Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 5 (Q.No. 2)
2.
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.
Answer: Option
Explanation:
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 .
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