Correct Answer - Option 4 : -2 ± j1
Given:
\(G(s)=\frac{1}{(s+2)^2}\) and H(s) = 1
Closed loop transfer function is given as:
\(CLTF=\frac{G(s)}{1+G(s)H(s)}\)
\(CLTF=\frac{\frac{1}{(s+2)^2}}{1+\frac{1}{(s+2)^2}}\)
\(CLTF=\frac{1}{s^2+4s+5}\)
∴ Closed-loop poles of this CLTF is given as:
s = -2 ± j1