# Interpolation on same number of samples after FIR filtration

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Lukas Poviser il 3 Mag 2022
Commentato: Star Strider il 3 Mag 2022
Hi,
I have made FIR filtration of my signal (sampled 25 Hz, 4500 samples) and after filtration that signal has about 200 samples less. The original signal is synced with another signal. And I need these two signals to work together and have same number of samples. What is best solution? How to interpolate filtrated signal to the same number of samples like before?
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### Risposta accettata

Star Strider il 3 Mag 2022
Filtering should not change the signal length.
I am not certain exactly what you are doing, how your designed your FIR filter, or what your signal is.
I always suggest using the filtfilt function to do the actual filtering.
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Star Strider il 3 Mag 2022
That is my filter. Before filtration the signal has 4538 samples and after it has just 4477 samples.
I cannot reproduce that behaviour —
y2 = randn(1,4538);
fs = 25;
fc = 0.65; % cut off frequency
Nfir = 124; % order of filter
N = length(y2);
wc=fc/(fs/2);
fil=fir1(Nfir,wc); % Filter design
filt_y_fir = filter(fil,1,y2); % filtering
filtered_signal_size = size(filt_y_fir)
filtered_signal_size = 1×2
1 4538
filt_y_fir = filtfilt(fil,1,y2); % filtering
filtered_signal_size = size(filt_y_fir)
filtered_signal_size = 1×2
1 4538
figure
freqz(fil,1,2^16,fs)
.

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### Più risposte (1)

Lukas Poviser il 3 Mag 2022
% y2 is vector with original data
fs = 25;
fc = 0.65; % cut off frequency
Nfir = 124; % order of filter
N = length(y2);
wc=fc/(fs/2);
fil=fir1(Nfir,wc); % Filter design
filt_y_fir = filter(fil,1,y2); % filtering
delay = mean(grpdelay(fil)); % removing time shift
tt = tt(1:end-delay);
sn = y2(1:end-delay);
sf = filt_y_fir;
sf(1:delay) = [];
That is my filter. Before filtration the signal has 4538 samples and after it has just 4477 samples.
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