c# - 从 WasapiLoopbackCapture 捕获音频，并转换为 muLaw

Question

我正在使用 WasapiLoopbackCapture 捕获音频

- format        = IeeeFloat- SampleRate    = 48000- BitsPerSample = 32

我需要将其转换为 muLaw(8Khz、8 位、单声道)- 最终它将通过 SIP 中继发送到电话。我已经尝试了 100 多个示例(其中大部分使用 NAudio)和解决方案，但仍然不知道如何执行此操作...

Answer 1

NAudio 中的 Mu-Law 工具有限，因此您可能需要自己动手。

您需要设置一系列 IWaveProvider 过滤器以转换为单声道、更改比特率和更改位深度。

waveBuffer = new BufferedWaveProvider(waveIn.WaveFormat);
waveBuffer.DiscardOnBufferOverflow = true;
waveBuffer.ReadFully = false;  // leave a buffer?

sampleStream = new WaveToSampleProvider(waveBuffer);

// Stereo to mono
monoStream = new StereoToMonoSampleProvider(sampleStream)
    {
        LeftVolume = 1f,
        RightVolume = 1f
    };

// Downsample to 8000
resamplingProvider = new WdlResamplingSampleProvider(monoStream, 8000);

// Convert to 16-bit in order to use ACM or MuLaw tools.
ieeeToPcm = new SampleToWaveProvider16(resamplingProvider);

然后为下一步创建自定义 IWaveProvider。

// In MuLawConversionProvider
public int Read(byte[] destinationBuffer, int offset, int readingCount)
{
    // Source buffer has twice as many items as the output array.
    var sizeOfPcmBuffer = readingCount * 2;
    _sourceBuffer = BufferHelpers.Ensure(_sourceBuffer, sizeOfPcmBuffer);
    var sourceBytesRead = _sourceProvider.Read(_sourceBuffer, offset * 2, sizeOfPcmBuffer);
    var samplesRead = sourceBytesRead / 2;

    var outIndex = 0;
    for (var n = 0; n < sizeOfPcmBuffer; n += 2)
    {
        destinationBuffer[outIndex++] = MuLawEncoder.LinearToMuLawSample(BitConverter.ToInt16(_sourceBuffer, offset + n));
    }

    return samplesRead * 2;
}

新的provider可以直接发送到WaveOut

outputStream = new MuLawConversionProvider(ieeeToPcm);
waveOut.Init(outputStream);
waveOut.Play();

这些过滤器以 BufferedWaveProvider 作为“根”保留在原位。每当您调用 BufferedWaveProvider.AddSamples() 时，数据都会通过所有这些过滤器。