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)

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

See similar textbooks

Similar questions

-Consider the following DT sequence, Is this signal as Power Signal or Energy signal? Determine it. * 0Snくs X [n ]= lo-n 55.n510 • ther wise Power Signal, P=85 Energy Signal, E=85 O Neither energy nor Power O Both, P=30 E=55 -Is the following signal is X(4) - Sin 3t + Sin 大t Periodic Aperiodic
Ererrises: IA platinum oil has a resifance of 3-14k r at go°C and 3.747JV at 100°C.- Fin'd the resistance at o'C ond the kmp -creff. of resifonce af 40°C 2. A pofential difference of r5oV is apphed to a field wnding at 15°C and the currend is SA. What will be the mean fempouafure of the winiding when currend has falen to 3.91A, apphedvrltage being constant Aasume s =p7.5. 3. A cril has a resi fance of 182 when its mean femperature is 20°C and 20v when ik mean femperature s 50°C. Frid is mean femperature rise when ik resotonce is 21JL and the sorrandinig femperature is i5°C. A cril of a relay takes a curre of 0.12 A when it is at the vem insbi o al 4. A jemperature of is'C ond canneckerdacms a oV supply If the minimum operating current of the relay is 0.1 A, colculate the femperature above ohich the relay will fail to operate when connected ti the same supply. Resi tncenfemp.coefficrenf of the cril makria] is o.0043 per °C at 0°C.
2-A resultant current is made of two components: a 10 A D.C. component and a sinusoidal component of maximum value 14.14. The average value of the resultant current is amperes (a) 0 (b) 24.14 (c) 10 (d) 4.14 and r.m.s. value is - amperes (e) 10 (f) 14.14 (g) 24.14 (h) 100
A certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage by
2_53528980697... -> For the circuit shown below, detemine the voltage of the -j10 2 capacitive reactance. -j5 0 ILOA(T) ==j102 jlon is Q 102 O 0.5L-90 52 A.
"If lab is equal to 595.8252306 and lb is 344296, the phase sequence is:" O POSITIVE O NEGATIVE
Power factor 90 to 100% 88 to <90% 85 to <88% 80 to <85% 75 to < 80% 70 to < 75% 65 to < 70% 60 to < 65% 55 to <60% 50 to <55% Less than 50% Surcharge None 2% 4% 9% 16% 24% 34% 44% 57% 72% 80% Use the table given for the following. a. What is the effective impedance? What is the average current drawn? What will be the surcharge? b. c. If a house with inductive loads has an avg consumption of 40kW and 26kVAR is supplied 230 Vrms source.
Required information The frequency of operation is 70.0 rad/s. Find a parallel equivalent of the series combination of a 2-0 resistor and a 100-mF capacitor. (Round the final answers to three decimal places.) The parallel equivalent of the series combination of a 2-0 resistor is The parallel equivalent of the series combination of a 100-mF capacitor is Q. mF.
HH 9 0btoin the tronsfer functle Vout Vin R. ) Whot is the C. Volue to place gero frequency to 206H2? Vout c) What is the Ce volue to place pole frequer to 500k Hz? R2=1k2 d) Plot the frogueney response.
C1 L1 R1 V1 Vout V1 contains the fundamental and its harmonics (fo, 2fo, 3fo, ...). The amplitude of the harmonics is inversely proportional to the harmonic number. V(fo) = 10 mV V(2fo) = 10mv/2 = 5 mV V(3fo) = 10mv/3 = 3.33 mV V(4fo) = 10mv/4 = 2.55 mV a) What is the minimum value of Q for the circuit that will attenuate the third harmonic (3fo) in Vout to be less than 1% of the fundamental? b) What is the amplitude of the fifth harmonic (5fo) as a percentage of the fundamental?
What is the form factor of 169sin 377t?
For the RL circuit shown, V1=600Vms, XL1=3002, and Rl=6002. What is the RMS current through R1? a. 0.707 Ams b. 1.0 Arms V1 c. 2.0 Arms d. 3.0 Arms R1 L1