我正在尝试用C语言编写离散傅里叶变换,以便处理32位浮点wav文件。每次读取2帧数据(每个声道一帧,但为了我的目的,我假设它们都相同,因此我只使用frame [0])。该代码旨在通过用频率20、40、60、......、10000来探测输入文件并输出幅度谱。我在输入帧上使用Hanning窗口。如果可以,我想避免使用复数。当我运行此代码时,它给我一些非常奇怪的振幅(其中大多数非常小,并且与正确频率不相关),这使我相信我在计算中犯了一个基本错误。有人能否提供关于这里发生了什么的见解?以下是我的代码:
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);
}
double
而不是float
,因为double
比float
快得多得多,而且不要使用pow(a,2)
,而是使用 (a*a)。pow
使用非常昂贵的指数函数调用,如果您只需要进行平方运算,则这是不必要的。 - Mark Lakata