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How to convert 16-bit audio created with Android's AudioRecord to 12-bit audio through bit shifting?

I am attempting to convert 16 bit audio into 12 bit audio. However, I am quite inexperienced with such conversions and believe my approach is possibly incorrect or flawed.
The use case, as context for the code snippets below, is an Android app which the user can speak into and that audio is transmitted to an IoT device for immediate playback. The IoT device expects audio in mono 12 bit, 8k sample rate, little endian, unsigned, with the data stored in the first twelve bits (0-11) and final four bits (12-15) are zeroes. Audio data needs to be received in packets of 1000 bytes.
The audio is being created in the Android app through the use of AudioRecord. The instantiation of which is as follows:
int bufferSize = 1000;
this.audioRecord = new AudioRecord(
MediaRecorder.AudioSource.MIC,
8000,
AudioFormat.CHANNEL_IN_MONO,
AudioFormat.ENCODING_PCM_16BIT,
bufferSize
);
In a while loop, the AudioRecord is being read from by 1000 byte packets and modified to the specifications in the use case. Not sure this part is relevant, but for completeness:
byte[] buffer = new byte[1000];
audioRecord.read(buffer, 0, buffer.length);
byte[] modifiedBytes = convert16BitTo12Bit(buffer);
Then the modifiedBytes are sent off to the device.
Here are the methods which modify the bytes. Basically, to conform to the specifications, I am shifting the bits in each 16 bit set (tossing the least significant 4) and adding zeroes to the final four spots. I do this through BitSet.
/**
* Takes a byte array presented as 16 bit audio and converts it to 12 bit audio through bit
* manipulation. Packets must be of 1000 bytes or no manipulation will occur and the input
* will be immediately returned.
*/
private byte[] convert16BitTo12Bit(byte[] input) {
if (input.length == 1000) {
for (int i = 0; i < input.length; i += 2) {
Log.d(TAG, "convert16BitTo12Bit: pass #" + (i / 2));
byte[] chunk = new byte[2];
System.arraycopy(input, i, chunk, 0, 2);
if (!isEmptyByteArray(chunk)) {
byte[] modifiedBytes = convertChunk(chunk);
System.arraycopy(
modifiedBytes,
0,
input,
i,
modifiedBytes.length
);
}
}
return input;
}
Log.d(TAG, "convert16BitTo12Bit: Failed - input is not 1000 in length; it is " + input.length);
return input;
}
/**
* Converts 2 bytes 16 bit audio into 12 bit audio. If the input is not 2 bytes, the input
* will be returned without manipulation.
*/
private byte[] convertChunk(byte[] chunk) {
if (chunk.length == 2) {
BitSet bitSet = BitSet.valueOf(chunk);
Log.d(TAG, "convertChunk: bitSet starts as " + bitSet.toString());
modifyBitSet(bitSet);
Log.d(TAG, "convertChunk: bitSet ends as " + bitSet.toString());
return bitSet.toByteArray();
}
Log.d(TAG, "convertChunk: Failed = chunk is not 2 in length; it is " + chunk.length);
return chunk;
}
/**
* Removes the first four bits and shifts the rest to leave the final four bits as 0.
*/
private void modifyBitSet(BitSet bitSet) {
for (int i = 4; i < bitSet.length(); i++) {
bitSet.set(i - 4, bitSet.get(i));
}
if (bitSet.length() > 8) {
bitSet.clear(12, 16);
} else {
bitSet.clear(4, 8);
}
}
/**
* Returns true if the byte array input contains all zero bits.
*/
private boolean isEmptyByteArray(byte[] input) {
BitSet bitSet = BitSet.valueOf(input);
return bitSet.isEmpty();
}
Unfortunately, this approach produces subpar results. The audio is quite noisy and it is difficult to make out what someone is saying (but you can hear that words are being spoken).
I also have been playing around with just saving the bytes to a file and playing it back on Android through AudioTrack. I noticed that if I just remove the first four bits and do not shift anything, the audio actually sounds good, as such:
private void modifyBitSet(BitSet bitSet) {
bitSet.clear(0, 4);
}
However, when played through the device, it sounds even worse, and I don't even think I can make out any words.
Clearly, my approach is not working here. Central question is how would one convert a 16 bit chunk into 12 bit audio and maintain audio quality given the requirement that the final four bits must be zero? Additionally, given my larger approach of using AudioRecord to obtain the audio, would such a solution for the prior question fit this use case?
Please let me know if there is anything more I can provide to clarify these questions and my intent.

Given that the audio is 16 bits but must be changed to 12 with four zeros at the end, four bits somewhere do have to be tossed.
Yes, of course and there is no other way, is there?
This is something quick that I can comeout with right now. Certainly not fully tested though. Only tested with input of 2 and 4 bytes. I'll leave it to you to test it.
//Reminder :: Convert as many as possible.
//Reminder :: To calculate the required size for store:
//if((bytes.length & 1) == 0) Math.round((bytes.length * 6) / 8F) : Math.round(((bytes.length - 1) * 6) / 8F).
//Return :: Amount of converted bytes.
public static final int convert16BitTo12Bit(final byte[] bytes, final byte[] store)
{
final int size = bytes.length;
int storeIndex = 0;
//Copy the first 2 bytes into store.
store[storeIndex++] = bytes[0]; store[storeIndex] = bytes[1];
if(size < 4) {
store[storeIndex] = (byte)(store[storeIndex] & 0xF0);
return 2;
}
final int result;
final byte tmp;
// 11111111 11110000 00000000 00000000
//+ 11111111 11110000 (<< 12)
//= 11111111 11111111 11111111 00000000 (1)
//-----------------------------------------
// 11111111 00000000 00000000 00000000 (1)
//+ 11111111 11110000 (<< 16)
//= 11111111 11111111 11110000 00000000 (2)
//-----------------------------------------
// 11110000 00000000 00000000 00000000 (2)
//+ 1111 11111111 0000 (<< 20)
//= 11111111 11111111 00000000 00000000 (3)
//-----------------------------------------
// 00000000 00000000 00000000 00000000 (3)
//+ 11111111 11110000 (<< 24)
//= 11111111 11110000 00000000 00000000
for(int i=2, shiftBits = 12; i < size; i += 2) {
if(shiftBits == 24) {
//Copy 2 bytes from bytes[] into store[] and move on.
store[storeIndex] = bytes[i];
//Never store byte 0 (Garbage).
tmp = (byte)(bytes[i + 1] & 0xF0); //Bit order: 11110000.
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 12; //Reset
} else if(shiftBits == 20) {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
store[storeIndex] = (byte)((result >> 24) & 0xFF);
tmp = (byte)((result >> 16) & 0xFF);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 24;
} else if(shiftBits == 16) {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 16) | ((bytes[i + 1] & 0xFF) << 8));
store[storeIndex] = (byte)((result >> 16) & 0xFF);
tmp = (byte)((result >> 8) & 0xF0);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 20;
} else {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 12) | ((bytes[i + 1] & 0xFF) << 4));
store[storeIndex] = (byte)((result >> 16) & 0xFF);
tmp = (byte)((result >> 8) & 0xFF);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 16;
}
}
return ++storeIndex;
}
Explanations
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
What this does is basically merge two integers into one.
((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
The first one is make an integer with same constant bit position.
(((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
The latter is for 2 current bytes with different bit positions.
(...) | (...)
Pipe or vertical bar at the middle is to merge these two integers we've just created into one.
Usage
To use this method is pretty straight forward.
byte[] buffer = new byte[1000];
byte[] store;
if((buffer.length & 1) == 0) { //Even.
store = new byte[Math.round((bytes.length * 6) / 8F)];
} else { //Odd.
store = new byte[Math.round(((bytes.length - 1) * 6) / 8F)];
}
audioRecord.read(buffer, 0, buffer.length);
int convertedByteSize = convert16BitTo12Bit(buffer, store);
System.out.println("size: " + convertedByteSize);

I have discovered a solution that produces clear audio. First, it is important to recount the requirements for the use case, which is 12 bit unsigned mono audio which will be read in little endian by the device in packets of 1000 bytes.
The initialization and configuration of the AudioRecord as described in the question is fine.
Once the 1000 bytes of audio is read from AudioRecord, it can be put into a ByteBuffer and defined as little endian for modification, and then put into a ShortBuffer to do manipulation on the 16 bit level:
// Audio specifications of device is in little endian.
ByteBuffer byteBuffer = ByteBuffer.wrap(input).order(ByteOrder.LITTLE_ENDIAN);
// Turn into a ShortBuffer so bitwise manipulation can occur on the 16 bit level.
ShortBuffer shortBuffer = byteBuffer.asShortBuffer();
Next, in a loop, take each short and modify it to 12 bit unsigned:
for (int i = 0; i < shortBuffer.capacity(); i++) {
short currentShort = shortBuffer.get(i);
shortBuffer.put(i, convertShortTo12Bit(currentShort));
}
This can be accomplished by shifting the 16 bits four spaces to the right to turn it into 12 bit signed. Then, to convert to unsigned, add 2048. For our purposes as a safety step, we also mask the least significant four bits as required by device, but given the shifting and adding, it shouldn't be the case that any bits actually remain there:
private static short convertShortTo12Bit(short input) {
int inputAsInt = input;
inputAsInt >>>= 4;
inputAsInt += 2048;
input = (short) (inputAsInt & 0B0000111111111111);
return input;
}
If one wishes to return 12 bits to 16 bits, do the reverse for each short (subtract 2048 and shift four spaces to the left).

None of the 3 bytes to integer examples work

Background
I am taking 8, 16, 24 or 32 bit audio data and converting them to integers, but BigInteger cannot be recycled and using it will waste lot of memory so I created this class to fix the memory consumption. And seems like ByteBuffer will do the job well, except if the input is 3 bytes long.
I have never done any bit or byte operations, so I am completely lost here.
Issue
None of the examples that I found on stackoverflow on 3 bytes to int do not give the wanted result. Check the bytesToInt3 method.
Question
Is there something obvious that I am doing completely wrong?
Is the return new BigInteger(byte[] data).intValue(); really the only solution to this?
Code
import java.math.BigInteger;
import java.nio.ByteBuffer;
class BytesToInt {
// HELP
private static int bytes3ToInt(byte[] data) {
// none below seem to work, even if I swap first and last bytes
// these examples are taken from stackoverflow
//return (data[2] & 0xFF) | ((data[1] & 0xFF) << 8) | ((data[0] & 0x0F) << 16);
//return ((data[2] & 0xF) << 16) | ((data[1] & 0xFF) << 8) | (data[0] & 0xFF);
//return ((data[2] << 28) >>> 12) | (data[1] << 8) | data[0];
//return (data[0] & 255) << 16 | (data[1] & 255) << 8 | (data[2] & 255);
return (data[2] & 255) << 16 | (data[1] & 255) << 8 | (data[0] & 255);
// Only thing that works, but wastes memory
//return new BigInteger(data).intValue();
}
public static void main(String[] args) {
// Test with -666 example number
byte[] negativeByteArray3 = new byte[] {(byte)0xff, (byte)0xfd, (byte)0x66};
testWithData(negativeByteArray3);
}
private static void testWithData(byte[] data) {
// Compare our converter to BigInteger
// Which we know gives wanted result
System.out.println("Converter = " + bytes3ToInt(data));
System.out.println("BigInteger = " + new BigInteger(data).intValue());
}
}
Output
Converter = 6749695
BigInteger = -666
full code here http://ideone.com/qu9Ulw

First of all, your indices are wrong. It's not 2, 1, 0 but 0, 1, 2.
Secondly the problem is that the sign isn't being extended, so even though it would work for positive values, negative values show wrong.
If you don't mask the highest (of 24bits) byte, it will sign extend properly, filling the highest (of 32bits) byte with 0x00 for positive values or 0xFF for negative values.
return (data[0] << 16) | (data[1] & 255) << 8 | (data[2] & 255);

Content of (DataBufferUShort).getData()

I can't understand how to get RGBA values from a short[] that I get from bufferedImage.getRaster().dataBuffer.getData() if dataBuffer is an instance of DataBufferUShort.
How to convert these values (that can be even -30000) to 0..255? If dataBuffer is an instance of DataBufferByte I can simply make something like this:
result[i] = (array[i] < 0) ?
array[i] + 256 :
array[i];
But what should I do with DataBufferUShort? Some PNG images has this type instead of expecting DataBufferByte.
getType() returns TYPE_CUSTOM. Here is the image: http://i.stack.imgur.com/YwmkO.png

A DataBufferUShort can hold multiple types of samples, so you first need to determine what data you have. Here are the most common ones:
If the image data represents 16 bit gray samples, all you need to do scale each value down to get an 8 bit gray value. You can do that by shifting the values 8 bits to the right.
DataBufferUShort dataBuffer;
short[] data = dataBuffer.getData();
byte[] gray = new byte[data.length];
for (int i = 0; i < data.length; i++) {
gray[i] = (byte) ((data[i] & 0xff00) >> 8);
}
If the image data represents 16 bits per sample (A)RGB values, you can do just the same as above, you will just have 3 (or 4 if there's alpha) samples or array elements per pixel instead of one.
If the data represents 16 bits per sample ARGB (as seems to be the case with your sample), you can also convert to int packed ARGB samples, like this:
DataBufferUShort dataBuffer;
short[] data = dataBuffer.getData();
int[] argb = new byte[data.length / 4];
for (int i = 0; i < data.length; += 4) {
int a = (data[i ] & 0xff00) >> 8;
int r = (data[i + 1] & 0xff00) >> 8;
int g = (data[i + 2] & 0xff00) >> 8;
int b = (data[i + 3] & 0xff00) >> 8;
argb[i / 4] = a << 24 | r << 16 | g << 8 | b;
}
If the image data represents 15/16 bit RGB (like the TYPE_USHORT_555_RGB or TYPE_USHORT_565_RGB) you'll have to scale the RGB values up to the full 8 bit/sample range. Something like:
DataBufferUShort dataBuffer;
short[] data = dataBuffer.getData();
byte[] rgb = new byte[data.length * 3];
for (int i = 0; i < data.length; i++) {
int shortRGB = data[i] & 0xffff;
// Assuming 5 bit R, 5 bit G, 5 bit B, using the lower 15 bits
rgb[i * 3 + 0] = ((((shortRGB & 0x7C00) >> 10) + 1) * 8) - 1;
rgb[i * 3 + 1] = ((((shortRGB & 0x03E0) >> 5) + 1) * 8) - 1;
rgb[i * 3 + 2] = ((((shortRGB & 0x001F) ) + 1) * 8) - 1;
}
For 565 RGB or even 4444 ARGB (as used in some Android devices), the procedure is very similar.

How to change 37 bits in a byte array to all complement of existing value

I have a byte array myByteArray[82]
I want to change exactly 37 bits in this array to the complement value. ie. if bit0 has a '1' I want to change it to '0'. I need to change first 37 such bits to introduce error in this byte array.
Please suggest how to do this

It's not entirely clear what you are trying to do. My best understanding is that you have an array of 82 bytes and you want to invert the lowest 37 bits of the array. Since a byte is 8 bits, so you can do this:
byte[] myByteArray = new byte[82];
// invert lowest 32 bits, 8 at a time
for (int i = 0; i < 4; ++i) {
myByteArray[i] = (byte)(~myByteArray[i]);
}
// invert next five bits
myByteArray[4] = (byte) (
(myByteArray[4] & 0xE0) // top 3 bits unchanged
|
((~myByteArray[4)) & 0x1F) // bottom 5 bits inverted
);

try
byte[] a82 = ...
Set<Integer> set = new HashSet<Integer>();
while (set.size() < 37) {
set.add((int) (Math.random() * 82));
}
for (int i : set) {
int ibyte = i / 8;
int ibit = i % 8;
int m = 1 << ibit;
a[ibyte] ^= m;
}

This works:
int nBits = 37;
int i = 0;
for (; i<nBits / 8; ++i)
myByteArray[i] = (byte)((byte) myByteArray[i] ^ 0xFF);
myByteArray[i] = (byte)(myByteArray[i] ^ ((0xFF >>> 5) ^ 0xFF));
Whether or not you do the final XOR with the 0xFF on the last line depends on whether you consider the most significant bit to be the first bit (then use it) or the last bit (then omit it);

To invert a specific 37 bits, randomly chosen:
// array of 82 bytes with 37 selected bits set to 1, all the rest zero
// you could generate this programmatically as well if you need a different
// set of bits each time, but your question implies you don't
byte[] mask = { 0x00, 0x01, 0x02, 0x80, .... 0x00 };
for (int i=0; i<myByteArray.length; i++)
{
myByteArray[i] ^= mask[i];
}
Use the bit-exclusive-or operator ^. The truth table for xor is
M a s k
| 0 | 1
D -+---+---
a 0| 0 | 1
t -+---+---
a 1| 1 | 0
Wherever there's a 1 in the mask the corresponding bit in the data will be flipped.

How to Convert a float array to a byte[2] array and back java

im applying DSP effects to my raw audio input which is in byte[2] array format.To apply DSP i need to convert the byte array to float array and back.To convert byte array to float array i use the following code:
private byte[] buffer;
/*
*
* Converts a byte[2] to a float, in LITTLE_ENDIAN format
*/
private float getFloat(byte argB1, byte argB2) {
return (float) (argB1 | (argB2 << 8));
}
for (int i = 0; i < N / 2; i++) {
curSample[i] = getFloat(buffer[i * 2],
buffer[i * 2 + 1]);}
I need to convert back curSample(which is a float array) to the byte[2] array.How to do that?

To convert byte array to float array, what you are doing does not consider the endianness.
int myInt = (byte[0] << 24) |
((byte[1] & 0xff) << 16) |
((byte[2] & 0xff) << 8) |
(byte[3] & 0xff);
or (for little-endian):
int myInt = (byte[3] << 24) |
((byte[2] & 0xff) << 16) |
((byte[1] & 0xff) << 8) |
(byte[0] & 0xff);
Then you can transform to a float using this:
float asFloat = Float.intBitsToFloat(asInt);
To convert it back to byte array
int j=0;
byte[] byteArray=new byte[4];
int data=Float.floatToIntBits(asFloat);
byteArray[j++]=(byte)(data>>>24);
byteArray[j++]=(byte)(data>>>16);
byteArray[j++]=(byte)(data>>>8);
byteArray[j++]=(byte)(data>>>0);
I also find some similar information here