Question

我能够使用来自https://code.google.com/p/musicg/的musicg库来可视化频谱图，但我发现了一些我不太懂的奇怪事物。我尝试使用采样率为22050的wav文件，并使用blackmann窗口使用1024个样本执行fft，重叠率为50％。计算结果是二维阵列（谱图[时间] [频率] =强度）。我的问题是如果第二维称为频率，为什么它的大小只有256？它与频率宽度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）。

然后我们应该将频谱图标准化吗？

spectrogram result

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结果区之间的间距是采样率除以FFT长度。对于以22050 sps的速率采样的数据，其馈送到1024的FFT长度，得到的频率仓间隔将约为21.5Hz。如果将FFT长度减小到512，那么较大的bin间距会导致光谱图垂直轴中的总二进制位数减少，然后才能达到采样率的一半以上。使用Blackman窗口（实际上是任何窗口），每个bin的带宽都会有一些重叠。

频谱图的频率范围

1 个答案: