please write this on step by step From the data,TF (W) =FactorIt jw1I+jWTF (∞) = 100 (1+ JW)10TF (W)+ jw50²35.71) radSecBy comparing 0%, zero is located at w/= 10 radsecAt W₁=10%and pole is located at W₂ = 50 rad/sec with order at 22Each zero contribute slope of +20dB/dec and+90° phase. where as each pole of order 2 can-18⁰⁰ phase.contribute -40dB/dec slope and43 dB100 (1 + 0.1 jw)1-0.0004 w² + jo.028=40dB+981+ (jw)² jw(50)2+⇒100 (₁ + jw)21+ (jw) ²wzgslope phase+2018/dec/+90°-40dB/des/-188Bode plot: Magnitude M= |TF(W) = 100 √ 1 + (0-10) 2(0.0284³²+ (1-80048)(20 log/100):: MdBM = 20 log (100) + 20 g √ [+ (6-1×10) ²)B35.71+ 2 & (jw)secMdB = 43 dB at W₁ = 10 rad+20dB/decMdB = Magnitude plotW₁=10phase plot (0)Resultant slope phase+2010/4/198-2018/407/-18dBW₁=10w=50|W₂=50-90°-20dB/dec>Wapproximate phase d19, Plot the Bode plot (magnitude and phase) of the following transfer function usinggraphical techniques.TF (w)-100(1+0.1jw)(1-0.0004w² + j0.028)

Answered: Image /qna-images/answer/b41bd62f-20a3-47ec-be31-ab3e554c9908.jpg

From the data, 100 (1 + 0.1 jw) 1-0.0004² + jo.028 TF (W) = 100 (1 + jw) 10 TF(W) = Factor It jw TF (W) 1 I+jW = By comparing 080, zero is located at w/= 10 rad sec 1+ (jw)² jw (50)2 + 100 (1 + jw) and pole is located at W₂ = = 2 + jw 50² 35.71 2 1+ (jw) ² + 2 & (jw) wzg Each zero contribute slope of +20dB/dec and +90° phase. where as each pole of order 2 can -18⁰⁰ phase. contribute -40dB/dec slope and 35.71 50 rad/sec with order of 2 slope phase +2010/460/+90° -4013/11/-188 Resultant slope phase 20 ds/dec/+90° + -20453/dc/-90 dB Bode plot: Magnitude M= |TF (W) = 100 √ 1 + (0-10)²2 0.028W²+(1-0.0004)

From the data, 100 (1 + 0.1 jw) 1-0.0004² + jo.028 TF (W) = 100 (1 + jw) 10 TF(W) = Factor It jw TF (W) 1 I+jW = By comparing 080, zero is located at w/= 10 rad sec 1+ (jw)² jw (50)2 + 100 (1 + jw) and pole is located at W₂ = = 2 + jw 50² 35.71 2 1+ (jw) ² + 2 & (jw) wzg Each zero contribute slope of +20dB/dec and +90° phase. where as each pole of order 2 can -18⁰⁰ phase. contribute -40dB/dec slope and 35.71 50 rad/sec with order of 2 slope phase +2010/460/+90° -4013/11/-188 Resultant slope phase 20 ds/dec/+90° + -20453/dc/-90 dB Bode plot: Magnitude M= |TF (W) = 100 √ 1 + (0-10)²2 0.028W²+(1-0.0004)

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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