Question

Consider an open-loop transfer function \(G\left( s \right) = \frac{{K\left( {s + 2} \right)}}{{\left( {s + 0.5} \right)\left( {s + 4} \right)}}\) with unity feedback. For K > 6, stability characteristic of the open-loop and closed-loop configurations of the system is:
1. unstable and stable
2. unstable and unstable
3. stable unstable
4. stable and stable

Answer 1

Correct Answer - Option 4 : stable and 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 lies on the right half of s plane = number of sign changes

Calculation:

Open-loop configuration:

Transfer function \(G\left( s \right) = \frac{{K\left( {s + 2} \right)}}{{\left( {s + 0.5} \right)\left( {s + 4} \right)}}\)

Poles: s = -0.5 and -4

All the poles are on the left half of the s-plane.

Therefore, the system is stable.

Closed-loop configuration:

The poles of the closed-loop configuration are the roots of the characteristic equation.

Characteristic equation: 1 + G(s) H(s) = 0

⇒ (s + 0.5) (s + 4) + K (s + 2) = 0

⇒ s² + (4.5 + K) s + (2 + 2K) = 0

\(\begin{array}{*{20}{c}} {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&{2 + 2K}\\ {4.5 + K}&0\\ {2 + 2K}&0 \end{array}} \right.\)

For K > 6, all the elements in the first column of the Routh array have the same sign.

Therefore, the system is stable.