Find the number of poles in the right-half plane (RHP) for the system as shown. Is the system stable?

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of the Routh array

The number of poles lie on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: \(1 + \frac{1}{{s\left( {2{s^4} + 3{s^3} + 2{s^2} + 3s + 2} \right)}} = 0\)

⇒ 2s⁵ + 3s⁴ + 2s³ + 3s² + 2s + 1 = 0

By applying Routh tabulation method,

\(\begin{array}{*{20}{c}} {{s^5}}\\ {{s^4}}\\ {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 2&2&2\\ 3&3&1\\ {0\left( \varepsilon \right)}&{\frac{4}{3}}&{}\\ {\left( {3 - \frac{4}{\varepsilon }} \right)}&1&{}\\ {\frac{4}{3}}&{}&{}\\ 1&{}&{} \end{array}} \right.\)

As ε is very small value, (3 – 4/ε) is a negative value and hence there are two sign changes.

Therefore, the number of right-half poles = 2

The system is unstable.