Selection of FFT sampling frequency and sampling points

Original signal ：$x(t)=2.5+sin(2\pi \ast 100t)+3sin(2\pi \ast 900t)$

1、 According to Nyquist sampling theorem, it is guaranteed not to overlap and can restore the original signal information , Determine the minimum sampling frequency $f_{smin}=1800Hz$.
2、 The maximum frequency resolution is 100Hz, Minimum total sampling time $L_{min}=1/100$.
3、 according to $L=N/f_s$, Get the minimum number of sampling points $N_{min}=L_{min}\times f_{smin}=1800/100=18$.
4、 The base 2-FFT Required points N by 2 Exponentially times （ Zero filling or truncation are not considered , The number of sampling points is directly equal to FFT Count points ）, Therefore, we can get more than $N_{min}$ Of N The valueable sequence of $N(n)=32,64,128,\mathrm{256....}$.
5、 To ensure the accuracy of frequency and amplitude （ Bloggers understand that the original signal period obtained in this sampling time is an integer multiple , Try not to leak the spectrum ）, Need to meet
$$m_1\times f_s/N=f_1=100,m_1\in Positive\; integer$$
$$m_2\times f_s/N=f_2=900,m_2\in Positive\; integer$$
6、 Combine the above two formulas to get $$\frac{m_2}{m_1}=9$$

• Yes $m_1、m_{2}andN$ Value in turn , Make up .
  Make $m_1=1,m_2=9,N=32$, obtain $f_s=3200$
  Make $m_1=1,m_2=9,N=64$, obtain $f_s=6400$
  Make $m_1=1,m_2=9,N=128$, obtain $f_s=12800$
  … wait
  Make $m_1=2,m_2=18,N=32$, obtain $f_s=1600<f_{smin}$ （ Give up ）
  Make $m_1=2,m_2=18,N=64$, obtain $f_s=3200$
  Make $m_1=2,m_2=18,N=128$, obtain $f_s=6400$
  Make $m_1=2,m_2=18,N=256$, obtain $f_s=12800$
  … wait
  Make $m_1=3,m_2=27,N=32$, obtain $f_s\approx 1066.7<f_{smin}$ （ Give up ）
  Make $m_1=3,m_2=27,N=64$, obtain $f_s\approx 2133.3$
  Make $m_1=3,m_2=27,N=128$, obtain $f_s\approx 4266.7$
  Make $m_1=3,m_2=27,N=256$, obtain $f_s\approx 8533.3$
  … wait

• Select one set of data ,$N=64,f_s=6400$ Analyze ：

%% [ Preprocessing ]
clc;   % Clear the command window 
clear; % Clear variables and functions in the workspace 
clf;   % Clears the current drawing 

%% [ Sampling parameters ]
fs=6400;   % sampling frequency  (Hz)  
ts=1/fs;   % Sampling period  (s)
N=64;      % Number of sampling points  ( individual )
n=0:N-1;   % Sample point index 
t=n*ts;    % Sampling timeline 

%% [ Sampled signal ] 
fa=100;           % The signal a The frequency of 
fb=900;           % The signal b The frequency of 
xn=2.5+sin(2*pi*fa*t)+3*sin(2*pi*fb*t); % Sampled signal = The signal a+ The signal b

%% [FFT Handle ]
M=N;           %FFT Count points 
X=fft(xn,M);   %FFT Output value 
X=[X(1)/M,X(2:M)*2/M]; % Amplitude axis 
k=0:M-1;       % Frequency point index 
f=fs*k/(M-1);  % Frequency axis 

%% [ Make a picture ]
stem(f,abs(X));  % Sequence diagram 
xlabel(' frequency ');  % Horizontal axis coordinates 
ylabel(' Range ');  % Vertical coordinates 
title({
    [' sampling frequency ',num2str(fs),' ,  Number of sampling points ',num2str(N)],['FFT Count points ',num2str(M)]});
grid;            % Show table lines 

• Constantly change the sampling frequency and sampling points , Get the picture below ：
  first line ：$m_1=1,m_2=9$
  The second line ：$m_1=2,m_2=18$
  The third line ：$m_1=3,m_2=27$
  
• Analyze the above figure ：
  In every line , The frequency resolution is gradually doubled from left to right .
  As shown in the two figures indicated by the double arrows in the figure below , The sampling frequency is different 、 The sampling points are the same 、FFT The result is the same , And with the increase of sampling points , The accuracy is gradually improved .
  
  7、 According to the law summarized above , According to the accuracy requirements of the actual project and the configuration of software and hardware , Select the appropriate number of sampling points and sampling frequency .