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In the active circuit shown below: R1 =10 k2, R2=100 ko, R3=50 k2, and R4=25 ka. The circuit can be used as a PI s+0.1 compensator G.(s) = K- when R C R3 vi(t) R1 R4 Vo(t) O a. K=10 and C= 100 uF O b. K=20 and C= 1 µF

In the active circuit shown below: R1 =10 k2, R2=100 ko, R3=50 k2, and R4=25 ka. The circuit can be used as a PI s+0.1 compensator G.(s) = K- when R C R3 vi(t) R1 R4 Vo(t) O a. K=10 and C= 100 uF O b. K=20 and C= 1 µF

Introductory Circuit Analysis (13th Edition)
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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In the active circuit shown below: R1 =10 k2, R2=100 ko, R3=50 k2, and R4=25 ka. The circuit can be used as a PI s+0.1 compensator G.(s) = K- when R C R3 vi(t) R1 R4 Vo(t) O a. K=10 and C= 100 uF O b. K=20 and C= 1 µF
Transcribed Image Text:In the active circuit shown below: R1 =10 k2, R2=100 ko, R3=50 k2, and R4=25 ka. The circuit can be used as a PI s+0.1 compensator G.(s) = K- when R C R3 vi(t) R1 R4 Vo(t) O a. K=10 and C= 100 uF O b. K=20 and C= 1 µF
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