勞斯陣列是劳斯–赫尔维茨稳定性判据 中,用來判斷系統是否穩定的方式,是透過系統的特徵多項式 係數所建立的陣列。勞斯陣列和勞斯–赫爾維茨理論 是古典控制理論的核心,結合了歐幾里得算法 和施图姆定理 來計算柯西指標 。
給定系統
f ( x ) = a 0 x n + a 1 x n − 1 + ⋯ + a n ( 1 ) = ( x − r 1 ) ( x − r 2 ) ⋯ ( x − r n ) ( 2 ) {\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 f ( x ) = 0 {\displaystyle f(x)=0} 的根都不在虛軸上,並且令
N {\displaystyle N} = 是 f ( x ) = 0 {\displaystyle f(x)=0} 的根的實部為負數的個數, P {\displaystyle P} = 是 f ( x ) = 0 {\displaystyle f(x)=0} 的根的實部為正數的個數, 因此可得
N + P = n ( 3 ) {\displaystyle N+P=n\quad (3)} 將 f ( x ) {\displaystyle f(x)} 以極座標型式表示,可得
f ( x ) = ρ ( x ) e j θ ( x ) ( 4 ) {\displaystyle f(x)=\rho (x)e^{j\theta (x)}\quad (4)} 其中
ρ ( x ) = R e 2 [ f ( x ) ] + I m 2 [ f ( x ) ] ( 5 ) {\displaystyle \rho (x)={\sqrt {{\mathfrak {Re}}^{2}[f(x)]+{\mathfrak {Im}}^{2}[f(x)]}}\quad (5)} 且
θ ( x ) = tan − 1 ( I m [ f ( x ) ] / R e [ f ( x ) ] ) ( 6 ) {\displaystyle \theta (x)=\tan ^{-1}{\big (}{\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]{\big )}\quad (6)} 根據(2)會發現
θ ( x ) = θ r 1 ( x ) + θ r 2 ( x ) + ⋯ + θ r n ( x ) ( 7 ) {\displaystyle \theta (x)=\theta _{r_{1}}(x)+\theta _{r_{2}}(x)+\cdots +\theta _{r_{n}}(x)\quad (7)} 其中
θ r i ( x ) = ∠ ( x − r i ) ( 8 ) {\displaystyle \theta _{r_{i}}(x)=\angle (x-r_{i})\quad (8)} 若 f ( x ) = 0 {\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 )
θ r i ( x ) | x = − j ∞ = ∠ ( x − r i ) | x = − j ∞ = ∠ ( 0 − R e [ r i ] , − ∞ − I m [ r i ] ) = ∠ ( − | R e [ r i ] | , − ∞ ) = π + lim ϕ → ∞ tan − 1 ϕ = 3 π 2 ( 9 ) {\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
θ r i ( x ) | x = j 0 = ∠ ( − | R e [ r i ] | , 0 ) = π − tan − 1 0 = π ( 10 ) {\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (-|{\mathfrak {Re}}[r_{i}]|,0)=\pi -\tan ^{-1}0=\pi \quad (10)} 且
θ r i ( x ) | x = j ∞ = ∠ ( − | R e [ r i ] | , ∞ ) = π − lim ϕ → ∞ tan − 1 ϕ = π 2 ( 11 ) {\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 f ( x ) = 0 {\displaystyle f(x)=0} 的第i個根的實部為負,
θ r i ( x ) | x = − j ∞ = ∠ ( x − r i ) | x = − j ∞ = ∠ ( 0 − R e [ r i ] , − ∞ − I m [ r i ] ) = ∠ ( | R e [ r i ] | , − ∞ ) = 0 − lim ϕ → ∞ tan 1 ϕ = − π 2 ( 2 ) {\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
θ r i ( x ) | x = j 0 = ∠ ( | R e [ r i ] | , 0 ) = tan − 1 0 = 0 ( 13 ) {\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (|{\mathfrak {Re}}[r_{i}]|,0)=\tan ^{-1}0=0\,\quad (13)} 且
θ r i ( x ) | x = j ∞ = ∠ ( | R e [ r i ] | , ∞ ) = lim ϕ → ∞ tan − 1 ϕ = π 2 ( 14 ) {\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 f ( x ) {\displaystyle f(x)} 的第i個根實部為正,則 θ r i ( x ) | x = − j ∞ x = j ∞ = − π {\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=-\pi } ,由(12)至(14)式可知,若 f ( x ) {\displaystyle f(x)} 的第i個根實部為負,則 θ r i ( x ) | x = − j ∞ x = j ∞ = π {\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\pi } 。因此
θ r i ( x ) | x = − j ∞ x = j ∞ = ∠ ( x − r 1 ) | x = − j ∞ x = j ∞ + ∠ ( x − r 2 ) | x = − j ∞ x = j ∞ + ⋯ + ∠ ( x − r n ) | x = − j ∞ x = j ∞ = π N − π P ( 15 ) {\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
Δ = 1 π θ ( x ) | − j ∞ j ∞ ( 16 ) {\displaystyle \Delta ={\frac {1}{\pi }}\theta (x){\Big |}_{-j\infty }^{j\infty }\quad (16)} 則可以得到以下的關係
N − P = Δ ( 17 ) {\displaystyle N-P=\Delta \quad (17)} 結合(3)式及(17)式可得
N = n + Δ 2 {\displaystyle N={\frac {n+\Delta }{2}}} 且 P = n − Δ 2 ( 18 ) {\displaystyle P={\frac {n-\Delta }{2}}\quad (18)} 因此,給定 n {\displaystyle n} 次的方程 f ( x ) {\displaystyle f(x)} ,只需要計算 Δ {\displaystyle \Delta } ,就可以得到根的實部為負的個數 N {\displaystyle N} ,以及根的實部為正的個數 P {\displaystyle P} 。
圖1 tan ( θ ) {\displaystyle \tan(\theta )} 相對 θ {\displaystyle \theta } 的圖
配合(6)式及圖1, tan ( θ ) {\displaystyle \tan(\theta )} 相對 θ {\displaystyle \theta } 的圖,將 x {\displaystyle x} 在區間(a,b)之間變化,其中 θ a = θ ( x ) | x = j a {\displaystyle \theta _{a}=\theta (x)|_{x=ja}} ,而 θ b = θ ( x ) | x = j b {\displaystyle \theta _{b}=\theta (x)|_{x=jb}} ,都是 π {\displaystyle \pi } 的整數倍,若此變化會使函數 θ ( x ) {\displaystyle \theta (x)} 增加 π {\displaystyle \pi } ,表示在從點a到點b的過程中, tan θ ( x ) = I m [ f ( x ) ] / R e [ f ( x ) ] {\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 θ ( x ) {\displaystyle \theta (x)} 減少 π {\displaystyle \pi } ,表示在從點a到點b的過程中, tan ( θ ) {\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, θ ( x ) | − j ∞ j ∞ {\displaystyle \theta (x){\Big |}_{-j\infty }^{j\infty }} 是 I m [ f ( x ) ] / R e [ f ( x ) ] {\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 π {\displaystyle \pi } 倍。假設在 x = ± j ∞ {\displaystyle x=\pm j\infty } 處, tan [ θ ( x ) ] {\displaystyle \tan[\theta (x)]} 有定義
圖2 − cot ( θ ) {\displaystyle -\cot(\theta )} 相對 θ {\displaystyle \theta } 的圖
若起始點是在不連續點( θ a = π / 2 ± i π {\displaystyle \theta _{a}=\pi /2\pm i\pi } , i = 0, 1, 2, ...),則因為公式(17)( N {\displaystyle N} 和 P {\displaystyle P} 都是整數,因此 Δ {\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 π / 2 {\displaystyle \pi /2} ,也就是在 θ {\displaystyle \theta } 上加 π / 2 {\displaystyle \pi /2} 。此時的指標函數在各種 f ( x ) {\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) = ( + j ∞ , − j ∞ ) {\displaystyle (+j\infty ,-j\infty )} 內計算 tan [ θ ] = I m [ f ( x ) ] / R e [ f ( x ) ] {\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]} ,再在起始點連續的區間,計算
tan [ θ ′ ( x ) ] = tan [ θ + π / 2 ] = − cot [ θ ( x ) ] = − R e [ f ( x ) ] / I m [ f ( x ) ] ( 19 ) {\displaystyle \tan[\theta '(x)]=\tan[\theta +\pi /2]=-\cot[\theta (x)]=-{\mathfrak {Re}}[f(x)]/{\mathfrak {Im}}[f(x)]\quad (19)} 差值 Δ {\displaystyle \Delta } 是 x {\displaystyle x} 從正跳躍和負跳躍的差值,若計算從 − j ∞ {\displaystyle -j\infty } 到 + j ∞ {\displaystyle +j\infty } 所產生的差值,即為相角正切的柯西指標 ,其相角為 θ ( x ) {\displaystyle \theta (x)} 或 θ ′ ( x ) {\displaystyle \theta '(x)} ,視 θ a {\displaystyle \theta _{a}} 是否是 π {\displaystyle \pi } 的整數倍而定。
為了要推導勞斯準則,會將 f ( x ) {\displaystyle f(x)} 的奇次方項和偶次方項分開來列:
f ( x ) = a 0 x n + b 0 x n − 1 + a 1 x n − 2 + b 1 x n − 3 + ⋯ ( 20 ) {\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
f ( j ω ) = a 0 ( j ω ) n + b 0 ( j ω ) n − 1 + a 1 ( j ω ) n − 2 + b 1 ( j ω ) n − 3 + ⋯ ( 21 ) = a 0 ( j ω ) n + a 1 ( j ω ) n − 2 + a 2 ( j ω ) n − 4 + ⋯ ( 22 ) + b 0 ( j ω ) n − 1 + b 1 ( j ω ) n − 3 + b 2 ( j ω ) n − 5 + ⋯ {\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 n {\displaystyle n} 為偶數:
f ( j ω ) = ( − 1 ) n / 2 [ a 0 ω n − a 1 ω n − 2 + a 2 ω n − 4 − ⋯ ] ( 23 ) + j ( − 1 ) ( n / 2 ) − 1 [ b 0 ω n − 1 − b 1 ω n − 3 + b 2 ω n − 5 − ⋯ ] {\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 n {\displaystyle n} 為奇數:
f ( j ω ) = j ( − 1 ) ( n − 1 ) / 2 [ a 0 ω n − a 1 ω n − 2 + a 2 ω n − 4 − ⋯ ] ( 24 ) + ( − 1 ) ( n − 1 ) / 2 [ b 0 ω n − 1 − b 1 ω n − 3 + b 2 ω n − 5 − ⋯ ] {\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 n {\displaystyle n} 為奇數,根據(3)式, N + P {\displaystyle N+P} 為奇數。若 N + P {\displaystyle N+P} 為奇數, N − P {\displaystyle N-P} 也是奇數。同樣的,若 n {\displaystyle n} 是偶數, N − P {\displaystyle N-P} 也是偶數。(15)式可以看出若 N − P {\displaystyle N-P} 是偶數, θ {\displaystyle \theta } 是 π {\displaystyle \pi } 的整數倍。因此在 n {\displaystyle n} 為偶數時, tan ( θ ) {\displaystyle \tan(\theta )} 有定義,是n為偶數時使用的正確指標,在而在 n {\displaystyle n} 為奇數時, tan ( θ ′ ) = tan ( θ + π ) = − cot ( θ ) {\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, n {\displaystyle n} 為偶數時:
Δ = I − ∞ + ∞ − I m [ f ( x ) ] R e [ f ( x ) ] = I − ∞ + ∞ b 0 ω n − 1 − b 1 ω n − 3 + ⋯ a 0 ω n − a 1 ω n − 2 + … ( 25 ) {\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, n {\displaystyle n} 為奇數時:
Δ = I − ∞ + ∞ R e [ f ( x ) ] I m [ f ( x ) ] = I − ∞ + ∞ b 0 ω n − 1 − b 1 ω n − 3 + … a 0 ω n − a 1 ω n − 2 + … ( 26 ) {\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:
Δ = I − ∞ + ∞ b 0 ω n − 1 − b 1 ω n − 3 + … a 0 ω n − a 1 ω n − 2 + … ( 27 ) {\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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