Frequency Response to IR calculation yields shallower Magnitude level?
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Michaela Warnecke
il 4 Ott 2023
Commentato: Mathieu NOE
il 9 Ott 2023
Hi all,
I have a frequency response curve of an audio recording. All I know is that is spans some range (see below) and was (most likely) sampled at fs = 48k.
I need to convert it to an IR to apply some specific filtering and am using the following code for that conversion:
FFT_length = 1024;
fs = 48e3;
%freq_axis = 1xN vector containing the range of frequencies in Hz ([71:20e3] in some octave steps)
%FR_data = 1xN vectora containing the magnitde values at each freq in freq_axis
%convert frequencies to normalized values to generate IR filters
filter_freqs = [0 freq_axis./(fs/2) 1];
%Create time-domain IRs
IR_sig = fir2(FFT_length, filter_freqs, [FR_data(1)-1 FR_data FR_data(end)+1]);
This creates an IR signal, which I then convert back into the frequency domain to validate against the original FR using the following:
cutoff_freq = fs/2;
HRTF_mag = abs(fft(IR_sig, FFT_length));
%Return only from 0 to pi
Mag = 20*log10(HRTF_mag(1:(FFT_length/2)+1));
f_vec = [0:(cutoff_freq*2)/FFT_length:cutoff_freq];
When I plot the original Frequency response curve (FR_data, left) and also plot the IR-converted, frequency domain signal (Mag, right), I can see a close match in the frequency domain, but the magnitude is entirely off. You can see that the original data (left) shows a difference of ~30 dB between the notch at 2.5k and the peak at 4.1k. The same notch and peak are only about 5dB apart in the deduced FR (Mag).
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1502214/image.png)
What am I missing/doing wrong?
2 Commenti
Paul
il 6 Ott 2023
Hi Michaela,
You're more likely to get help if you upload the data needed to run your code and recreate your results. Save freq_axis and FR_data (if that's all that's needed) to a .mat file and add the file to your question using the paperclip icon in the Insert ribbon. Also, can you add the plotting commands used to make the plots.
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Mathieu NOE
il 6 Ott 2023
hi
I have done something similar in the past - see below FYI
I start with a given FIR filter then computation of the transfer function and backward to the FIR ( = impulse response of a TF)
%
Fs = 1e3;
coef_fs = 2; % 2 for Nyquist (exact match) , measurement factor can be 2.56
Freq = linspace(0,Fs/coef_fs,200); % make sure freq vector goes up to Fs/2
b = fir1(48,[0.2 0.3]); % Window-based FIR filter design
frf = freqz(b,1,Freq,Fs);
%% FIR obtained with ifft method
if mod(length(frf),2)==0 % iseven
frf_sym = conj(frf(end:-1:2));
else
frf_sym = conj(frf(end-1:-1:2));
end
fir = real(ifft([frf frf_sym]));
% resample fir (impulse response) if frf data is not provided up to Nyquits frequency (Fs/2)
corfactor = 2/coef_fs;
x_og = (1:length(fir));
x_cor = linspace(1,length(fir)*corfactor,length(fir));
fir = interp1(x_og,fir,x_cor);
fir = fir(1:50); % truncation is possible if fir decays enough
frfid = freqz(fir,1,Freq,Fs);
% alternative with invfreqz
ITER = 1e3;
% FIR filter design
NB = 50; %
NA = 0;
W = 2*pi*Freq/Fs;
Wt = ones(size(W));
TOL = 1e-2;
[fir2,A] = invfreqz(frf,W,NB,NA,Wt,ITER,TOL);
frfid2 = freqz(fir2,1,Freq,Fs);
figure(1)
subplot(311),plot(1:length(b),b,'-*',1:length(fir),fir,1:length(fir2),fir2);
legend('FIR model','identified FIR (fft)','identified FIR (invfreqz)');
xlabel('samples');
ylabel('amplitude');
subplot(312),plot(Freq,20*log10(abs(frf)),Freq,20*log10(abs(frfid)),Freq,20*log10(abs(frfid2)),'*');
legend('input model','identified FIR model (fft)','identified FIR model (invfreqz)');
xlabel('frequency (Hz)');
ylabel('FRF modulus (dB)');
subplot(313),plot(Freq,180/pi*angle(frf),Freq,180/pi*angle(frfid),Freq,180/pi*angle(frfid),'*');
legend('input model','identified FIR model (fft)','identified FIR model (invfreqz)');
xlabel('frequency (Hz)');
ylabel('FRF angle (°)');
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