NEED to design a FIR LPF with cuff frequency at 1 Hz

fir1(31,0.5) length of the filter is 32 and cuttoff frequency is 1 Hz please some body help me what should be there in place of 0.5

Answer 1

[Wayne King]

Nobody can tell you the answer to this question if you do not tell us your sampling frequency.

Solve for Wn in the following

Wn = (2*1)/Fs;

where Fs is your sampling frequency. Then

   b = fir1(31,Wn);

You can view the frequency response with:

fvtool(b,1,'Fs',Fs);  % again where Fs is your sampling frequency

[suresh]

sir, my input has sampling period of 30s (for every 30s im getting a sample) then when it is converted into sampling frequency it is 0.03333Hz as per above formula 2*1/ Fs gives me 60 but Wn must be in between 0 to 1

Answer 2

[Wayne King]

The problem is that if you are really sampling only every 30 seconds, then you cannot accurately reproduce data at 1 Hz.

If your sampling frequency is really 0.0333 Hz, then your Nyquist frequency is 0.0333/2 . The highest frequency sine wave you can accurately represent will have a period of 60 seconds, NOT a period of 1 second.