如何从信号中提取频率 [英] How to extract a frequency from a signal

问题描述

有没有一种简单的方法来从信号中提取主频率/周期（不借助FFT）？

对于我的要求，这可能导致主频率（例如3Hz）的值或表示目标频率的强度的值。例如，在下面的一维信号中，频率大约是4Hz，假设采样率是50ms。

这怎么能通过编程的方式从数据中提取出来呢？

p>

10
2
1
2
8
10
8
2
1
1
8
10
7
1
1
2
7
10
5
1

解决方案

使用自动关联功能！

将采样率转换为赫兹

fs = 1 /（50/1000）％result = 20hz

vector = [10 2 1 2 8 10 8 2 1 1 8 10 7 1 1 2 7 10 5 1];

R = xcorr（vector）;

[pks，locs] = findpeaks（R）;

％导致​​赫兹

fs./(diff(locs））

ans =

3.3333 4.0000 3.3333 3.3333 4.0000 3.3333

max（fs./diff（locs））

ans =

4

• 应用自相关的信号，你可以在网上找到很多
  的源代码，用不同的语言来做自相关，一个伪代码：

    TotalSamples = length（signal）
   for z = 1：TotalSamples 
   sum = 0; 
   
   for i = 1：TotalSamples 
   
   sum = sum +（signal（i）* signal（i + pos））; 
   
   end 
   
   Xcorre（z）= Xcorre（z）+ sum; 
   end 
    




• 从自相关结果中找出所有局部峰值




• 计算局部峰值之间的差异 locs [k + 1] - locs [k]




• 将帧频除以局部峰值之间的差值




• 频率是最大值

Is there a simple way to extract the main frequency/period from a signal (without resorting to the FFT)?

For my requirements, this can result in either a value for the main frequency (e.g. 3Hz) or a value representing the strength of a target frequency. For example, in the following 1-D signal the frequency is about 4Hz, assuming the sampling rate is 50ms.

How can this be extracted from the data programmatically?

10 2 1 2 8 10 8 2 1 1 8 10 7 1 1 2 7 10 5 1

解决方案

Use Auto Correlation !

%using Matlab

%convert sample rate to hertz

fs = 1/(50/1000) % result = 20hz

vector = [10 2 1 2 8 10 8 2 1 1 8 10 7 1 1 2 7 10 5 1];

R = xcorr(vector);

[pks,locs]=findpeaks(R);

%result in hertz

fs./(diff(locs))

ans =

3.3333    4.0000    3.3333    3.3333    4.0000    3.3333

max(fs./diff(locs))

ans =

4

• Apply Autocorrelation on the signal, you can find a lot of source code in the web in defferent languages to do autocorrelation, a pseudo code:

  TotalSamples = length(signal)
  for z=1:TotalSamples
   sum = 0;
  
   for i=1:TotalSamples
  
             sum = sum + (signal(i)*signal(i + pos));
  
   end
  
   Xcorre(z) = Xcorre(z) + sum;
  end
  

• Find all local peaks from result of autocorrelation

• Compute the difference between local peaks locs[k+1] - locs[k]

• Divide your frame rate by the difference between local peaks

• The Frequency is the Maximum value

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