# Designing a fir filter with Parks-McClellan method

18 views (last 30 days)
Nediljko Ivisic on 5 Jul 2020
Commented: Sai Sri Pathuri on 9 Jul 2020
Hello, I have a task to designs a bandstop fir filter using parks-mcclellan method, apply it to a signal and show the results. The cutoff frequencys are fp1=0.15 and fp2=0.3 and the signal is x(n)=sin(2*pi*0.2*n)+sin(2*pi*0.1*n), 0<n<128. So far I have wrote this:
clear all;
fs = 10;
fp1 = 0.15;
fp2 = 0.3;
f1 = 0.2;
f2 = 0.1;
t = 128;
n = 0:1/fs:t;
x = sin(2*pi*f1*n) + sin(2*pi*f2*n);
F = [0 0.05 0.15 0.3 0.4 1];
A = [1 1 0 0 1 1];
B = firpm(10, F, A);
[H, w] = freqz(B);
y = filtfilt(B, 1, x);
figure(1)
subplot(2,1,1)
plot(n, x);
subplot(2,1,2)
plot(n, y);
dB = mag2db(abs(H));
figure(2)
subplot(2,1,1)
plot(F, A ,w, abs(H));
subplot(2,1,2)
plot(w, dB);
But I am not getting the output that I want. If anyone knows what I am doing wrong any help would be greatly appreciated.

Sai Sri Pathuri on 8 Jul 2020
I assume you are expecting a graph like this You may obtain this by changing w to w/pi while plotting
figure(2)
subplot(2,1,1)
plot(F, A ,w/pi, abs(H));
subplot(2,1,2)
plot(w/pi, dB);
##### 2 CommentsShowHide 1 older comment
Sai Sri Pathuri on 9 Jul 2020
When you use the frequencies f1 = 0.2 and f2 = 0.1,
normFreq = freq/(fs/2))
then the corresponding normal frequencies are 0.04 and 0.02 respectively. Due to this, when the bandstop filter filters the 0.2 frequency, there is no change in filtered signal because no such frequency existed in your signal.
Try changing the f1, f2 values to
f1 = 1;
f2 = 0.5;