How to calculate passband and stopband coefficients for signal filtering?

Hi All

using the following code, I need that my passband frequency be 250 Hz , and the Stop Band at 512 Hz. how do I calculate Wp and Ws to insert in the code ?

Ts = 0.001;                                                     % Sampling Interval (s)
Fs = 1/Ts;                                                      % Sampling Frequency (Hz)
Fn = Fs/2;                                                      % Nyquist Frequency (Hz)
Wp = 0.001;                                                     % Passband Frequency For Lowpass Filter (Hz)
Ws = 0.0012;                                                    % Stopband Frequency For Lowpass Filter (Hz)
Rp =  1;                                                        % Passband Ripple For Lowpass Filter (dB)
Rs = 50;                                                        % Stopband Ripple (Attenuation) For Lowpass Filter (dB)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs);                                 % Calculate Filter Order
[z,p,k] = ellip(n,Rp,Rs,Wp);                                    % Calculate Filter
[sos,g] = zp2sos(z,p,k);                                        % Second-Order-Section For Stability
% A_Filt = filtfilt(sos,g,A);                                     % Filter Signal
figure
freqz(sos, 2^14, Fs)                                            % Filter Bode Plot (Check Performance)
set(subplot(2,1,1), 'Xlim',[0 1])
set(subplot(2,1,2), 'Xlim',[0 1])

Answer 1

[Star Strider]

Specifying both passband and stopband frequencies (and without other arguments to the ellip function) means that you are designing a bandpass filter. I appreciate your quoting my code, however I also posted several examples of elliptical filters using bandpass filters.

For your filter, replace ‘Wp’ with:

Wp = [250  512]/Fn;                                             % Passband Frequencies For Bandpass Filter (Hz)

and ‘Ws’ with:

Ws = [0.99 1.01]*Wp;                                            % Stopband Frequencies For Bandpass Filter (Hz)

or define it yourself, similarly to the way I defined ‘Wp’ here. Note that for a bandpass filter, the stopbands must be outside the passbands.

[farzad]

thank you very much

given that I have a double type variable for time :

class(t) = double ; size(t) = n x 1

How should I define T ?

is it :

T = t(end)

or

T = t

cause in both cases I have trouble with Dimensioning of Wn and Wp

[Star Strider]

As always, my pleasure!

The ‘Ts’ (sampling interval) value is ideally calculated as:

Ts = t(2)-t(1);

or alternatively:

Ts = mean(diff(t));

where ‘t’ is the time vector. (If it is not uniformly sampled, it is necessary to use the resample function to crate a regularly-sampled version of the signal and time vector so that the filters will work correctly.)

With ‘Ts’ defined, the frequencies and related variables will then be calculated correctly.

[farzad]

Thank you !

well with this Wp will be

Wp = 250

But Ws can't

Error using  * 
Inner matrix dimensions must agree.

Error in filtsignal (line 9)
Ws = [0.99 1.01]*Wp;                                                     % Stopband Frequency For
Lowpass Filter (Hz)

[Star Strider]

As always, my pleasure!

I initially thought that you wanted to design a bandpass filter. If you want to design a lowpass filter with those frequencies:

Wp = 250 / Fn;
Ws = 512 / Fn;

That should do what you want. ;

[farzad]

dear Star !

I am not sure then what should I do with the ellipord and next function , because in this way, I mean, the low pass band, the filtering function will not work :

Error using ellipord (line 72)
The cutoff frequencies must be within the interval of (0,1).

Error in filtsignal (line 12)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs);                                 % Calculate Filter Order

[Star Strider]

Your sampling frequency (Fs) must be greater than 1024 Hz (Ts less than 0.0009765625 s). The Nyquist frequency (Fs/2) is the limiting factor. All frequencies that you specify for the filter must be less than the Nyquist frequency, so the normalised frequencies are all less than 1.

[farzad]

Indeed they are :

my sampling Frequency is 512 Hz , and I am interested in keeping the frequencies under 256 Hz.

so I have :

>> Ts= t(2)-t(1);
Ts = 0.002
Fs = 512
Fn = 256
Wp =1
Ws =2
>> Rp = 1
Rp =    1
Rs =   50
>> [n,Wp] = ellipord(Wp,Ws,Rp,Rs)
Error using ellipord (line 72)
The cutoff frequencies must be within the interval of (0,1).

So Still not resolved

[Star Strider]

The ‘Wp’ and ‘Ws’ parameters are normalised frequencies such that the passdband and stopband frequencies (in Hz) are normalised with respect to the Nyquist frequency. If you only want to keep the frequencies below 256 Hz, then there is nothing further that you need to do.

If the ADC (analog-digital converter) you used to acquire those signals had a hardware Bessel lowpass filter front-end with a cutoff frequency approximately equal to the Nyquist frequency of the ADC (to prevent aliasing) that in this instance was 256 Hz, then your entire signal already meets your requirements, and no further filtering is necessary.

[farzad]

Well, I think the sampling frequency is 512 , and I think the filter is set to 512 Hz, or else : why do you think it's set on 256 Hz ? just because I get the error ?

Given that my X is a double, How can I know the present frequencies in my signal ?

[Star Strider]

The sampling frequency of 512 Hz produces a Nyquist frequency of half that, or 256 Hz. The Nyquist frequency is the highest uniquely resolvable frequency in a sampled signal. That is the reason the anti-aliasing Bessel filter is necessary, and if your instrumentation had one set to reject all signals above the Nyquist frequency of the ADC, and you only wanted signals in that spectrum, then no further filtering is necessary. The hardware Bessel filter ahead of the ADC already did all of that for you.

Unless you want to reject signals less than 256 Hz, do not do any more filtering.

[farzad]

I mean ! Thank you ! It took a day for me to really understand this ! and finally ! well I wish I have also learnt a bit about filter design.

also one thing : that's right that my frequency of interest is 256 Hz. but : when I look at my signal, I feel it's too noisy ! what is your comment on this ? the signal is the blue one above

[Star Strider]

You need to do a Fourier transform of your signal with the fft function. If the noise is band-limited, a frequency-selective filter can eliminate much of it. If it is broadband noise, a frequency-selective filter will have little effect.

[farzad]

How do I use a frequency selective filter ? shall you please help me with that ?

[Star Strider]

This entire discussion has been about frequency-selective filters.

The point is that if there is broadband noise (noise at almost every frequency in the Fourier transform of the signal), a frequency-selective filter is not going to be of much use.

If there is noise that begins at a particular frequency (for example above the frequency band of the signal), a frequency-selective filter can eliminate much of it without affecting the parts of the signal with the deisred information.

Wavelets can be used for removing at least some broadband noise, however they are not always successful.

[Danish khan]

Can you tell us a way on how to calculate maximum stopband attenuation?

[Pallove Nautiyal]

please let me know the same @Danish khan if you got to know how to calculate stopband attenuation too.

Answer 2

[Maria Amr]

Hello Star Strider,

I have a signal, that is a well log Acoustic Impedance log and I need to apply a Antialias filter in coarser samoling that means a bandpass filter.

Sample frequency is 6562 per kilometer (28000 sample/4.27km). I do need to identify cutt off frequencies correctly. I applied your passband code into my dataset but my proffessors says: apply an anti-alias filter that has a flat spectrum over the band of interest. Would you please direct me how I can identify cutoff frequencies? Appreciated!