问题
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] = butter(8,fnorm,'low');
filteredData = filter(b,a,rawData); % I want to write this myself
Now I don't want to directly use the filter-function given in Matlab but write it myself. In the Matlab documentation it's described as follows:
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.
So I've already tried to write the function like that:
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
But that's not working. Can you please help me? What am I doing wrong?
Sincerely, Cerdo
PS: the filter documentation can be foud here: http://www.mathworks.de/de/help/matlab/ref/filter.html#f83-1015962 when expanding More About -> Algorithms
回答1:
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
回答2:
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!!
回答3:
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
回答4:
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.
回答5:
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...
来源：https://stackoverflow.com/questions/17506343/how-can-i-write-the-matlab-filter-function-myself