reading a raw audio file as Matlab does in C

Question

I have the following small script that I want to write in C :

`%% getting the spectgrum 
clear, clc ;

 fileName ='M0.raw'
   [x,fs] = audioread(fileName);
   [xPSD,f] = pwelch(x,hanning(8192),0,8192*4 ,fs);
   plot(f,10*log10(abs(xPSD)));
   xlim([0 22e3]);
   absxPSD = abs(xPSD);
   save('absXPSD.txt','absxPSD','-ascii');
   save('xPSD.txt','xPSD','-ascii');
   save('xValues.txt','x','-ascii');
   save('frequency.txt','f','-ascii');`

without goning in details, I have a problem getting the correct result, when I checked I figured out that the data that I read is wrong here's sample that read the raw file to compare it with what Matalb reads :

#include <stdio.h>
#include <stdlib.h>
int main (){

FILE* inp =NULL;
FILE* oup =NULL;
double value =0;

inp = fopen("M0.raw","r");
oup = fopen("checks.txt","w+");

UPDATE after LoPiTaL's answer I've tried to jump over the RIFF header which is 44Byte length0 using fseek

fseek (inp,352,SEEK_SET);// that didn't help getting the right result !! 

if( inp == NULL || oup==NULL){
    printf(" error at file opning \n");
    return -1;
}

while (!(feof(inp))){

    fread(&value,sizeof(double),1,inp);
    printf(" %f \n ",value);
    fprintf(oup,"%f\n",value);
}
fclose(inp);
fclose(oup);
return 0;

}

and the result that I get is :

    -28083683309813134333858080554409220100578902032859386180468433149049781495379346137536863936326139303879846829175766826833343673613788446579155215033623707200818670767132304934425064429529496303287641688697019947073799877821581901737052884168025721481955133510652655692037990001524306465271815108431928360960.000000
0.000000
0.000000
0.000000
0.000000
0.000000
-20701636078248669570005757343846586744027511881225108933223144646890577802102653022204406730988428912367583701134782419138464527797567258583836429190479797597328678189654150340845........................................................................

and my aim is to get those value :

  -1.0162354e-02
  -9.3688965e-03
  -7.5073242e-03
  -1.9531250e-03
   3.7231445e-03
   1.3549805e-02
   2.3223877e-02
   3.2867432e-02
   4.4830322e-02
   5.5114746e-02
   6.7291260e-02
   7.7636719e-02
   8.8562012e-02
   9.5794678e-02
   1.0055542e-01
   1.0415649e-01
   1.0351563e-01
   1.0235596e-01
   9.8785400e-02
   9.1796875e-02
   8.3648682e-02
   7.1594238e-02

the audio file is mono an is 16bit resolution , any idea how can solve this ? thanks for any help

score 2 · Answer 1 · answered Jul 08 '14 at 11:53

You must open the file in binary mode, for starters. Otherwise you get text mode, which can do translations of line endings, for instance. Not good with binary data.

Binary mode:

inp = fopen("M0.raw", "rb");
                        ^
                        |
                       muy
                   importante

LoPiTaL · Accepted Answer · 2014-07-08T13:28:10.973

1

Sure, you cannot read an audio file as is and hope that the data is as you think it is.

Ignoring any coded audio file, which of course you have to decode prior to read it, lets focus in the RAW audio files:

RAW audio files usually are WAV files. WAV files have a .RIFF header at the beginning of the file, which obviously you would have to ignore before reading audio data.

http://en.wikipedia.org/wiki/Resource_Interchange_File_Format

After you have removed the RIFF header, then the data starts. As you stated, the data is encoded as 16 bit resolution. 16 bit resolution means that 0x0000 is 0.0 and 0xFFFF is 1.0, and the size of the data is only two bytes! So you have to read two bytes at a time (i.e. with a signed short) and then do the conversion to the range 0 to 1:

signed short ss;
double value;
FILE* inp =NULL;
inp = fopen("M0.raw","rb");  //As stated in other answer, use binary mode!
fseek (inp,44,SEEK_SET);     // Only 44 bytes!!
//We already have discarded the header here....

while (fread(&ss, sizeof(signed short) ,1 , inp) == 1){
    //Now we have to convert from signed short to double:
    value=((double)ss)/(unsigned)0xFFFF;

    //Print the results:
    printf(" %f \n ",value);
    fprintf(oup,"%f\n",value);
}

Of course, the function "audioread" from Matlab already does all of this for you, so you don't have to care about the encoding, as in your example, your particular data is in 16 bit, but if you use any other file, it could be 8, 16, 24 or 32, even could be differential or be encoded despite being a WAV file (see the RIFF header for more information).

edited Jul 08 '14 at 13:28

answered Jul 08 '14 at 11:03

LoPiTaL

2,495
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thanks for your answer ! I have to figure out how to access the data directly ! thanks again – Engine Jul 08 '14 at 11:26
one question how big is this header after reading this https://ccrma.stanford.edu/courses/422/projects/WaveFormat/ it appears to 44Byte meaning 352 bits I've tried this but it didn't helps to ! any idea ? – Engine Jul 08 '14 at 11:58
1

@Engine Yes, it seems to be 44 bytes (although it is not necessary to be always like this, but lets supose in your file it is), why you need the size in bits??? Read the documentation on fseek: http://www.cplusplus.com/reference/cstdio/fseek/, and you will see that the offset is in BYTES, not bits I have updated my answer to include the seek to jump the header and the opening mode to be binary mode, instead of text mode, as stated by unwind in his answer – LoPiTaL Jul 08 '14 at 13:21
thanks for replying I tried this to and it still don't work , I guess that the header is after all not 44 Bytes ! any Idea how to figure out how long it is ? – Engine Jul 08 '14 at 13:34
Well, you can watch directly the file using a binary reader (i.e.hexEdit http://www.hexedit.com/), so you can see what you have to read and what you actually read. – LoPiTaL Jul 08 '14 at 13:39
After re-reading your update, I see that you haven't updated all the code, so have you changed what I told you in my answer? Have you changed the fread and have added the division? Sorry if you already have, but since I didn't see it in your update, I have to ask... :D – LoPiTaL Jul 08 '14 at 13:41

reading a raw audio file as Matlab does in C

2 Answers2