Convert .csv to .wav audio

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Debagnik
Debagnik 2020 年 4 月 16 日
回答済み: Walter Roberson 2023 年 8 月 10 日
I am fairly new to matlab
I wanted to convert a comma separated value (.csv) file to a Wave (.wav) audio file while also plotting it.
I saw this previously answered Here on matlab answers but I had a few questions and some errors, for the sake of simplicity I have attached the code and the csv file here
error shown is
Undefined function or variable 'y'.
Error in Untitled (line 17)
audiowrite('audio_output.wav',y,Fs)
and I did not understand why in the code the variable 'y' and 'Fs' are there in code
clc;
clear all;
close all;
data=csvread("output.csv");
time=data(: ,1);
output=data(: ,2);
plot(time,output);
save('aec.mat','time','output');
load aec.mat
filename='audio_output.wav';
Fs=6000;
audiowrite('audio_output.wav',y,Fs)
clear y Fs
[y,Fs]=audioread(filename);
sound(y,Fs);

採用された回答

Geoff Hayes
Geoff Hayes 2020 年 4 月 16 日
編集済み: Geoff Hayes 2020 年 4 月 16 日
Debagnik - the y variable represents the audio samples. In your case, that would be output. Just change your code to the following
time = data(: ,1);
y = data(: ,2);
plot(time, y);
I see that you have set the sampling rate, Fs, to be 6000...just like in the link that you have provided. If your time data is in seconds, then it looks like you have 2.5 seconds (time(end) - time(1)) worth of data. Since there are 108569 (length(output)) samples, then this would mean your Fs could be
Fs = length(output) / (time(end) - time(1)); % 43427.6
Is this correct? Or is your sampling freqency really 6000?
  4 件のコメント
Debagnik
Debagnik 2020 年 4 月 17 日
another followup
as per the first error in the previous error that they expected the sample rate to be integer so I used round() function in that line
Fs=round(length(data)/(time(end)-time(1)); %43428
hence mostly all my probles are solved
except this warning.
Warning: Data clipped when writing
file.
> In audiowrite>clipInputData (line 396)
In audiowrite (line 176)
In Untitled (line 18)
Geoff Hayes
Geoff Hayes 2020 年 4 月 17 日
From audiowrite y parameter, The valid range for the data in y depends on the data type of y. For data types double (your input) the range is -1.0 to +1.0. Your input is between -6.5 and 6.0. I suppose that you could divide your data y by the maximum absolute value:
time = data(: ,1);
y = data(: ,2);
y = y / max(abs(min(y)),max(y));

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その他の回答 (2 件)

Mehreen Jabbeen
Mehreen Jabbeen 2021 年 4 月 25 日
clc;
clear all;
close all;
y=csvread("original.csv");
display(y)
filename='yes.wav';
Fs=16000;
audiowrite(filename,y,Fs)
clear y Fs
[y,Fs]=audioread(filename);
display(y)
display(Fs)
sound(y,Fs);

Walter Roberson
Walter Roberson 2023 年 8 月 10 日
The time in the file is not uniformly sampled. The data needs to be resampled before being written to a file. Also, the file contains duplicate times with different data.
From previous discussion and the evidence of the file, we know that the file is intended to represent 2.5 seconds. The number of entries in the file is about 7 1/2 times greater than 2.5 seconds * 6000 samples/second .
We cannot just resample because the input is not regular timesteps. We cannot just use fft methods either for the same reason. We cannot use interp1 because of the duplicate times.
I experimented with nufft but the results I got back were suspicious, and I do not trust them yet.
data = csvread("output.csv");
time = data(: ,1);
output = data(: ,2);
filename='audio_output.wav';
Fs=6000;
TT = timetable(output, 'RowTimes', seconds(time));
TT = retime(TT, 'regular', 'mean', 'TimeStep', seconds(1/Fs));
t = TT.Properties.RowTimes;
y = rescale(TT.output, -1, 1);
audiowrite('audio_output.wav', y, Fs)
plot(t, y)
sound(y, Fs);

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