Question

For a second order system settling time is T_s = 7 s and peak time is T_p = 3 s. The location of poles are
1. – 0.97 ± j0.69
2. – 0.69 ± j0.97
3. – 1.047 ± j0.571
4. – 0.571 ± j1.047

Answer 1

Correct Answer - Option 4 : – 0.571 ± j1.047
\(\begin{array}{l} \zeta {\omega _n} = \frac{4}{{{T_s}}} = 0.571\\ {\omega _n}\sqrt {1 - {\zeta ^2}} = \frac{\pi }{{{T_p}}} = 1.047\\ poles = - 0.571 \pm j1.047 \end{array}\)