均方根RMS:通常用于噪声分析 Calculate decibels

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https://stackoverflow.com/questions/4152201/calculate-decibels

“Root Mean Square”的缩写。

均方根值也称有效值，它可以指示信号发送功率的能力。不管什么波形，具有相同均方眼值的信号发送到阻性负载上的功率是相同的。

https://baike.baidu.com/item/均方根/5374543

When measuring the level of a sound signal, you should calculate the dB from the RMS value. In your sample you are looking at the absolute peak level. A single (peak) sample value determines your dB value, even when all other samples are exactly 0.

try this:

For 'instantaneous' dB levels you would normally calculate the RMS over a segment of 20-50 ms. Note that the calculated dB value is relative to full-scale. For sound the dB value should be related to 20 uPa, and you will need to calibrate your signal to find the proper conversion from digital values to pressure values.

在数据统计分析中，将所有值平方求和，求其均值，再开平方，就得到均方根值，常用均方根值来分析噪声，也是定义AC波的有效电压或电流的一种最普遍的数学方法。。

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And by calibrate you mean that the each client would have to find their zero...because each device and environment will be different? For instance mine, seems to be -40 while everything is silent....would I calibrate that to zero? – Ryan Eastabrook Nov 11 '10 at 15:58
Normally you would use a microphone calibrator for that. The calibrator delivers a signal with a very precise known level, say 98 dB. You then measure/record this signal and derive a scale factor (to be multiplied with each sample value) such that the decibel value you calculate is 98 dB. – Han Nov 11 '10 at 17:55
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The 98dB of the calibrator is relative to 20uPa, so the actual rms pressure level would be 20*10E-6 * 10^(98/20) = 1.59 Pascal. Sometimes you can find the microphone sensitivity in mV/Pa. Then you would only need to know the relation between the voltage on the ADC input and the digital value the ADC delivers. This would allow you to us a known voltage source (or a voltmeter) to calibrate the circuit behind the microphone, and use the microphone sensitivity to get the calibration scale factor. – Han Nov 11 '10 at 18:09
Although I don't fully understand all of the algorithms you're describing, I'm convinced you know much more about this than me. ;) – Ryan Eastabrook Nov 12 '10 at 1:07
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Shouldn't it be: Math.Sqrt(sum / (_buffer.length/2));