这就是我现在正在使用的内容：

for (int i = 0, numSamples = soundBytes.length / 2; i < numSamples; i += 2)
{
    // Get the samples.
    int sample1 = ((soundBytes[i] & 0xFF) << 8) | (soundBytes[i + 1] & 0xFF);   // Automatically converts to unsigned int 0...65535                                 
    int sample2 = ((outputBytes[i] & 0xFF) << 8) | (outputBytes[i + 1] & 0xFF); // Automatically converts to unsigned int 0...65535

    // Normalize for simplicity.
    float normalizedSample1 = sample1 / 65535.0f;
    float normalizedSample2 = sample2 / 65535.0f;

    float normalizedMixedSample = 0.0f;

    // Apply the algorithm.
    if (normalizedSample1 < 0.5f && normalizedSample2 < 0.5f)
        normalizedMixedSample = 2.0f * normalizedSample1 * normalizedSample2;
    else
        normalizedMixedSample = 2.0f * (normalizedSample1 + normalizedSample2) - (2.0f * normalizedSample1 * normalizedSample2) - 1.0f;

    int mixedSample = (int)(normalizedMixedSample * 65535);

    // Replace the sample in soundBytes array with this mixed sample.
    soundBytes[i] = (byte)((mixedSample >> 8) & 0xFF);
    soundBytes[i + 1] = (byte)(mixedSample & 0xFF);
}

据我所知,它是本页定义的算法的准确表示：http://www.vttoth.com/CMS/index.php/technical-notes/68

然而,只是混合声音和静音(全0)会产生非常明显听起来不正确的声音,也许最好将其描述为更高音和更响亮.

非常感谢帮助确定我是否正确地实现了算法,或者我是否只需要以不同的方式(不同的算法/方法)来实现它？

解决方法:

在链接的文章中,作者假设A和B代表整个音频流.更具体地说,X表示流X中所有样本的最大abs值 – 其中X是A或B.所以他的算法所做的是扫描两个流的全部以计算每个流的最大abs样本然后进行比例缩放输出理论上达到峰值1.0.您需要对数据进行多次传递才能实现此算法,如果您的数据是流式传输,则它将无法正常工作.

这是我认为算法如何工作的一个例子.它假定样本已经转换为浮点到侧面步骤,转换代码的问题是错误的.我稍后会解释它有什么问题：

 double[] samplesA = ConvertToDoubles(samples1);
 double[] samplesB = ConvertToDoubles(samples2);
 double A = ComputeMax(samplesA);
 double B = ComputeMax(samplesB);

 // Z always equals 1 which is an un-useful bit of information.
 double Z = A+B-A*B;

 // really need to find a value x such that xA+xB=1, which I think is:
 double x = 1 / (Math.sqrt(A) * Math.sqrt(B));

 // Now mix and scale the samples
 double[] samples = MixAndScale(samplesA, samplesB, x);

混合和缩放：

 double[] MixAndScale(double[] samplesA, double[] samplesB, double scalingFactor)
 {
     double[] result = new double[samplesA.length];
     for (int i = 0; i < samplesA.length; i++)
         result[i] = scalingFactor * (samplesA[i] + samplesB[i]);
 }

计算最大峰值：

double ComputeMaxPeak(double[] samples)
{
    double max = 0;
    for (int i = 0; i < samples.length; i++)
    {
        double x = Math.abs(samples[i]);
        if (x > max)
            max = x;
    }
    return max;
}

和转换.注意我是如何使用short以便正确维护符号位：

double[] ConvertToDouble(byte[] bytes)
{
    double[] samples = new double[bytes.length/2];
    for (int i = 0; i < samples.length; i++)
    {
        short tmp = ((short)bytes[i*2])<<8 + ((short)(bytes[i*2+1]);
        samples[i] = tmp / 32767.0;
    }
    return samples;
}