混合8位8000 Hz PCM_MULAW样本的最简单方法是什么?

时间:2010-10-25 13:34:32

标签: java audio voip

我想在Java中将多个ULAW样本混合在一起。有没有办法在不转换为PCM的情况下这样做?我试图做一个8000Hz 8位VoIP应用程序。我使用ULAW编码测试AU文件(在goldwave中创建)。我目前的实施是:

[代码]

    AudioFormat f = new AudioFormat(AudioFormat.Encoding.PCM_SIGNED, 8000, 16, 1, 2, 8000, false);
    SourceDataLine sdl = AudioSystem.getSourceDataLine(f);
    sdl.open(f);
    sdl.start();

    File file1 = new File("C:\\Scream3.au");
    AudioInputStream ais1 = AudioSystem.getAudioInputStream(file1);
    AudioInputStream aisTarget1 = AudioSystem.getAudioInputStream(f, ais1);

    File file2 = new File("C:\\Blackout3.au");
    AudioInputStream ais2 = AudioSystem.getAudioInputStream(file2);
    AudioInputStream aisTarget2 = AudioSystem.getAudioInputStream(f, ais2);

    byte[] data = new byte[10000];
    int[] calc = new int[5000];
    AudioInputStream[] streams = {aisTarget1, aisTarget2};
    int count = streams.length + 1;
    while (true) {
        int r = -1;
        for (int i = 0; i < streams.length; i++) {
            r = streams[i].read(data, 0, data.length);
            if (r == -1) break;
            for (int j = 0; j < calc.length; j++) {
                int tempVal = ((data[j * 2 + 1] << 8) | (data[j * 2] & 0xFF));
                calc[j] += tempVal;
            }
        }
        for (int i = 0; i < calc.length; i++) {
            calc[i] /= count;
            data[i * 2 + 0] = (byte) (calc[i] & 0xFF);
            data[i * 2 + 1] = (byte) (calc[i] >> 8);
        }
        if (r == -1) break;
        sdl.write(data, 0, data.length);
    }

[/代码]

如果不能直接混合ulaw样本并且我必须转换为PCM,我该如何转换PCM格式(AudioFormat.Encoding.PCM_SIGNED,8000 Hz,16bits,1通道,2字节帧大小,8000帧速率,小端) 到ULAW(8bit 8000Hz)。

我做的事情是这样的: 1)将WAVE标头写入字节流 2)将PCM数据写入字节流 3)使用AudioSystem.getAudioInputStream(字节流)获取PCM AIS 4)使用AudioSystem.getAudioInputStream(ulawFormat,PCM AIS)获取ULAW目标AIS

感谢任何帮助。

编辑: 试图从Mu-LAW转换为PCM(试图使用http://en.wikipedia.org/wiki/Ulaw中的等式):

[CODE]

    AudioFormat f = new AudioFormat(AudioFormat.Encoding.PCM_SIGNED, 8000, 8, 1, 1, 8000, false);
    SourceDataLine sdl = AudioSystem.getSourceDataLine(f);
    sdl.open(f);
    sdl.start();
    File file1 = new File("C:\\scream3.au");
    FileInputStream fis = new FileInputStream(file1);
    // Skip header
    fis.skip(24);
    byte[] data = new byte[8196];

    while (true) {
        int r = fis.read(data, 0, data.length);
        if (r == -1) break;
        for (int i = 0; i < r; i++) {
            float y = ((float) data[i] / Byte.MAX_VALUE);
            float sample = -1.0f * (float) (Math.signum(y) * (1.0f / 255.0f) * (Math.pow(1 + 255, Math.abs(y)) - 1.0f));
            data[i] = (byte) (sample * Byte.MAX_VALUE);
        }
        sdl.write(data, 0, data.length);
    }
    sdl.drain();
    sdl.stop();
    sdl.close();
    fis.close();

[/ CODE]

1 个答案:

答案 0 :(得分:1)

我会使用Wikipedia:μ-law algorithm中的公式,手动转换/到线性编码,一次一个样本。

示例代码(基于OP的代码):

float sample1 = Math.signum(y1)/255.0*(Math.pow(256, Math.abs(y1))-1);
float sample2 = Math.signum(y2)/255.0*(Math.pow(256, Math.abs(y2))-1);
float combined = somethig * sample1 + (1-something) * sample2;
float result = Math.signum(combined)*Math.log(1+255*Math.abs(combined))/Math.log(256);