在Java中生成正弦波时背景噪声

Question

当我运行以下代码时,我在后台得到轻微的失真(听起来像嗡嗡声).由于其微妙的性质,它使得相信存在字节转换的某种混叠.

AudioFormat = PCM_SIGNED 44100.0 Hz,16位,立体声,4字节/帧,big-endian

注意:代码假定(现在)数据是大端.

public static void playFreq(AudioFormat audioFormat, double frequency, SourceDataLine sourceDataLine)
{
    System.out.println(audioFormat);
    double sampleRate = audioFormat.getSampleRate();
    int sampleSizeInBytes = audioFormat.getSampleSizeInBits() / 8;
    int channels = audioFormat.getChannels();

    byte audioBuffer[] = new byte[(int)Math.pow(2.0, 19.0) * channels * sampleSizeInBytes];

    for ( int i = 0; i < audioBuffer.length; i+=sampleSizeInBytes*channels )
    {
        int wave = (int) (127.0 * Math.sin( 2.0 * Math.PI * frequency * i / (sampleRate * sampleSizeInBytes * channels) )  );

        //wave = (wave > 0 ? 127 : -127);

        if ( channels == 1 )
        {
            if ( sampleSizeInBytes == 1 )
            {
                audioBuffer[i] = (byte) (wave);
            }

            else if ( sampleSizeInBytes == 2 )
            {
                audioBuffer[i] = (byte) (wave);
                audioBuffer[i+1] = (byte)(wave >>> 8);
            }
        }

        else if ( channels == 2 )
        {
            if ( sampleSizeInBytes == 1 )
            {
                audioBuffer[i] = (byte) (wave);
                audioBuffer[i+1] = (byte) (wave);
            }

            else if ( sampleSizeInBytes == 2 )
            {
                audioBuffer[i] = (byte) (wave);
                audioBuffer[i+1] = (byte)(wave >>> 8);

                audioBuffer[i+2] = (byte) (wave);
                audioBuffer[i+3] = (byte)(wave >>> 8);
            }
        }
    }

    sourceDataLine.write(audioBuffer, 0, audioBuffer.length);
}

Answer 1

您的评论说该代码假定为big-endian.

从技术上讲,你实际上是以little-endian输出,但它似乎并不重要,因为通过一个幸运的怪癖,你最重要的字节总是0.

编辑:进一步解释 - 当你的值达到最大值127时,你应该写(0x00,0x7f),但你的代码的实际输出是(0x7f,0x00),这是32512.这恰好接近适当的16位最大值32767,但最低8位全为零.最好始终使用32767作为最大值,然后根据需要丢弃底部的8位.

这意味着即使您输出的是16位数据,有效分辨率也只有8位.这似乎是缺乏音质的原因.

我已经制作了一个代码版本,它只是将原始数据转储到一个文件中,并且看不到任何与位移本身有关的错误.符号或丢失位没有意外的变化,但有一个与8位样本质量一致的嗡嗡声.

此外,如果您根据样本计数计算波动方程,那么您的数学运算会更容易,然后分别担心字节偏移:

int samples = 2 << 19;
byte audioBuffer[] = new byte[samples * channels * sampleSizeInBytes];

for ( int i = 0, j = 0; i < samples; ++i )
{
    int wave = (int)(32767.0 * Math.sin(2.0 * Math.PI * frequency * i / sampleRate));
    byte msb = (byte)(wave >>> 8);
    byte lsb = (byte) wave;

    for (int c = 0; c < channels; ++c) {
        audioBuffer[j++] = msb;
        if (sampleSizeInBytes > 1) {
            audioBuffer[j++] = lsb;
        }
    }
 }