You are told to find the transfer functionof the following circuit. You know that thetransfer function will take the form H(w)=jA/(R+jB), where A and B are variablesthat depend on component values andthe input signal frequency. Find the valueof B, given the following values: R = 1 kQ,L = 6 mH, C = 773 μµF, w = 1,445 rad/s.Please enter the answer to 3 significantfigures.VinCRwwwVout

Find the transfer function in the form of Hence circuit in frequency domain:

Answered: You are told to find the transfer…

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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Question

You are told to find the transfer function
of the following circuit. You know that the
transfer function will take the form H(w)=
jA/(R+jB), where A and B are variables
that depend on component values and
the input signal frequency. Find the value
of B, given the following values: R = 1 kQ,
L = 6 mH, C = 773 μµF, w = 1,445 rad/s.
Please enter the answer to 3 significant
figures.
Vin
C
R
www
Vout

Expert Solution

Step 1: Question analysis:

Find the transfer function in the form of $straight H open parentheses straight omega close parentheses equals fraction numerator jA over denominator straight R plus jB end fraction space space space space..... left parenthesis straight i right parenthesis$

$Given space values colon straight R equals 1 straight k straight capital omega straight L equals 6 mH straight C equals 773 straight mu straight F straight omega equals 1445 space rad divided by sec$

$Inductive space impedance colon straight Z subscript straight L equals straight j straight omega straight L equals straight j cross times 1445 cross times 6 cross times 10 to the power of negative 3 end exponent straight capital omega equals j 8.67 straight capital omega Capacitive space impedance colon space straight Z subscript straight C equals negative fraction numerator straight j over denominator straight omega straight C end fraction equals negative fraction numerator straight j over denominator 1445 cross times 773 cross times 10 to the power of negative 6 end exponent end fraction straight capital omega equals negative straight j 0.859 straight capital omega$

Hence circuit in frequency domain:

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