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How did he get the value inside the circle? It is possible to explain step by step and how he got the output, the indicator in green Y(F) = 12000 S(F +3000) + 12000 6(f-3000) +D 000 S(f+2010)+ 20000 S (f-2o0) * YE = 24000 SCF-3000) +24000 S(F+3000) + 2. y0 000 S(F-2000) + 40000 S(f+2000) 2. 2 > YF) = > YE)* 24000 S(F-3000) + SIf+3000) + %3D 2 [s(^200) + 6(f+200) 400 00 S(F+2000) Step 2 of 2:) we know that Ac cos(27TfcE) < Foudier TransFom Ac SCF-fc)+ S(F+F) APply Inverse T8omSFOom to 4(F) we get y (t) = 24000 Cos (21TX 3000 t) + 40000 COs (2T X 200ot) => yE) = 24000cos (G000TTE) + 4000 CoS (4000 TTE) %3D

How did he get the value inside the circle? It is possible to explain step by step and how he got the output, the indicator in green Y(F) = 12000 S(F +3000) + 12000 6(f-3000) +D 000 S(f+2010)+ 20000 S (f-2o0) * YE = 24000 SCF-3000) +24000 S(F+3000) + 2. y0 000 S(F-2000) + 40000 S(f+2000) 2. 2 > YF) = > YE)* 24000 S(F-3000) + SIf+3000) + %3D 2 [s(^200) + 6(f+200) 400 00 S(F+2000) Step 2 of 2:) we know that Ac cos(27TfcE) < Foudier TransFom Ac SCF-fc)+ S(F+F) APply Inverse T8omSFOom to 4(F) we get y (t) = 24000 Cos (21TX 3000 t) + 40000 COs (2T X 200ot) => yE) = 24000cos (G000TTE) + 4000 CoS (4000 TTE) %3D

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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How did he get the value inside the circle? It is possible to explain step by step and how he got the output, the indicator in green Y(F) = 12000 S(F +3000) + 12000 6(f-3000) +D 000 S(F+2010)+ 20000 8 (f-2o0) * YE 24000 SCF-3000) +24000 S(f+3000) + 2. y0 000 S(F-2000) + 40000 S(f+2000) 2. 2 > YF) = > YE)* 24000 S(F-3000) + SIf+3000) + %3D 2 [s(^2000) + 6(f+200) 400 00 S(f+2000) Step 2 of 2:) we know that Ac cos(27TFct) < Foudier TransFom Ac SCF-fc)+ S(F+F) APply Inverse T8omSFOom to 4(F) we get y (t) = 24000 Cos (21TX 3000 t) + 40000 COs(2T X 200ot) => yE) = 24000cos (G000TTE) + 4000 CoS (4000 TTE) %3D
Transcribed Image Text:How did he get the value inside the circle? It is possible to explain step by step and how he got the output, the indicator in green Y(F) = 12000 S(F +3000) + 12000 6(f-3000) +D 000 S(F+2010)+ 20000 8 (f-2o0) * YE 24000 SCF-3000) +24000 S(f+3000) + 2. y0 000 S(F-2000) + 40000 S(f+2000) 2. 2 > YF) = > YE)* 24000 S(F-3000) + SIf+3000) + %3D 2 [s(^2000) + 6(f+200) 400 00 S(f+2000) Step 2 of 2:) we know that Ac cos(27TFct) < Foudier TransFom Ac SCF-fc)+ S(F+F) APply Inverse T8omSFOom to 4(F) we get y (t) = 24000 Cos (21TX 3000 t) + 40000 COs(2T X 200ot) => yE) = 24000cos (G000TTE) + 4000 CoS (4000 TTE) %3D
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