Routh–Hurwitz formulas

For a causal LTI system with rational, proper $T(s)$ (numerator degree $\le$ denominator degree), with poles counted after cancelling common factors. If $T(s)$ is proper but not strictly proper, $h(t)$ also contains an impulse $D\delta (t)$, which does not affect BIBO stability. An improper $T(s)$, such as the pure differentiator $T(s)=s$, is never BIBO stable. A single pole on the $j\omega$-axis, even a simple one, is enough to make the system not BIBO stable.

Here $G=N_G/D_G$ and $H=N_H/D_H$, written without cancelling any factor. The roots are the closed-loop poles. For positive feedback, $T(s)=\frac{G(s)}{1-G(s)H(s)}$ and the characteristic equation is $1-G(s)H(s)=0$, that is $D_G{D}_H-N_G{N}_H=0$.

Necessary for every order $n$. It is necessary and sufficient only for $n=1$ and $n=2$. A missing term or a sign change proves that not every root is in the LHP, but it does not tell you how many roots are in the RHP or whether some lie on the $j\omega$-axis.

General rule: with rows $i-2$ and $i-1$ directly above row $i$, the entry in column $j$ is $x_{i,j}=\frac{x_{i-1,1}{x}_{i-2,j+1}-x_{i-2,1}{x}_{i-1,j+1}}{x_{i-1,1}}$. The divisor is always the first-column entry of the row directly above. Take any missing entry at the right-hand end as 0. In an array with no special case, the single $s^0$ entry equals $a_n$ (or a positive multiple of it if you rescaled a row), which is a quick arithmetic check.

Applies when no first-column entry is zero. With $a_0>0$, the system is stable if and only if every first-column entry is positive. If the array is completed with no zero in the first column, no root lies on the $j\omega$-axis, so all the remaining roots are in the LHP.

Assumes $a_0>0$. This comes from the $s^1$ entry $(a_1{a}_2-a_0{a}_3)/a_1$. If $a_1{a}_2=a_0{a}_3$ with every coefficient positive, the polynomial factors as $(a_0{s}^2+a_2)\left(s+\frac{a_1}{a_0}\right)$. That gives a pair of roots at $s=\pm j\sqrt{a_2/a_0}$ (equivalently $\pm j\sqrt{a_3/a_1}$) and a third root at $-a_1/a_0$, which is marginal stability.

Assumes $a_0>0$. This comes from requiring the $s^2$ entry $(a_1{a}_2-a_0{a}_3)/a_1$ and the $s^1$ entry to be positive. When every coefficient is positive, this condition already implies $a_1{a}_2>a_0{a}_3$.

Assumes $a_0>0$. Row $i$, column $j$ of $H$ is $a_{2j-i}$, with $a_k=0$ for $k<0$ or $k>n$. ${\Delta }_i$ is the determinant of the top-left $i\times i$ block, so ${\Delta }_1=a_1$, ${\Delta }_2=a_1{a}_2-a_0{a}_3$ and ${\Delta }_n=a_n{\Delta }_{n-1}$. When no ${\Delta }_i$ is zero, the Routh first column is $a_0,{\Delta }_1,\frac{{\Delta }_2}{{\Delta }_1},\frac{{\Delta }_3}{{\Delta }_2},\dots ,\frac{{\Delta }_n}{{\Delta }_{n-1}}$, so the two tests always agree.

$c_1,c_2,\dots$ are the entries of row $s^k$, directly above the zero row (a positive multiple of that row works equally well). $A(s)$ divides the characteristic polynomial exactly.

$k$ is the degree of $A(s)$ and $p$ is the number of sign changes from the $A(s)$ row down to the $s^0$ row. If $p=0$, all $k$ roots of $A(s)$ lie on the $j\omega$-axis. If a second row of zeros appears lower down, $A(s)$ has repeated roots; see the last tip in this section.

$\alpha$ and $\beta$ are the $s^2$-row entries evaluated at the $K$ that makes the $s^1$ row zero. Both must be positive for the roots to lie on the $j\omega$-axis. If the limit is set by the $s^0$ entry going to zero instead, the boundary root is at $s=0$ and there is no oscillation.

Unity negative feedback, real $a,b,c\ge 0$. Put $c=0$ for $G(s)=\frac{K}{s(s+a)(s+b)}$, which gives $K_{\text{mar}}=ab(a+b)$ and $\omega =\sqrt{ab}$. The characteristic equation is $s^3+(a+b+c)s^2+(ab+bc+ca)s+abc+K=0$. The third-order shortcut, with $(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)$, gives stability for $-abc<K<(a+b)(b+c)(c+a)$. At $K_{\text{mar}}$ the third root is $s=-(a+b+c)$. For any other form of $G(s)$, build the array.

Substitute into the characteristic polynomial, expand in $z$ and apply Routh. The count holds when no special case arises. No sign change and no zero in the first column means every root lies strictly to the left of the line $s=-\sigma$; a row of zeros means some roots lie on the line.