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Vin(t) = 5 cos(250t) + 5cos(2000t) + 5cos(20,000rt) R = 1000 = 318.3 N TT + C = 1 µF Vin(1) Vout(t) +

Vin(t) = 5 cos(250t) + 5cos(2000t) + 5cos(20,000rt) R = 1000 = 318.3 N TT + C = 1 µF Vin(1) Vout(t) +

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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1. Find the half power frequency and use the transfuntion to find the out for the given signal

2. Use the bode plots for the low pass filter to find the filter response for the signal

Vin (t) = 5 cos(250rt) + 5cos(2000rtt) + 5cos(20,000rt) R- 1000 – 318.3 . TT + Vin(t) C = 1 µF Vour(t) +
Transcribed Image Text:Vin (t) = 5 cos(250rt) + 5cos(2000rtt) + 5cos(20,000rt) R- 1000 – 318.3 . TT + Vin(t) C = 1 µF Vour(t) +
10 Approximate Actual .Approximate Actual -10 -20 -10 -30 -40 -20 -50 -30 -60 -70 40 -80 -50 -90 fB 10-1 fB 10fB 100fg fB fB fB 10fB 100fB 100 10 100 Phase angle, deg
Transcribed Image Text:10 Approximate Actual .Approximate Actual -10 -20 -10 -30 -40 -20 -50 -30 -60 -70 40 -80 -50 -90 fB 10-1 fB 10fB 100fg fB fB fB 10fB 100fB 100 10 100 Phase angle, deg
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