A lead compensator is realized using ideal OpAmps and linear circuit components C₁=100μF 100 R₂ 10 C₂=100μF R₁=C=100μF| 2π 2π C₂=100μF (a) Demonstrate that the transfer function of the circuit is given by: +1 H(f) = jf +1 1000 V out (b) Sketch the bode plot of this transfer function. Make sure to label the magnitude and phase plots and slope of the segments, the two break points and the highest value on the phase plot
A lead compensator is realized using ideal OpAmps and linear circuit components C₁=100μF 100 R₂ 10 C₂=100μF R₁=C=100μF| 2π 2π C₂=100μF (a) Demonstrate that the transfer function of the circuit is given by: +1 H(f) = jf +1 1000 V out (b) Sketch the bode plot of this transfer function. Make sure to label the magnitude and phase plots and slope of the segments, the two break points and the highest value on the phase plot
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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
Transcribed Image Text:A lead compensator is realized using ideal OpAmps and linear circuit components
C₁=100μF
100
R₂
10
C₂=100μF
R₁=C=100μF|
2π
2π
C₂=100μF
(a) Demonstrate that the transfer function of the circuit is given by:
+1
H(f) = jf
+1
1000
V
out
(b) Sketch the bode plot of this transfer function. Make sure to label the magnitude and
phase plots and slope of the segments, the two break points and the highest value
on the phase plot
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