我使用DFT的此实现:
/*
Direct fourier transform
*/
int DFT(int dir,int m,double *x1,double *y1)
{
long i,k;
double arg;
double cosarg,sinarg;
double *x2=NULL,*y2=NULL;
x2 = malloc(m*sizeof(double));
y2 = malloc(m*sizeof(double));
if (x2 == NULL || y2 == NULL)
return(FALSE);
for (i=0;i<m;i++) {
x2[i] = 0;
y2[i] = 0;
arg = - dir * 2.0 * 3.141592654 * (double)i / (double)m;
for (k=0;k<m;k++) {
cosarg = cos(k * arg);
sinarg = sin(k * arg);
x2[i] += (x1[k] * cosarg - y1[k] * sinarg);
y2[i] += (x1[k] * sinarg + y1[k] * cosarg);
}
}
/* Copy the data back */
if (dir == 1) {
for (i=0;i<m;i++) {
x1[i] = x2[i] / (double)m;
y1[i] = y2[i] / (double)m;
}
} else {
for (i=0;i<m;i++) {
x1[i] = x2[i];
y1[i] = y2[i];
}
}
free(x2);
free(y2);
return(TRUE);
}
放置在这里http://paulbourke.net/miscellaneous/dft/
第一个问题是为什么在应用直接transform(dir=1
)之后我们应该缩放值?我阅读了有关DFT实施的一些想法,但没有发现任何相关信息。
作为输入,我使用具有1024采样频率的cos
#define SAMPLES 2048
#define ZEROES_NUMBER 512
double step = PI_2/(SAMPLES-2*ZEROES_NUMBER);
for(int i=0; i<SAMPLES; i++)
{
/*
* Fill in the beginning and end with zeroes
*/
if(i<ZEROES_NUMBER || i > SAMPLES-ZEROES_NUMBER)
{
samplesReal[i] = 0;
samplesImag[i] = 0;
}
/*
* Generate one period cos with 1024 samples
*/
else
{
samplesReal[i] = cos(step*(double)(i-ZEROES_NUMBER));
samplesImag[i] = 0;
}
}
对于绘图,我删除了上面我询问的缩放比例,因为输出值变得非常小,无法绘制图形。
如您所见,相位始终为0,幅度频谱相反。为什么?
以下是我更易读的版本,没有进行缩放,但产生的结果相同:
void DFT_transform(double complex* samples, int num, double complex* res)
{
for(int k=0; k<num; k++)
{
res[k] = 0;
for(int n=0; n<num; n++)
{
double complex Wkn = cos(PI_2*(double)k*(double)n/(double)num) -
I*sin(PI_2*(double)k*(double)n/(double)num);
res[k] += samples[n]*Wkn;
}
}
}
Ok, guys. I'm glad to say that this implementation works. Problem was the wrong way of plotting graphs and a lack of understanding formulas.
As you can see there is k
variable which is used to vary the frequency. So frequency is ν = k / T where T is a period of time taken to get samples. T = N/S where S is your sampling frequency. Then you can find your frequency as v = S*k/N
So when you get your result you should calculate frequencies for each point and remove everything that above S/2 and only then plot graph Magnitude = Magnitude(Frequency). This is what I didn't understand before. Hope it will be helpful to someone.
Sin 100HZ.
如您所见,显示了频率和相关幅度。缩放存在问题,但是我们是否要确定信号中显示的频率并不重要。
感谢@PaulR
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