Question

我正在使用pusleMeter项目测量脉冲信号。

所以，我使用递归最小均方的自适应滤波来从脉冲信号中去除噪声（类似于ECG信号中存在的运动伪影的运动伪像），并且我修改了来自RLS <的RLMS代码/ p>

1）因为只有一个有噪声的脉冲信号所以第一次没有参考，所以试图使用第一个脉冲信号（500个样本）的低通滤波器版本作为RLS滤波器的参考

2）在第一个脉冲信号（接下来的500个样本）之后，它将RLSfitler输出作为参考信号。

我像这样实施

 clear all;
 clc;
 P=load('Noise_Pulse_signal.mat');
a2=P.a1;
for i=1:500:length(a2)
    if(i<499)
 For the first pulse signal(500samples) the reference signal is approximated as the LPfilter version of the first pulse signal.
     x1=a2(i:i+499,:);
        d = fdesign.lowpass('Fp,Fst,Ap,Ast',2.4,8,1,40,500);
        Hd1 = design(d);
        LP = filtfilt(Hd1.Numerator,1,x1);
          plot(LP,'-r');
        pause(0.2);
        e=LP';
      else
         x=e;
        x1=a2(i:i+499,:); 
        x1=x1';
        [e,w] = RLS_1(x,x1,500);
        pause(0.2);
    end;
end

RLS_1就是这个

function [e,w] = RLS_1(n,x,fs)
  % Filter Parameters
  p       = 4;                % filter order
  lambda  = 1.0;              % forgetting factor
  laminv  = 1/lambda;
  delta   = 1.0;              % initialization parameter
  % Filter Initialization
  w       = zeros(p,1);       % filter coefficients
  P       = delta*eye(p);     % inverse correlation matrix
  %e       = x*0;              error signal I'm not using this as the LPfilter output will be used as the e(reference signal).
    for m = p:length(x)    
      y = n(m:-1:m-p+1);% Acquire chunk of data
      e(m) = x(m)-w'*y;% Error signal equation
      Pi = P*y;% Parameters for efficiency
      k = (Pi)/(lambda+y'*Pi);% Filter gain vector update  
      P = (P - k*y'*P)*laminv; % Inverse correlation matrix update   
      w = w + k*e(m);% Filter coefficients adaption   
  end
  toc
    t = linspace(0,length(x)/fs,length(x));
  figure;
  plot(t,x,t,e);
  title('Result of RLS Filter')
  xlabel('Time (s)');
  legend('Reference', 'Filtered', 'Location', 'NorthEast');
  title('Comparison of Filtered Signal to Reference Input');
  return

但它正在给出这样的错误

        Error in RLS_1 (line 15)
     e(m) = x(m)-w'*y;% Error signal equation
         Error in Motion (line 18)
       [e,w] = RLS_1(x,x1,500);

任何人都可以解释为什么会产生错误，如果这是上述要求的正确实现方式，因为我需要一个有效的方法，以便可以在微控制器上实现。三江源。

递归最小均方的自适应滤波

0 个答案: