How did they get wn= sqrt 50 in the question?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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How did they get wn= sqrt 50 in the question?
![det Kal = derivative gain
Kp = 96
dem þing ratio z= o-8
Now, Gels)
Kp t kd s
%D
1+ (kp+ Kq s)
284s+4
C·E =
= 0;>
25+S+4 + Kd S + Kp = 0
st K(+ Kd) + 4+kp =0
s*+ x (1+ Kd] + 4+96 > s+ k(i+kd) +50=0
Jão rad/sec
: Wn =
and =
KlIt kd) =
0.8
2 150
It kd =
4150 x 0.8
22.6274
Kd = 21. 6274](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a214cd6-f30d-411d-a1ad-f5c9f3ceb081%2Fbc8bbe67-15b1-4fc4-b9b1-e7da202dbfa1%2F7e9ik1h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:det Kal = derivative gain
Kp = 96
dem þing ratio z= o-8
Now, Gels)
Kp t kd s
%D
1+ (kp+ Kq s)
284s+4
C·E =
= 0;>
25+S+4 + Kd S + Kp = 0
st K(+ Kd) + 4+kp =0
s*+ x (1+ Kd] + 4+96 > s+ k(i+kd) +50=0
Jão rad/sec
: Wn =
and =
KlIt kd) =
0.8
2 150
It kd =
4150 x 0.8
22.6274
Kd = 21. 6274
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