Stability and natural responses.

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

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

Given

\frac{N(s)}{a_4s^4 + a_3s^3 + a_2s^2 + a_1s + a_0}

The characteristic equation is $a_4s^4 + a_3s^3 + a_2s^2 + a_1s + a_0 = 0$

Create a basic Routh table

\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

Étiquette