c语言实现4096个点的FFT程序
时间: 2024-05-08 12:17:20 浏览: 185
由于FFT算法比较复杂,实现4096点的FFT需要一定的数学和编程知识。以下是一个简单的C语言实现4096点FFT的程序:
```c
#include <stdio.h>
#include <math.h>
#define N 4096
void fft(double *real, double *imag, int n, int inverse) {
int i, j, k, m;
double tempreal, tempimag, wreal, wimag, treal, timag;
double pi = inverse ? -acos(-1.0) : acos(-1.0);
for (i = 0; i < n; i++) {
j = 0;
for (k = 0; k < log2(n); k++) {
j |= ((i >> k) & 1) << (log2(n) - 1 - k);
}
if (j > i) {
tempreal = real[i];
real[i] = real[j];
real[j] = tempreal;
tempimag = imag[i];
imag[i] = imag[j];
imag[j] = tempimag;
}
}
for (m = 2; m <= n; m <<= 1) {
wreal = cos(2.0 * pi / m);
wimag = sin(2.0 * pi / m);
for (k = 0; k < n; k += m) {
treal = 1.0;
timag = 0.0;
for (i = k; i < k + (m >> 1); i++) {
tempreal = treal * real[i + (m >> 1)] - timag * imag[i + (m >> 1)];
tempimag = treal * imag[i + (m >> 1)] + timag * real[i + (m >> 1)];
real[i + (m >> 1)] = real[i] - tempreal;
imag[i + (m >> 1)] = imag[i] - tempimag;
real[i] += tempreal;
imag[i] += tempimag;
tempreal = treal * wreal - timag * wimag;
timag = treal * wimag + timag * wreal;
treal = tempreal;
}
}
}
if (inverse) {
for (i = 0; i < n; i++) {
real[i] /= n;
imag[i] /= n;
}
}
}
int main() {
double real[N], imag[N];
int i;
// 生成输入信号
for (i = 0; i < N; i++) {
real[i] = sin(2.0 * acos(-1.0) * i / N);
imag[i] = 0.0;
}
// 进行FFT变换
fft(real, imag, N, 0);
// 输出结果
for (i = 0; i < N; i++) {
printf("%f + %fi\n", real[i], imag[i]);
}
return 0;
}
```
该程序通过调用fft函数实现了对4096个点进行FFT变换,其中real和imag分别为输入实数和虚数数组,n为变换点数,inverse为0表示正变换,为1表示逆变换。程序中使用了蝴蝶运算和位逆序置换两种常用的实现FFT的方法。最后,程序输出了变换后的实部和虚部。
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