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)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.1MCQ: The rms value of v(t)=Vmaxcos(t+) is given by a. Vmax b. Vmax/2 c. 2Vmax d. 2Vmax
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