共振算法来检测间距 [英] Resonance algorithm to detect pitches

问题描述

我一直在寻找的检测盛入麦克风声音的音调不同的方法。

I have been looking at different methods of detecting the pitch of a tone sung into the microphone.

看到，因为我想找到它与特定的间距类多么紧密的共鸣，我不知道如果我可以做某种形式的基于物理学的共振算法。

Seeing as I want to find how closely it resonates with a particular pitch class, I wonder if I could do some sort of physics-based resonance algorithm.

如果你按住维持在钢琴踏板，唱一个调子进去，（如果你足够接近其现有球场的）一个音符将产生共鸣同情。

If you hold down to sustain pedal on the piano, and sing a tone into it, (and if you are close enough to one of its existing pitches) a note will resonate sympathetically.

我很想能够模拟这种行为。但如何将我去的任务是什么？谁能帮我搬这个前进？

I would love to be able to model this behaviour. But how would I go about the task? Can anyone help me move this forward?

推荐答案

一个有趣的解决方案，我发现仅仅是喂养麦克风输入到Karplus强大的算法。

One interesting solution I found is simply feeding the microphone input into a Karplus Strong algorithm.

所以Karplus强通过模拟弹拨：

So Karplus Strong simulates a plucked string by:

• 创建一个循环缓冲区（如果我们在取样44.1千赫，我们希望来模拟中间一个，即A4是440Hz的，那么我们的缓冲区大小为〜101元）
• 填充它充满了静态-1到1之间
• 走圈，每次的当前值设置为的previous 2的平均值，（和发射电流值至扬声器）
• 在一个恒定的阻尼可以加入

现在，如果我们加上麦克风流到这个过程，所以：

Now if we add the microphone stream into this process, so:

x = ( ringBuf[prev] + ring theBuf[prev2] ) * 0.5 * 0.998;
micOut[moPtr++] = x;
ringBuf[curr] = x + micIn[miPtr++];

它实际上是模仿唱歌变成吉他真的准确。如果你在得到你的语气点，它真实地哀号。

It actually simulates singing into a guitar really accurately. if you get your tone spot on, it truly wails.

但是有一个严重的问题，这种方法：考虑以100元的缓冲器产生的间距，并通过101元的缓冲区产生。没有办法以产生这两个值之间的任何间距。我们仅限于离散的工作设置间距。而这将是pretty的准确低音符（A2将有〜400缓冲区长度），更高才好更多的错误：A7将有〜12.5缓冲区长度。该错误可能是在一个半音。

But there is a severe problem with this approach: consider the pitch generated by a buffer of 100 elements, and that generated by a buffer of 101 elements. there is no way to generate any pitch in between these two values. we are limited to a discrete working set Of pitches. while this is going to be pretty accurate for low notes (A2 would have a buffer length of ~400), the higher we go the more the error: A7 would have a buffer length of ~12.5. That error is probably over a semitone.

我看不出有什么办法对付这个问题。我认为，这种方法已被删除。

I cannot see any way of countering this problem. I think the approach has to be dropped.

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