det kd = derivative gain' Kp = 96 dem ping ratio &= 0.8 Now, Gel@) = Kp t Kds %3D C•E = 1+ kp+ Kq s) 254s+4 o;→ 25²+5+4t Kelst Kp =O s+ K(L+ Kd) + 4+kp =0 2. s+ (1+ K4) + 4+96 > s+ k(itkd) +5o=0 ķ (itkd) +50=0 : wn= J50 rad/pec and & = klt kd) - o-8 2 150 It kd 450 x 0-8 22.6274 %3D Kd = 21• 6274 %3D 介介
det kd = derivative gain' Kp = 96 dem ping ratio &= 0.8 Now, Gel@) = Kp t Kds %3D C•E = 1+ kp+ Kq s) 254s+4 o;→ 25²+5+4t Kelst Kp =O s+ K(L+ Kd) + 4+kp =0 2. s+ (1+ K4) + 4+96 > s+ k(itkd) +5o=0 ķ (itkd) +50=0 : wn= J50 rad/pec and & = klt kd) - o-8 2 150 It kd 450 x 0-8 22.6274 %3D Kd = 21• 6274 %3D 介介
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
How did they get square root of 50?
![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%2F35baf419-d907-444a-bf3c-36b51550b1ba%2F358261a5-f1a2-4aa6-a957-98f1d7d2f95a%2Fy7edl39_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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