Given an audio file, how can I construct a histogram of all bipolar pulses?

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AZ0 2023-11-14

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https://ww2.mathworks.cn/matlabcentral/answers/2046907-given-an-audio-file-how-can-i-construct-a-histogram-of-all-bipolar-pulses

评论： Star Strider 2023-11-27

Hi! I'm looking to take an audio file (.flac) and append each instance of a bipolar pulse to a histogram vector. Is there a version of the "findpeaks" function that only looks for bipolar signals (ie: looks for peaks that are adjacent to a negative peak)? If not, how would I go about constructing a signal template and processing through my file to search for signals of a similar shape? I'm not sure what key words I should be Googling (maybe wavelet or templating?) so any help is appreciated. Thanks!

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Answer 1

Star Strider 2023-11-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2046907-given-an-audio-file-how-can-i-construct-a-histogram-of-all-bipolar-pulses#answer_1351907

在 MATLAB Online 中打开

The findpeaks function can easily detect negative peaks.

Simply negate the original signal to get them —

t = linspace(0, 50, 125);

s = tan(2*pi*t);

[pks,plocs] = findpeaks(s, 'MinPeakProminence',10); % Peaks

[vys,vlocs] = findpeaks(-s, 'MinPeakProminence',10); % Valleys

Pulse = plocs(plocs-vlocs <= 10)+[-5; 10]; % Pulses

figure

plot(t, s)

hold on

plot(t(plocs), pks, '^r')

plot(t(vlocs), -vys, 'vg')

plot(t(Pulse), [10; 10], '-k', 'LineWidth',2)

hold off

grid

axis('padded')

I do not understand the histogram reference.

Make appropriate changes to get the result you want.

.

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Star Strider 2023-11-15

在 MATLAB Online 中打开

My pleasure!

First, this line:

Pulse = plocs(plocs-vlocs <= 10)+[-5; 10]; % Pulses

defines each pulse (according to my criteria) and is then used to plot a horizonotal balck line identifying them. It identifies the beginning of a pulse if the difference between the peak and valley are less than 10 index units. Once you have identified the pulses, you could then use the maximum in that signal range (or the peak-to-peak value) to create a vector for the histogram counts function to use.

That might go something like this —

t = linspace(0, 1500, 1250);

s = tan(2*pi*t);

[pks,plocs] = findpeaks(s, 'MinPeakProminence',50); % Peaks

[vys,vlocs] = findpeaks(-s, 'MinPeakProminence',50); % Valleys

Pulse = plocs(plocs-vlocs <= 10)+[-5; 10] % Pulses

Pulse = 2×12

47 149 256 358 460 567 669 776 878 980 1087 1189 62 164 271 373 475 582 684 791 893 995 1102 1204

for k = 1:size(Pulse,2)

Pulsev{k,:} = Pulse(1,k) : Pulse(2,k);

maxPulse(k) = max(s(Pulsev{k}));

p2pPulse(k) = max(s(Pulsev{k})) - min(s(Pulsev{k}));

end

maxPulse

maxPulse = 1×12

159.0255 53.0029 795.1377 72.2807 37.8549 113.5882 46.7657 265.0448 61.1590 34.5616 88.3449 41.8414

p2pPulse

p2pPulse = 1×12

200.8669 141.3478 829.6993 133.4397 302.8997 160.3539 160.3539 302.8997 133.4397 829.6993 141.3478 200.8669

figure

plot(t, s)

hold on

plot(t(plocs), pks, '^r')

plot(t(vlocs), -vys, 'vg')

plot(t(Pulse), [1; 1]*50, '-k', 'LineWidth',2)

hold off

grid

axis('padded')

xlabel('Time')

ylabel('Amplitude (mV)')

figure

histogram(maxPulse, 10)

xlabel('Pulse Maximum')

ylabel('Counts')

title('Pulse Height')

figure

histogram(p2pPulse, 10)

xlabel('Pulse Magnitude')

ylabel('Counts')

title('Pulse Peak-To-Peak')

Thjat seems to work acceptably well with my synthetic data. You will likely have to adjust the 'MinPeakProminence' values (individually for the peaks and valleys) to fit your actual data

.

Star Strider 2023-11-24

在 MATLAB Online 中打开

It would be helpful to have the signal (at least a representative sample) to work with.

I did my best to simulate the situation you describe, setting the 'MinPeakProminence' in the ‘Valleys’ detection to produce fewer of them than there are ‘Peaks’.

This works with my simulated data —

t = linspace(0, 1500, 1250).';

s = tan(2*pi*t);

[pks,plocs] = findpeaks(s, 'MinPeakProminence',50); % Peaks

[vys,vlocs] = findpeaks(-s, 'MinPeakProminence',125); % Valleys

for k = 1:numel(vlocs)

[dist,ix] = min(abs(plocs-vlocs(k)));

Pulse(k,:) = [NaN NaN];

if dist <= 10

Pulse(k,:) = vlocs(k)+ix+[-5; 10]; % Pulses

end

Pulse

Pulse = 5×2

156 171 470 485 681 696 995 1010 1206 1221

for k = 1:size(Pulse,1)

Pulsev{k,:} = Pulse(k,1) : Pulse(k,2);

maxPulse(k) = max(s(Pulsev{k}));

p2pPulse(k) = max(s(Pulsev{k})) - min(s(Pulsev{k}));

end

maxPulse

maxPulse = 1×5

1.6511 1.6234 1.7592 1.7287 1.8783

p2pPulse

p2pPulse = 1×5

89.9960 266.6681 27.3958 33.5237 16.8587

figure

hp{1} = plot(t, s, 'DisplayName','Signal');

hold on

hp{2} = plot(t(plocs), pks, '^r', 'DisplayName','Peaks');

hp{3} = plot(t(vlocs), -vys, 'vg', 'DisplayName','Valleys');

hp{4} = plot(t(Pulse)+[-10 10], [1 1]*200, '-k', 'LineWidth',5, 'DisplayName','Identified Biploar Pulses');

hold off

grid

axis('padded')

xlabel('Time')

ylabel('Amplitude (mV)')

legend([hp{1},hp{2},hp{3},hp{4}(1)],'Location','best')

figure

histogram(maxPulse, 10)

xlabel('Pulse Maximum')

ylabel('Counts')

title('Pulse Height')

figure

histogram(p2pPulse, 10)

xlabel('Pulse Magnitude')

ylabel('Counts')

title('Pulse Peak-To-Peak')

This iterates through the ‘Valleys’ (since there are fewer of them), and then uses the minimum peak distance from each to define a bipolar pulse. I included a distance test for ‘Pulse’ and set the entries that do not meet that test to NaN. Set the distance interval (here 10) to whatever you want.

Also, my code here works equally well with row or column vectors for ‘t’ and ‘s’. (I tested it to be sure.)

Thank you! I do, and I did. I hope you did, too!

.

Star Strider 2023-11-26

My pleasure.

I prefer not to download files from external sites because I cannot download them to here and I do not want to download them to my computer and then upload them here. (Ever since the Run utility — the right-facing green arrow in the top toolstrip — was added here, I have used it or MATLAB Online almost exclusively to develop and run code on MATLAB Answers.) Use the ‘paperclip’ icon to the right of the Σ to upload them here. If they have an unrecognised extension (such as .dat), use the zip function to enclose it in a .zip file and then upload the .zip file.

I will wait for the file to appear here.

The ‘Pulse’ loop iterates through ‘vlocs’ because you mentioned that there are fewer valleys than peaks. The ‘Pulse’ assignment begins with the ‘vlocs’ index, adds ‘ix’ to it because the min function returns the index (‘ix’) relative to the beginning of the difference between the ‘plocs’ vector and the indexed ‘vlocs’ value, so adding ‘ix’ produces the absolute index (for the entire vector). The loop counter ‘k’ is the ‘vlocs’ index.

The size-of-Pulse loop uses the ‘[-5; 10]’ offset vector result in ‘Pulse’ to create a range to determine the maximum and peak-to-peak values within that range. (I chose the ‘[-5; 10]’ range because it works with my simulated data. It may need to be changed to work with real data.)

.

Star Strider 2023-11-26

在 MATLAB Online 中打开

matlab sample audio.flac.zip

My pleasure!

After a bit of tweaking of my original code, this works —

% t = linspace(0, 1500, 1250).';

% s = tan(2*pi*t);

Uz = unzip('matlab sample audio.flac.zip')

Uz = 1×1 cell array

{'matlab sample audio.flac'}

[s, Fs] = audioread(Uz{1})

s = 2128109×2

-0.0085 0.0378 -0.0055 0.0339 -0.0032 0.0305 0.0019 0.0240 0.0073 0.0196 0.0109 0.0227 0.0109 0.0289 0.0156 0.0273 0.0191 0.0294 0.0172 0.0304

Fs = 4410

L = size(s,1);

t = linspace(0, L-1, L).'/Fs;

figure

plot(t, s(:,1), '-', 'DisplayName','Left Channel')

hold on

plot(t, s(:,2), 'DisplayName','Left Channel')

hold off

grid

xlabel('Time (s)')

ylabel('Amplitude (Units)')

title('Signals Plot')

legend('Location','best')

SelectIdx = [6483 6512];

SelectT = t(SelectIdx);

SelectWidth = diff([1556 1606])

SelectWidth = 50

figure

plot(t, s(:,1), '.-', 'DisplayName','Left Channel')

hold on

plot(t, s(:,2), 'DisplayName','Left Channel')

hold off

grid

xlabel('Time (s)')

ylabel('Amplitude (Units)')

xlim(SelectT+[-0.015; 0.015])

xline(SelectT(1), '--c', 'DisplayName','Valley')

xline(SelectT(2), '--g', 'DisplayName','Peak')

legend('Location','best')

title('Arbitrary ‘Zoom’ To Examine Signals Detail')

s = s(:,1); % Choose Left Channel

[pks,plocs] = findpeaks(s, 'MinPeakProminence',0.2); % Peaks

[vys,vlocs] = findpeaks(-s, 'MinPeakProminence',0.2); % Valleys

ofst = [-50 50];

for k = 1:numel(vlocs)

[dist,ix] = min(abs(plocs-vlocs(k)));

Pulse(k,:) = [NaN NaN];

if dist <= 100

idxrng = vlocs(k)+ix+ofst;

if idxrng(1) < 1

idxrng(1) = 1;

end

if idxrng(2) > L

idxrng(2) = L;

end

Pulse(k,:) = idxrng; % Pulses

end

NrNaN = nnz(isnan(Pulse(:,1)))

NrNaN = 105

% Pulse

for k = 1:size(Pulse,1)

maxPulse(k,:) = NaN;

p2pPulse(k,:) = NaN;

if ~all(isnan(Pulse(k,:)),2) & ~all(isempty(Pulse(k,:)),2)

Pulsev = Pulse(k,1) : Pulse(k,2);

if numel(Pulsev) < 2 % Check For Empty 'Pulsev'

break

end

maxPulse(k,:) = max(s(Pulsev));

p2pPulse(k,:) = max(s(Pulsev)) - min(s(Pulsev));

end

% maxPulse

% p2pPulse

figure

hp{1} = plot(t, s, 'DisplayName','Signal');

hold on

hp{2} = plot(t(plocs), pks, '^r', 'DisplayName','Peaks');

hp{3} = plot(t(vlocs), -vys, 'vg', 'DisplayName','Valleys');

% hp{4} = plot(t(Pulse)+[-10 10], [1 1]*200, '-k', 'LineWidth',5, 'DisplayName','Identified Biploar Pulses');

hold off

grid

axis('padded')

xlabel('Time')

ylabel('Amplitude (mV)')

% legend([hp{1},hp{2},hp{3},hp{4}(1)],'Location','best')

legend([hp{1},hp{2},hp{3}],'Location','best')

figure

histogram(maxPulse, 5E+2, 'EdgeColor','none')

xlabel('Pulse Maximum')

ylabel('Counts')

title('Pulse Height')

figure

histogram(p2pPulse, 5E+2, 'EdgeColor','none')

xlabel('Pulse Magnitude')

ylabel('Counts')

title('Pulse Peak-To-Peak')

Not much needed tweaking, since my original code accounted for most of it. I had to change the ‘ofst’ index range to accommodate the sampling intervals with respect to the peaks, and adjusted the detection limits in the findpeaks calls. (Those use my favourite name-value pair argument —'MinPeakProminence' — however you can choose any criterion and detection level level that works best for you.) Some other changes were necessary with respect to the indexing (it took a few minutes to figure out where that problem — indexing beyond the ranges of the vectors — actually was, fixing it involved an if block to check ‘Pulsev’) and with those tweaks, it works. It should do what you requested.

.

Star Strider 2023-11-27

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

Star Strider 2023-11-27

Thank you!

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Given an audio file, how can I construct a histogram of all bipolar pulses?

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Given an audio file, how can I construct a histogram of all bipolar pulses?

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