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Exam Questions Papers - Engineering

Q1:

Find the value of K and velocity constant Kv so that the maximum overshoot in the unit step response is 0.2 and the peak time is 1 sec.

A 12.37, 0.175

B 12.86, 2.175

C 11.36, 1.175

D 1.86, 0.175

ANS:A - 12.37, 0.175

ζ2 = 0.259(1 - ζ2) = 0.259 - 0.259ζ2 1.259ζ2 = 0.259 ζ2 = 0.206 ζ = 0.45 ω2n = K ∴ ωn = K and 2ζωn = 1 + KKv ∴ ωn = K K = ω2n K = 12.37 1 + 12.37Kv = 2 x 0.45 x 3.52 Kv = 0.175 .