我如何可以比较两个捕获，看看哪一个更响亮？ [英] How can I compare two captures to see which one is louder?

问题描述

由于从麦克风捕获的数据的两个字节数组，我怎么能确定哪一个有更多的噪音尖峰？我会假设有一个算法，我可以适用于数据，但我不知道从哪里开始。

Given two byte arrays of data captured from a microphone, how can I determine which one has more spikes in noise? I would assume there is an algorithm I can apply to the data, but I have no idea where to start.

获取到它，我需要能够确定何时宝宝哭闹VS在室内环境噪音。

Getting down to it, I need to be able to determine when a baby is crying vs ambient noise in the room.

如果有帮助，我现在用的是Microsoft.Xna.Framework.Audio.Microphone类来捕捉声音。

If it helps, I am using the Microsoft.Xna.Framework.Audio.Microphone class to capture the sound.

推荐答案

您可以将每个样本应用下面的公式转换（标准化为1.0〜-1.0）成分贝的评价

you can convert each sample (normalised to a range 1.0 to -1.0) into a decibel rating by applying the formula

分贝= 20 *日志基10（采样值）

dB = 20 * log-base-10 (sample-value)

要诚实，只要你不介意偶尔的假阳性，并且您的麦克风设置好了，你应该没有问题，告诉一个婴儿的啼哭和环境背景噪声之间的差异，而不需要通过麻烦中做一个FFT。

To be honest, so long as you don't mind the occasional false positive, and your microphone is set up OK, you should have no problem telling the difference between a baby crying and ambient background noise, without going through the hassle of doing an FFT.

我建议你在看看源$ C ​​$ C为噪声门，这的确pretty的太多你是什么后，可配置的发作次数和放大器;阈值。

I'd recommend you having a look at the source code for a noise gate, which does pretty much what you are after, with configurable attack times & thresholds.

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