How can I write the Matlab “filter”-function myself?

Question

I would like to use a Butterworth filter on a 1D-Signal. In Matlab the script would look like this:

 f=100;
 f_cutoff = 20; 
 fnorm =f_cutoff/(f/2);
 [b,a] = but         


        

Answer 1

Check my Answer

filter

    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;
    }

conv

    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;
    }

deconv

    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;
    }

deconvRes

    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;
    }

Example Using

Example Using

  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));

OUTPUT

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]

Please give me your feedbacks!!

Answer 2

I implemented filter function used by Matlab in Java :

The filter function is implemented as a direct form II transposed structure,

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)

where n-1 is the filter order, which handles both FIR and IIR filters [1], na is the feedback filter order, and nb is the feedforward filter order.

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);
    }

and Here is an example :

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);

and output was :

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

as in Matlab output

That's it

Answer 3

BlackEagle's solution does not reproduce the same results as MATLAB with other arrays. For example:

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

Gives you completely other results. Be careful.

Answer 4

I found my mistake. Here's the working code (as a function):

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

Now the filter works nearly fine, but at the first 40 values i've got divergent results. I'll have to figure that out...

Answer 5

I have found a text described the Direct Form II Transposed used in the Matlab filter function and it works perfectly. See script below. Other implementations are also available but with error of around 1e-15, you'll see this by running the script yourself.

%% 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