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.
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.
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...
Related questions
Question
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.
Transcribed Image Text: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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,