DSP PRACTICAL
Q:Design an low pass fir filter using
hanning window of order 11 having following specification:
pass band cut off frequency=250 Hz
sampling frequency=1000 Hz
fp=250 Hz
Fs=1000 Hz
PROGRAM:
clc
clear all;
close all;
syms w
H=input('enter value of H');
N=input('order of filter');
a=input('enter lower limit');
b=input('enter upper limit');
hd=0;
for n=1:N
hd(n)=(1/(2*pi))*int(H*exp(j*w*(n-1)),a,b)
end
h=hd.*rishhann(N)
figure(1)
plot(h)
figure(2)
freqz(h)
pause
a=6
FUNCTION WINDOW:
PROGRAM1:
function [w] =
rishhamm(N)
w=0;
for n=1:N
w(n)=.54-.46*cos(2*pi*(n-1)/(N-1))
end
end
PROGRAM 2:
function [w] = rishhann(N)
w=0;
for n=1:N
w(n)=.5-.5*cos(2*pi*(n-1)/(N-1))
end
end
PROGRAM 3:
function [w] = rishrect(N)
w=0;
for n=1:N
w(n)=1
end
end
OUTPUT:
enter value of Hexp(-5*j*w)
order of filter11
enter lower limit-1.57
enter upper limit1.57
hd =
0.0637
hd =
0.0637 -0.0003
hd =
0.0637 -0.0003
-0.1061
hd =
0.0637 -0.0003
-0.1061 0.0003
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997 0.3183
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997 0.3183 0.0003
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997 0.3183 0.0003
-0.1061
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997 0.3183 0.0003
-0.1061 -0.0003
hd =
0.0637 -0.0003
-0.1061 0.0003 0.3183
0.4997 0.3183 0.0003
-0.1061 -0.0003 0.0637
w =
0
w =
0 0.0955
w =
0 0.0955 0.3455
w =
0 0.0955 0.3455
0.6545
w =
0 0.0955 0.3455
0.6545 0.9045
w =
0 0.0955 0.3455
0.6545 0.9045 1.0000
w =
0 0.0955 0.3455
0.6545 0.9045 1.0000
0.9045
w =
0 0.0955 0.3455
0.6545 0.9045 1.0000
0.9045 0.6545
w =
0 0.0955 0.3455
0.6545 0.9045 1.0000
0.9045 0.6545 0.3455
w =
0
0.0955 0.3455 0.6545
0.9045 1.0000 0.9045
0.6545 0.3455 0.0955
w =
0 0.0955 0.3455
0.6545 0.9045 1.0000
0.9045 0.6545 0.3455
0.0955 0
h =
0 -0.0000 -0.0367
0.0002 0.2879 0.4997
0.2879 0.0002 -0.0367
-0.0000 0
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