Java中可靠快速的FFT[闭合]

http://www.fftw.org/download.html

晚到聚会-这里作为一个纯java的解决方案,当JNI不是一个选项。 JTransforms

我用Java为FFT编写了一个函数: http://www.wikijava.org/wiki/The_Fast_Fourier_Transform_in_Java_%28part_1%29

它在公共领域,所以你可以在任何地方使用这些功能(个人或商业项目也一样)。只要引用我的信用证和发送给我只是一个链接,你的工作,你没事。

它是完全可靠的。我已经对照Mathematica的FFT检查了它的输出,它们一直正确到第15位小数。我认为这是一个非常好的FFT实现。我在j2se1.6版本上编写了它,并在j2se1.5-1.6版本上进行了测试。

让我知道它是否有用,并告诉我任何你喜欢的评论。

我在这里共享相同的代码:

/**
* @author Orlando Selenu
*
*/
public class FFTbase {
/**
 * The Fast Fourier Transform (generic version, with NO optimizations).
 *
 * @param inputReal
 *            an array of length n, the real part
 * @param inputImag
 *            an array of length n, the imaginary part
 * @param DIRECT
 *            TRUE = direct transform, FALSE = inverse transform
 * @return a new array of length 2n
 */
public static double[] fft(final double[] inputReal, double[] inputImag,
                           boolean DIRECT) {
    // - n is the dimension of the problem
    // - nu is its logarithm in base e
    int n = inputReal.length;

    // If n is a power of 2, then ld is an integer (_without_ decimals)
    double ld = Math.log(n) / Math.log(2.0);

    // Here I check if n is a power of 2. If exist decimals in ld, I quit
    // from the function returning null.
    if (((int) ld) - ld != 0) {
        System.out.println("The number of elements is not a power of 2.");
        return null;
    }

    // Declaration and initialization of the variables
    // ld should be an integer, actually, so I don't lose any information in
    // the cast
    int nu = (int) ld;
    int n2 = n / 2;
    int nu1 = nu - 1;
    double[] xReal = new double[n];
    double[] xImag = new double[n];
    double tReal, tImag, p, arg, c, s;

    // Here I check if I'm going to do the direct transform or the inverse
    // transform.
    double constant;
    if (DIRECT)
        constant = -2 * Math.PI;
    else
        constant = 2 * Math.PI;

    // I don't want to overwrite the input arrays, so here I copy them. This
    // choice adds \Theta(2n) to the complexity.
    for (int i = 0; i < n; i++) {
        xReal[i] = inputReal[i];
        xImag[i] = inputImag[i];
    }

    // First phase - calculation
    int k = 0;
    for (int l = 1; l <= nu; l++) {
        while (k < n) {
            for (int i = 1; i <= n2; i++) {
                p = bitreverseReference(k >> nu1, nu);
                // direct FFT or inverse FFT
                arg = constant * p / n;
                c = Math.cos(arg);
                s = Math.sin(arg);
                tReal = xReal[k + n2] * c + xImag[k + n2] * s;
                tImag = xImag[k + n2] * c - xReal[k + n2] * s;
                xReal[k + n2] = xReal[k] - tReal;
                xImag[k + n2] = xImag[k] - tImag;
                xReal[k] += tReal;
                xImag[k] += tImag;
                k++;
            }
            k += n2;
        }
        k = 0;
        nu1--;
        n2 /= 2;
    }

    // Second phase - recombination
    k = 0;
    int r;
    while (k < n) {
        r = bitreverseReference(k, nu);
        if (r > k) {
            tReal = xReal[k];
            tImag = xImag[k];
            xReal[k] = xReal[r];
            xImag[k] = xImag[r];
            xReal[r] = tReal;
            xImag[r] = tImag;
        }
        k++;
    }

    // Here I have to mix xReal and xImag to have an array (yes, it should
    // be possible to do this stuff in the earlier parts of the code, but
    // it's here to readibility).
    double[] newArray = new double[xReal.length * 2];
    double radice = 1 / Math.sqrt(n);
    for (int i = 0; i < newArray.length; i += 2) {
        int i2 = i / 2;
        // I used Stephen Wolfram's Mathematica as a reference so I'm going
        // to normalize the output while I'm copying the elements.
        newArray[i] = xReal[i2] * radice;
        newArray[i + 1] = xImag[i2] * radice;
    }
    return newArray;
}

/**
 * The reference bitreverse function.
 */
private static int bitreverseReference(int j, int nu) {
    int j2;
    int j1 = j;
    int k = 0;
    for (int i = 1; i <= nu; i++) {
        j2 = j1 / 2;
        k = 2 * k + j1 - 2 * j2;
        j1 = j2;
    }
    return k;
  }
}

我想这取决于你在处理什么。如果你在一个大的持续时间内计算FFT,你可能会发现它确实需要一段时间,这取决于你想要多少个频点。然而,在大多数情况下,对于音频,它被认为是非平稳的(即信号的均值和方差随时间变化很大),因此采用一个大的FFT( Periodogram PSD 估计)不是准确的表示。或者可以使用短时傅里叶变换,将信号分解为更小的帧并计算FFT。帧的大小取决于统计数据变化的速度,对于语音,通常是20-40ms,对于音乐,我认为稍高一些。

KissFFT libraries in Android Speech Enhancement for Android .

SSTJ 用于Java中的FFT。它可以通过JNI重定向到 FFTW 如果库可用,或者如果不可用,将使用纯Java实现。