Question

以下是什么算法？

我从以下源代码中了解到：

dir是FFT的方向：forward = 1，inverse = -1。
x是真实的部分
y是想象中的部分

这里m是什么？

如果x = {1,1,1,1,0,0,0,0}，y = {0,0,0,0,0,0,0 ，0,0}，m的价值是什么？

        //Inplace 1D FFT
        public static void FFT1D(int dir, int m, ref double[] x, ref double[] y)
        {
            long nn, i, i1, j, k, i2, l, l1, l2;
            double c1, c2, tx, ty, t1, t2, u1, u2, z;
            /* Calculate the number of points */
            nn = 1;
            for (i = 0; i < m; i++)
                nn *= 2;
            /* Do the bit reversal */
            i2 = nn >> 1;
            j = 0;
            for (i = 0; i < nn - 1; i++)
            {
                if (i < j)
                {
                    tx = x[i];
                    ty = y[i];
                    x[i] = x[j];
                    y[i] = y[j];
                    x[j] = tx;
                    y[j] = ty;
                }
                k = i2;
                while (k <= j)
                {
                    j -= k;
                    k >>= 1;
                }
                j += k;
            }
            /* Compute the FFT */
            c1 = -1.0;
            c2 = 0.0;
            l2 = 1;
            for (l = 0; l < m; l++)
            {
                l1 = l2;
                l2 <<= 1;
                u1 = 1.0;
                u2 = 0.0;
                for (j = 0; j < l1; j++)
                {
                    for (i = j; i < nn; i += l2)
                    {
                        i1 = i + l1;
                        t1 = u1 * x[i1] - u2 * y[i1];
                        t2 = u1 * y[i1] + u2 * x[i1];
                        x[i1] = x[i] - t1;
                        y[i1] = y[i] - t2;
                        x[i] += t1;
                        y[i] += t2;
                    }
                    z = u1 * c1 - u2 * c2;
                    u2 = u1 * c2 + u2 * c1;
                    u1 = z;
                }
                c2 = Math.Sqrt((1.0 - c1) / 2.0);
                if (dir == 1)
                    c2 = -c2;
                c1 = Math.Sqrt((1.0 + c1) / 2.0);
            }
            /* Scaling for forward transform */
            if (dir == 1)
            {
                for (i = 0; i < nn; i++)
                {
                    x[i] /= (double)nn;
                    y[i] /= (double)nn;
                }
            }
        }

Answer 1

您发布的FFT的实施仅限于大小为2 ^m的输入。因此m间接指定FFT块大小的大小。因此，对于x = {1,1,1,1,0,0,0,0}和y={1,1,1,1,0,0,0,0}为大小为8 = 2 ³的数组的示例，m将等于3.

请注意，没有对数组x和y的大小进行额外检查，因此请确保它们至少为该大小。

1D快速傅里叶变换

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