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Use Filter Constants to Hard Code Filter

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Donald Hume
Donald Hume on 20 Jun 2011
Commented: Han Zerui on 5 Mar 2021
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);
This is the mathematical representation as far as I can gather from the doc:
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.

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);
z = z(1:n - 1);
[EDITED]: A C-Mex implementation which handles arrays also: FEX: FilterM.
Han Zerui
Han Zerui on 5 Mar 2021
@Jan Thank you for the answer. That makes so much more sense.

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More Answers (2)

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
The meaning of z is explained in the documentation: doc filter.
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.

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Yves on 10 May 2018
Can someone please comment on whether this z/z1 or zi/zf - initial/final condition (delay) of the digital filter is equivalent to the state variables in state-space model (ABCD matrix) of the filter?

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