在Android中实现FFT算法时出现问题。我们假设我有一个长度为8.000字节的wav文件。 我知道你必须选择FFT算法的大小(也必须是2的幂)。我的问题是,我现在不确定如何进一步开展工作。假设我已经选择了N = 1024的FFT大小。 我基本上在我的脑海选项: 1)直接应用FFT算法的8.000字节 2)整个阵列划分为1024个字节的块的8000字节数组wav文件(与0's的最后一块补直到有8个确切的块), 然后将fft应用于每个块,最后再次将所有不同的块重新整理成单个字节数组来表示。 8000 * 2 * 1秒= 8192FFT阵列的实现
我觉得it's选项2,但我不能完全肯定。
这里是FFT阵列我使用:
package com.example.acoustics;
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);
}
}
/***************************************************************
* fft.c
* Douglas L. Jones
* University of Illinois at Urbana-Champaign
* January 19, 1992
* http://cnx.rice.edu/content/m12016/latest/
*
* fft: in-place radix-2 DIT DFT of a complex input
*
* input:
* n: length of FFT: must be a power of two
* m: n = 2**m
* input/output
* x: double array of length n with real part of data
* y: double array of length n with imag part of data
*
* Permission to copy and use this program is granted
* as long as this header is included.
****************************************************************/
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;
}
}
}
}
}
你想要对FFT结果做什么?对于大多数实际应用,您可以将数据阵列分成重叠块,应用窗口,进行FFT,进行滤波/修改,进行逆FFT并使用相同的重叠结构添加结果。窗口可以被安排为统一的分割,以便在没有修改的情况下通过这些操作获得原始阵列。 – LutzL
我试图绘制这样的一个FFT频谱:[剧情](http://www.mathworks.com/help/signal/ug/periodogram_psd.png) – paviflo