希尔伯特变换C语言实现
时间: 2023-08-31 21:12:29 浏览: 225
希尔伯特变换(Hilbert Transform)可以通过快速傅里叶变换(FFT)实现,其C语言实现如下:
```c
#include <stdio.h>
#include <math.h>
#define N 1024
void hilbert(float *x, float *y) {
int i, j, k;
float h[N], re[N], im[N], mag[N];
// FFT
for (i = 0; i < N; i++) {
h[i] = x[i];
re[i] = x[i];
im[i] = 0.0;
}
fft(re, im, N);
// Hilbert Transform
for (i = 1; i < N/2; i++) {
re[i] *= 2.0;
im[i] *= 2.0;
}
for (i = N/2; i < N; i++) {
re[i] = 0.0;
im[i] = 0.0;
}
ifft(re, im, N);
// Magnitude and Phase Calculation
for (i = 0; i < N; i++) {
mag[i] = sqrt(re[i]*re[i] + im[i]*im[i]);
y[i] = atan2(im[i], re[i]);
}
}
void fft(float *re, float *im, int n) {
int i, j, k;
float tr, ti, ur, ui, sr, si;
j = 0;
for (i = 0; i < n; i++) {
if (j > i) {
tr = re[j];
ti = im[j];
re[j] = re[i];
im[j] = im[i];
re[i] = tr;
im[i] = ti;
}
k = n/2;
while (k <= j) {
j -= k;
k /= 2;
}
j += k;
}
for (k = 1; k < n; k *= 2) {
sr = cos(M_PI/k);
si = -sin(M_PI/k);
ur = 1.0;
ui = 0.0;
for (j = 0; j < k; j++) {
for (i = j; i < n; i += 2*k) {
tr = ur*re[i+k] - ui*im[i+k];
ti = ur*im[i+k] + ui*re[i+k];
re[i+k] = re[i] - tr;
im[i+k] = im[i] - ti;
re[i] += tr;
im[i] += ti;
}
tr = ur;
ur = tr*sr - ui*si;
ui = tr*si + ui*sr;
}
}
}
void ifft(float *re, float *im, int n) {
int i;
for (i = 0; i < n; i++) {
im[i] = -im[i];
}
fft(re, im, n);
for (i = 0; i < n; i++) {
re[i] /= n;
im[i] = -im[i]/n;
}
}
int main() {
float x[N], y[N];
int i;
// Generate Input Signal
for (i = 0; i < N; i++) {
x[i] = sin(2*M_PI*50*i/1000) + sin(2*M_PI*150*i/1000);
}
// Hilbert Transform
hilbert(x, y);
// Output Results
for (i = 0; i < N; i++) {
printf("%f %f\n", x[i], y[i]);
}
return 0;
}
```
其中,`hilbert` 函数实现了希尔伯特变换,`fft` 和 `ifft` 函数实现了快速傅里叶变换和逆变换。在 `main` 函数中,首先生成一个包含两个正弦波的输入信号,然后调用 `hilbert` 函数进行希尔伯特变换,最后输出结果。