你好，我已经复制了这个 FFT 实现，但它说没有像 System. Windows我怎样才能使这个代码工作？已经得到了答案，我只是想编辑这篇文章，所以它现在对某人有用。
资料来源： http: //gerrybeauregard.wordpress.com/2011/04/01/an-fft-in-c/
代码：：
using System;
using System.Net;
namespace FFT {
public class FFT2 {
// Element for linked list in which we store the
// input/output data. We use a linked list because
// for sequential access it's faster than array index.
class FFTElement {
public double re = 0.0; // Real component
public double im = 0.0; // Imaginary component
public FFTElement next; // Next element in linked list
public uint revTgt; // Target position post bit-reversal
}
private uint m_logN = 0; // log2 of FFT size
private uint m_N = 0; // FFT size
private FFTElement[] m_X; // Vector of linked list elements
/**
*
*/
public FFT2() {
}
/**
* Initialize class to perform FFT of specified size.
*
* @param logN Log2 of FFT length. e.g. for 512 pt FFT, logN = 9.
*/
public void init(
uint logN) {
m_logN = logN;
m_N = (uint)(1 << (int)m_logN);
// Allocate elements for linked list of complex numbers.
m_X = new FFTElement[m_N];
for (uint k = 0; k < m_N; k++)
m_X[k] = new FFTElement();
// Set up "next" pointers.
for (uint k = 0; k < m_N - 1; k++)
m_X[k].next = m_X[k + 1];
// Specify target for bit reversal re-ordering.
for (uint k = 0; k < m_N; k++)
m_X[k].revTgt = BitReverse(k, logN);
}
/**
* Performs in-place complex FFT.
*
* @param xRe Real part of input/output
* @param xIm Imaginary part of input/output
* @param inverse If true, do an inverse FFT
*/
public void run(
double[] xRe,
double[] xIm,
bool inverse = false) {
uint numFlies = m_N >> 1; // Number of butterflies per sub-FFT
uint span = m_N >> 1; // Width of the butterfly
uint spacing = m_N; // Distance between start of sub-FFTs
uint wIndexStep = 1; // Increment for twiddle table index
// Copy data into linked complex number objects
// If it's an IFFT, we divide by N while we're at it
FFTElement x = m_X[0];
uint k = 0;
double scale = inverse ? 1.0 / m_N : 1.0;
while (x != null) {
x.re = scale * xRe[k];
x.im = scale * xIm[k];
x = x.next;
k++;
}
// For each stage of the FFT
for (uint stage = 0; stage < m_logN; stage++) {
// Compute a multiplier factor for the "twiddle factors".
// The twiddle factors are complex unit vectors spaced at
// regular angular intervals. The angle by which the twiddle
// factor advances depends on the FFT stage. In many FFT
// implementations the twiddle factors are cached, but because
// array lookup is relatively slow in C#, it's just
// as fast to compute them on the fly.
double wAngleInc = wIndexStep * 2.0 * Math.PI / m_N;
if (inverse == false)
wAngleInc *= -1;
double wMulRe = Math.Cos(wAngleInc);
double wMulIm = Math.Sin(wAngleInc);
for (uint start = 0; start < m_N; start += spacing) {
FFTElement xTop = m_X[start];
FFTElement xBot = m_X[start + span];
double wRe = 1.0;
double wIm = 0.0;
// For each butterfly in this stage
for (uint flyCount = 0; flyCount < numFlies; ++flyCount) {
// Get the top & bottom values
double xTopRe = xTop.re;
double xTopIm = xTop.im;
double xBotRe = xBot.re;
double xBotIm = xBot.im;
// Top branch of butterfly has addition
xTop.re = xTopRe + xBotRe;
xTop.im = xTopIm + xBotIm;
// Bottom branch of butterly has subtraction,
// followed by multiplication by twiddle factor
xBotRe = xTopRe - xBotRe;
xBotIm = xTopIm - xBotIm;
xBot.re = xBotRe * wRe - xBotIm * wIm;
xBot.im = xBotRe * wIm + xBotIm * wRe;
// Advance butterfly to next top & bottom positions
xTop = xTop.next;
xBot = xBot.next;
// Update the twiddle factor, via complex multiply
// by unit vector with the appropriate angle
// (wRe + j wIm) = (wRe + j wIm) x (wMulRe + j wMulIm)
double tRe = wRe;
wRe = wRe * wMulRe - wIm * wMulIm;
wIm = tRe * wMulIm + wIm * wMulRe;
}
}
numFlies >>= 1; // Divide by 2 by right shift
span >>= 1;
spacing >>= 1;
wIndexStep <<= 1; // Multiply by 2 by left shift
}
// The algorithm leaves the result in a scrambled order.
// Unscramble while copying values from the complex
// linked list elements back to the input/output vectors.
x = m_X[0];
while (x != null) {
uint target = x.revTgt;
xRe[target] = x.re;
xIm[target] = x.im;
x = x.next;
}
}
/**
* Do bit reversal of specified number of places of an int
* For example, 1101 bit-reversed is 1011
*
* @param x Number to be bit-reverse.
* @param numBits Number of bits in the number.
*/
private uint BitReverse(
uint x,
uint numBits) {
uint y = 0;
for (uint i = 0; i < numBits; i++) {
y <<= 1;
y |= x & 0x0001;
x >>= 1;
}
return y;
}
}
}