Matlab没有绘制确切的傅立叶信号

`fft`函数采用单位时间步长.为了校正非单位时间步长，您需要根据奈奎斯特速率定义频率分量.以下代码以正确的频率轴绘制`fft`的大小.

``````clc;
clear all;
close all;
x=1:0.001:10;
%     ^ this is your sampling time step
f1=sin(2*pi*10*x);
f2=sin(2*pi*15*x);
f3=sin(2*pi*30*x);

% bounds of fourier transform based on sampling rate
Fs = 1/0.001;
ff = linspace(-Fs/2,Fs/2,numel(x));

F1 = fftshift(fft(f1)/numel(x));
F2 = fftshift(fft(f2)/numel(x));
F3 = fftshift(fft(f3)/numel(x));

figure();
plot(ff,abs(F1),'-r'); hold on;
plot(ff,abs(F2),'-b');
plot(ff,abs(F3),'-k');
``````

Edit2:归一化项(`/numel(x)`)对于使用离散傅立叶变换估计连续时间傅立叶变换是必需的.我不记得确切的数学原因，而是关于MIT OpenCourseWare的讲座.

I'm trying to plot a simple signal in fourier domain using Matlab. It's not plotting the correct signal. Here is my code:

``````clc;
clear all;
close all;
x=1:0.001:10;
f1=sin(2*pi*10*x);
f2=sin(2*pi*15*x);
f3=sin(2*pi*30*x);
f=f1+f2+f3;
plot(2*pi*x,fft(f1));
figure
plot(x,fft(f1));
``````

I've expected a peak at 10 since the frequency is 10. But it is giving a peak at some other point

Here are the two plot images:

This is the image for `plot(x,fft(f1))`

This is the image for `plot(2*pi*x,fft(f1))`

It is not showing the peak at 10.I even tried using abs(fft(f1)). No luck :/

Isn't it the correct way to plot signal in fourier domain?

The `fft` function assumes unit time step. In order to correct for non unit time step you need to define the frequency component based on the nyquist rate. The following code plots the magnitude of the `fft` with the correct frequency axis.

``````clc;
clear all;
close all;
x=1:0.001:10;
%     ^ this is your sampling time step
f1=sin(2*pi*10*x);
f2=sin(2*pi*15*x);
f3=sin(2*pi*30*x);

% bounds of fourier transform based on sampling rate
Fs = 1/0.001;
ff = linspace(-Fs/2,Fs/2,numel(x));

F1 = fftshift(fft(f1)/numel(x));
F2 = fftshift(fft(f2)/numel(x));
F3 = fftshift(fft(f3)/numel(x));

figure();
plot(ff,abs(F1),'-r'); hold on;
plot(ff,abs(F2),'-b');
plot(ff,abs(F3),'-k');
``````

Edit: To answer OPs question in the comment.

Speaking in normalized frequency units (assuming sampling rate of 1). The `fft` function returns the frequency response from `0` to `2*pi` radians, but due to some signal processing properties and the way that discrete signals are interpreted when performing an FFT, the signal is actually periodic so the `pi` to `2*pi` section is identical to the `-pi` to `0` section. To display the plot with the DC component (0 frequency) in the center we use `fftshift` which does a circular shift equal to 1/2 the length of the signal on the data returned by `fft`. Before you take the `ifft` make sure you use `ifftshift` to put it back in the right place.

Edit2: The normalization term (`/numel(x)`) is necessary to estimate the continuous time fourier transform using the discrete fourier transform. I don't remember the precise mathematical reason off the top of my head but the examples in the MATLAB documentation also imply the necessity of this normalization.

Edit 3: The original link that I had is down. I may come back to add a more detailed answer but in the mean time I definitely recommend that anyone interested in understanding the relationship between the fundamentals of the FS, FT, DTFT, and DFT watch Professor Oppenheim's hilariously old, but amazingly informative and straightforward lectures on MIT OpenCourseWare.