EXAMPLE 9.3.1 Suppose that we would like to design an FIR linear phase low-pass filter according to the following specifications: 0.99 H(e) ≤1.01 H(e) 0.01 0❘w❘ ≤0.19 0.21 ≤ For a stopband attenuation of 20 log(0.01) = -40 dB, we may use a Hanning window. Although we could also use a Hamming or a Blackman window, these windows would overdesign the filter and produce a larger stopband attenuation at the expense of an increase in the transition width. Because the specification calls for a transition width of Aww, wp = 0.02, or Af = 0.01, with NAS 3.1 for a Hanning window (see Table 9.2), an estimate of the required filter order is 3.1 N = 310 Af The last step is to find the unit sample response of the ideal low-pass filter that is to be windowed. With a cutoff frequency of w₁ = (w, + wp)/2 = 0.2, and a delay of α = N/2 = 155, the unit sample response is sin[0.2π (n-155)] ha(n) = (n-155)
EXAMPLE 9.3.1 Suppose that we would like to design an FIR linear phase low-pass filter according to the following specifications: 0.99 H(e) ≤1.01 H(e) 0.01 0❘w❘ ≤0.19 0.21 ≤ For a stopband attenuation of 20 log(0.01) = -40 dB, we may use a Hanning window. Although we could also use a Hamming or a Blackman window, these windows would overdesign the filter and produce a larger stopband attenuation at the expense of an increase in the transition width. Because the specification calls for a transition width of Aww, wp = 0.02, or Af = 0.01, with NAS 3.1 for a Hanning window (see Table 9.2), an estimate of the required filter order is 3.1 N = 310 Af The last step is to find the unit sample response of the ideal low-pass filter that is to be windowed. With a cutoff frequency of w₁ = (w, + wp)/2 = 0.2, and a delay of α = N/2 = 155, the unit sample response is sin[0.2π (n-155)] ha(n) = (n-155)
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
![EXAMPLE 9.3.1 Suppose that we would like to design an FIR linear phase low-pass filter according to the following
specifications:
0.99 H(e) ≤1.01
H(e) 0.01
0❘w❘ ≤0.19
0.21 ≤
For a stopband attenuation of 20 log(0.01) = -40 dB, we may use a Hanning window. Although we could also use a Hamming
or a Blackman window, these windows would overdesign the filter and produce a larger stopband attenuation at the expense
of an increase in the transition width. Because the specification calls for a transition width of Aww, wp = 0.02, or
Af = 0.01, with
NAS 3.1
for a Hanning window (see Table 9.2), an estimate of the required filter order is
3.1
N =
310
Af
The last step is to find the unit sample response of the ideal low-pass filter that is to be windowed. With a cutoff frequency
of w₁ = (w, + wp)/2 = 0.2, and a delay of α = N/2 = 155, the unit sample response is
sin[0.2π (n-155)]
ha(n) =
(n-155)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85133c3c-d87c-414d-89fc-102819ff3f2b%2F69ed80a7-e3e7-4b0b-8719-b80f9b7cc7de%2Fbrob7l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:EXAMPLE 9.3.1 Suppose that we would like to design an FIR linear phase low-pass filter according to the following
specifications:
0.99 H(e) ≤1.01
H(e) 0.01
0❘w❘ ≤0.19
0.21 ≤
For a stopband attenuation of 20 log(0.01) = -40 dB, we may use a Hanning window. Although we could also use a Hamming
or a Blackman window, these windows would overdesign the filter and produce a larger stopband attenuation at the expense
of an increase in the transition width. Because the specification calls for a transition width of Aww, wp = 0.02, or
Af = 0.01, with
NAS 3.1
for a Hanning window (see Table 9.2), an estimate of the required filter order is
3.1
N =
310
Af
The last step is to find the unit sample response of the ideal low-pass filter that is to be windowed. With a cutoff frequency
of w₁ = (w, + wp)/2 = 0.2, and a delay of α = N/2 = 155, the unit sample response is
sin[0.2π (n-155)]
ha(n) =
(n-155)
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 2 steps with 3 images

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,