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Low Pass Filter in Software question

  1. May 7, 2012 #1
    Hi Everyone,
    I read an article by Barry L. Dorr about how to implement a Low Pass Filter in software, however, the equation that he gave confused me as I didn't see any sort of convergence to the value desired.

    So here is the equation:
    y(n) = (1 - 2^(-k)) * y(n-1) + x(n)
    x is the input, y is the output, n is the index, and k is the filter strength

    So I understand how k works as it just makes it so that previous values affect the output less. But where I run into problems is the + x(n) term.

    So taking the example where k = 4, I have the equation:
    y(n) = 0.9375 * y(n-1) + x(n)

    Then I will just create an arbitrary array with which I want to low-pass and smooth out.
    So lets say, x(n) = {1, 5, 3, 2}

    Then that means,
    y(3) = .9375 * y(2) + x(3) = 8.031 + 2 = 10.031
    y(2) = .9375 * y(1) + x(2) = 5.5664 + 3 = 8.5664
    y(1) = .9375 * y(0) + x(1) = .9375 + 5 = 5.9375
    y(0) = x(0) = 1

    So at this point, I can see the the output array is not even converging to getting larger as time goes on, even though the maximum input to low-pass filter was 5.

    Any insight on how this works would be great.
  2. jcsd
  3. May 7, 2012 #2


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    Science Advisor
    Homework Helper

    You mean ths article? http://www.edn.com/article/465351-A...filter_suits_embedded_system_applications.php

    He "forgot" to tell you that his simple filter has a gain at low frequences. If you see what happens when you filter a constant input signal with all the x(n) values equal to 1, the output values are
    1 + .9375
    1 + .9375 + .9375^2
    1 + .9375 + .9375^2 + 9375^3
    This is a geometric series. The output will rise "slowly" and then level off at the value of
    1 / (1 - .9375) = 16.

    Note: you will need to get about 50 output values to see the complete picture of what happens.

    If you want the filter to have a gain of 1 at low frequencies, use
    y(n) = 2^(-k)(1 - 2^(-k)) * y(n-1) + 2^(-k)x(n)
    Last edited: May 7, 2012
  4. May 7, 2012 #3
    Ah yes,
    I did notice that when I was running it in my actual software. It would level off at around 16 if I passed in a large array. I will try implementing your low-pass filter with gain of 1 and see how it works out. Thanks.
  5. May 7, 2012 #4
    Okay I just implemented the low-pass filter and it does have the affect of "smoothing" out my data, now I want to implement a high pass filter for my data. My main question is,
    What affect does a high pass filter have on my dataset?
    For example, I know that the low pass smooths out data and passes low frequencies (changes?). So applying this logic, does the high pass filter converge more quickly to my data, (ie. pass large changes in data but not small?)
  6. May 8, 2012 #5


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    Science Advisor

    Short answer: high pass leaves you with the part that low pass removes. If your data are the sum of a low-frequency "signal" and some higher-frequency "noise" then low-pass will tend to filter out the noise, while high-pass will leave you with the noise. (Of course, one person's noise is another person's data, etc., etc...)

    The time derivative (changes) is weighted toward high frequencies in a sort of "ramp" rather than providing good "high pass."
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