在Android SDK中FFT库 [英] FFT library in android Sdk

问题描述

我与Android project.I工作需要FFT算法来处理Android SDK中提供了Android加速度计data.Is有FFT库？

I am working with android project.I need FFT algorithm to process the android accelerometer data.Is there FFT library available in android sdk?

推荐答案

您可以使用这个类，这是足够快的实时音频分析

You can use this class, which is fast enough for real time audio analysis

public class FFT {

  int n, m;

  // Lookup tables. Only need to recompute when size of FFT changes.
  double[] cos;
  double[] sin;

  public FFT(int n) {
      this.n = n;
      this.m = (int) (Math.log(n) / Math.log(2));

      // Make sure n is a power of 2
      if (n != (1 << m))
          throw new RuntimeException("FFT length must be power of 2");

      // precompute tables
      cos = new double[n / 2];
      sin = new double[n / 2];

      for (int i = 0; i < n / 2; i++) {
          cos[i] = Math.cos(-2 * Math.PI * i / n);
          sin[i] = Math.sin(-2 * Math.PI * i / n);
      }

  }

  public void fft(double[] x, double[] y) {
      int i, j, k, n1, n2, a;
      double c, s, t1, t2;

      // Bit-reverse
      j = 0;
      n2 = n / 2;
      for (i = 1; i < n - 1; i++) {
          n1 = n2;
          while (j >= n1) {
              j = j - n1;
              n1 = n1 / 2;
          }
          j = j + n1;

          if (i < j) {
              t1 = x[i];
              x[i] = x[j];
              x[j] = t1;
              t1 = y[i];
              y[i] = y[j];
              y[j] = t1;
          }
      }

      // FFT
      n1 = 0;
      n2 = 1;

      for (i = 0; i < m; i++) {
          n1 = n2;
          n2 = n2 + n2;
          a = 0;

          for (j = 0; j < n1; j++) {
              c = cos[a];
              s = sin[a];
              a += 1 << (m - i - 1);

              for (k = j; k < n; k = k + n2) {
                  t1 = c * x[k + n1] - s * y[k + n1];
                  t2 = s * x[k + n1] + c * y[k + n1];
                  x[k + n1] = x[k] - t1;
                  y[k + n1] = y[k] - t2;
                  x[k] = x[k] + t1;
                  y[k] = y[k] + t2;
              }
          }
      }
  }
}

警告：此code似乎是从的此处，并拥有了GPLv2许可证。

Warning: this code appears to be derived from here, and has a GPLv2 license.

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