Electronics and Communication Engineering - Exam Questions Papers - Discussion

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.

[A]. 12.37, 0.175
[B]. 12.86, 2.175
[C]. 11.36, 1.175
[D]. 1.86, 0.175

Answer: Option A

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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