java - 频谱图的频率范围

Question

我能够使用 https://code.google.com/p/musicg/ 中的音乐库来可视化频谱图。但我发现了一些我不太理解的奇怪的事情。我尝试使用采样率为 22050 的 wav 文件，并使用 Blackmann 窗口使用 1024 个样本(重叠 50%)执行 fft。计算结果是二维数组(频谱图[时间][频率]=强度)。我的问题是，如果第二个维度称为频率，为什么它的大小只有256？与频率宽度bin有关吗？那么我如何确定频率？当我尝试使用 512 个样本时，大小减少到一半(128)。

那么我们应该对频谱图进行归一化吗？

这是我从 musicg 获得的代码

short[] amplitudes=wave.getSampleAmplitudes();

    int numSamples = amplitudes.length;
    int pointer=0;
    // overlapping
    if (overlapFactor>1){

        int numOverlappedSamples=numSamples*overlapFactor;
        int backSamples=fftSampleSize*(overlapFactor-1)/overlapFactor;
        int fftSampleSize_1=fftSampleSize-1;
        short[] overlapAmp= new short[numOverlappedSamples];
        pointer=0;
        for (int i=0; i<amplitudes.length; i++){
            overlapAmp[pointer++]=amplitudes[i];
            if (pointer%fftSampleSize==fftSampleSize_1){
                // overlap
                i-=backSamples;
            }
        }
        numSamples=numOverlappedSamples;
        amplitudes=overlapAmp;
    }
    // end overlapping

    numFrames=numSamples/fftSampleSize;
    framesPerSecond=(int)(numFrames/wave.length()); 

    // set signals for fft (windowing)
    WindowFunction window = new WindowFunction();
    window.setWindowType("BLACKMAN");
    double[] win=window.generate(fftSampleSize);

    double[][] signals=new double[numFrames][];
    for(int f=0; f<numFrames; f++) {
        signals[f]=new double[fftSampleSize];

        int startSample=f*fftSampleSize;
        for (int n=0; n<fftSampleSize; n++){

            signals[f][n]=amplitudes[startSample+n]*win[n];                         
        }
    }
    // end set signals for fft

    absoluteSpectrogram=new double[numFrames][];
    // for each frame in signals, do fft on it
    FastFourierTransform fft = new FastFourierTransform();
    for (int i=0; i<numFrames; i++){            
        absoluteSpectrogram[i]=fft.getMagnitudes(signals[i]);
    }

    if (absoluteSpectrogram.length>0){

        numFrequencyUnit=absoluteSpectrogram[0].length;
        unitFrequency=(double)wave.getWaveHeader().getSampleRate()/2/numFrequencyUnit;  // frequency could be caught within the half of nSamples according to Nyquist theory

        // normalization of absoluteSpectrogram
        spectrogram=new double[numFrames][numFrequencyUnit];

        // set max and min amplitudes
        double maxAmp=Double.MIN_VALUE;
        double minAmp=Double.MAX_VALUE; 
        for (int i=0; i<numFrames; i++){
            for (int j=0; j<numFrequencyUnit; j++){
                if (absoluteSpectrogram[i][j]>maxAmp){
                    maxAmp=absoluteSpectrogram[i][j];
                }
                else if(absoluteSpectrogram[i][j]<minAmp){
                    minAmp=absoluteSpectrogram[i][j];
                }
            }
        }

谢谢

Answer 1

每个 FFT 结果箱之间的间距是采样率除以 FFT 长度。对于以 22050 sps 的速率采样并馈送到 1024 FFT 长度的数据，所得频率仓间隔约为 21.5 Hz。如果将 FFT 长度减少到 512，则在达到不超过采样率一半之前，较大的箱间距会导致频谱图垂直轴中的总箱数减少。使用布莱克曼窗口(实际上是任何窗口)，每个 bin 的带宽都会有一些重叠。