What does interleaved stereo PCM linear Int16 big endian audio look like?

Question

I know that there are a lot of resources online explaining how to deinterleave PCM data. In the course of my current project I have looked at most of them...but I have no background in audio processing and I have had a very hard time finding a detailed explanation of how exactly this common form of audio is stored.

I do understand that my audio will have two channels and thus the samples will be stored in the format [left][right][left][right]... What I don't understand is what exactly this means. I have also read that each sample is stored in the format [left MSB][left LSB][right MSB][right LSB]. Does this mean the each 16 bit integer actually encodes two 8 bit frames, or is each 16 bit integer its own frame destined for either the left or right channel?

Thank you everyone. Any help is appreciated.

Edit: If you choose to give examples please refer to the following.

Method Context

Specifically what I have to do is convert an interleaved short[] to two float[]'s each representing the left or right channel. I will be implementing this in Java.

public static float[][] deinterleaveAudioData(short[] interleavedData) {
    //initialize the channel arrays
    float[] left = new float[interleavedData.length / 2];
    float[] right = new float[interleavedData.length / 2];
    //iterate through the buffer
    for (int i = 0; i < interleavedData.length; i++) {
        //THIS IS WHERE I DON'T KNOW WHAT TO DO
    }
    //return the separated left and right channels
    return new float[][]{left, right};
}

My Current Implementation

I have tried playing the audio that results from this. It's very close, close enough that you could understand the words of a song, but is still clearly not the correct method.

public static float[][] deinterleaveAudioData(short[] interleavedData) {
    //initialize the channel arrays
    float[] left = new float[interleavedData.length / 2];
    float[] right = new float[interleavedData.length / 2];
    //iterate through the buffer
    for (int i = 0; i < left.length; i++) {
        left[i] = (float) interleavedData[2 * i];
        right[i] = (float) interleavedData[2 * i + 1];
    }
    //return the separated left and right channels
    return new float[][]{left, right};
}

Format

If anyone would like more information about the format of the audio the following is everything I have.

• Format is PCM 2 channel interleaved big endian linear int16
• Sample rate is 44100
• Number of shorts per short[] buffer is 2048
• Number of frames per short[] buffer is 1024
• Frames per packet is 1

Answer 1

I do understand that my audio will have two channels and thus the samples will be stored in the format [left][right][left][right]... What I don't understand is what exactly this means.

Interleaved PCM data is stored one sample per channel, in channel order before going on to the next sample. A PCM frame is made up of a group of samples for each channel. If you have stereo audio with left and right channels, then one sample from each together make a frame.

• Frame 0: [left sample][right sample]
• Frame 1: [left sample][right sample]
• Frame 2: [left sample][right sample]
• Frame 3: [left sample][right sample]
• etc...

Each sample is a measurement and digital quantization of pressure at an instantaneous point in time. That is, if you have 8 bits per sample, you have 256 possible levels of precision that the pressure can be sampled at. Knowing that sound waves are... waves... with peaks and valleys, we are going to want to be able to measure distance from the center. So, we can define center at 127 or so and subtract and add from there (0 to 255, unsigned) or we can treat those 8 bits as signed (same values, just different interpretation of them) and go from -128 to 127.

Using 8 bits per sample with single channel (mono) audio, we use one byte per sample meaning one second of audio sampled at 44.1kHz uses exactly 44,100 bytes of storage.

Now, let's assume 8 bits per sample, but in stereo at 44.1.kHz. Every other byte is going to be for the left, and every other is going to be for the R.

LRLRLRLRLRLRLRLRLRLRLR...

Scale it up to 16 bits, and you have two bytes per sample (samples set up with brackets [ and ], spaces indicate frame boundaries)

[LL][RR] [LL][RR] [LL][RR] [LL][RR] [LL][RR] [LL][RR]...

I have also read that each sample is stored in the format [left MSB][left LSB][right MSB][right LSB].

Not necessarily. The audio can be stored in any endianness. Little endian is the most common, but that isn't a magic rule. I do think though that all channels go in order always, and front left would be channel 0 in most cases.

Does this mean the each 16 bit integer actually encodes two 8 bit frames, or is each 16 bit integer its own frame destined for either the left or right channel?

Each value (16-bit integer in this case) is destined for a single channel. Never would you have two multi-byte values smashed into each other.

I hope that's helpful. I can't run your code but given your description, I suspect you have an endian problem and that your samples aren't actual big endian.

Answer 2

Let's start by getting some terminology out of the way

• A channel is a monaural stream of samples. The term does not necessarily imply that the samples are contiguous in the data stream.
• A frame is a set of co-incident samples. For stereo audio (e.g. L & R channels) a frame contains two samples.
• A packet is 1 or more frames, and is typically the minimun number of frames that can be processed by a system at once. For PCM Audio, a packet often contains 1 frame, but for compressed audio it will be larger.
• Interleaving is a term typically used for stereo audio, in which the data stream consists of consecutive frames of audio. The stream therefore looks like L1R1L2R2L3R3......LnRn

Both big and little endian audio formats exist, and depend on the use-case. However, it's generally ever an issue when exchanging data between systems - you'll always use native byte-order when processing or interfacing with operating system audio components.

You don't say whether you're using a little or big endian system, but I suspect it's probably the former. In which case you need to byte-reverse the samples.

Although not set in stone, when using floating point samples are usually in the range -1.0<x<+1.0, so you want to divide the samples by 1<<15. When 16-bit linear types are used, they are typically signed.

Taking care of byte-swapping and format conversions:

int s = (int) interleavedData[2 * i];
short revS = (short) (((s & 0xff) << 8) | ((s >> 8) & 0xff)) 
left[i] = ((float) revS) / 32767.0f;