所以我尝试用 C 语言编写离散傅里叶变换来处理真正的 32 位浮点 wav 文件。它一次读取 2 个帧(每个通道一个,但出于我的目的,我假设它们都是相同的,所以我使用帧 [0])。该代码应该通过使用频率 20,40,60,...,10000 探测输入文件来写出输入文件的幅度谱。我在输入帧上使用汉宁窗。如果可以的话,我想避免使用复数。当我运行这个程序时,它给了我一些非常奇怪的幅度(其中大多数都非常小,并且与正确的频率无关),这让我相信我在计算中犯了一个根本性错误。有人可以提供一些关于这里发生的事情的见解吗?这是我的代码:
int windowSize = 2205;
int probe[500];
float hann[2205];
int j, n;
// initialize probes to 20,40,60,...,10000
for (j=0; j< len(probe); j++) {
probe[j] = j*20 + 20;
fprintf(f, "%d\n", probe[j]);
}
fprintf(f, "-1\n");
// setup the Hann window
for (n=0; n< len(hann); n++) {
hann[n] = 0.5*(cos((2*M_PI*n/(float)windowSize) + M_PI))+0.5;
}
float angle = 0.0;
float w = 0.0; // windowed sample
float realSum[len(probe)]; // stores the real part of the probe[j] within a window
float imagSum[len(probe)]; // stores the imaginary part of probe[j] within window
float mag[len(probe)]; // stores the calculated amplitude of probe[j] within a window
for (j=0; j<len(probe);j++) {
realSum[j] = 0.0;
imagSum[j] = 0.0;
mag[j] = 0.0;
}
n=0; //count number of samples within current window
framesread = psf_sndReadFloatFrames(ifd,frame,1);
totalread = 0;
while (framesread == 1){
totalread++;
// window the frame with hann value at current sample
w = frame[0]*hann[n];
// determine both real and imag product values at sample n for all probe freqs times the windowed signal
for (j=0; j<len(probe);j++) {
angle = (2.0 * M_PI * probe[j] * n) / windowSize;
realSum[j] = realSum[j] + (w * cos(angle));
imagSum[j] = imagSum[j] + (w * sin(angle));
}
n++;
// checks to see if current window has ended
if (totalread % windowSize == 0) {
fprintf(f, "B(%f)\n", totalread/44100.0);
printf("%f breakpoint written\n", totalread/44100.0);
for (j=0; j < len(mag); j++) { // print out the amplitudes
realSum[j] = realSum[j]/windowSize;
imagSum[j] = imagSum[j]/windowSize;
mag[j] = sqrt(pow((double)realSum[j],2)+pow((double)imagSum[j],2))/windowSize;
fprintf(f, "%d\t%f\n", probe[j], mag[j]);
realSum[j] = 0.0;
imagSum[j] = 0.0;
}
n=0;
}
framesread = psf_sndReadFloatFrames(ifd,frame,1);
}
我认为错误在于角度的计算。每个样本的角度增量取决于采样频率。 像这样的东西(你似乎有44100Hz):
angle = (2.0 * M_PI * probe[j] * n) / 44100;
您的样本窗口将包含最低探测频率 20Hz 的一个完整周期。如果将 n 循环到 2205,则该角度将为 2*M_PI。 您看到的可能是混叠,因为您的参考频率为 2205Hz,所有高于 1102Hz 的频率都混叠为较低频率。
使用下面的代码 - 仅稍微重新组织以编译和创建假样本,我确实不得到全零。我已将输出调用更改为最后:
fprintf(f, "%d\t%f\n", probe[j], mag[j] );
到
if (mag[j] > 1e-7)
fprintf(f, "%d\t%f\n", probe[j], mag[j] * 10000);
这只是让查看非零数据变得更容易。也许唯一的问题是理解比例因子?请注意我如何伪造输入来生成纯音作为测试用例。
#include <math.h>
#include <stdio.h>
#define M_PI 3.1415926535
#define SAMPLE_RATE 44100.0f
#define len(array) (sizeof array/sizeof *array)
unsigned psf_sndReadFloatFrames(FILE* inFile,float* frame,int framesToRead)
{
static float counter = 0;
float frequency = 1000;
float time = counter++;
float phase = time/SAMPLE_RATE*frequency;
*frame = (float)sin(phase);
return counter < SAMPLE_RATE;
}
void discreteFourier(FILE* f)
{
FILE* ifd = 0;
float frame[1];
int windowSize = 2205;
int probe[500];
float hann[2205];
float angle = 0.0;
float w = 0.0; // windowed sample
float realSum[len(probe)]; // stores the real part of the probe[j] within a window
float imagSum[len(probe)]; // stores the imaginary part of probe[j] within window
float mag[len(probe)]; // stores the calculated amplitude of probe[j] within a window
int j, n;
unsigned framesread = 0;
unsigned totalread = 0;
for (j=0; j<len(probe);j++) {
realSum[j] = 0.0;
imagSum[j] = 0.0;
mag[j] = 0.0;
}
// initialize probes to 20,40,60,...,10000
for (j=0; j< len(probe); j++) {
probe[j] = j*20 + 20;
fprintf(f, "%d\n", probe[j]);
}
fprintf(f, "-1\n");
// setup the Hann window
for (n=0; n< len(hann); n++)
{
hann[n] = 0.5*(cos((2*M_PI*n/(float)windowSize) + M_PI))+0.5;
}
n=0; //count number of samples within current window
framesread = psf_sndReadFloatFrames(ifd,frame,1);
totalread = 0;
while (framesread == 1){
totalread++;
// window the frame with hann value at current sample
w = frame[0]*hann[n];
// determine both real and imag product values at sample n for all probe freqs times the windowed signal
for (j=0; j<len(probe);j++) {
angle = (2.0 * M_PI * probe[j] * n) / windowSize;
realSum[j] = realSum[j] + (w * cos(angle));
imagSum[j] = imagSum[j] + (w * sin(angle));
}
n++;
// checks to see if current window has ended
if (totalread % windowSize == 0) {
fprintf(f, "B(%f)\n", totalread/SAMPLE_RATE);
printf("%f breakpoint written\n", totalread/SAMPLE_RATE);
for (j=0; j < len(mag); j++) { // print out the amplitudes
realSum[j] = realSum[j]/windowSize;
imagSum[j] = imagSum[j]/windowSize;
mag[j] = sqrt(pow((double)realSum[j],2)+pow((double)imagSum[j],2))/windowSize;
if (mag[j] > 1e-7)
fprintf(f, "%d\t%f\n", probe[j], mag[j] * 10000);
realSum[j] = 0.0;
imagSum[j] = 0.0;
}
n=0;
}
framesread = psf_sndReadFloatFrames(ifd,frame,1);
}
}
在 C 中进行真正的 DFT 最简单的方法是矩阵乘法。这是可以对音频块执行此操作的函数示例。
#define SAMPLING_RATE 44100
#define FRAMES_PER_SECOND 30
#define N (SAMPLING_RATE / FRAMES_PER_SECOND)
float dft_matrix[N][N];
void build_dft_matrix(void) {
for (int k = 0; k < N; k++)
for (int n = 0; n < N; n++)
dft_matrix[k][n] = cosf(2 * M_PIf * k * n / N);
}
void dft(float input[N], float output[N]) {
for (int k = 0; k < N; k++) {
output[k] = 0;
for (int n = 0; n < N; n++)
output[k] = fmaf(dft_matrix[k][n], input[n], output[k]);
}
}
您应该从
build_dft_matrix()
函数调用 main()
来初始化系数矩阵。如果您想让它运行得更快,您可以使用 matmul 库,或者您可以将 dft() 函数中的两个循环展开为 3x4 块(适用于 avx2+),这可以让您利用指令级并行性。