# Use Filter Constants to Hard Code Filter

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Hey,

I am trying to implement a real-time filter so am using MATLAB's butter() function to generate the needed [b,a] vectors

[b,a] = butter(4,4/35,'low');

Just to be clear I have used these generated vectors with the filter(a,b,data) function to successfully filter my data and it looks quite desirable. But as I am in the end trying to create a real time filter, I am trying to code this into a for-loop (for testing purposes). My code is as follows:

for n=5:1:length(x)

y(n) = b(1)*x(n)+b(2)*x(n-1)+b(3)*x(n-2)+b(4)*x(n-3)+b(5)*x(n-4)-a(2)*y(n-1)-a(3)*y(n-2)+a(4)*y(n-3)+a(5)*y(n-4);

end

This is the mathematical representation as far as I can gather from the doc: http://www.mathworks.com/help/techdoc/ref/filter.html

Can anyone tell me how I am incorrectly modeling the filter() command? I have also switched the a, b, column vectors (in case that was an issue). The above method just goes to infinity, and with a<->b the data just seems to be amplified.

Thanks for the help in advance.

##### 0 Comments

### Accepted Answer

Jan
on 20 Jun 2011

Edited: Jan
on 26 Oct 2014

The difference equation looks ok, but you do not show how e.g. "y(n-4)" is initialized.

Matlab's FILTER uses the "direct form II transposed" implementation, which is more efficient. Together with inital and final conditions:

function [Y, z] = myFilter(b, a, X, z)

% Author: Jan Simon, Heidelberg, (C) 2011

n = length(a);

z(n) = 0; % Creates zeros if input z is omitted

b = b / a(1); % [Edited, Jan, 26-Oct-2014, normalize parameters]

a = a / a(1);

Y = zeros(size(X));

for m = 1:length(Y)

Y(m) = b(1) * X(m) + z(1);

for i = 2:n

z(i - 1) = b(i) * X(m) + z(i) - a(i) * Y(m);

end

end

z = z(1:n - 1);

##### 19 Comments

Jan
on 8 Jun 2022

With some more features:

function [Y, z] = myFilter(b, a, X, z)

% Author: Jan, Heidelberg, (C) 2022

na = numel(a); % Equilize size of a and b

n = numel(b);

if na > n

b(na) = 0;

n = na;

elseif na < n

a(n) = 0;

end

z(n) = 0; % Creates zeros if input z is omitted

b = b / a(1); % Normalize a and b

a = a / a(1);

Y = zeros(size(X));

if na > 1 % IIR filter

for m = 1:length(Y)

Y(m) = b(1) * X(m) + z(1);

for i = 2:n-1

z(i - 1) = b(i) * X(m) + z(i) - a(i) * Y(m);

end

z(n - 1) = b(n) * X(m) - a(n) * Y(m); % Omit z(n), which is 0

end

else % FIR filter: a(2:n) = 0

for m = 1:length(Y)

Y(m) = b(1) * X(m) + z(1);

for i = 2:n-1

z(i - 1) = b(i) * X(m) + z(i);

end

z(n - 1) = b(n) * X(m); % Omit z(n), which is 0

end

end

z = z(1:n - 1);

end

### More Answers (3)

khatereh
on 6 Jan 2012

Hi, I want to use your function instead of matlab filter function. I calculated the filter coefficient in b matrix and it is FIR filter so all the a values are 1. What should be the value for z in my case? I am confused how should I use z.

Thanks so much. Regards, KHatereh

##### 1 Comment

Jan
on 26 Oct 2014

Edited: Jan
on 26 Oct 2014

The initial conditions for the internal state of the filter can be set such, that the transient effects are damped. Look into the code of the filtfilt function for a method to do this automatically.

Set z to zero, if you do not have any information about the signal.

For me the meaning of z got clear, when I examined this example: Imagine a long signal X, which is divided in 2 parts X1 and X2. Now the complete signal X is filtered with certain parameters and the initial settings z=0 (this means zeros(1,n-1) with n is the length of the filter parameters):

z = zeros(1, length(b) - 1);

Y = filter(b, a, X, z);

Now we do this for the first part:

z = zeros(1, length(b) - 1);

[Y1, z1] = filter(b, a, X1, z);

Now the output z1 is the internal state of the filter, a kind of history over the last elements. If we use the output z1 of the 1st part as input of the 2nd, we get exactly the same outpt as for the full signal:

Y2 = filter(b, a, X2, z1);

isequal(Y, [Y1, Y2]) % TRUE

But if we omit z1 as input for filtering X2, there is a small difference mostly at the start of Y2 due to the transient effects.

In this case, we do have some knowledge about the history of the internal filter state for X2, but for X1 this state is not defined and zeros are a fair guess, but not necessarily smart.

Yves
on 10 May 2018

##### 0 Comments

Michal
on 21 Feb 2022

Probably fully functional naive version of filter:

function [Y, z] = myFilter(b, a, X, z)

% a and b should have same order

na = length(a);

nb = length(b);

if na > nb

b = [b,zeros(1,na-nb)];

n = na;

elseif na < nb

a = [a,zeros(1,nb-na)];

n = nb;

else

n = na;

end

% naive filter implementation

z(n-1) = 0; % Creates zeros if input z is omitted

b = b / a(1); % normalize parameters

a = a / a(1);

Y = zeros(size(X));

for m = 1:length(Y)

Xm = X(m);

Y(m) = b(1) * Xm + z(1);

Ym = Y(m);

for i = 2:n-1

z(i - 1) = b(i) * Xm + z(i) - a(i) * Ym;

end

z(n - 1) = b(n) * Xm - a(n) * Ym;

end

end

##### 3 Comments

Jan
on 17 Sep 2022 at 9:53

@Michal: No, z(n)=0 does not break the compatibility with Matlab's filter, if it is applied inside the function and cropped finally: z = z(1:n - 1) as in my implementation above. The error message appears, if you expand z before calling filter, but this was not suggested.

Your method to expand z by the line:

z(n-1) = 0;

is a bug, because it sets the last element of z to 0. If z was [1,2,3] initially (then a and b have 4 elements), your code sets z to [1,2,0]. Simply try it:

x = rand(1, 100);

B = [0.000416546, 0.00124964, 0.00124964, 0.000416546];

A = [1, -2.68616, 2.41966, -0.730165]; % Lowpass Butterworth

z = [1,2,3];

y1 = filter(B, A, x, z);

y2 = michaelsFilter(B, A, x, z);

plot(y1, 'r')

hold('on')

plot(y2, 'b') % Obviously different

function [Y, z] = michaelsFilter(b, a, X, z)

% a and b should have same order

na = length(a);

nb = length(b);

if na > nb

b = [b,zeros(1,na-nb)];

n = na;

elseif na < nb

a = [a,zeros(1,nb-na)];

n = nb;

else

n = na;

end

% naive filter implementation

z(n - 1) = 0; % BUG! Must be z(n)=0, then crop z(n) after the loop.

% Or:

% if length(z) < n-1

% z(n - 1) = 0;

% end

b = b / a(1); % normalize parameters

a = a / a(1);

Y = zeros(size(X));

for m = 1:length(Y)

Xm = X(m);

Y(m) = b(1) * Xm + z(1);

Ym = Y(m);

for i = 2:n-1

z(i - 1) = b(i) * Xm + z(i) - a(i) * Ym;

end

z(n - 1) = b(n) * Xm - a(n) * Ym;

end

end

Do you see it now?

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