FIR digiral filter Using FREQUENCY SAMPLING
9(a) LPF
M=63;
Wp=0.5*pi;%the number of samples and the passband cutoff frequency
m=0:(M+1)/2;
Wm=2*pi*m./(M+1);%the sampling points and the stopband cutoff frequency
mtr=floor(Wp*(M+1)/(2*pi))+2;%round to negative part,i.e.floor(3.5)=3;floor
(-3.2)=-4
Ad=[Wm<=Wp];
Ad(mtr)=0.38;
Hd=Ad.*exp(-j*0.5*M*Wm);%define frequency-domain sampling vector H(k)
Hd=[Hd conj(fliplr(Hd(2:(M+1)/2)))];
%fliplr is to realize the fliplr of matrix and conj
h=real(ifft(Hd));% h(n)=IDFT[H(k)
w=linspace(0,pi,1000);%get 1000 row vectors between 0 and pi
H=freqz(h,[1],w);%the amplitude -frequency characteristic diagram of the
filter
figure(1)
plot(w/pi,20*log10(abs(H)));%parameters are respectively the normalized
frequency and amplitude
xlabel('the normailzed frequency');
ylabel('gian/dB');
title('The gain response of lowpass filter');
axis([0 1 -50 0.5]);
9(b) HPF
M=32;%the number of samples
Wp=0.6*pi;%passband cutoff frequency
m=0:M/2;%the sampling points
Wm=2*pi*m./(M+1);%stopband cutoff frequency
mtr=ceil(Wp*(M+1)/(2*pi));%round to positive part,i.e.ceil(3.5)=4;ceil
(-3.2)=-3;
Ad=[Wm>=Wp];
Ad(mtr)=0.38;
Hd=Ad.*exp(-j*0.5*M*Wm);%define frequency-domain sampling vector H(k))
Hd=[Hd conj(fliplr(Hd(2:M/2+1)))];
%fliplr is to realize the fliplr of matrix and conj is the conjugate
h=real(ifft(Hd));%h(n)=IDFT[H(k)]
w=linspace(0,pi,1000);%get 1000 row vectors between 0 and pi
H=freqz(h,[1],w);%the amplitude -frequency characteristic diagram of the
filter
figure(1)
plot(w/pi,20*log10(abs(H)));%parameters are respectively the noemalized
frequency and amplitude
xlabel('the normailzed frequency');
ylabel('gian/dB');
title('The gain response of highpass filter');
axis([0 1 -50 0]);
