Answer the following wuth illustration and solutions.1.) A second-order band pass filter is to be constructed using RC components that will only allow a range of frequencies to pass above 500Hz and below 1.5kHz. Assuming that both the resistor have values of 1kohms, calculate the values of the two capacitors required. Find also the center frequency of the two frequencies.This is the example/guide that might help. Example: Band-pass Filter CircuitA second-order band pass filter is to be constructed usingRC components that will only allow a range of frequenciesto pass above 1kHz and below 30kHz. Assuming that boththe resistors have values of 10k, calculate the values ofthe two capacitors required. Find also the centerfrequency of the two frequencies.For each filter capacitance:1C₁ =2πf₁R1C₂ =2πfHRFor the center frequency:1= 15.9nF2π(1kHz) (10kn)1= 530pF2π (30kHz) (10kn)fr=√√fLx fH = √1kHz × 30kHz = 5.48kHz

Given data, Lower frequency fL=500 Hz Higher frequency fH=1.5 kHz = 1500 Hz Value of resistance R= 1…

Answered: Answer the following wuth illustration…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Answer the following wuth illustration and solutions.

1.) A second-order band pass filter is to be constructed using RC components that will only allow a range of frequencies to pass above 500Hz and below 1.5kHz. Assuming that both the resistor have values of 1kohms, calculate the values of the two capacitors required. Find also the center frequency of the two frequencies.

This is the example/guide that might help.

Example: Band-pass Filter Circuit
A second-order band pass filter is to be constructed using
RC components that will only allow a range of frequencies
to pass above 1kHz and below 30kHz. Assuming that both
the resistors have values of 10k, calculate the values of
the two capacitors required. Find also the center
frequency of the two frequencies.
For each filter capacitance:
1
C₁ =
2πf₁R
1
C₂ =
2πfHR
For the center frequency:
1
= 15.9nF
2π(1kHz) (10kn)
1
= 530pF
2π (30kHz) (10kn)
fr=√√fLx fH = √1kHz × 30kHz = 5.48kHz

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