目前，我正在使用JTransforms对我从MediaCodecAudioRenderer复制的音频样本执行FFT，并使用RTMP {{1 }}。
我得到的是4096（或4608 ...是的，有些MP3奇怪地具有非2幂的样本大小，我不知道为什么）长度为MediaSource。那是我必须放入ByteBuffer或FloatFFT_1D对象中的内容，对吗？
现在我的代码如下：
DoubleFFT_1D我注释掉了通道拆分代码，因为我不确定是将字节数组批发还是将其按通道拆分，然后将其中一个通道的数据放入。
但是最终的crf.setHook((dupe, format) -> {
if(currentMediaSource == mMediaSourceAudio) {
byte[] data = new byte[dupe.limit()];
dupe.position(0);
dupe.get(data);
if(format != null) {
new FFTTask(data, format).execute();
}
Log.i("straight_from_renderer", data.length+" "+format);
}
});
...
...
private class FFTTask extends AsyncTask<Void, Void, float[]> {
byte[] bufferContents;
MediaFormat format;
FFTTask(byte[] samples, MediaFormat format) {
this.bufferContents = samples;
this.format = format;
}
float[] floatMe(short[] pcms) {
float[] floaters = new float[pcms.length];
for (int i = 0; i < pcms.length; i++) {
floaters[i] = pcms[i];
}
return floaters;
}
short[] shortMe(byte[] bytes) {
short[] out = new short[bytes.length / 2]; // will drop last byte if odd number
ByteBuffer bb = ByteBuffer.wrap(bytes);
for (int i = 0; i < out.length; i++) {
out[i] = bb.getShort();
}
return out;
}
float[] directFloatMe(byte[] bytes) {
float[] out = new float[bytes.length / 2]; // will drop last byte if odd number
ByteBuffer bb = ByteBuffer.wrap(bytes);
for (int i = 0; i < out.length; i++) {
out[i] = bb.getFloat();
}
return out;
}
private double db2(double r, double i, double maxSquared) {
return 5.0 * Math.log10((r * r + i * i) / maxSquared);
}
double[] convertToDb(double[] data, double maxSquared) {
data[0] = db2(data[0], 0.0, maxSquared);
int j = 1;
for (int i=1; i < data.length - 1; i+=2, j++) {
data[j] = db2(data[i], data[i+1], maxSquared);
}
data[j] = data[0];
return data;
}
@Override
protected float[] doInBackground(Void... voids) {
//WARNING: bufferContents is from a 2-channel 48k bitrate audio, so convert to mono first?
/*
byte[] oneChannel = new byte[bufferContents.length/2];
for(int i = 0; i < oneChannel.length; i+=2) {
oneChannel[i] = bufferContents[i*2+2];
oneChannel[i+1] = bufferContents[i*2+3];
}
*/
float[] dataAsFloats = floatMe(shortMe(bufferContents));
int fftLen = dataAsFloats.length/2;
fft = new FloatFFT_1D(fftLen);
fft.complexForward(dataAsFloats);
String log = "";
float[] magnitudes = new float[dataAsFloats.length/2];
float magMax = 0;
int maxIndex = 0;
float dominantFreq = 0;
for(int i = 0; i < dataAsFloats.length/2; i++) {
float re = dataAsFloats[2*i];
float im = dataAsFloats[2*i+1];
magnitudes[i] = (float)(Math.sqrt(re * re + im * im) / 1e7);
//log += re+" "+im+" "+magnitudes[i]+"\n";
if(magnitudes[i] > magMax) {
magMax = (float)(magnitudes[i]);
maxIndex = i;
}
}
dominantFreq = format.getInteger(MediaFormat.KEY_SAMPLE_RATE) * maxIndex / fftLen;
Log.i("fft_results", magMax+" "+dominantFreq);
return magnitudes;
}
@Override
protected void onPostExecute(float[] res) {
super.onPostExecute(res);
//fftListener.onFFTResultsAvailable(res);
caView.feedFFTMagnitudes(res);
}
}
值确实很嘈杂-幅值混乱，并且没有显示出通常的音频分析图像中应该出现的明显模式-而是紧密的曲折，高于20000 Hz的频率值。
我在做什么错了？
答案 0 :(得分：0)
事实证明，我做错的是字节到短的转换。现在，我只是使用此函数将两个字节转换为短字节：
private short getSixteenBitSample(byte high, byte low) {
return (short)((high << 8) | (low & 0xff));
}
然后将短路转换为转换数组中的浮点数。然后将转换数组通过FFT。