See also slides

Stable if natural response tend to zero as tt \to \infty.

BIBO stability (bounded-input, bounded-output)

A system is BIBO stable if the output is bounded for any bounded input.

Stability and Poles

stable if all poles are strictly in the left side of the complex plane.

unstable if any pole is in the right side of the complex plane.

marginally stable e if no pole is on the right hand side, and its poles on the imaginary axis are of multiplicity one

Necessary and sufficient condition for stability

to have all roots in open left hand plane is to have all coefficients of polynomial to be present and have same sign.

Given

N(s)a4s4+a3s3+a2s2+a1s+a0\frac{N(s)}{a_4s^4 + a_3s^3 + a_2s^2 + a_1s + a_0}

The characteristic equation is a4s4+a3s3+a2s2+a1s+a0=0a_4s^4 + a_3s^3 + a_2s^2 + a_1s + a_0 = 0

Create a basic Routh table

s4a4a2a0s3a3a10s2a4a2a3a1a3=b1a4a0a30a3=b2a40a30a3=0s1a3a1b1b2b1=c1a30b10b1=0a30b10b1=0s0b1b2c10c1=d1b10c10c1=0b10c10c1=0\begin{array}{c|c|c|c} s^4 & a_4 & a_2 & a_0 \\ \hline s^3 & a_3 & a_1 & 0 \\ \hline s^2 & \frac{\begin{vmatrix} -a_4 & a_2 \\ -a_3 & a_1 \\ \end{vmatrix}}{a_{3}} = b_1 & \frac{\begin{vmatrix} -a_4 & a_0 \\ -a_3 & 0 \\ \end{vmatrix}}{a_{3}} = b_2 & \frac{\begin{vmatrix} -a_4 & 0 \\ -a_3 & 0 \\ \end{vmatrix}}{a_{3}} = 0 \\ \hline s^1 & \frac{\begin{vmatrix} -a_3 & a_1 \\ b_1 & b_2 \\ \end{vmatrix}}{b_{1}} = c_1 & \frac{\begin{vmatrix} -a_3 & 0 \\ b_1 & 0 \\ \end{vmatrix}}{b_{1}} = 0 & \frac{\begin{vmatrix} -a_3 & 0 \\ b_1 & 0 \\ \end{vmatrix}}{b_{1}} = 0 \\ \hline s^0 & \frac{\begin{vmatrix} -b_1 & b_2 \\ c_1 & 0 \\ \end{vmatrix}}{c_1} = d_1 & \frac{\begin{vmatrix} -b_1 & 0 \\ c_1 & 0 \\ \end{vmatrix}}{c_{1}} = 0 & \frac{\begin{vmatrix} -b_1 & 0 \\ c_1 & 0 \\ \end{vmatrix}}{c_{1}} = 0 \\ \end{array}

states that the number of poles in the right half plane is equal to the number of sign changes in the first coefficient column of the table

stability

System is deemed Stable if there are no sign changes in the first column