# Stability using Routh Stability Criterion

Determine the stability of the system whose characteristics equation is: .

# Solution

All coefficients are positive and non-zero; therefore, the necessary condition for stability is satisfied. Let’s write the Routh array: At this stage, we see that the top row corresponding to can be divided by two to make the calculation a little bit easier. So, we go ahead and divide that row by two:

*** QuickLaTeX cannot compile formula:
\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

*** Error message:
File ended while scanning use of \end .
Emergency stop.



Let’s continue writing the Routh table:   *** QuickLaTeX cannot compile formula:
\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 $\frac{3 \times 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}}\end{array} *** Error message: Missing$ inserted.
leading text: ...mes 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}}
Missing $inserted. leading text: ...mes 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}} Missing$ inserted.
leading text: ...mes 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}}
Extra }, or forgotten $. leading text: ...mes 2 -0 \times -\frac{1}{2}}{-\frac{1}{2}} Missing } inserted. leading text: ...imes -\frac{1}{2}}{-\frac{1}{2}}\end{array} Extra }, or forgotten$.
Missing } inserted.
Extra }, or forgotten $. leading text: ...imes -\frac{1}{2}}{-\frac{1}{2}}\end{array} Missing } inserted. leading text: ...imes -\frac{1}{2}}{-\frac{1}{2}}\end{array} Extra }, or forgotten$. 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.

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1. 1 Comment

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