计算FFT相关系数

static Complex[] CrossCorrelation(Complex[] ffta, Complex[] fftb)
{
    var conj = ffta.Select(i => new Complex(i.Re, -i.Im)).ToArray();
    for (int a = 0; a < conj.Length; a++)
        conj[a] = Complex.Multiply(conj[a], fftb[a]);
    FourierTransform.FFT(conj, FourierTransform.Direction.Backward);
    return conj;
}
static double CorrelationCoefficient(Complex[] ffta, Complex[] fftb)
{
    var correlation = CrossCorrelation(ffta, fftb);
    var a = CrossCorrelation(ffta, ffta);
    var b = CrossCorrelation(fftb, fftb);
    // Not sure if this part is correct..
    var numerator = correlation.Select(i => i.SquaredMagnitude).Max();
    var denominatora = a.Select(i => i.Magnitude).Max();
    var denominatorb = b.Select(i => i.Magnitude).Max();
    return numerator / (denominatora * denominatorb);
}

与fft和Afrog进行交叉相关:

• 用零填充信号:根据AForge的fft文档:该方法只接受2n大小的数据数组，其中n可以在[1,14]范围内变化。

所以你需要确保输入的长度被正确填充为2的幂，并且在指定的范围内:考虑到至少一半的波是"空白"(零)

裁判:

https://dsp.stackexchange.com/questions/741/why-should-i-zero-pad-a-signal-before-taking-the-fourier-transform

https://dsp.stackexchange.com/questions/1919/efficiently-calculating-autocorrelation-using-ffts

• 取两个信号的FFT
• 将其中一个与另一个的共轭相乘(逐元素乘法)
• 做反向FFT
• 取最大值作为相关系数，其指数作为延迟(信号滞后)

为什么结果IFFT的最大值:

互相关对于确定两个信号之间的时间延迟是有用的，例如，用于确定声学信号在麦克风阵列上传播的时间延迟。2[3][澄清需要]在计算两个信号之间的互相关后，互相关函数的最大值(如果信号是负相关的，则为最小值)表示信号最佳排列的时间点，即两个信号之间的时间延迟由最大的参数或互相关的arg max决定，如

参考:https://math.stackexchange.com/questions/1080709/why-is-the-maximum-value-of-cross-correlation-achieved-at-similar-section

基于以上几点，交叉计算使用以下代码计算:

  //same as OP
  public Complex[] CrossCorrelation(Complex[] ffta, Complex[] fftb)
  {
    var conj = ffta.Select(i => new Complex(i.Re, -i.Im)).ToArray();
    conj = conj.Zip(fftb, (v1, v2) => Complex.Multiply(v1, v2)).ToArray();
    FourierTransform.FFT(conj, FourierTransform.Direction.Backward);
    //get that data and plot in Excel, to show the max peak 
    Console.WriteLine("---------rr[i]---------");
    double[] rr = new double[conj.Length];
    rr = conj.Select(i => i.Magnitude).ToArray();
    for (int i = 0; i < conj.Length; i++)
        Console.WriteLine(rr[i]);
    Console.WriteLine("----end-----");
    return conj;
  } 
 //tuble.Item1: Cor. Coefficient
 //tuble.Item2: Delay of signal (Lag)
 public Tuple<double, int> CorrelationCoefficient(Complex[] ffta, Complex[] fftb)
{
    Tuple<double, int> tuble;
    var correlation = CrossCorrelation(ffta, fftb);
    var seq = correlation.Select(i => i.Magnitude);
    var maxCoeff = seq.Max();
    int maxIndex = seq.ToList().IndexOf(maxCoeff);
    Console.WriteLine("max: {0}", maxIndex);
    tuble = new Tuple<double, int>(maxCoeff, maxIndex);
    return tuble;
}
  // Pad signal with zeros up to 2^n and convert to complex 
  public List<Complex> ToComplexWithPadding(List<double> sample, int padding = 1)
    {
        //As per AForge documentation:
        //    The method accepts data array of 2n size only, where n may vary in the [1, 14] range
        //So you would need to make sure the input size is correctly padded to a length that is a power of 2, and in the specified range:
        double logLength = Math.Ceiling(Math.Log(sample.Count * padding, 2.0));
        int paddedLength = (int)Math.Pow(2.0, Math.Min(Math.Max(1.0, logLength), 14.0));
        Complex[] complex = new Complex[paddedLength];
        var samples = sample.ToArray();
        // copy all input samples
        int i = 0;
        for (; i < sample.Count; i++)
        {
            complex[i] = new Complex(samples[i], 0);
            Console.WriteLine(complex[i].Re);
        }
        // pad with zeros
        for (; i < paddedLength; i++)
        {
            complex[i] = new Complex(0, 0);
            Console.WriteLine(complex[i].Re);
        }
        return complex.ToList();
    }
    // extra for signal generation for testing. You can find in the link of the life demo.

您可以使用延迟11产生的两个信号的样本运行寿命演示结果与信号的实际延时

吻合。

生成两个信号的生活演示

输出结果:

 correlation Coef: 0.33867796353274 | Delay: 11| actual delay: 11