如何测试声级rms算法

Question

我的应用。正在计算输入声音的噪声水平和频率峰值。我使用快速傅立叶变换得到shorts[]缓冲区的数组，代码如下: bufferSize = 1024，sampleRate = 44100

 int bufferSize = AudioRecord.getMinBufferSize(sapleRate,
                channelConfiguration, audioEncoding);
        AudioRecord audioRecord = new AudioRecord(
                MediaRecorder.AudioSource.DEFAULT, sapleRate,
                channelConfiguration, audioEncoding, bufferSize);

这是转换代码：

short[] buffer = new short[blockSize];
        try {
            audioRecord.startRecording();
        } catch (IllegalStateException e) {
            Log.e("Recording failed", e.toString());
        }
        while (started) {
            int bufferReadResult = audioRecord.read(buffer, 0, blockSize);

            /*
             * Noise level meter begins here
             */
            // Compute the RMS value. (Note that this does not remove DC).
            double rms = 0;
            for (int i = 0; i < buffer.length; i++) {
                rms += buffer[i] * buffer[i];
            }
            rms = Math.sqrt(rms / buffer.length);
            mAlpha = 0.9;   mGain = 0.0044;
            /*Compute a smoothed version for less flickering of the
            // display.*/
            mRmsSmoothed = mRmsSmoothed * mAlpha + (1 - mAlpha) * rms;
            double rmsdB = 20.0 * Math.log10(mGain * mRmsSmoothed);

现在我想知道这个算法是否正确工作，或者我遗漏了什么？我想知道它是否正确并且我在手机上显示了dB格式的声音，如何测试它？我需要任何帮助，请提前谢谢:)

Answer 1

代码看起来是正确的，但您可能应该处理缓冲区最初包含零的情况，这可能会导致Math.log10失败，例如更改：

        double rmsdB = 20.0 * Math.log10(mGain * mRmsSmoothed);

至：

        double rmsdB = mGain * mRmsSmoothed >.0 0 ?
                           20.0 * Math.log10(mGain * mRmsSmoothed) :
                           -999.99;  // choose some appropriate large negative value here for case where you have no input signal