\infty }$「跳到」${\displaystyle -\infty }$的次數比從${\displaystyle -\infty }$「跳到」${\displaystyle +\infty }$的次數少一次。

因此，${\displaystyle \theta (x){\Big |}_{-j\infty }^{j\infty }}$是${\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$從${\displaystyle -\infty }$跳到${\displaystyle +\infty }$的次數，減掉同函數從${\displaystyle +\infty }$跳到${\displaystyle -\infty }$的次數，兩者差的${\displaystyle \pi }$倍。假設在${\displaystyle x=\pm j\infty }$處，${\displaystyle \tan[\theta (x)]}$有定義

圖2

${\displaystyle -\cot(\theta )}$相對${\displaystyle \theta }$的圖

若起始點是在不連續點（${\displaystyle \theta _{a}=\pi /2\pm i\pi }$, i = 0, 1, 2, ...），則因為公式(17)（${\displaystyle N}$和${\displaystyle P}$都是整數，因此${\displaystyle \Delta }$也是整數），其結束點也會在不連續點。此時可以調整指標函數（正跳躍和負跳躍的差值）的計算方式，將正切函數的X軸移動${\displaystyle \pi /2}$，也就是在${\displaystyle \theta }$上加${\displaystyle \pi /2}$。此時的指標函數在各種${\displaystyle f(x)}$的係數組合下都有定義，就是在起始點（及結束點）連續的區間(a,b) = ${\displaystyle (+j\infty ,-j\infty )}$內計算${\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}$，再在起始點連續的區間，計算

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

差值${\displaystyle \Delta }$是${\displaystyle x}$從正跳躍和負跳躍的差值，若計算從${\displaystyle -j\infty }$到${\displaystyle +j\infty }$所產生的差值，即為相角正切的柯西指標，其相角為${\displaystyle \theta (x)}$或${\displaystyle \theta '(x)}$，視${\displaystyle \theta _{a}}$是否是${\displaystyle \pi }$的整數倍而定。

為了要推導勞斯準則，會將${\displaystyle f(x)}$的奇次方項和偶次方項分開來列：

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

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${\displaystyle n}$為偶數：

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

若${\displaystyle n}$為奇數：

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

可以看出若${\displaystyle n}$為奇數，根據(3)式，${\displaystyle N+P}$為奇數。若${\displaystyle N+P}$為奇數，${\displaystyle N-P}$也是奇數。同樣的，若 ${\displaystyle n}$是偶數，${\displaystyle N-P}$也是偶數。(15)式可以看出若${\displaystyle N-P}$是偶數，${\displaystyle \theta }$是${\displaystyle \pi }$的整數倍。因此在${\displaystyle n}$為偶數時，${\displaystyle \tan(\theta )}$有定義，是n為偶數時使用的正確指標，在而在${\displaystyle n}$為奇數時，${\displaystyle \tan(\theta ')=\tan(\theta +\pi )=-\cot(\theta )}$有定義，也是n為奇數時使用的正確指標。

因此，根據(6)式及(23)式，${\displaystyle n}$為偶數時：

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

因此，根據(19)式及(24)式，${\displaystyle n}$為奇數時：

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

因此可以計算相同的柯西指標：

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

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