Stability

$H(s)$ such that,

H (s) = \frac{N (s)}{D (s)}

where

\begin{matrix} \begin{aligned} D (s) & = s^{n} + a_{n - 1} s^{n - 1} + . . . + a_{1} s + a_{0} \\ = (s + p_{1}) (s + p_{2}) . . . (s + p_{n}) \end{aligned} \end{matrix}

$D(s)$ $(p_1,p_2\:...\:p_n)$ are in the open left half plane. In order words,

R e (p_{i}) < 0, for all 0 < i \leq n

Necessary Condition for Stability

$H(s)$ $D(s)$ $(a_0,a_1\:...\:a_{n-1})$ are positive. In other words, if

H (s) = \frac{N (s)}{D (s)}

and

\begin{matrix} \begin{aligned} D (s) & = s^{n} + a_{n - 1} s^{n - 1} + . . . + a_{1} s + a_{0} \\ = (s + p_{1}) (s + p_{2}) . . . (s + p_{n}) \end{aligned} \end{matrix}

Then,

a_{i} > 0 for all 0 < i \leq n ⟹ the system is stable

Routh-Hurwitz Test

$D(s)$ such that,

D (s) = a_{n} s^{n} + a_{n - 1} s^{n - 1} + . . . + a_{1} s + a_{0}

We can construct a Routh array:

$\Large s^n$	$\Large a^n$	$\Large a^{n - 2}$	$\Large a^{n - 4}$
$\Large s_{n - 1}$	$\Large a_{n - 1}$	$\Large a_{n - 3}$	$\Large a_{n - 5}$
$\Large s_{n - 2}$	$\Large b_1$	$\Large b_2$	$\Large b_3$
$\Large s_{n - 3}$	$\Large c_1$	$\Large c_2$	$\Large c_3$
...	...	...	...
$\Large s^2$	...	...	...
$\Large s$	...	...	...
$\Large 1$	...	...	...

where,

\begin{matrix} b_{i} = \frac{1}{a_{n - 1}} (a_{n - 1} a_{n - i - 1} - a_{n} a_{n - 2 i - 1}) \\ c_{i} = \frac{1}{b_{1}} (b_{1} a_{n - i - 2} - a_{n - 1} b_{i + 1}) \end{matrix}

The number of sign changes in the second column of the table indicates the number of roots that are in the open left hand plane. All roots are in the open left hand plane if and only if there are no sign changes in the second column (i.e. either all entries are positive, or all entries are negative). In other words, the system described above is stable if and only if

\begin{matrix} a_{n} > 0, a_{n - 1} > 0, b_{1} > 0, c_{1} > 0, . . . \\ OR \\ a_{n} < 0, a_{n - 1} < 0, b_{1} < 0, c_{1} < 0, . . . \end{matrix}