3/(10) Marks: The Bode diagram of a compensator network is shown in Fig.3.10i-Describe the compensator networki-Find Ge(s).Magnitude (dB)0Lag - Leadbc (s) = kc (J₁S+1) (J₂S+1)(25+1) (4₂52 +1 8laglead2²₂ =-=-=-=-=> 32 = -=-=-15 = 6.675.00011510.3d=LS D15-109010-2110-10-152,91.510⁰Zied1013010²"Phiedi10aust

The Given is Compensator Network. Additional structures called compensation networks are sometimes…

3/(10) Marks: The Bode diagram of a compensator network is shown in Fig.3. 10 i-Describe the compensator network ii-Find Ge(s). Lag - Lead = (s) = kc (J.S+1) (J₂S+1) (5+1) (4₂52 +1 Z lead o lag 29 = = = = = 525 —- = 6-675 J₂ 0115 Magnitude (dB) Phase (deg) 0 -10 0 90

3/(10) Marks: The Bode diagram of a compensator network is shown in Fig.3. 10 i-Describe the compensator network ii-Find Ge(s). Lag - Lead = (s) = kc (J.S+1) (J₂S+1) (5+1) (4₂52 +1 Z lead o lag 29 = = = = = 525 —- = 6-675 J₂ 0115 Magnitude (dB) Phase (deg) 0 -10 0 90

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

3/(10) Marks: The Bode diagram of a compensator network is shown in Fig.3.
10
i-Describe the compensator network
i-Find Ge(s).
Magnitude (dB)
0
Lag - Lead
bc (s) = kc (J₁S+1) (J₂S+1)
(25+1) (4₂52 +1 8
lag
lead
2²₂ =-=-=-=-=> 32 = -=-=-15 = 6.675.00
0115
10.3
d=LS D
15
-10
90
10-2
1
10-10-15
2,9
1.5
10⁰
Zied
10130
10²
"Phiedi
10
aust

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