Android:我是语音处理的新手,我使用JTransform库创建了带有sampleRate 8000Hz的AudioRecord对象我试图过滤频率,下面的代码中有几件事我不明白我的问题如下
Q.1为什么我们要转换" toTransform [i] =(double)buffer [i] / 32768.0; //签名16位"缓冲到16位值?
Q.2 Rite now audioRecord read data array is short array如果我读取字节数组怎么把它转换成烧焦的16位?
Q.3我想用双数组显示以Hz为单位的声音频率如何计算声音频率?
Q.4我写过滤器方法filterAudio()但它不是过滤频率范围?
请帮帮我,我脑子里有很多问题
/ *代码如下* /
private final int[] mSampleRates = new int[] { 8000, 11025, 22050, 44100 };
final AudioRecord audioRecord = findAudioRecord();
if(audioRecord == null){
return null;
}
final short[] buffer = new short[blockSize];
final double[] toTransform = new double[blockSize];
audioRecord.startRecording();
while (started) {
Thread.sleep(100);
final int bufferReadResult = audioRecord.read(buffer, 0, blockSize);
for (int i = 0; i < blockSize && i < bufferReadResult; i++) {
toTransform[i] = (double) buffer[i] / 32768.0; // signed 16 bit
}
//Audio Filter passing frequency of mSampleRates[3]
filterAudio(bufferReadResult, toTransform, mSampleRates[3]);
transformer.realForward(toTransform);
publishProgress(toTransform);
}
audioRecord.stop();
audioRecord.release();
public static void filterAudio(int bufferSize, double[] audioBuffer, float sampleRate ){
//it is assumed that a float array audioBuffer exists with even length = to
//the capture size of your audio buffer
//float frequency=0F;
//The size of the FFT will be the size of your audioBuffer / 2
int FFT_SIZE = bufferSize / 2;
//RealDoubleFFT mFFT = new RealDoubleFFT(FFT_SIZE);
DoubleFFT_1D mFFT = new DoubleFFT_1D(FFT_SIZE); //this is a jTransforms type
//Take the FFT
mFFT.realForward(audioBuffer);
//mFFT.ft(audioBuffer);
//The first 1/2 of audioBuffer now contains bins that represent the frequency
//of your wave, in a way. To get the actual frequency from the bin:
//frequency_of_bin = bin_index * sample_rate / FFT_SIZE
//assuming the length of audioBuffer is even, the real and imaginary parts will be
//stored as follows
//audioBuffer[2*k] = Re[k], 0<=k<n/2
//audioBuffer[2*k+1] = Im[k], 0<k<n/2
//Define the frequencies of interest
float freqMin = 14400;
float freqMax = 14500;
//Loop through the fft bins and filter frequencies
for(int fftBin = 0; fftBin < FFT_SIZE; fftBin++){
//Calculate the frequency of this bin assuming a sampling rate of 44,100 Hz
float frequency = (float)fftBin * sampleRate / (float)FFT_SIZE;
//Now filter the audio, I'm assuming you wanted to keep the
//frequencies of interest rather than discard them.
if(frequency < freqMin || frequency > freqMax){
//Calculate the index where the real and imaginary parts are stored
int real = 2 * fftBin;
int imaginary = 2 * fftBin + 1;
//zero out this frequency
audioBuffer[real] = 0;
audioBuffer[imaginary] = 0;
}
}
//Take the inverse FFT to convert signal from frequency to time domain
mFFT.realInverse(audioBuffer, false);
}
final AudioRecord findAudioRecord() {
for (int rate : mSampleRates) {
for (short audioFormat : new short[] { AudioFormat.ENCODING_PCM_8BIT, AudioFormat.ENCODING_PCM_16BIT }) {
for (short channelConfig : new short[] { AudioFormat.CHANNEL_CONFIGURATION_MONO , AudioFormat.CHANNEL_CONFIGURATION_STEREO }) {
try {
bufferSize = AudioRecord.getMinBufferSize(rate, channelConfig, audioFormat);
if (bufferSize != AudioRecord.ERROR_BAD_VALUE) {
// check if we can instantiate and have a success
AudioRecord recorder = new AudioRecord(AudioSource.DEFAULT, rate, channelConfig, audioFormat, bufferSize);
if (recorder.getState() == AudioRecord.STATE_INITIALIZED){
Log.d(TAG, "Attempting rate " + rate + "Hz, bits: " + audioFormat + ", channel: "
+ channelConfig);
return recorder;
}
}
} catch (Exception e) {
Log.e(TAG, rate + "Exception, keep trying.",e);
}
}
}
}
return null;
}
答案 0 :(得分:4)
Q.1为什么我们要将“toTransform [i] =(double)buffer [i] / 32768.0; //签名的16位”缓冲区转换为16位值?
32768是最大值,我认为该行代码正在对数据进行规范化,使其介于0和1之间。
Q.2现在audioRecord读取数据数组是短数组,如果我读取字节数组,我将如何将其转换为烧结的16位?
为什么要将它读入byte
数组?如果您这样做,那么您必须将两个bytes
组合在一起以获得16位值。而只需读入short
数组即可获取数据。 Here is an example如何处理短数组。 This code读取它。
Q.3我想用双数组显示以Hz为单位的声音频率如何计算声音频率?
This是估算频率的一种不太准确的方法。存在其他方式,但更复杂