Instant Notes: 5. Stability Analysis
5.1 Routh Stability Analysis
For the characteristic equation (with \(a_0\) as positive) \[a_0s^n + a_1s_{n1}+\cdots+a_{n1}s + a_n = 0\] First test: If any of the coefficients of the
characteristic equation \(a_1,a_2,\ldots,a_{n1},a_n\) is negative, then there is at least one root of the characteristic equation which has a positive real part, and the corresponding system is unstable. No further
analysis is needed.
Second test: If all coefficients \(a_1,a_2,\ldots,a_{n1},a_n\) are positive, then the following analysis by using Routh array is to be made.
The elements of Routh array are written as
Row 1  \(\quad a_0\)  \(a_2\)  \(a_4\)  \(a_6\)  \(\qquad \cdots\) 
2  \(\quad a_1\)  \(a_3\)  \(a_5\)  \(a_7\)  \(\qquad \cdots\) 
3  \(\quad A_1\)  \(A_2\)  \(A_3\)  \(\cdot\)  \(\qquad \cdots\) 
4  \(\quad B_1\)  \(B_2\)  \(B_3\)  \(\cdot\)  \(\qquad \cdots\) 
5  \(\quad C_1\)  \(C_2\)  \(C_3\)  \(\cdot\)  \(\qquad \cdots\) 
\(\cdot\)  \(\quad \cdot\)  \(\cdot\)  \(\cdot\)  \(\cdot\)  \(\qquad \cdots\) 
\(\cdot\)  \(\quad \cdot\)  \(\cdot\)  \(\cdot\)  \(\cdot\)  \(\qquad \cdots\) 
\(n+1\)  \(\quad V_1\)  \(V_2\)  \(\cdot\)  \(\cdot\)  \(\qquad \cdots\) 
where
\(\displaystyle A_1=\frac{a_1a_2a_0a_3}{a_1}\)  \(\displaystyle \quad A_2=\frac{a_1a_4a_0a_5}{a_1}\)  \(\displaystyle \quad A_3=\frac{a_1a_6a_0a_7}{a_1} \quad \cdots\) 
\(\displaystyle B_1=\frac{A_1a_3a_1A_2}{A_1}\)  \(\displaystyle \quad B_2=\frac{A_1a_5a_1A_3}{A_1}\)  \(\displaystyle \cdots\) 
\(\displaystyle C_1=\frac{B_1A_2A_1B_2}{B_1}\)  \(\displaystyle \quad C_2=\frac{B_1A_3A_1B_3}{B_1}\)  \(\displaystyle \cdots\) 
etc. 
Examine the elements of the first column of the Routh array: \[a_0 \quad a_1 \quad A_1 \quad B_1 \quad C_1 \quad \ldots \quad V_1\]

If any of these elements is negative, we have atleast one root to the right of the imaginary axis and the system is unstable.

The number of sign changes in the elements of the first column is equal to the number of roots to the the right of the imaginary axis.
Therefore, a system is stable if all the elements in the first column of the Routh array are positive.
Limitations of Routh Test:

Can’t be used for deadtime systems.

Only indicates stability; doesn’t tell about location of roots.