Example

• Let ${\displaystyle f(z)=az^{2}+bz+c}$ (for the sake of simplicity we take real coefficients) where ${\displaystyle c\neq 0}$ (to avoid a root in zero so that we can use the Routh–Hurwitz theorem). First, we have to calculate the real polynomials ${\displaystyle P_{0}(y)}$ and ${\displaystyle P_{1}(y)}$:

${\displaystyle f(iy)=-ay^{2}+iby+c=P_{0}(y)+iP_{1}(y)=-ay^{2}+c+i(by).}$

Next, we divide those polynomials to obtain the generalized Sturm chain:

• ${\displaystyle P_{0}(y)=((-a/b)y)P_{1}(y)+c,}$ yields ${\displaystyle P_{2}(y)=-c,}$
• ${\displaystyle P_{1}(y)=((-b/c)y)P_{2}(y),}$ yields ${\displaystyle P_{3}(y)=0}$ and the Euclidean division stops.

Notice that we had to suppose b different from zero in the first division. The generalized Sturm chain is in this case ${\displaystyle (P_{0}(y),P_{1}(y),P_{2}(y))=(c-ay^{2},by,-c)}$. Putting ${\displaystyle y=+\infty }$, the sign of ${\displaystyle (c-ay^{2})}$ is the opposite sign of a and the sign of by is the sign of b. When we put ${\displaystyle (y=-\infty )}$, the sign of the first element of the chain is again the opposite sign of a and the sign of by is the opposite sign of b. Finally, -c has always the opposite sign of c.

Suppose now that f is Hurwitz-stable. This means that ${\displaystyle w(+\infty )-w(-\infty )=2}$ (the degree of f). By the properties of the function w, this is the same as ${\displaystyle w(+\infty )=2}$ and ${\displaystyle w(-\infty )=0}$. Thus, a, b and c must have the same sign. We have thus found the necessary condition of stability for polynomials of degree 2.

Routh–Hurwitz criterion for second, third and fourth-order polynomials

• The second-degree polynomial ${\displaystyle P(s)=s^{2}+a_{1}s+a_{0}}$ has both roots with negative real part (and the system with characteristic equation ${\displaystyle P(s)=0}$ is stable) if and only if both coefficients satisfy ${\displaystyle a_{i}>0}$.
• The third-order polynomial ${\displaystyle P(s)=s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}$ has all roots in the open left half-plane if and only if ${\displaystyle a_{2}}$, ${\displaystyle a_{1}}$, and ${\displaystyle a_{0}}$ are positive and ${\displaystyle a_{2}a_{1}>a_{0}.}$
• For a fourth-order polynomial ${\displaystyle P(s)=s^{4}+a_{3}s^{3}+a_{2}s^{2}+a_{1}s+a_{0}=0}$, all the coefficients must satisfy ${\displaystyle a_{i}>0}$, and ${\displaystyle a_{3}a_{2}a_{1}>a_{1}^{2}+a_{3}^{2}a_{0}.}$^[4] (When this is derived you do not know all coefficients should be positive, and you add ${\displaystyle a_{3}a_{2}>a_{1}}$.)
• In general the Routh stability criterion states a polynomial has all roots in the open left half-plane if and only if all first-column elements of the Routh array have the same sign.
• All coefficients being positive (or all negative) is necessary for all roots to be located in the open left half-plane. That is why here ${\displaystyle a_{n}}$ is fixed to 1, which is positive. When this is assumed, we can remove ${\displaystyle a_{3}a_{2}>a_{1}}$ from fourth-order polynomial, and conditions for fifth- and sixth-order can be simplified. For fifth-order we only need to check that ${\displaystyle \Delta _{2}>0,\Delta _{4}>0}$ and for sixth-order we only need to check ${\displaystyle \Delta _{3}>0,\Delta _{5}>0}$ and this is further optimised in Liénard–Chipart criterion.^[5] Indeed, some coefficients being positive is not independent with principal minors being positive, like ${\displaystyle a_{2}>0}$ check can be removed for third-order polynomial.

Higher-order example

A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an nth-degree polynomial

• ${\displaystyle D(s)=a_{n}s^{n}+a_{n-1}s^{n-1}+\cdots +a_{1}s+a_{0}}$

the table has n + 1 rows and the following structure:

${\displaystyle a_{n}}$	${\displaystyle a_{n-2}}$	${\displaystyle a_{n-4}}$	${\displaystyle \dots }$
${\displaystyle a_{n-1}}$	${\displaystyle a_{n-3}}$	${\displaystyle a_{n-5}}$	${\displaystyle \dots }$
${\displaystyle b_{1}}$	${\displaystyle b_{2}}$	${\displaystyle b_{3}}$	${\displaystyle \dots }$
${\displaystyle c_{1}}$	${\displaystyle c_{2}}$	${\displaystyle c_{3}}$	${\displaystyle \dots }$
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \ddots }$

where the elements ${\displaystyle b_{i}}$ and ${\displaystyle c_{i}}$ can be computed as follows:

• ${\displaystyle b_{i}={\frac {a_{n-1}\times {a_{n-2i}}-a_{n}\times {a_{n-(2i+1)}}}{a_{n-1}}}.}$
• ${\displaystyle c_{i}={\frac {b_{1}\times {a_{n-(2i+1)}}-a_{n-1}\times {b_{i+1}}}{b_{1}}}.}$

When completed, the number of sign changes in the first column will be the number of non-negative roots.

In the first column, there are two sign changes (0.75 → −3, and −3 → 3), thus there are two non-negative roots where the system is unstable.

The characteristic equation of a servo system is given by:^[6]

• ${\displaystyle b_{0}s^{4}+b_{1}s^{3}+b_{2}s^{2}+b_{3}s+b_{4}=0}$

${\displaystyle b_{0}}$	${\displaystyle b_{2}}$	${\displaystyle b_{4}}$	0
${\displaystyle b_{1}}$	${\displaystyle b_{3}}$	0	0
${\displaystyle b_{1}b_{2}-b_{0}b_{3} \over b_{1}}$	${\displaystyle b_{1}b_{4}-b_{0}\times 0 \over b_{1}}$=${\displaystyle b_{4}}$	0	0
${\displaystyle (b_{1}b_{2}-b_{0}b_{3})b_{3}-b_{1}^{2}b_{4} \over b_{1}b_{2}-b_{0}b_{3}}$	0	0	0
${\displaystyle b_{4}}$	0	0	0

for stability, all the elements in the first column of the Routh array must be positive. So the conditions that must be satisfied for stability of the given system as follows:^[6]

${\displaystyle b_{1}>0,b_{1}b_{2}-b_{0}b_{3}>0,(b_{1}b_{2}-b_{0}b_{3})b_{3}-b_{1}^{2}b_{4}>0,b_{4}>0}$^[6]

We see that if

${\displaystyle (b_{1}b_{2}-b_{0}b_{3})b_{3}-b_{1}^{2}b_{4}\geq 0}$

then

${\displaystyle b_{1}b_{2}-b_{0}b_{3}>0}$

Is satisfied.

• ${\displaystyle s^{4}+6s^{3}+11s^{2}+6s+200=0}$^[7]

We have the following table :

there are two sign changes. The system is unstable, since it has two right-half-plane poles and two left-half-plane poles. The system cannot have jω poles since a row of zeros did not appear in the Routh table.^[7]

Sometimes the presence of poles on the imaginary axis creates a situation of marginal stability. In that case the coefficients of the "Routh array" in a whole row become zero and thus further solution of the polynomial for finding changes in sign is not possible. Then another approach comes into play. The row of polynomial which is just above the row containing the zeroes is called the "auxiliary polynomial".

• ${\displaystyle s^{6}+2s^{5}+8s^{4}+12s^{3}+20s^{2}+16s+16=0.\,}$

We have the following table:

In such a case the auxiliary polynomial is ${\displaystyle A(s)=2s^{4}+12s^{2}+16\,}$ which is again equal to zero. The next step is to differentiate the above equation which yields the polynomial ${\displaystyle B(s)=8s^{3}+24s^{1}}$. The coefficients of the row containing zero now become "8" and "24". The process of Routh array is proceeded using these values which yield two points on the imaginary axis. These two points on the imaginary axis are the prime cause of marginal stability.^[8]