Trying to plot the fft of a sinc function

Question

I am trying to plot the fft of a set of data I have. These data form a nearly perfect sinc function. Here is the data of which I am trying to plot the fft: .

I know the fft of a sinc function should look like kind of a step function. However, the results I get are nowhere near that. Finding the fft in itself is super easy, but I think my mistake is when I try to compute the frequency axis. I have found several methods online, but so far I have not been able to make it work. Here is my code:

sampleRate = (max(xdata) - min(xdata))/length(xdata);
sampleN = length(xdata);
y = fft(ydata, sampleN);
Y = y.*conj(y)/sampleN;
freq = (0:1:length(Y)-1)*sampleRate/sampleNumber;
plot(freq, Y)

I have found pretty much all of that online and I understand pretty much none of it (which might be why it's not working...)

Zoom on what I get using that code:

It now seems to be working! This is what I get when I subtract the mean:

Answer 1

你在这里看到的是零频率远远超过其他所有东西。使用plot(freq,Y,'o-')绘图以证明您看到的形状只是两个样本之间的线性插值。

零频率是所有样本的总和。因为信号的平均值比幅度大很多，零频率使其他所有信息相形见绌。而且由于你正在绘制功率（DFT的绝对平方），这种差异会进一步增强。

有两个简单的解决方案：

1. 使用对数y轴进行绘图：

   plot(freq, Y)
   set(gca,'yscale','log')
   

2. 从信号中减去均值，删除零频率或缩放y轴（这些都或多或少等价）：

   y = fft(ydata-mean(ydata), sampleN);
   

   或

   y(1) = 0;
   

   或

   plot(freq, Y)
   set(gca,'ylim',[0,max(Y(2:end))]);