Android AudioRecord频率过滤器一步一步

Question

Android:我是语音处理的新手，我使用 JTransform 库创建了 sampleRate 8000Hz 的 AudioRecord 对象我正在尝试过滤频率以下代码中有几件事我不明白我的问题如下

Q.1 为什么我们要将“toTransform[i] = (double) buffer[i]/32768.0;//signed 16 bit”缓冲区转换为 16 位值？

Q.2 Rite 现在audioRecord 读取的数据数组是short array 如果我读取byte 数组我如何将它转换成singed 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 我写了filter方法filterAudio()但是不是过滤频率范围？

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;
    }

Answer 1

Q.1 为什么我们要将“toTransform[i] = (double) buffer[i]/32768.0;//signed 16 bit”缓冲区转换为 16 位值？

32768 是最大值，我认为那行代码正在规范化数据，使其介于 0 和 1 之间。

Q.2 现在audioRecord读取的数据数组是短数组，如果我读取字节数组，我如何将它转换成单16位？

为什么要将它读入byte 数组？如果这样做，则必须将两个 bytes 组合在一起以获得 16 位值。相反，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阅读它。

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是一种不太准确的估计频率的方法。存在其他方法，但更复杂