The Routh array is a tabular method permitting one to establish the stability of a system using only the coefficients of the characteristic polynomial. Central to the field of control systems design, the Routh–Hurwitz theorem and Routh array emerge by using the Euclidean algorithm and Sturm's theorem in evaluating Cauchy indices.
The Cauchy index Given the system:
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}}}
Assuming no roots of f ( x ) = 0 {\displaystyle f(x)=0} lie on the imaginary axis, and letting
N {\displaystyle N} = The number of roots of f ( x ) = 0 {\displaystyle f(x)=0} with negative real parts, and
P {\displaystyle P} = The number of roots of f ( x ) = 0 {\displaystyle f(x)=0} with positive real parts then we have
N + P = n ( 3 ) {\displaystyle N+P=n\quad (3)}
Expressing f ( x ) {\displaystyle f(x)} in polar form, we have
f ( x ) = ρ ( x ) e j θ ( x ) ( 4 ) {\displaystyle f(x)=\rho (x)e^{j\theta (x)}\quad (4)}
where
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)}
and
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)}
from (2) note that
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)}
where
θ r i ( x ) = ∠ ( x − r i ) ( 8 ) {\displaystyle \theta _{r_{i}}(x)=\angle (x-r_{i})\quad (8)}
Now if the ith root of f ( x ) = 0 {\displaystyle f(x)=0} has a positive real part, then (using the notation y=(RE[y],IM[y]))
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}}}
and
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)}
and
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)}
Similarly, if the ith root of f ( x ) = 0 {\displaystyle f(x)=0} has a negative real part,
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 ( 12 ) {\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 (12)\\\end{aligned}}}
and
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)}
and
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)}
From (9) to (11) we find that θ r i ( x ) | x = − j ∞ x = j ∞ = − π {\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=-\pi } when the ith root of f ( x ) {\displaystyle f(x)} has a positive real part, and from (12) to (14) we find that θ r i ( x ) | x = − j ∞ x = j ∞ = π {\displaystyle \theta _{r_{i}}(x){\Big |}_{x=-j\infty }^{x=j\infty }=\pi } when the ith root of f ( x ) {\displaystyle f(x)} has a negative real part. Thus,
θ ( 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 (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)}
So, if we define
Δ = 1 π θ ( x ) | − j ∞ j ∞ ( 16 ) {\displaystyle \Delta ={\frac {1}{\pi }}\theta (x){\Big |}_{-j\infty }^{j\infty }\quad (16)}
then we have the relationship
N − P = Δ ( 17 ) {\displaystyle N-P=\Delta \quad (17)}
and combining (3) and (17) gives us
N = n + Δ 2 {\displaystyle N={\frac {n+\Delta }{2}}} and P = n − Δ 2 ( 18 ) {\displaystyle P={\frac {n-\Delta }{2}}\quad (18)}
Therefore, given an equation of f ( x ) {\displaystyle f(x)} of degree n {\displaystyle n} we need only evaluate this function Δ {\displaystyle \Delta } to determine N {\displaystyle N} , the number of roots with negative real parts and P {\displaystyle P} , the number of roots with positive real parts.
In accordance with (6) and Figure 1, the graph of tan ( θ ) {\displaystyle \tan(\theta )} vs θ {\displaystyle \theta } , varying x {\displaystyle x} over an interval (a,b) where θ a = θ ( x ) | x = j a {\displaystyle \theta _{a}=\theta (x)|_{x=ja}} and θ b = θ ( x ) | x = j b {\displaystyle \theta _{b}=\theta (x)|_{x=jb}} are integer multiples of π {\displaystyle \pi } , this variation causing the function θ ( x ) {\displaystyle \theta (x)} to have increased by π {\displaystyle \pi } , indicates that in the course of travelling from point a to point b, θ {\displaystyle \theta } has "jumped" from + ∞ {\displaystyle +\infty } to − ∞ {\displaystyle -\infty } one more time than it has jumped from − ∞ {\displaystyle -\infty } to + ∞ {\displaystyle +\infty } . Similarly, if we vary x {\displaystyle x} over an interval (a,b) this variation causing θ ( x ) {\displaystyle \theta (x)} to have decreased by π {\displaystyle \pi } , where again θ {\displaystyle \theta } is a multiple of π {\displaystyle \pi } at both x = j a {\displaystyle x=ja} and x = j b {\displaystyle x=jb} , implies that tan θ ( x ) = I m [ f ( x ) ] / R e [ f ( x ) ] {\displaystyle \tan \theta (x)={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]} has jumped from − ∞ {\displaystyle -\infty } to + ∞ {\displaystyle +\infty } one more time than it has jumped from + ∞ {\displaystyle +\infty } to − ∞ {\displaystyle -\infty } as x {\displaystyle x} was varied over the said interval. Thus, θ ( x ) | − j ∞ j ∞ {\displaystyle \theta (x){\Big |}_{-j\infty }^{j\infty }} is π {\displaystyle \pi } times the difference between the number of points at which I m [ f ( x ) ] / R e [ f ( x ) ] {\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]} jumps from − ∞ {\displaystyle -\infty } to + ∞ {\displaystyle +\infty } and the number of points at which I m [ f ( x ) ] / R e [ f ( x ) ] {\displaystyle {\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]} jumps from + ∞ {\displaystyle +\infty } to − ∞ {\displaystyle -\infty } as x {\displaystyle x} ranges over the interval ( − j ∞ , + j ∞ ) {\displaystyle (-j\infty ,+j\infty \,)} provided that at x = ± j ∞ {\displaystyle x=\pm j\infty } , tan [ θ ( x ) ] {\displaystyle \tan[\theta (x)]} is defined.
In the case where the starting point is on an incongruity (i.e. θ a = π / 2 ± i π {\displaystyle \theta _{a}=\pi /2\pm i\pi } , i = 0, 1, 2, ...) the ending point will be on an incongruity as well, by equation (17) (since N {\displaystyle N} is an integer and P {\displaystyle P} is an integer, Δ {\displaystyle \Delta } will be an integer). In this case, we can achieve this same index (difference in positive and negative jumps) by shifting the axes of the tangent function by π / 2 {\displaystyle \pi /2} , through adding π / 2 {\displaystyle \pi /2} to θ {\displaystyle \theta } . Thus, our index is now fully defined for any combination of coefficients in f ( x ) {\displaystyle f(x)} by evaluating tan [ θ ] = I m [ f ( x ) ] / R e [ f ( x ) ] {\displaystyle \tan[\theta ]={\mathfrak {Im}}[f(x)]/{\mathfrak {Re}}[f(x)]} over the interval (a,b) = ( + j ∞ , − j ∞ ) {\displaystyle (+j\infty ,-j\infty )} when our starting (and thus ending) point is not an incongruity, and by evaluating
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)}
over said interval when our starting point is at an incongruity. This difference, Δ {\displaystyle \Delta } , of negative and positive jumping incongruities encountered while traversing x {\displaystyle x} from − j ∞ {\displaystyle -j\infty } to + j ∞ {\displaystyle +j\infty } is called the Cauchy Index of the tangent of the phase angle, the phase angle being θ ( x ) {\displaystyle \theta (x)} or θ ′ ( x ) {\displaystyle \theta '(x)} , depending as θ a {\displaystyle \theta _{a}} is an integer multiple of π {\displaystyle \pi } or not.
The Routh criterion To derive Routh's criterion, first we'll use a different notation to differentiate between the even and odd terms of 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)}
Now we have:
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}}}
Therefore, if n {\displaystyle n} is even,
f ( j ω ) = ( − 1 ) n / 2 [ a 0 ω n − a 1 ω