I wrote a java code to read .wav file into a byte array.
.WAV file are 44100Hz (sample per second), 16 bits depth
For audio with length 1 min, I expect to get a byte array with length 60*44100*2. (1 min = 60 sec, each second has 44100 samples, each sample contains 16bits = 2bytes)
However, the array length seems to be doubled. Is there any explanation ?
Assuming this is a normal recording it will be Stereo - 2 channels. So you get all this data for each channel.
I got the answer !! The .WAV file has channel as stereo. So got right and left channel.
Specific question related to the DASH Standard.
A "moof" is followed by "mdat" in MPEG-DASH standard in Segment Templates.
For ex
Segment 1 - duration 2 seconds - moof + mdat
Segment 2 - duration 2 seconds - moof + mdat
Is the "moof" length across segments a constant? Can this be considered a constant to arrive at the mdat offset in subsequent segments.
You can not assume the "moof" length to be constant across segments - it could be, but it does not necessarily have to be.
Why would you assume this? The "moof", like each box in an MP4 file, contains the box size at the very beginning of a box, which you can easily parse.
What do you want to achieve?
When I store the data in a .wav file into a byte array, what do these values mean?
I've read that they are in two-byte representations, but what exactly is contained in these two-byte values?
You will have heard, that audio signals are represented by some kind of wave. If you have ever seen this wave diagrams with a line going up and down -- that's basically what's inside those files. Take a look at this file picture from http://en.wikipedia.org/wiki/Sampling_rate
You see your audio wave (the gray line). The current value of that wave is repeatedly measured and given as a number. That's the numbers in those bytes. There are two different things that can be adjusted with this: The number of measurements you take per second (that's the sampling rate, given in Hz -- that's how many per second you grab). The other adjustment is how exact you measure. In the 2-byte case, you take two bytes for one measurement (that's values from -32768 to 32767 normally). So with those numbers given there, you can recreate the original wave (up to a limited quality, of course, but that's always so when storing stuff digitally). And recreating the original wave is what your speaker is trying to do on playback.
There are some more things you need to know. First, since it's two bytes, you need to know the byte order (big endian, little endian) to recreate the numbers correctly. Second, you need to know how many channels you have, and how they are stored. Typically you would have mono (one channel) or stereo (two), but more is possible. If you have more than one channel, you need to know, how they are stored. Often you would have them interleaved, that means you get one value for each channel for every point in time, and after that all values for the next point in time.
To illustrate: If you have data of 8 bytes for two channels and 16-bit number:
abcdefgh
Here a and b would make up the first 16bit number that's the first value for channel 1, c and d would be the first number for channel 2. e and f are the second value of channel 1, g and h the second value for channel 2. You wouldn't hear much there because that would not come close to a second of data...
If you take together all that information you have, you can calculate the bit rate you have, that's how many bits of information is generated by the recorder per second. In our example, you generate 2 bytes per channel on every sample. With two channels, that would be 4 bytes. You need about 44000 samples per second to represent the sounds a human beeing can normally hear. So you'll end up with 176000 bytes per second, which is 1408000 bits per second.
And of course, it is not 2-bit values, but two 2 byte values there, or you would have a really bad quality.
The first 44 bytes are commonly a standard RIFF header, as described here:
http://tiny.systems/software/soundProgrammer/WavFormatDocs.pdf
and here: http://www.topherlee.com/software/pcm-tut-wavformat.html
Apple/OSX/macOS/iOS created .wav files might add an 'FLLR' padding chunk to the header and thus increase the size of the initial header RIFF from 44 bytes to 4k bytes (perhaps for better disk or storage block alignment of the raw sample data).
The rest is very often 16-bit linear PCM in signed 2's-complement little-endian format, representing arbitrarily scaled samples at a rate of 44100 Hz.
The WAVE (.wav) file contain a header, which indicates the formatting information of the audio file's data. Following the header is the actual audio raw data. You can check their exact meaning below.
Positions Typical Value Description
1 - 4 "RIFF" Marks the file as a RIFF multimedia file.
Characters are each 1 byte long.
5 - 8 (integer) The overall file size in bytes (32-bit integer)
minus 8 bytes. Typically, you'd fill this in after
file creation is complete.
9 - 12 "WAVE" RIFF file format header. For our purposes, it
always equals "WAVE".
13-16 "fmt " Format sub-chunk marker. Includes trailing null.
17-20 16 Length of the rest of the format sub-chunk below.
21-22 1 Audio format code, a 2 byte (16 bit) integer.
1 = PCM (pulse code modulation).
23-24 2 Number of channels as a 2 byte (16 bit) integer.
1 = mono, 2 = stereo, etc.
25-28 44100 Sample rate as a 4 byte (32 bit) integer. Common
values are 44100 (CD), 48000 (DAT). Sample rate =
number of samples per second, or Hertz.
29-32 176400 (SampleRate * BitsPerSample * Channels) / 8
This is the Byte rate.
33-34 4 (BitsPerSample * Channels) / 8
1 = 8 bit mono, 2 = 8 bit stereo or 16 bit mono, 4
= 16 bit stereo.
35-36 16 Bits per sample.
37-40 "data" Data sub-chunk header. Marks the beginning of the
raw data section.
41-44 (integer) The number of bytes of the data section below this
point. Also equal to (#ofSamples * #ofChannels *
BitsPerSample) / 8
45+ The raw audio data.
I copied all of these from http://www.topherlee.com/software/pcm-tut-wavformat.html here
As others have pointed out, there's metadata in the wav file, but I think your question may be, specifically, what do the bytes (of data, not metadata) mean? If that's true, the bytes represent the value of the signal that was recorded.
What does that mean? Well, if you extract the two bytes (say) that represent each sample (assume a mono recording, meaning only one channel of sound was recorded), then you've got a 16-bit value. In WAV, 16-bit is (always?) signed and little-endian (AIFF, Mac OS's answer to WAV, is big-endian, by the way). So if you take the value of that 16-bit sample and divide it by 2^16 (or 2^15, I guess, if it's signed data), you'll end up with a sample that is normalized to be within the range -1 to 1. Do this for all samples and plot them versus time (and time is determined by how many samples/second is in the recording; e.g. 44.1KHz means 44.1 samples/millisecond, so the first sample value will be plotted at t=0, the 44th at t=1ms, etc) and you've got a signal that roughly represents what was originally recorded.
I suppose your question is "What do the bytes in data block of .wav file represent?" Let us know everything systematically.
Prelude:
Let us say we play a 5KHz sine wave using some device and record it in a file called 'sine.wav', and recording is done on a single channel (mono). Now you already know what the header in that file represents.
Let us go through some important definitions:
Sample: A sample of any signal means the amplitude of that signal at the point where sample is taken.
Sampling rate: Many such samples can be taken within a given interval of time. Suppose we take 10 samples of our sine wave within 1 second. Each sample is spaced by 0.1 second. So we have 10 samples per second, thus the sampling rate is 10Hz. Bytes 25th to 28th in the header denote sampling rate.
Now coming to the answer of your question:
It is not possible practically to write the whole sine wave to the file because there are infinite points on a sine wave. Instead, we fix a sampling rate and start sampling the wave at those intervals and record the amplitudes. (The sampling rate is chosen such that the signal can be reconstructed with minimal distortion, using the samples we are going to take. The distortion in the reconstructed signal because of the insufficient number of samples is called 'aliasing'.)
To avoid aliasing, the sampling rate is chosen to be more than twice the frequency of our sine wave (5kHz)(This is called 'sampling theorem' and the rate twice the frequency is called 'nyquist rate'). Thus we decide to go with sampling rate of 12kHz which means we will sample our sine wave, 12000 times in one second.
Once we start recording, if we record the signal, which is sine wave of 5kHz frequency, we will have 12000*5 samples(values). We take these 60000 values and put it in an array. Then we create the proper header to reflect our metadata and then we convert these samples, which we have noted in decimal, to their hexadecimal equivalents. These values are then written in the data bytes of our .wav files.
Plot plotted on : http://fooplot.com
Two bit audio wouldn't sound very good :) Most commonly, they represent sample values as 16-bit signed numbers that represent the audio waveform sampled at a frequency such as 44.1kHz.
In anime, does frame means number of scene per second? Each scene can consist of several layer background, hero, object, etc. I think this is the reason why I am confused.
In wave (raw audio) file,
Does one frame contain data for one side (left or right) only?
Does bit sampling precision refer to a single side/channel?
With audio, do frames represent changes in loudness?
One frame can consist of left and right?
I.e. stereo 8 bit sampling depth => 1 frame => 2 bytes?
I do not know whether a formal definition of a frame exists, but when referring to an audio frame we usually mean a single time sample of a number of channels. So 2 audio channels # 8 bits per channel results in 2 bytes per frame. 4 channels # 16 bit per sample is 8 bytes.
I am wondering on the relationship between a block of samples and its time equivalent. Given my rough idea so far:
Number of samples played per second = total filesize / duration.
So say, I have a 1.02MB file and a duration of 12 sec (avg), I will have about 89,300 samples played per second. Is this right?
Is there other ways on how to compute this? For example, how can I know how much a byte[1024] array is equivalent to in time?
Generally speaking for PCM samples you can divide the total length (in bytes) by the duration (in seconds) to get the number of bytes per second (for WAV files there will be some inaccuracy to account for the header). How these translate into samples depends on
the sample rate
bits used per sample, i.e. commonly
used is 16 bits = 2 bytes
number of channels, i.e. for stereo
this is 2
If you know 2) and 3) you can determine 1)
In your example 89300 bytes/second, assuming stereo and 16 bits per sample would be 89300 / 4 ~= 22Khz sample rate
In addition to #BrokenGlass's very good answer, I'll just add that for uncompressed audio with a fixed sample rate, number of channels and bits per sample, the arithmetic is fairly straightforward. E.g. for "CD quality" audio we have a 44.1 kHz sample rate, 16 bits per sample, 2 channels (stereo), therefore the data rate is:
44100 * 16 * 2
= 1,411,200 bits / sec
= 176,400 bytes / sec
= 10 MB / minute (approx)