coefficients of a difference equation?

Hello everybody!

How to find the coefficients of this difference equation in matlab in order to filter a music signal?

Thanks in advance!

[Mathieu NOE]

helo

this looks like an echo flter ... but the answer is in the question , you have written all coefficients of the equation - !

pretty much what is done here :

infile='DirectGuitar.wav';
outfile='out_echo.wav';
% read the sample waveform
[x,Fs] = audioread(infile);

% normalize x to +/- 1 amplitude
x = x ./ (max(abs(x)));

% parameters 
N_delay=20; % delay in samples
C = 0.7; % amplitude of direct sound

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

y = zeros(length(x),1);       % create empty out vector
y(1:N_delay)=x(1:N_delay); % to avoid referencing of negative samples

% for each sample > N_delay
for i = (N_delay+1):length(x)
    y(i) = C*x(i) + (1-C)*(x(i-N_delay));   % add delayed sample
    
end

% write output
% normalize y to +/- 1 amplitude
y = y ./ (max(abs(y)));
audiowrite(outfile, y, Fs);

figure(1)
hold on
plot(x,'r');
plot(y,'b');
title('Echoed and original Signal');
sound(y,Fs);

[RoBoTBoY]

Yes, it is an echo filter and I want to delay 0.15 sec from my original signal

[RoBoTBoY]

This is the original equation

where: c = 0.6 but i don't know the P in order to delay my signal

[Mathieu NOE]

the delay in samples = round( Fs x 0.15 sec)

do you know Fs ?

[RoBoTBoY]

Yes it is 44100Hz

[RoBoTBoY]

Thank so much!! You were very helpful!

Do you know how to implement a reverb filter?

I know that implement with three echo filter in series, but I don't know how to do that in matlab

[Mathieu NOE]

here is it : 3 filters in series :

infile='DirectGuitar.wav';
outfile='out_reverb.wav';
% read the sample waveform
[x,Fs] = audioread(infile);

% normalize x to +/- 1 amplitude
x = x ./ (max(abs(x)));

% parameters (3 delay filters in series)
N1_delay=15; % delay in samples
C1 = 0.7; % amplitude of direct sound

N2_delay=20; % delay in samples
C2 = 0.6; % amplitude of direct sound

N3_delay=50; % delay in samples
C3 = 0.5; % amplitude of direct sound

N_delay = max([N1_delay N2_delay N3_delay]);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% initialization %
y1 = zeros(length(x),1);       % create empty out vector
y2 = y1;
y3 = y1;
y1(1:N_delay)=x(1:N_delay); % to avoid referencing of negative samples
y2(1:N_delay)=x(1:N_delay); % to avoid referencing of negative samples
y3(1:N_delay)=x(1:N_delay); % to avoid referencing of negative samples


% for each sample > N_delay
for i = (N_delay+1):length(x)
    y1(i) = C1*x(i) + (1-C1)*(x(i-N1_delay));   % add delayed sample
    y2(i) = C2*y1(i) + (1-C2)*(y1(i-N2_delay));   % add delayed sample
    y3(i) = C3*y2(i) + (1-C3)*(y2(i-N3_delay));   % add delayed sample
    
end

% write output
% normalize y to +/- 1 amplitude
y3 = y3 ./ (max(abs(y3)));
audiowrite(outfile, y3, Fs);

figure(1)
hold on
plot(x,'r');
plot(y3,'b');
title('Echoed and original Signal');
sound(y3,Fs);

[RoBoTBoY]

Thank you very much!

Do you know about that?

https://www.mathworks.com/matlabcentral/answers/752674-is-this-signal-periodic?s_tid=srchtitle

[Mathieu NOE]

have you tried to do a fft of it ?

example below :

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%% data
[data,Fs] = audioread('test_voice.wav');

channel = 1;
signal = data(:,channel);
samples = length(signal);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 4096;    % 
OVERLAP = 0.75;

% spectrogram dB scale
spectrogram_dB_scale = 80;  % dB range scale (means , the lowest displayed level is XX dB below the max level)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums 

% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;

%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 4;
if decim>1
    signal = decimate(signal,decim);
    Fs = Fs/decim;
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[freq, sensor_spectrum] = myfft_peak(signal,Fs,NFFT,OVERLAP);

% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);

% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(freq);
    sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
    my_ylabel = ('Amplitude (dB (A))');
else
    my_ylabel = ('Amplitude (dB (L))');
end


figure(1),plot(freq,sensor_spectrum_dB,'b');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[sg,fsg,tsg] = specgram(signal,NFFT,Fs,hanning(NFFT),floor(NFFT*OVERLAP));  

% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg));     % NB : X=fft(x.*hanning(N))*4/N; % hanning only

% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(fsg);
    sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
    my_title = ('Spectrogram (dB (A))');
else
    my_title = ('Spectrogram (dB (L))');
end

% saturation of the dB range : 
% saturation_dB = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;

% plots spectrogram
figure(2);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');


function pondA_dB = pondA_function(f)
	% dB (A) weighting curve
	n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
	r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
	pondA = n./r;
	pondA_dB = 20*log10(pondA(:));
end

function  [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
% FFT peak spectrum of signal  (example sinus amplitude 1   = 0 dB after fft).
% Linear averaging
%   signal - input signal, 
%   Fs - Sampling frequency (Hz).
%   nfft - FFT window size
%   Overlap - buffer overlap % (between 0 and 0.95)
signal = signal(:);
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
    s_tmp = zeros(nfft,1);
    s_tmp((1:samples)) = signal;
    signal = s_tmp;
    samples = nfft;
end

% window : hanning
window = hanning(nfft);
window = window(:);

%    compute fft with overlap 
 offset = fix((1-Overlap)*nfft);
 spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
%     % for info is equivalent to : 
%     noverlap = Overlap*nfft;
%     spectnum = fix((samples-noverlap)/(nfft-noverlap));	% Number of windows

    % main loop
    fft_spectrum = 0;
    for i=1:spectnum
        start = (i-1)*offset;
        sw = signal((1+start):(start+nfft)).*window;
        fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft);     % X=fft(x.*hanning(N))*4/N; % hanning only 
    end
    fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling

% one sidded fft spectrum  % Select first half 
    if rem(nfft,2)    % nfft odd
        select = (1:(nfft+1)/2)';
    else
        select = (1:nfft/2+1)';
    end
fft_spectrum = fft_spectrum(select);
freq_vector = (select - 1)*Fs/nfft;
end 

[RoBoTBoY]

What exactly does this example do?

About periodic of signal?

[Mathieu NOE]

this will do averaged fft and spectrogram analysis of any signal , and in case it has some periodic content, it will show up as a peak in the spectrum (at the dominant frequency)

otherwise , if you are sure that your signal is basically a decaying (damped) sinus wave, you can also simply compute the time intervals (= period of oscillation) between consecutives crossing points (positive or negative slope of signal) , given a certain threshold , see example on steady sinus wave attached

[RoBoTBoY]

Hello again!

I have these coefficients from a filter for dereverberation:

b = [6 0 0 0 0 0 0];
a = [1 0 2.45 0 2 0 0.55];

and this is the difference equation:

dereverb(n) = 6*reverb(n)-2.45*dereverb(n-2)-2*dereverb(n-4)-0.55*dereverb(n-6);

also I have these coefficients from a filter for reverb:

b = [0.1664 0 0.4084 0 0.3341 0 0.0911];
a = [1 0 0 0 0 0 0];

and this is the difference equation:

y(n) = 0.1664*x(n)+0.4084*x(n-2)+0.3341*x(n-4)+0.0911*x(n-6);

Have I calculated the dereverberation coefficients correctly? If yes, how to filter a reverb signal through dereverberation filter?

Thanks

[Mathieu NOE]

hello

IMO, your dereverb filter is incorrect. I understand that you take the reverb filter (a FIR filter) and inverse numerator / denominator to create the dereverb filter. But that does not generate a causal , stable and realizable dereverb filter.

It is unstable as you can see by generating the impulse response :

b = [6 0 0 0 0 0 0];
a = [1 0 2.45 0 2 0 0.55];
dimpulse(b,a)

[Mathieu NOE]

the usual technique used in acoustic room compensations , is to make a time delayed version of the flipped FIR filter

you can do that by creating the flipped FIR :

b_flipped = flipud(b')'

then you will see that putting the two filters in series lead to a perfectly flat tranfer function, that is , the impulse response of both filters in series is a sharp pulse (Dirac)

plot(conv(flipud(b')',b))

for your dereverb simulation, you only need to rewritte the difference equation based on b_flipped coefficients (as for the reverb case)

[RoBoTBoY]

I wrote this:

b_dereverb = [4.6297 0 0 0 0 0 0];
a_dereverb = [1 0 2 0 1.3333 0 4];
flipped_b_dereverb = flipud(b_dereverb')';
flipped_a_dereverb = flipud(a_dereverb')';
y_derevereration = filter(flipped_b_dereverb,flipped_a_dereverb,reverb_piano);
y_derevereration = y_derevereration./(max(abs(y_derevereration))); % normalization of dereverb piano
figure(41);
plot(y_derevereration);
title('Dereverb Signal');
xlabel('Time');
pause(3);
sound(y_derevereration,Fs_piano);

But I don't have the expectantly results. I got the same reverb signal.

Why that?

[Mathieu NOE]

hello

the "flip" technique applies only on FIR filter , not IIR; you can see that in many applications dealing with room acoustcs compesntion (electronic equalization of loudspeakers)

infile='DirectGuitar.wav';
outfile1='out_reverb2.wav';
outfile2='out_reverb_derev.wav';

% read the sample waveform
[x,Fs] = audioread(infile);

% normalize x to +/- 1 amplitude
x = x ./ (max(abs(x)));

%%%%%% reverb using FIR filter %%%%%%%%%%%%%%
% b = [0.1664 0 0.4084 0 0.3341 0 0.0911];
% a = [1 0 0 0 0 0 0];
% and this is the difference equation:
y = x; 
for n = 7:length(x)
    y(n) = 0.1664*x(n)+0.4084*x(n-2)+0.3341*x(n-4)+0.0911*x(n-6);
end

% write output
% normalize y to +/- 1 amplitude
y = y ./ (max(abs(y)));
audiowrite(outfile1, y, Fs);

%%%%%% dereverb using flipped FIR filter %%%%%%%%%%%%%%
% b_flipped = flipud(b')' 
% b = [0.0911         0    0.3341         0    0.4084         0    0.1664];
% a = [1 0 0 0 0 0 0];
% and this is the difference equation:
y2 = y; 
for n = 7:length(y)
    y2(n) = 0.0911*y(n)+0.3341*y(n-2)+0.4084*y(n-4)+0.1664*y(n-6);
end

% write output
% normalize y to +/- 1 amplitude
y2 = y2 ./ (max(abs(y2)));
audiowrite(outfile2, y2, Fs);


figure(1)
hold on
plot(x,'r');
plot(y2,'b');
title('Echoed and original Signal');
sound(y2,Fs);

[RoBoTBoY]

So the above code will work?

[Mathieu NOE]

should be working, I tested it on another wav file and there was some improvements, even though the reverb + dereverb process was creating some masking effects - the final wav file sound is not as clean as the original one.

maybe there are more advanced methods...as attached papers