2

我想在一维信号上使用巴特沃斯滤波器。在 Matlab 中，脚本如下所示：

 f=100;
 f_cutoff = 20; 
 fnorm =f_cutoff/(f/2);
 [b,a] = butter(8,fnorm,'low');
 filteredData = filter(b,a,rawData); % I want to write this myself

现在我不想直接使用 Matlab 中给出的filter函数，而是自己编写。在 Matlab 文档中描述如下：

滤波器函数实现为直接形式 II 转置结构，

y(n) = b(1)*x(n) + b(2)*x(n-1) + ... + b(nb+1)*x(n-nb) - a(2)* y(n-1) - ... - a(na+1)*y(n-na)

其中 n-1 是滤波器阶数，它同时处理 FIR 和 IIR 滤波器 [1]，na 是反馈滤波器阶数，nb 是前馈滤波器阶数。

所以我已经尝试过编写这样的函数：

f=100;
f_cutoff = 20; 
fnorm =f_cutoff/(f/2);
[b,a] = butter(8,fnorm,'low');
for n = 9:size(rawData,1)
    filteredData(n,1) = b(1)*n + b(2)*(n-1) + b(3)*(n-2) + b(4)*(n-3) + b(5)*(n-4) ...
                      - a(2)*rawData(n-1,1) - a(3)*rawData(n-2,1) - a(4)*rawData(n-3,1) - a(5)*accel(n-4,1);
end

但这行不通。你能帮我么？我究竟做错了什么？

真诚的， 塞尔多

PS：过滤器文档可以在这里找到：http: //www.mathworks.de/de/help/matlab/ref/filter.html#f83-1015962展开更多关于 -> 算法时

4

5 回答 5

8

检查我的答案

筛选

    public static double[] filter(double[] b, double[] a, double[] x) {
      double[] filter = null;
      double[] a1 = getRealArrayScalarDiv(a,a[0]);
      double[] b1 = getRealArrayScalarDiv(b,a[0]);
      int sx = x.length;
      filter = new double[sx];
      filter[0] = b1[0]*x[0];
      for (int i = 1; i < sx; i++) {
        filter[i] = 0.0;
        for (int j = 0; j <= i; j++) {
          int k = i-j;
          if (j > 0) {
            if ((k < b1.length) && (j < x.length)) {
              filter[i] += b1[k]*x[j];
            }
            if ((k < filter.length) && (j < a1.length)) {
              filter[i] -= a1[j]*filter[k];
            }
          } else {
            if ((k < b1.length) && (j < x.length)) {
              filter[i] += (b1[k]*x[j]);
            }
          }
        }
      }
      return filter;
    }

转化率 

    public static double[] conv(double[] a, double[] b) {
      double[] c = null;
      int na = a.length;
      int nb = b.length;
      if (na > nb) {
        if (nb > 1) {
          c = new double[na+nb-1];
          for (int i = 0; i < c.length; i++) {
            if (i < a.length) {
              c[i] = a[i];
            } else {
              c[i] = 0.0;
            }
          }
          a = c;
        }
        c = filter(b, new double [] {1.0} , a);
      } else {
        if (na > 1) {
          c = new double[na+nb-1];
          for (int i = 0; i < c.length; i++) {
            if (i < b.length) {
              c[i] = b[i];
            } else {
              c[i] = 0.0;
            }
          }
          b = c;
        }
        c = filter(a, new double [] {1.0}, b);
      }
      return c;
    }

反卷积 

    public static double[] deconv(double[] b, double[] a) {
      double[] q = null;
      int sb = b.length;
      int sa = a.length;
      if (sa > sb) {
        return q;
      }
      double[] zeros = new double[sb - sa +1];
      for (int i =1; i < zeros.length; i++){
        zeros[i] = 0.0;
      }
      zeros[0] = 1.0;
      q = filter(b,a,zeros);
      return q;
    }

反卷积 

    public static double[] deconvRes(double[] b, double[] a) {
      double[] r = null;
      r = getRealArraySub(b,conv(a,deconv(b,a)));
      return r;
    }

getRealArraySub 

    public static double[] getRealArraySub(double[] dSub0, double[] dSub1) {
      double[] dSub = null;
      if ((dSub0 == null) || (dSub1 == null)) { 
        throw new IllegalArgumentException("The array must be defined or diferent to null"); 
      }
      if (dSub0.length != dSub1.length) { 
        throw new IllegalArgumentException("Arrays must be the same size"); 
      }
      dSub = new double[dSub1.length];
      for (int i = 0; i < dSub.length; i++) {
        dSub[i] = dSub0[i] - dSub1[i];
      }
      return dSub;
    }

getRealArrayScalarDiv 

    public static double[] getRealArrayScalarDiv(double[] dDividend, double dDivisor) {
      if (dDividend == null) {
        throw new IllegalArgumentException("The array must be defined or diferent to null");
      }
      if (dDividend.length == 0) {
        throw new IllegalArgumentException("The size array must be greater than Zero");
      }
      double[] dQuotient = new double[dDividend.length];

      for (int i = 0; i < dDividend.length; i++) {
        if (!(dDivisor == 0.0)) {
          dQuotient[i] = dDividend[i]/dDivisor;
        } else {
          if (dDividend[i] > 0.0) {
            dQuotient[i] = Double.POSITIVE_INFINITY;
          }
          if (dDividend[i] == 0.0) {
            dQuotient[i] = Double.NaN;
          }
          if (dDividend[i] < 0.0) {
            dQuotient[i] = Double.NEGATIVE_INFINITY;
          }
        }
      }
      return dQuotient;
    }

示例使用

示例使用

  double[] a, b, q, u, v, w, r, z, input, outputVector;


  u = new double [] {1,1,1};
  v = new double [] {1, 1, 0, 0, 0, 1, 1};
  w = conv(u,v);
  System.out.println("w=\n"+Arrays.toString(w));

  a = new double [] {1, 2, 3, 4};
  b = new double [] {10, 40, 100, 160, 170, 120};
  q = deconv(b,a);
  System.out.println("q=\n"+Arrays.toString(q));

  r = deconvRes(b,a);
  System.out.println("r=\n"+Arrays.toString(r));

  a = new double [] {2, -2.5, 1};
  b = new double [] {0.1, 0.1};
  u = new double[31];
  for (int i = 1; i < u.length; i++) {
    u[i] = 0.0;
  }
  u[0] = 1.0;
  z = filter(b, a, u);
  System.out.println("z=\n"+Arrays.toString(z));

  a = new double [] {1.0000,-3.518576748255174,4.687508888099475,-2.809828793526308,0.641351538057564};
  b = new double [] { 0.020083365564211,0,-0.040166731128422,0,0.020083365564211};
  input = new double[]{1,2,3,4,5,6,7,8,9};

  outputVector = filter(b, a, input);
  System.out.println("outputVector=\n"+Arrays.toString(outputVector));

输出

w=
[1.0, 2.0, 2.0, 1.0, 0.0, 1.0, 2.0, 2.0, 1.0]
q=
[10.0, 20.0, 30.0]
r=
[0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
z=
[0.05, 0.1125, 0.115625, 0.08828125, 0.0525390625, 0.021533203124999997, 6.469726562499979E-4, -0.009957885742187502, -0.012770843505859377, -0.010984611511230471, -0.007345342636108401, -0.003689372539520266, -9.390443563461318E-4, 6.708808243274683E-4, 0.0013081232085824014, 0.0012997135985642675, 9.705803939141337E-4, 5.633686931105333E-4, 2.189206694310998E-4, -8.033509766391922E-6, -1.195022219235398E-4, -1.453610225212288E-4, -1.219501671897661E-4, -7.975719772659323E-5, -3.8721413563358476E-5, -8.523168090901481E-6, 8.706746668052387E-6, 1.5145017380516224E-5, 1.4577898391619086E-5, 1.0649864299265747E-5, 6.023381178272641E-6]
outputVector=
[0.020083365564211, 0.11083159422936348, 0.31591188140651166, 0.648466936215357, 1.0993782391344866, 1.6451284697769106, 2.25463601232057, 2.8947248889603028, 3.534126758562552]

请给我你的反馈！！

于 2015-12-25T01:25:36.377 回答
2

我找到了一篇描述 Matlab 过滤器函数中使用的 Direct Form II Transposed 的文本，它工作得很好。请参阅下面的脚本。其他实现也可用，但错误约为 1e-15，您将通过自己运行脚本来看到这一点。

%% Specification of the Linear Chebysev filters
clc;clear all;close all
ord = 5; %System order (from 1 to 5)
[bq,aq] = cheby1(ord,2,0.2);theta = [bq aq(2:end)]';
figure;zplane(bq,aq); % Z-Pole/Zeros
u = [ones(40,1); zeros(40,1)];
%% Naive implementation of the basic algorithm
y0 = filter(bq,aq,u); % Built-in filter
b = fliplr(bq);a = fliplr(aq);a(end) = [];
y1 = zeros(40,1);pad = zeros (ord,1);
yp = [pad; y1(:)];up = [pad; u(:)];
for i = 1:length(u)
    yp(i+ord) = sum(b(:).*up(i:i+ord))-sum(a(:).*yp(i:i+ord-1));
end
y1 = yp(ord+1:end); % Naive implementation
err = y0(:)-y1(:);
figure
plot(y0,'r')
hold on
plot(y1,'*g')
xlabel('Time')
ylabel('Response')
legend('My code','Built-in filter')
figure
plot(err)
xlabel('Time')
ylabel('Error')
%% Direct Form II Transposed  
% Direct realization of rational transfer functions
% trps: 0 for direct realization, 1 for transposed realisation 
% b,a: Numerator and denominator
% x:   Input sequence
% y:   Output sequence
% u:   Internal states buffer

trps = 1;
b=theta(1:ord+1);
a=theta(ord+2:end);
y2=zeros(size(u));
x=zeros(ord,1);
%%
if trps==1
    for i=1:length(u)
        y2(i)=b(1)*u(i)+x(1);
        x=[x(2:ord);0];
        x=x+b(2:end)*u(i)-a*y2(i);
    end
else
    for i=1:length(u)
        xnew=u(i)-sum(x(1:ord).*a);
        x=[xnew,x];
        y2(i)=sum(x(1:ord+1).*b);
        x=x(1:ord);
    end
end
%%
err = y2 - filter(bq,aq,u);
figure
plot(y0,'r')
hold on
plot(y2,'*g')
xlabel('Time')
ylabel('Response')
legend('Form II Transposed','Built-in filter')
figure
plot(err)
xlabel('Time')
ylabel('Error')
% end
于 2014-01-14T08:09:36.163 回答
1

我在 Java 中实现了 Matlab 使用的过滤器功能：

滤波器函数实现为直接形式 II 转置结构，

y(n) = b(1)*x(n) + b(2)*x(n-1) + ... + b(nb+1)*x(n-nb) - a(2)* y(n-1) - ... - a(na+1)*y(n-na)

其中 n-1 是滤波器阶数，它同时处理 FIR 和 IIR 滤波器 [1]，na 是反馈滤波器阶数，nb 是前馈滤波器阶数。

public void filter(double [] b,double [] a, ArrayList<Double> inputVector,ArrayList<Double> outputVector){
        double rOutputY = 0.0;    
        int j = 0;
        for (int i = 0; i < inputVector.size(); i++) {
            if(j < b.length){
                rOutputY += b[j]*inputVector.get(inputVector.size() - i - 1);
            }
            j++;
        }
        j = 1;
        for (int i = 0; i < outputVector.size(); i++) {
            if(j < a.length){
                rOutputY -= a[j]*outputVector.get(outputVector.size() - i - 1);
            }
            j++;   
        }
        outputVector.add(rOutputY);
    }

这是一个例子：

ArrayList<Double>inputVector = new ArrayList<Double>();
ArrayList<Double>outputVector = new ArrayList<Double>();
double [] a = new double [] {1.0000,-3.518576748255174,4.687508888099475,-2.809828793526308,0.641351538057564};
double [] b = new double [] { 0.020083365564211,0,-0.040166731128422,0,0.020083365564211};
double  []input = new double[]{1,2,3,4,5,6,7,8,9};
for (int i = 0; i < input.length; i++) {
    inputVector.add(input[i]);
    filter(b, a, inputVector, outputVector);
}
System.out.println(outputVector);

输出是：

[0.020083365564211, 0.11083159422936348, 0.31591188140651166, 0.6484669362153569, 1.099378239134486, 1.6451284697769086, 2.254636012320566, 2.894724888960297, 3.534126758562545]

如在 Matlab 输出中

就是这样

于 2015-02-24T05:23:04.960 回答
1

BlackEagle 的解决方案无法重现与 MATLAB 与其他数组相同的结果。例如：

b = [0.1 0.1]
a = [2 -2.5 1]
u = [1, zeros(1, 30)];
z = filter(b, a, u)

给你完全不同的结果。当心。

于 2015-10-16T10:34:50.187 回答
0

我发现了我的错误。这是工作代码（作为函数）：

function filtered = myFilter(b, a, raw)

filtered = zeros(size(raw));
for c = 1:3
    for n = 9:size(raw,1)

        filtered(n,c) = b(1)* raw(n,c)   + b(2)* raw(n-1,c) + b(3)* raw(n-2,c) ...
                      + b(4)* raw(n-3,c) + b(5)* raw(n-4,c) + b(6)* raw(n-5,c) ...
                      + b(7)* raw(n-6,c) + b(8)* raw(n-7,c) + b(9)* raw(n-8,c) ...
                      - a(1)*filtered(n,c)   - a(2)*filtered(n-1,c) - a(3)*filtered(n-2,c) ...
                      - a(4)*filtered(n-3,c) - a(5)*filtered(n-4,c) - a(6)*filtered(n-5,c) ...
                      - a(7)*filtered(n-6,c) - a(8)*filtered(n-7,c) - a(9)*filtered(n-8,c);
    end
end

现在过滤器几乎可以正常工作，但在前 40 个值中，我得到了不同的结果。我得想办法...

于 2013-07-07T08:45:06.037 回答