# Stability using Routh Stability Criterion

Determine the stability of the system whose characteristics equation is:

$a(s) = 2s^5 + 3s^4 + 2s^3 + s^2 + 2s + 2$.

# Solution

All coefficients are positive and non-zero; therefore, the necessary condition for stability is satisfied. Let's write the Routh array:

$\begin{array}{l | c c c} s^5 & 2 & 2 & 2 \\ s^4 & 3 & 1 & 2 \\ s^3 & & & \\ s^2 & & & \\ s^1 & & & \\ s^0 & & &\end{array}$

At this stage, we see that the top row corresponding to $S^5$ can be divided by two to make the calculation a little bit easier. So, we go ahead and divide that row by two:

$\begin{array}{l | c c c} s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\\ s^2 &\\s^1 &\\s^0 \end{array}$

Let's continue writing the Routh table:

$\begin{array}{l | c c c}s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\frac{3 \times 1 - 1 \times 1}{3} &\frac{3 \times 1 - 2 \times 1}{3} & \\ s^2 &\\s^1 &\\s^0 \end{array}$

$\begin{array}{l | c c c}s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\frac{2}{3} &\frac{1}{3} & \\ s^2 &\frac{\frac{2}{3} \times 1 - \frac{1}{3} \times 3}{\frac{2}{3}} & \frac{\frac{2}{3} \times 2 - 0 \times 3}{\frac{2}{3}} & \\s^1 &\\s^0 \end{array}$

$\begin{array}{l | c c c}s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\frac{2}{3} &\frac{1}{3} & \\ s^2 &-\frac{1}{2} & 2 & \\s^1 &\frac{-\frac{1}{2} \times \frac{1}{3} - 2 \times \frac{2}{3}}{-\frac{1}{2}} &\\s^0 \end{array}$

$\begin{array}{l | c c c}s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\frac{2}{3} &\frac{1}{3} & \\ s^2 &-\frac{1}{2} & 2 & \\s^1 &3 & \\s^0 \frac{3 \times 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}}\end{array}$

$\begin{array}{l | c c c}s^5 &1 & 1 & 1 \\ s^4 &3 & 1 & 2 \\ s^3 &\frac{2}{3} &\frac{1}{3} & \\ s^2 &-\frac{1}{2} & 2 & \\s^1 &3 & \\s^0 & 2 \end{array}$

Since there are two sign changes in the first column, the characteristic equation has two roots with negative real parts. Therefore, the system is unstable.

## One thought on “Stability using Routh Stability Criterion”

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