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Generated filter reduces signal time

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farzad on 18 Dec 2020
Commented: Star Strider on 18 Dec 2020
Hi All
I desgined a filter with the following data, but it reduced the signal time to half , after filtering. is there a reason ? how to resolve?
function y = myFilter(x)
persistent Hd;
if isempty(Hd)
N = 3; % Order
Fstop1 = 55; % First Stopband Frequency
Fpass1 = 65; % First Passband Frequency
Fpass2 = 9998; % Second Passband Frequency
Fstop2 = 10000; % Second Stopband Frequency
Fs = 256000; % Sampling Frequency
h = fdesign.bandpass('n,fst1,fp1,fp2,fst2', N, Fstop1, Fpass1, Fpass2, ...
Fstop2, Fs);
Hd = design(h, 'equiripple');
y = filter(Hd,x);


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Answers (1)

Star Strider
Star Strider on 18 Dec 2020
The ‘y’ output should be the same length as the ‘x’ input. The filter should not change that.
It is important not to confuse time duration with frequency. The frequency displayed will only be up to the Nyquist frequency, half the original sampling frequency. (The Nyquist frequency is the highest frequency that can be uniquely resolvable in a sampled signal.)


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Star Strider
Star Strider on 18 Dec 2020
My pleasure!
If the filter is doing what you want it to do, I wouldn’t change anything.
However, since it’s a FIR filter, an order of 3 is likely much too short. I would start with an order of about 48, and then increase that until you get what you want, unless it’s working optimally as a 48-order filter. Longer filters are less efficient, and with this filter, an order of 256 or less will likely do what you want it to do. I wouldn’t increase it much beyond that unless it isn’t.
farzad on 18 Dec 2020
Thank you very much! seems a good filtering. well there is a disturbing sinusoidal wave inside the signal, possibly from the electricty that feeds the accelerometer, I want to filter it out. it should be around 50 to 60Hz
Star Strider
Star Strider on 18 Dec 2020
If you want to filter out the mains frequency noise, the easiest way would be to use the bandstop function, introduced in R2018a. If you have an earlier version, it’s easy to design a IIR bandstop filter:
Fs = 256000; % Sampling Frequency
Fn = Fs/2; % Nyquist Frequency
Ws = [48 62]/Fn; % Stopband Frequency (Normalised)
Wp = [0.99 1.01].*Ws; % Passband Frequency (Normalised)
Rp = 1; % Passband Ripple
Rs = 90; % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs); % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp,'stop'); % Elliptic Filter Design: Zero-Pole-Gain
[sos,g] = zp2sos(z,p,k); % Second-Order Section For Stability
freqz(sos, 2^20, Fs) % Filter Bode Plot
set(subplot(2,1,1), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
set(subplot(2,1,2), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
Use filtfilt to do the actual filtering:
y = filtfilt(sos,g,x);
Experiment with the ‘Ws’ frequencies to get the result you want. (The ‘Wp’ frequencies are calculated automatically from them.)

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