沃梦达 / 编程问答 / 前端问题 / 正文

发声的一个问题：如何用傅立叶系数发声

前端问题
2022-01-01

A problem with sound producing: How to make sound with Fourier coefficients(发声的一个问题：如何用傅立叶系数发声)

本文介绍了发声的一个问题：如何用傅立叶系数发声的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在尝试使用傅立叶系数创建声音。

首先，请让我说明一下我是如何得到傅里叶系数的。

(1)我从麦克风声音拍摄了波形快照。

  • 获取麦克风：getUserMedia()
  • 获取麦克风声音：MediaStreamAudioSourceNode
  • 获取波形数据：AnalyserNode.getByteTimeDomainData()

数据如下：(我字符串Uint8Array，这是getByteTimeDomainData()的返回值，并添加了length属性，以便稍后将此对象更改为Array)

const raw = '{"length": 512,"0":126,"1":121,"2":121,"3":124,"4":129,"5":135,"6":140,"7":147,"8":153,"9":156,"10":152,"11":141,"12":125,"13":112,"14":106,"15":108,"16":113,"17":120,"18":127,"19":132,"20":138,"21":142,"22":141,"23":136,"24":126,"25":115,"26":106,"27":103,"28":105,"29":111,"30":117,"31":121,"32":123,"33":124,"34":124,"35":120,"36":112,"37":103,"38":97,"39":95,"40":96,"41":98,"42":101,"43":106,"44":112,"45":117,"46":117,"47":113,"48":105,"49":98,"50":93,"51":91,"52":91,"53":92,"54":93,"55":95,"56":97,"57":101,"58":105,"59":108,"60":106,"61":101,"62":96,"63":95,"64":97,"65":100,"66":100,"67":97,"68":94,"69":94,"70":99,"71":104,"72":106,"73":105,"74":104,"75":105,"76":108,"77":111,"78":112,"79":110,"80":108,"81":105,"82":105,"83":107,"84":110,"85":113,"86":114,"87":115,"88":116,"89":120,"90":123,"91":125,"92":124,"93":121,"94":120,"95":121,"96":123,"97":124,"98":124,"99":126,"100":128,"101":131,"102":133,"103":134,"104":134,"105":134,"106":134,"107":134,"108":134,"109":133,"110":132,"111":131,"112":131,"113":134,"114":137,"115":139,"116":141,"117":142,"118":143,"119":142,"120":142,"121":139,"122":136,"123":131,"124":128,"125":128,"126":131,"127":134,"128":137,"129":139,"130":140,"131":141,"132":142,"133":141,"134":137,"135":132,"136":126,"137":122,"138":123,"139":127,"140":132,"141":135,"142":135,"143":134,"144":134,"145":135,"146":134,"147":130,"148":125,"149":121,"150":120,"151":121,"152":124,"153":129,"154":132,"155":134,"156":134,"157":133,"158":131,"159":129,"160":128,"161":127,"162":125,"163":124,"164":123,"165":124,"166":125,"167":128,"168":130,"169":131,"170":132,"171":132,"172":131,"173":129,"174":129,"175":129,"176":130,"177":129,"178":129,"179":128,"180":129,"181":132,"182":134,"183":135,"184":134,"185":133,"186":131,"187":131,"188":131,"189":132,"190":134,"191":134,"192":134,"193":134,"194":137,"195":140,"196":142,"197":142,"198":141,"199":138,"200":136,"201":135,"202":137,"203":138,"204":137,"205":135,"206":134,"207":137,"208":142,"209":147,"210":148,"211":147,"212":146,"213":144,"214":144,"215":144,"216":144,"217":142,"218":138,"219":136,"220":137,"221":141,"222":145,"223":149,"224":150,"225":150,"226":150,"227":150,"228":150,"229":148,"230":145,"231":142,"232":142,"233":144,"234":146,"235":146,"236":146,"237":147,"238":150,"239":153,"240":153,"241":149,"242":145,"243":143,"244":141,"245":141,"246":142,"247":143,"248":143,"249":142,"250":144,"251":148,"252":153,"253":152,"254":142,"255":130,"256":123,"257":123,"258":127,"259":130,"260":132,"261":134,"262":139,"263":147,"264":154,"265":155,"266":148,"267":134,"268":119,"269":108,"270":106,"271":110,"272":115,"273":119,"274":124,"275":129,"276":136,"277":141,"278":141,"279":135,"280":125,"281":115,"282":108,"283":105,"284":105,"285":108,"286":111,"287":115,"288":119,"289":122,"290":121,"291":116,"292":110,"293":106,"294":104,"295":101,"296":98,"297":96,"298":98,"299":103,"300":110,"301":115,"302":116,"303":112,"304":104,"305":98,"306":95,"307":95,"308":94,"309":91,"310":88,"311":88,"312":94,"313":101,"314":107,"315":110,"316":107,"317":103,"318":100,"319":99,"320":99,"321":98,"322":95,"323":89,"324":87,"325":89,"326":96,"327":103,"328":107,"329":109,"330":110,"331":111,"332":113,"333":113,"334":110,"335":105,"336":102,"337":102,"338":104,"339":105,"340":107,"341":110,"342":115,"343":120,"344":123,"345":123,"346":122,"347":120,"348":120,"349":121,"350":123,"351":124,"352":123,"353":122,"354":122,"355":126,"356":133,"357":137,"358":136,"359":132,"360":128,"361":129,"362":134,"363":139,"364":139,"365":135,"366":131,"367":131,"368":135,"369":141,"370":144,"371":143,"372":140,"373":138,"374":138,"375":140,"376":142,"377":140,"378":136,"379":131,"380":130,"381":133,"382":138,"383":141,"384":141,"385":140,"386":140,"387":140,"388":139,"389":136,"390":132,"391":129,"392":128,"393":128,"394":129,"395":131,"396":133,"397":135,"398":136,"399":136,"400":135,"401":132,"402":129,"403":125,"404":123,"405":123,"406":125,"407":126,"408":126,"409":126,"410":128,"411":131,"412":133,"413":133,"414":130,"415":127,"416":125,"417":125,"418":125,"419":125,"420":125,"421":125,"422":125,"423":126,"424":129,"425":131,"426":132,"427":131,"428":128,"429":126,"430":126,"431":128,"432":129,"433":130,"434":130,"435":130,"436":132,"437":134,"438":136,"439":135,"440":133,"441":131,"442":129,"443":128,"444":129,"445":130,"446":132,"447":134,"448":136,"449":138,"450":140,"451":142,"452":143,"453":142,"454":140,"455":137,"456":135,"457":134,"458":134,"459":134,"460":134,"461":135,"462":137,"463":139,"464":143,"465":147,"466":148,"467":147,"468":146,"469":145,"470":144,"471":141,"472":139,"473":137,"474":136,"475":137,"476":139,"477":142,"478":145,"479":149,"480":150,"481":151,"482":152,"483":152,"484":151,"485":146,"486":141,"487":138,"488":140,"489":145,"490":147,"491":146,"492":145,"493":147,"494":152,"495":157,"496":156,"497":151,"498":145,"499":140,"500":137,"501":139,"502":143,"503":147,"504":147,"505":144,"506":143,"507":146,"508":152,"509":152,"510":143,"511":129}';

※如果我们将此数据绘制到Canvas，我们可以看到以下内容：(这是元音‘i’(发音像‘ee’))

似乎捕获了2个周期的波。由于长度为512，我们可以猜测一个周期的数据位于索引0~255。

(2)我处理了数据。

const parsed = JSON.parse(raw);
const arrayfied = Array.from(parsed);
const sliced = arrayfied.slice(0, 256);
const refined = [];

// According to the Web Audio API specification,
// "The values stored in the unsigned byte array are computed in the following way.
// Let x[k] be the time-domain data. Then the byte value, b[k], is
// b[k]=⌊128(1+x[k])⌋." So, I manipulate the array like the following:

for (let i = 0; i < sliced.length; i++) {
  refined[i] = (sliced[i] / 128) - 1;
}

(3)我计算了傅立叶系数。

// This function calculates Riemann sum (area approximation using rectangles)
// fn: function to be calculated
// initial: calculation start point
// final: calculation end point
// division: number of rectangles to use
// nth: used for an, bn (please see below)
function numerical_integration(fn, initial, final, division, nth = null) {
  let accumulation = 0;
  const STEP = (final - initial) / division;

  for (let i = initial; i <= final; i++) {
    // calculate an area of a rectangle and add
    accumulation += fn(i, initial, final, nth) * STEP;
  }

  return accumulation;
}

// This is f(t)
function f0(t) {
  const result = refined[t];

  return result;
}

// This is f(t) * cos(nwt)
// ※ w = 2 * Math.PI / period
function fc(t, i, f, n) {
  const result = f0(t) * Math.cos(n * 2 * Math.PI * t / (f - i));

  return result;
}

// This is f(t) * sin(nwt)
function fs(t, i, f, n) {
  const result = f0(t) * Math.sin(n * 2 * Math.PI * t / (f - i));

  return result;
}

// This function returns a0 value
// period is 256 (0 ~ 255) and the last element of array refined is at index 255,
// so I subtract one.
function getA0(period) {
  const result = numerical_integration(f0, 0, period - 1, 100) / period;

  return result;
}

// This function returns an values
function getAn(period) {
  const result = [];

  for (let i = 1; i <= 49; i++) {
    result.push(numerical_integration(fc, 0, period - 1, 100, i) * 2 / period);
  }

  return result;
}

// This function returns bn values
function getBn(period) {
  const result = [];

  for (let i = 1; i <= 49; i++) {
    result.push(numerical_integration(fs, 0, period - 1, 100, i) * 2 / period);
  }

  return result;
}

到目前为止还不错！现在，我们可以通过使用系数制作波函数并将其绘制到画布来检查我们的傅立叶系数是否计算得很好！

const a0 = getA0(refined.length);
const an = getAn(refined.length);
const bn = getBn(refined.length);

// returns y coordinate
function getY(t) {
  let anSum = 0;
  let bnSum = 0;

  for (let i = 0; i <= 48; i++) {
    anSum += an[i] * Math.cos((i + 1) * 2 * Math.PI * t / refined.length);
    bnSum += bn[i] * Math.sin((i + 1) * 2 * Math.PI * t / refined.length);
  }

  const result = a0 + anSum + bnSum;

  return result;
}

// draw
canvasContext.lineTo(x, getY(t));

哇！干得好！与原始波形几乎相同！

然后您可能会问&那么，您的问题是什么？&因此，我将提出我的问题：如何使用傅立叶系数再现声音？(我对Web Audio API和数字声音了解不多)

我想的是三件事：

  • 可能是AudioBuffer？
  • AudioWorklet？
  • 周期波和振荡器节点？

我试过AudioWorklet，但听起来像是饱和的(？)A4(也许)有"滴答声"的爆裂声。AudioWorklet代码如下：

class IWaveProducer extends AudioWorkletProcessor {
  constructor() {
    super();

    this.t = 0;
  }

  process(inputs, outputs, parameters) {
    const output = outputs[0];

    output.forEach(channel => {
      for (let i = 0; i < channel.length; i++) {
        channel[i] = getY(this.t);
      }
    });

    this.t++;

    return true;
  }
}

registerProcessor('i-wave-producer', IWaveProducer);

这是图表：

所以这次我尝试了PeriodicWave和OscillatorNode，但也失败了。代码如下：

const real = new Float32Array(50);
const imag = new Float32Array(50);

real[0] = a0;
imag[0] = 0;

for (let i = 1; i <= 48; i++) {
  real[i + 1] = an[i];
  imag[i + 1] = bn[i];
}

const wave = new PeriodicWave(audioCtx, { real, imag, disableNormalization: false });
const osc = new OscillatorNode(audioCtx, { periodicWave: wave });

osc.connect(analyser)
   .connect(audioCtx.destination);
osc.start();

这是图表：

听起来(可能)像是A4锯齿波。另外，有趣的是，似乎所有的数据都被正确插入，因为波形的形式与上图非常相似(请参见"波形近似"图)。(其格局：一高一小山)

.但这完全不是我想要的！我想要的是重现一个元音"i"！我怎样才能实现我的目标呢？如果你知道什么，请告诉我。我们将不胜感激。我好奇得要死。请帮帮我，ㅠㅠ。非常感谢您阅读这个长长的问题。

还是不可能用Web Audio API制作"语音"？但我以前见过使用JavaScript制作语音的库。例如：

  • https://github.com/kripken/speak.js
  • https://www.masswerk.at/mespeak/
  • https://github.com/usdivad/mesing

又来了！我想我找到答案了！答案是."AudioBuffer"。我真的快要哭了……欣喜若狂……不管怎样，这是代码！

// Since the length of wave is 256
// and I guess (maybe wrong) that it means
// this wave lasts for 256 / 44100 seconds (= 0.0058).
// Thus, in order to make it longer,
// multiply 1000. So this sound will exist for 5.8 seconds.
// (Since sampling rate is 44100 per sec,
// the formula results in the length of this buffer--256000.)
const audioBuffer = new AudioBuffer({ numberOfChannels: 1, length: 1000 * audioCtx.sampleRate * 256 / 44100, sampleRate: audioCtx.sampleRate });

const buffering = audioBuffer.getChannelData(0);
let count = 0;

for (let i = 0; i < audioBuffer.length; i++) {
  buffering[i] = refined[count];

  if (count === 255) {
    count = 0;
  } else {
    count++;
  }
}

const source = new AudioBufferSourceNode(audioCtx, { buffer: audioBuffer });
  
source.connect(analyser)
      .connect(audioCtx.destination);

source.start();

由此产生的声音有点滑稽！但我觉得它听起来肯定像"i(Ee)"。它听起来也像"fa"音符。为什么会这样呢？让我们一起考虑一下吧。首先，我想我们需要计算一下波的频率。因为波的一个周期是0.0058秒，所以频率是1/0.0058，也就是172.4138赫兹。

下一步，A4是440 Hz。因此，A3是220赫兹。下面的四个音符是F3(A3、G#4、G4、F#3、F3)。则F3的频率为220×2^(-4/12)=174.6141 Hz。

172与174几乎相同。！这绝对说得通！秘密现在解开了。这就是为什么那个声音听起来像fa。

感谢您阅读我的艰难，但同时也是与Web Audio API的美丽战斗故事。再见！

又见面了！我刚刚发现PeriodicWave和OscillatorNode也可以解决问题！

const osc = new OscillatorNode(audioCtx, { periodicWave: wave, frequency: 174 });

设置频率参数是关键！再次再见！

那么唯一剩下的就是AudioWorklet。这是否也是答案呢？这让我很好奇。

推荐答案

在golang中，我采用了一个数组arr1，它表示时间序列(可以是音频，在我的情况下可以是图像)，其中该时域数组的每个元素都是浮点值，表示原始音频曲线在摆动时的高度.然后，我将这个浮点数组馈送到一个FFT调用中，该调用根据频域中的定义返回一个新数组ARR2，其中该数组的每个元素都是单个复数，其中实部和虚部都是浮点.然后，当我将此数组送入反向FFT调用(IFFT)时，它返回了时域中的浮点数组ARR3……一次近似ARR3匹配ARR1.不用说，如果我取ARR3并将其送入FFT调用，它的输出ARR4将与ARR2匹配……实际上，您有这个time_domain_array-->；FFT调用-->；FREQUENCY_DOMAIN_ARRAY-->；InverseFFT调用-->；TIME_DOMAIN_ARRAY.冲洗N个重复

我知道Web Audio API有一个FFT调用.不知道它是否有IFFT API调用，但是如果没有IFFT(逆FFT)，你可以编写自己的这样的函数，这里是如何…遍历ARR2并为每个元素计算该频率的幅度(ARR2的每个元素表示一个频率，在文献中您将看到ARR2被称为频率仓，其简单地表示数组的每个元素保存一个复数，并且当您遍历数组时，每个连续的元素表示从元素0到存储频率0的不同频率，并且每个后续的数组元素将表示通过将incr_freq添加到前一个数组元素的频率来定义的频率)

ARR2的每个索引代表一个频率，其中元素0是DC偏置，它是输入ARR1曲线的零偏移偏置，如果它以零交叉点为中心，则该值通常为零，元素0可以忽略...ARR2的每个元素之间的频率差是一个恒定的频率增量，可以使用

incr_freq := sample_rate / number_of_samples  //  with sample_rate of 44100 samples per second, and one second worth of samples ( 44100 )
                                              //  this gives you a frequency increment resolution of 1 Hertz ... IE each freq bin is 1 Hertz apart

对于ARR2的给定元素，您需要使用

计算震级 
curr_real = real(curr_complex) // pluck out real portion of imaginary number
curr_imag = imag(curr_complex) // ditto for im

curr_mag = 2.0 * math.Sqrt(curr_real*curr_real+curr_imag*curr_imag) / number_of_samples  // magnitude of this freq

curr_theta = math.Atan2(curr_imag, curr_real)  //  phase of this frequency
此时，您正在迭代ARR2，并且在此循环中，您正在为ARR2的每个元素进行上面的幅度和相位计算……现在简单地为每个频率合成一条正弦曲线，因为你已经知道了幅度和相位……请记住，当您遍历ARR2时，您将按上述incr_freq递增...然后，只需将这些新合成的正弦曲线中的每一条合并为一条输出曲线，并将其存储到新的ARR3中，该ARR3将与您的原始源ARR1匹配

玩得开心，祝你好运！

更新窗口的概念是指一组音频样本...假设您有一个输入wav格式的文件，您可以从该文件打开并迭代其原始音频曲线。如果这个WAV文件有一首10分钟的歌曲，如果你将整个音频输入到一个FFT调用中，结果将代表整个文件，然而，如果你将整个音频样本集(这些只是原始音频曲线上的点)切成几个窗口，其中每个窗口只有1024个这样的音频样本，并将这些样本的每个窗口都输入到FFT调用中，则FFT调用的每个输出将仅特定于歌曲的该部分...当您使用Audacity之类的工具播放一首歌曲时，同时查看其实时FFT频谱，该频谱会显示当前窗口的刷新视图，因为Audacity会将输入音频每隔1024个左右的音频样本切换到一个新窗口

这篇关于发声的一个问题：如何用傅立叶系数发声的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！

本文标题为：发声的一个问题：如何用傅立叶系数发声

上一篇： DialogFlow：如何处理具有多个路由的NodeJS服务器？
下一篇： 通过Google Apps Script-POST和JSON编辑JIRA问题

基础教程推荐

学习HTML
学习jQuery
学习Laravel
学习CSS3
学习Vue.js
学习Bootstrap5
学习ThinkPHP
学习AJAX
猜你喜欢
最新文章
热门文章
编程基础
热门标签
网站首页 HTML/CSS 菜单 Layui Yii2