A second order control system has the settling time of 5 seconds (for $\pm 2$ ) tolerance ) and the peak time of 2 seconds. The locations of poles in the s-plane are

(- 0.8 \pm j 3.14) r a d / s e c

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(- 2 \pm j 0.628) r a d / s e c

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(- 0.8 \pm j 1.57) r a d / s e c

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(- 2 \pm j 5) r a d / s e c

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Solution

The correct option is C
$(- 0.8 \pm j 1.57) r a d / s e c$
Settling time $t_{s} = \frac{4}{ζ ω_{π}} 5 = \frac{4}{ζ w_{n}} ζ w_{n} = \frac{4}{5} = 0.8 r a d / s e c$
Peak time $t_{p} = \frac{π}{ω_{d}} 2 = \frac{π}{w_{d}} w_{d} = \frac{π}{2} = 1.57 r a d / s e c$

For second order system, the location of poles $= ζ w_{n} \pm j w_{d} = (- 0.8 \pm j 1.57) r a d / s e c$

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A second order control system has the settling time of 5 seconds (for ± 2) tolerance ) and the peak time of 2 seconds. The locations of poles in the s-plane are

The correct option is C (−0.8± j 1.57) rad/secSettling time ts=4ζωπ5=4ζ wnζwn=45=0.8 rad/sec Peak time tp=πωd2=πwdwd=π2=1.57 rad/sec For second order system, the location of poles =ζwn±jwd=(−0.8±j1.57)rad/sec

A second order control system has the settling time of 5 seconds (for $\pm 2$ ) tolerance ) and the peak time of 2 seconds. The locations of poles in the s-plane are