Use frequency sampling to design a filter b[n] that approximates the "batman" frequency response B(e) in the figure below. What are the fewest number of samples needed to get a good representation of the filter? The value of B(e) for any w can be found using the function batman.m on the Canvas homepage. (Note that batman.m provides the actual gain, not the gain in dB). JU Amplitude 3 2.5 2 1.5 0.5 0 -1 -0.8 -0.6 -0.4 0 | 0.2 -0.2 Normalized Frequency 0.4 0.6 0.8 1
Use frequency sampling to design a filter b[n] that approximates the "batman" frequency response B(e) in the figure below. What are the fewest number of samples needed to get a good representation of the filter? The value of B(e) for any w can be found using the function batman.m on the Canvas homepage. (Note that batman.m provides the actual gain, not the gain in dB). JU Amplitude 3 2.5 2 1.5 0.5 0 -1 -0.8 -0.6 -0.4 0 | 0.2 -0.2 Normalized Frequency 0.4 0.6 0.8 1
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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![function [gain]=batman (f)
%function [gain]=batman (f)
%
% produces the batman function
% f - normalized frequencies between -1 and 1
% gain value of the function at each value of f
x=8*f(:);
Nx=length(x);
for k=1: Nx
if x(k) < -7
y(k)=0;
elseif x(k) < -3
y(k)=3*sqrt(1-(x(k)/7)^2);
elseif x(k) < -1
y(k)=6*sqrt(10)/7+1.5-abs(x(k))/2-6*sqrt(10)*sqrt(4-(abs(x(k))-1)^2)/14;
elseif x(k) < -0.8
y(k)=10* (x(k)+1)+1;
elseif x(k) < -0.5
end
gain=y;
y(k)=-0.8* (x(k)+0.8)/0.3+3;
elseif x(k) < 0.5
y (k)=2.2;
elseif x(k) < 0.8
y(k)=0.8* (x(k)-0.5)/0.3+2.2;
elseif x(k) < 1
y(k)=10*(-x(k)+1)+1;
elseif x(k) < 3
y(k)=6*sqrt (10)/7+1.5-abs(x(k))/2-6*sqrt(10)*sqrt(4-(abs(x(k))-1)^2)/14;
elseif x(k) < 7
y(k)=3*sqrt(1-(x(k)/7)^2);
else
end
y(k)=0;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F232308f4-caae-4107-95ff-d4fc12180a88%2Fac830596-3e72-4f36-8264-6c29cd5d6e26%2F23fjt8_processed.png&w=3840&q=75)
Transcribed Image Text:function [gain]=batman (f)
%function [gain]=batman (f)
%
% produces the batman function
% f - normalized frequencies between -1 and 1
% gain value of the function at each value of f
x=8*f(:);
Nx=length(x);
for k=1: Nx
if x(k) < -7
y(k)=0;
elseif x(k) < -3
y(k)=3*sqrt(1-(x(k)/7)^2);
elseif x(k) < -1
y(k)=6*sqrt(10)/7+1.5-abs(x(k))/2-6*sqrt(10)*sqrt(4-(abs(x(k))-1)^2)/14;
elseif x(k) < -0.8
y(k)=10* (x(k)+1)+1;
elseif x(k) < -0.5
end
gain=y;
y(k)=-0.8* (x(k)+0.8)/0.3+3;
elseif x(k) < 0.5
y (k)=2.2;
elseif x(k) < 0.8
y(k)=0.8* (x(k)-0.5)/0.3+2.2;
elseif x(k) < 1
y(k)=10*(-x(k)+1)+1;
elseif x(k) < 3
y(k)=6*sqrt (10)/7+1.5-abs(x(k))/2-6*sqrt(10)*sqrt(4-(abs(x(k))-1)^2)/14;
elseif x(k) < 7
y(k)=3*sqrt(1-(x(k)/7)^2);
else
end
y(k)=0;
![2. Use frequency sampling to design a filter b[n] that approximates the "batman" frequency response
B(e) in the figure below. What are the fewest number of samples needed to get a good representation
of the filter? The value of B(e) for any w can be found using the function batman.m on the Canvas
homepage. (Note that batman.m provides the actual gain, not the gain in dB).
3
tois
Amplitude
2.5
2
0.5
0
-1
-0.8
-0.6
-0.4
-0.2
0.2
Normalized Frequency
0
0.4
0.6
0.8
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F232308f4-caae-4107-95ff-d4fc12180a88%2Fac830596-3e72-4f36-8264-6c29cd5d6e26%2Fiydlr8l_processed.png&w=3840&q=75)
Transcribed Image Text:2. Use frequency sampling to design a filter b[n] that approximates the "batman" frequency response
B(e) in the figure below. What are the fewest number of samples needed to get a good representation
of the filter? The value of B(e) for any w can be found using the function batman.m on the Canvas
homepage. (Note that batman.m provides the actual gain, not the gain in dB).
3
tois
Amplitude
2.5
2
0.5
0
-1
-0.8
-0.6
-0.4
-0.2
0.2
Normalized Frequency
0
0.4
0.6
0.8
1
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