audio - 音符合成、和声( fiddle 、钢琴、吉他、贝司)、频率、MIDI

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我想找到笔记是如何构建的。
以乐器( fiddle 或钢琴)为例，Note LA4 (A4) 具有 440Hz 的主(或中心)频率 FC，具有特定的 AC 振幅，但它也必须具有其他频率(谐波？)FH，具有其他振幅 AH。

谐波具有依赖于主频率的其他频率，其幅度(几乎)小于主频率的幅度。

成型(建筑)注意事项

我想知道笔记是如何形成(建立)的(不考虑时间)。

例子:
A4 = AC(FC) + AH1(FH1)+ AH2(FH2) + AH3(FH3) + AH4(FH4)....AHn(FHn)
也许，FH1 = 2*FC，FH2 = 3*FC，FH3 = 4*FC，等等......

乐器比较( fiddle 和钢琴)

对于钢琴，Note LA4 (A4) 的主频 FC 为 440Hz，并且
也许，FC(钢琴)= FC( fiddle )，FH1(钢琴)= FH1( fiddle )，FH2(钢琴)= FH2( fiddle )，等等......

但是，AC(Piano) != AC(Violin), AH1(Piano) != AH1(Violin), AH2(Piano) != AH2(Violin) 等等....

我的问题的例子是:
http://www.phys.unsw.edu.au/jw/sound.spectrum.html

我想弹奏这个音符，避免使用 MIDI 格式，这可以在 Java/C#(或其他语言编程)中实现，并且可以更好地控制我的声音。

谢谢你。

安娜

Answer 1

我有这个...

    int iTone = 40;   //Tone to be interpreted
    iSmplRate = 32000;  //Sample Rate
    int NumBytesPerSample = 16;  // 8 or 16
    int NumChannels = 2;  //1 Mono, 2 Stereo
    double Duration = 6.5; //Seconds performing
    Short sAmplit = 1200;
    int iNumSmpl = (int)(SampleRate*Duration);
    NumTotalBytes = (int)(SampleRate*Duration*NumBytesPerSample*NumChannels);
    ByteBuffer bbWav = ByteBuffer.allocate(NumTotalBytes);


    double dMaxInstr = (double)Short.MIN_VALUE;
    double dMinInstr = (double)Short.MAX_VALUE;


    //Amplitude for violin's armonics 
    double[] violAmps = {1.0, 0.286699025, 0.150079537, 0.042909002, 
                        0.203797365, 0.229228698, 0.156931925, 
                        0.115470898, 0.0, 0.097401803, 0.087653465, 
                        0.052331036, 0.052922462, 0.038850593, 
                        0.053554676, 0.053697434, 0.022270261, 
                        0.013072562, 0.008585879, 0.005771505,
                        0.004343925, 0.002141371, 0.005343231, 
                        0.000530244, 0.004711017, 0.009014153};

    //Amplitude for piano's armonics 
    double[] pianAmps = {1.0, 0.399064778, 0.229404484, 0.151836061,
                        0.196754229, 0.093742264, 0.060871957,
                        0.138605419, 0.010535002, 0.071021868,
                        0.029954614, 0.051299684, 0.055948288,
                        0.066208224, 0.010067391, 0.00753679,
                        0.008196947, 0.012955577, 0.007316738,
                        0.006216476, 0.005116215, 0.006243983,
                        0.002860679, 0.002558108, 0.0, 0.001650392};
    double[] operator = {1.0};
    if (instrument.equals("violin")) {
      operator = violAmps;
    }
    if (instrument.equals("piano")) {
      operator = pianAmps;
    }
    double dFreq = 440.0*Math.pow(2.0, (iTone-69)/12.0;

    double dFreqRel = iSmplRate/dFreq;
    Integer iSampleInstrument = null;
    double PI2 = 2*Math.PI;

    int[] iSamplesInstr = new int[iNumSmpl];
    for (int i = 0;i < iNumSmpl; i++) {
      Double Angle = i*PI2/dFreqRel;
      Double dInstrument = 0.0;
      for (int a = 1; a <=operator.length; a++) {
        dInstrument += operator[a-1]*Math.sin((double)a*Angle);
      }

      dMaxInstr = (dInstrument>dMaxInstr)?dInstrument:dMaxInstr;
      dMinInstr = (dInstrument<dMinInstr)?dInstrument:dMinInstr;

      iSampleInstrument = (int)(sAmplit*dInstrument);

      if (instrument.equals("violin")) {
        double FreqEnvV = iSmplRate/6.0;
        double FracEnvV = 35.0;
        double dEnvViolin = sAmplit*DStepperExt(Math.sin(1.0*i*PI2/FreqEnvV),4)/FracEnvV;
        iSampleInstrument = (int)(iSampleInstrument+dEnvViolin);
      }
      if (instrument.equals("piano")) {
        double FracEnvP = 8.0/10.0;
        double AngP = (double)i/(iSmplRate*FracEnvP);
        double EnvPiano = 1.0/Math.exp(AngP);
        iSampleInstrument = (int)(iSampleInstrument*EnvPiano);
      }
      dMxSmplInstr = (iSampleInstrument>dMxSmplInstr)?iSampleInstrument:dMxSmplInstr;
      dMnSmplInstr = (iSampleInstrument<dMnSmplInstr)?iSampleInstrument:dMnSmplInstr;
      iSamplesInstr[i] = iSampleInstrument;
    }

    double dMaxAbs = 
            (Math.abs(dMaxInstr)>Math.abs(dMinInstr))?Math.abs(dMaxInstr):Math.abs(dMinInstr);
    double dMxAbsSmpl = 
            (Math.abs(dMxSmplInstr)>Math.abs(dMnSmplInstr))?Math.abs(dMxSmplInstr):Math.abs(dMnSmplInstr);
    double dNormal = 1.0;
    if (dMxAbsSmpl > 32768.0) {
      dNormal = 32768.0/dMxAbsSmpl;
    }

    for (int i = 0;i < iNumSmpl; i++) {
      short sSampleInst = (short)(iSamplesInstr[i]*dNormal);
      try {
        if (iNumByteSmpl == 2) {
          bbWav.put((byte)((sSampleInst >> 0) & 0xFF));
          bbWav.put((byte)((sSampleInst >> 8) & 0xFF));
          if (iNumChnnls == 2) {
            bbWav.put((byte)((sSampleInst >> 0) & 0xFF));
            bbWav.put((byte)((sSampleInst >> 8) & 0xFF));
          }
        } else {
          byte ByteSample = (byte)((sSampleInst >> 8) & 0xFF);
          short ShrtSample = (short)(ByteSample & 0xFF);
          ShrtSample += 128;
          bbWav.put((byte)(ShrtSample & 0xFF));
          if (iNumChnnls == 2) {
            bbWav.put((byte)(ShrtSample & 0xFF));
          }
        }
      } catch (Exception e) {
        System.out.println(e.getMessage());
      }

此代码用于 fiddle 乐器:

  private Double DStepperExt(Double Val, Integer Steps) {
    //Return a value inside in range defined by step
    //Divide [-1.0,1.0]/(Steps-1), retorning the value according to the range
    //The value must be between 0.0 and 1.0
    if (Steps <= 0.0) { 
      return 0.0;
    }
    if (Val != -1.0 && Val != 1.0) {
      Val = Val - Val.intValue();
    }
    Double sDouble = new Double(Steps-1);
    Double bdStep = 2.0/sDouble;
    Double bdRef = bdStep/2.0;
    bdRef = bdRef - 1.0;
    Double bdInit = -1.0;

    Double bdRet = null;
    for (int c = 0; c<=sDouble;c++) {
      if (Val < bdRef) {
        bdRet = bdInit;
        break;
      } else {
        bdInit = bdInit+bdStep;
        bdRef = bdRef+bdStep;
      }
    }
    return Math.min(bdRet.doubleValue(),1.0);
  }

试试这个代码，我的声音并不完美，但非常相似。