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CE105 DIGITAL SIGNAL PROCESSING
ASSIGNMENT 5: FIR Filter Design
Guide: Students are expected to submit MATLAB code and report
the results in form of
either MS Word or PDF format. All is zipped in a single file as
[StudentID].zip.
Send email to instructors with title [CE105.E12-AS1]
[StudentID].
For example, the student with ID 12345 should send the file
named 12345.zip and the
email title should be [CE105.E12-AS1] 12345
1.1 Solution:
Transition band:
2 / 2 .1850 / 8000 0.4625p p sF F
2 / 2 .2150 / 8000 0.5375s stop sF F
| | 0.075p s
Base on Table in Slide 32 Lecture 7, selecting the rectangular
window will result in a
passband ripple of 0.75dB and a stopband attenuation 21dB. Thus,
this window selection
would satisfy the design requirement for the passband ripple of
1dB and stopband
attenuation of 20dB.
Next, we determine the length of the filter as
1.824L
L
The cutoff frequency is determined by:
0.52
p s
c
1.2 Solution:
Based on the specifications, the Hann window will do the job,
since it has a passband
ripple of 0.055dB and a stopband attenuation of 44dB. Then
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2
| | 2500 1500 0.258000
p s
6.225L
Hence, we choose 25 filter coefficients using the Hann window
method. The cutoff
frequency is 0.5 .
Matlab Program:
function B=firwd(N, Ftype, WnL, WnH, Wtype)
%B=firwd(N,Ftype,WnL,WnH,Wtype)
% FIR filter design using the window function method.
% Input parameters:
% N: the number of the FIR filter taps.
% Note: It must be an odd number.
% Ftype: the filter type
%1. Lowpass filter;
%2. Highpass filter;
%3. Bandpass filter;
%4. Band reject filter;
% WnL: lower cutoff frequency in radians. Set WnL=0 for the
highpass filter.
% WnH: upper cutoff frequency in radians. Set WnH=0 for the
lowpass filter.
% Wtypw: window function type
%1. Rectangular window;
%2. Triangular window;
%3. Hanning window;
%4. Hamming window;
%5. Blackman window;
% Output:
% B: FIR filter coefficients.
M=(N-1)/2;
hH = sin(WnH*[-M:1:-1])./([-M:1:-1]*pi);
hH(M+1)=WnH/pi;
hH(M+2:1:N)=hH(M:-1:1);
hL=sin (WnL*[-M:1:-1])./([-M:1:-1]*pi);
hL(M+1)=WnL/pi;
hL(M+2:1:N)=hL(M:-1:1);
if Ftype==1
h(1:N)=hL(1:N);
end
if Ftype==2
h(1:N)=-hH(1:N);
h(M+1)=1+h(M+1);
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end
if Ftype==3
h(1:N)=hH(1:N)-hL(1:N);
end
if Ftype==4
h(1:N)=hL(1:N)-hH(1:N);
h(M+1)=1+h(M+1);
end
% window functions;
if Wtype==1
w(1:N)=ones(1,N);
end
if Wtype==2
w=1-abs([-M:1:M])/M;
end
if Wtype==3
w = 0.5 + 0.5*cos([-M:1:M]*pi/M);
end
if Wtype==4
w = 0.54 + 0.46*cos([-M:1:M]*pi/M);
end
if Wtype==5
w = 0.42 + 0.5*cos([-M:1:M]*pi/M) + 0.08*cos(2*[-
M:1:M]*pi/M);
end
B = h.*w
Program P1.2
% MATLAB program to create Figure in P1.2
%
N=25; Ftype=2; WnL=0; WnH=0.5*pi; Wtype=3;fs=8000;
Bhan=firwd(N,Ftype,WnL,WnH,Wtype);
freqz(Bhan,1,512,fs);
axis([0 fs/2 -120 10]);
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1.3 Solution:
1 1 12
| | 1600 500 0.2758000
p s
2 2 22
| | 3500 2300 0.38000
p s
Based on the specifications, the Hamming window will do the
job.
1
2
246.6
22
L
LL
Choosing 25L filter coefficients using the Hamming window
method:
1 11
1600 5002. 0.2625
2 8000 2
p s
c
2 22
3500 23002. 0.725
2 8000 2
p s
c
0 500 1000 1500 2000 2500 3000 3500 4000-1500
-1000
-500
0
500
Frequency (Hz)
Phase (
degre
es)
0 500 1000 1500 2000 2500 3000 3500 4000
-100
-50
0
Frequency (Hz)
Magnitude (
dB
)
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Program
% MATLAB program to create Figure in P1.3
%
N=25;Ftype=3;WnL=0.2625*pi;WnH=0.725*pi;Wtype=4;fs=8000;
Bham=firwd(N,Ftype,WnL,WnH,Wtype);
freqz(Bham,1,512,fs);
axis([0 fs/2 -130 10]);
1.4 Solution:
The normalized transition widths:
1
15002 0.375
8000 , and 2
13002 0.325
8000
The filter lengths are determined using the Blackman window
as:
1
2
29.3311
33.8
L
LL
0 500 1000 1500 2000 2500 3000 3500 4000-2000
-1000
0
1000
Frequency (Hz)
Phase (
degre
es)
0 500 1000 1500 2000 2500 3000 3500 4000
-100
-50
0
Frequency (Hz)
Magnitude (
dB
)
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We choose an odd number 35L . The normalized lower and upper
cutoff frequencies are
calculated as:
2 .12500.3125
8000cL
2 .28500.7125
8000cH
Program P1.4
% MATLAB program to create Figure in P1.4
%
N=35;Ftype=4;WnL=0.3125*pi;WnH=0.7125*pi;Wtype=5;fs=8000;
Bblack=firwd(N,Ftype,WnL,WnH,Wtype);
freqz(Bblack,1,512,fs);
axis([0 fs/2 -120 10]);
0 500 1000 1500 2000 2500 3000 3500 4000-3000
-2000
-1000
0
Frequency (Hz)
Phase (
degre
es)
0 500 1000 1500 2000 2500 3000 3500 4000
-100
-50
0
Frequency (Hz)
Magnitude (
dB
)
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Example:
a) Design a 3-tap (L=3) FIR lowpass with cutoff frequency of
800Hz and
sampling rate of 8000Hz using Hamming window function.
b) Determine the transfer function and difference equation of
the designed FIR
system.
c) Compute and plot the magnitude frequency response for 0, / 4,
/ 2,3 / 4
and radians.
