Answer the following show illustration and solution. 2.) Design a basic wide-band, RC band stop filter with a lower cut-off frequency of 250 Hz and a higher cut-off frequency of 1.2kHz. Find its center frequency, bandwidth and Q of the circuit assuming the capacitor for both filter sections is 0.1 microfarad.

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...
icon
Related questions
icon
Concept explainers
Question

Answer the following show illustration and solution.

2.) Design a basic wide-band, RC band stop filter with a lower cut-off frequency of 250 Hz and a higher cut-off frequency of 1.2kHz. Find its center frequency, bandwidth and Q of the circuit assuming the capacitor for both filter sections is 0.1 microfarad.

This is the example that might help.

 

Example: Band-pass Filter Circuit
Design a basic wide-band, RC band stop filter with a lower
cut-off frequency of 200HZ and a higher cut-off
frequency of 800Hz. Find the value of the resistor for
both frequencies, its center frequency, bandwidth and Q
of the circuit assuming the capacitor for both filter
sections is 0.1uF
For resistor values of the filters:
1
= 7.95kN
R, =
2nf.C
1
RH =
2nfµC¯ 2n(800HZ)(0.1µF)
%3D
2п (200H2) (0.1иF)
= 1.99kN
For the center frequency:
fr = fi × fu = /(200H2)(800H2) = 400HZ
For the bandwidth:
BW = fµ – f.
BW = 800HZ – 200HZ
For the quiescent point (operating point):
fr
Q =
400HZ
= 0.67
%3D
BW
600HZ
BW = 600HZ
Transcribed Image Text:Example: Band-pass Filter Circuit Design a basic wide-band, RC band stop filter with a lower cut-off frequency of 200HZ and a higher cut-off frequency of 800Hz. Find the value of the resistor for both frequencies, its center frequency, bandwidth and Q of the circuit assuming the capacitor for both filter sections is 0.1uF For resistor values of the filters: 1 = 7.95kN R, = 2nf.C 1 RH = 2nfµC¯ 2n(800HZ)(0.1µF) %3D 2п (200H2) (0.1иF) = 1.99kN For the center frequency: fr = fi × fu = /(200H2)(800H2) = 400HZ For the bandwidth: BW = fµ – f. BW = 800HZ – 200HZ For the quiescent point (operating point): fr Q = 400HZ = 0.67 %3D BW 600HZ BW = 600HZ
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Types of Filter and Its Properties
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.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,