Convert .csv to .wav audio
38 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Debagnik
am 16 Apr. 2020
Beantwortet: Walter Roberson
am 10 Aug. 2023
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);
0 Kommentare
Akzeptierte Antwort
Geoff Hayes
am 16 Apr. 2020
Bearbeitet: Geoff Hayes
am 16 Apr. 2020
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 Kommentare
Geoff Hayes
am 17 Apr. 2020
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));
Weitere Antworten (2)
Mehreen Jabbeen
am 25 Apr. 2021
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);
0 Kommentare
Walter Roberson
am 10 Aug. 2023
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.
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);
0 Kommentare
Siehe auch
Kategorien
Mehr zu Transforms finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
