勞斯陣列是勞斯–赫爾維茨穩定性判據 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 歐幾里得算法 和施圖姆定理 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}
N
{\displaystyle N}
f
(
x
)
=
0
{\displaystyle f(x)=0}
P
{\displaystyle P}
f
(
x
)
=
0
{\displaystyle f(x)=0}
因此可得
N
+
P
=
n
(
3
)
{\displaystyle N+P=n\quad (3)}
將
f
(
x
)
{\displaystyle f(x)}
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}
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}
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)}
θ
r
i
(
x
)
|
x
=
−
j
∞
x
=
j
∞
=
−
π
{\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=-\pi }
f
(
x
)
{\displaystyle f(x)}
θ
r
i
(
x
)
|
x
=
−
j
∞
x
=
j
∞
=
π
{\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\pi }
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}}}
P
=
n
−
Δ
2
(
18
)
{\displaystyle P={\frac {n-\Delta }{2}}\quad (18)}
因此，給定
n
{\displaystyle n}
f
(
x
)
{\displaystyle f(x)}
Δ
{\displaystyle \Delta }
N
{\displaystyle N}
P
{\displaystyle P}
圖1
tan
(
θ
)
{\displaystyle \tan(\theta )}
θ
{\displaystyle \theta }
tan
(
θ
)
{\displaystyle \tan(\theta )}
θ
{\displaystyle \theta }
x
{\displaystyle x}
θ
a
=
θ
(
x
)
|
x
=
j
a
{\displaystyle \theta _{a}=\theta (x)|_{x=ja}}
θ
b
=
θ
(
x
)
|
x
=
j
b
{\displaystyle \theta _{b}=\theta (x)|_{x=jb}}
π
{\displaystyle \pi }
θ
(
x
)
{\displaystyle \theta (x)}
π
{\displaystyle \pi }
tan
θ
(
x
)
=
I
m
[
f
(
x
)
]
/
R
e
[
f
(
x
)
]
{\displaystyle \tan \theta (x)={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}
+
∞
{\displaystyle +\infty }
−
∞
{\displaystyle -\infty }
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
θ
(
x
)
{\displaystyle \theta (x)}
π
{\displaystyle \pi }
tan
(
θ
)
{\displaystyle \tan(\theta )}
+
∞
{\displaystyle +\infty }
−
∞
{\displaystyle -\infty }
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
因此，
θ
(
x
)
|
−
j
∞
j
∞
{\displaystyle \theta (x){\Big |}_{-j\infty }^{j\infty }}
I
m
[
f
(
x
)
]
/
R
e
[
f
(
x
)
]
{\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}
−
∞
{\displaystyle -\infty }
+
∞
{\displaystyle +\infty }
+
∞
{\displaystyle +\infty }
−
∞
{\displaystyle -\infty }
π
{\displaystyle \pi }
x
=
±
j
∞
{\displaystyle x=\pm j\infty }
tan
[
θ
(
x
)
]
{\displaystyle \tan[\theta (x)]}
圖2
−
cot
(
θ
)
{\displaystyle -\cot(\theta )}
θ
{\displaystyle \theta }
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}
P
{\displaystyle P}
Δ
{\displaystyle \Delta }
π
/
2
{\displaystyle \pi /2}
θ
{\displaystyle \theta }
π
/
2
{\displaystyle \pi /2}
f
(
x
)
{\displaystyle f(x)}
(
+
j
∞
,
−
j
∞
)
{\displaystyle (+j\infty ,-j\infty )}
tan
[
θ
]
=
I
m
[
f
(
x
)
]
/
R
e
[
f
(
x
)
]
{\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}
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 }
x
{\displaystyle x}
−
j
∞
{\displaystyle -j\infty }
+
j
∞
{\displaystyle +j\infty }
柯西指標
θ
(
x
)
{\displaystyle \theta (x)}
θ
′
(
x
)
{\displaystyle \theta '(x)}
θ
a
{\displaystyle \theta _{a}}
π
{\displaystyle \pi }
為了要推導勞斯準則，會將
f
(
x
)
{\displaystyle f(x)}
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}
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}
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}
N
+
P
{\displaystyle N+P}
N
+
P
{\displaystyle N+P}
N
−
P
{\displaystyle N-P}
n
{\displaystyle n}
N
−
P
{\displaystyle N-P}
N
−
P
{\displaystyle N-P}
θ
{\displaystyle \theta }
π
{\displaystyle \pi }
n
{\displaystyle n}
tan
(
θ
)
{\displaystyle \tan(\theta )}
n
{\displaystyle n}
tan
(
θ
′
)
=
tan
(
θ
+
π
)
=
−
cot
(
θ
)
{\displaystyle \tan(\theta ')=\tan(\theta +\pi )=-\cot(\theta )}
因此，根據(6)式及(23)式，
n
{\displaystyle n}
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}
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)}
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)}
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. 
![{\displaystyle \rho (x)={\sqrt {{\mathfrak {Re}}^{2}[f(x)]+{\mathfrak {Im}}^{2}[f(x)]}}\quad (5)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3342273646dc0f1235561b387b7af52e776938a)
![{\displaystyle \theta (x)=\tan ^{-1}{\big (}{\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]{\big )}\quad (6)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7b2086266668256c2065a93ae151bb2aba594d0)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b5b7395910dafa111cef24840a04b6f678dde27)
![{\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (-|{\mathfrak {Re}}[r_{i}]|,0)=\pi -\tan ^{-1}0=\pi \quad (10)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8a52bccf4aeb16d032d5a102c1ae9bc8b6bcd1f)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/838372f37cf13c3c184a2477039b676104cb7a5f)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de58d21c449a38a7a60691d0083d7d1d5a467d67)
![{\displaystyle \theta _{r_{i}}(x){\big |}_{x=j0}=\angle (|{\mathfrak {Re}}[r_{i}]|,0)=\tan ^{-1}0=0\,\quad (13)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0af975bb9b398ec5e8cf3b476787d1db575c52d)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec04aab88fc87619bb9b6771313b6b3e63ccb771)
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![{\displaystyle \tan \theta (x)={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/517bdc0afc4cdf73846d5681bfd6527a3189b43f)
![{\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f620a737220c91bbf20684bf6b26abd7231b22c)
![{\displaystyle \tan[\theta (x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1add4cb96e2e9fde725301edaec88e5603cbf8ef)
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![{\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ac74db804df37ddb6ef8ff4d6f721bd086850a8)
![{\displaystyle \tan[\theta '(x)]=\tan[\theta +\pi /2]=-\cot[\theta (x)]=-{\mathfrak {Re}}[f(x)]/{\mathfrak {Im}}[f(x)]\quad (19)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e718d1233f9c08d6cb4344dae637f97bc9804db)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e495902015ba0b7f57501e6d3d82ee9ab5ef62a)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2031e504dd4c89445e158080035a39a2e0e17c2b)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9829781584a0ac5eac8c828f3b075583bb496da)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d8786a9be2f8a0f8cff61b436453fa922966c5d)
