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  • Low pass filter order

    Hi All,

    I am trying to ascertain the correct way to name a low pass filter. If I follow the examples as detailed in Robertson and Dowling (reference below), then a single pass filter is a second order, a dual pass is a fourth order, 2 dual passes is an 8th order etc.

    However, recent discussions with a colleague have indicated this may not be accurate. My colleague has indicated these are simply second order filters with ‘x’ number of bidirectional passes.

    While I have found many texts that provide a description of poles and zeros and patterns of change with different orders and number of bidirectional passes, I cannot find any other references that link the two.

    Please can anyone in the group shed any light – and ideally a reference, explaining the correct way to describe filters?

    Many thanks
    Dan

    Robertson, D. G. E., & Dowling, J. J. (2003). Design and responses of Butterworth and critically damped digital filters. Journal of electromyography and kinesiology : official journal of the International Society of Electrophysiological Kinesiology, 13(6), 569–73.

  • #2
    Re: Low pass filter order

    The dual pass second order filter is indeed a 4th order filter because at high frequencies, the response drops off proportional to frequency to the power -4.

    I am not familiar with the critically damped filters of Robertson & Dowling, but if each pass is a Butterworth filter, the dual pass filter is not a 4th order Butterworth filter. A 4th order BW filter would have a frequency response that is not equal to the square of the 2nd order BW filter response. The difference is small at very low or very high frequencies.

    Also the cutoff frequency (defined as the frequency where the signal is attenuated by a factor sqrt(2)) is going to be lowered due to having two passes. This is all clearly explained in David Winter's book.

    See also http://en.wikipedia.org/wiki/Butterworth_filter. You can square the frequency response and quickly see that it is not the same as doubling the order.

    Ton van den Bogert

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    • #3
      Re: Low pass filter order

      Hi Ton,

      Thanks for getting back to me. If I understand your post correctly, a Butterworth filter with one bidirectional pass is a 4th order filter, though a Butterworth filter with 2 bidirectional passes is not an 8th order filter?

      I looked over the link you posted; while it provides a good overview of Butterworth filters it does not explain exactly what defines filter orders. I have Winter’s book and have re-read the section on filters. Winter clearly states that one bidirectional pass = a 4th order filter, but does not explain what happens if I apply 2 bidirectional passes?

      Can you shed some light on exactly what a Butterworth filter with 2 bidirectional passes is? i.e. is it simply a 4th order filter applied twice or is it actually a different order? If so how do I establish which order it is?

      Thank you for your help.

      Comment


      • #4
        Re: Low pass filter order

        Dan,

        The order of the filter describes its response as frequency goes to infinity. If the amplitude response is 1/f^n, the filter is of order n.

        This explains why two passes of a nth order filter will produce a 2n-th order filter. So 4 passes of the 2nd order filter will give an 8th order filter.

        But, there are many ways to make a 4th order filter. All will converge to the same high-frequency response but they can be very different near the corner frequency. This is where the double 2nd order filter (two passes of the 2nd order filter) is different from a 4th order butterworth filter.

        If I had more time I would produce some plots of the frequency responses, this is not hard with the formulas from the Wikipedia page.

        I hope this helps.

        Ton van den Bogert

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        • #5
          Re: Low pass filter order

          Hi Ton,

          Thanks for getting back to me again; I am beginning to get an idea of what you are saying. Though I am still struggling to get my head the exact ways I can assess filters.

          I note that despite a lack of replies this post is one of the most viewed out of the recent posts; therefore I wonder if others are in a similar situation to me? What concerns me most is that I am applying filters without honestly understanding the filters.

          Do you know of a good learning resource (textbook/web page/ technical article) that covers the areas of filters discussed? I have followed the link you previously posted, though I am not sure how I would use this to establish the order of a filter I have applied in Microsoft excel or in software such as Visual3D (where I can set the cut-off frequency and the number of bidirectional passes but not the filter order)?

          Thanks again
          Dan

          Comment


          • #6
            Re: Low pass filter order

            You can try a textbook on digital signal processing, but that is probably not needed and not a good use of your time. Winter's "Biomechanics and Motor Control of Human Movement" describes the Butterworth filter that is commonly used in biomechanics.

            If you already have a filter (in Excel or Visual3D), the filter order should be known because that was part of the filter design process. But, you can indeed determine the filter order experimentally. Create a sine wave with frequency far above the cutoff frequency. Maybe twice the cutoff frequency at least. Run it through the filter and measure the amplitude A of the sine wave that comes out of the filter. Now make a sine wave with exactly twice the frequency of the first sine wave. Run it through the filter and measure the amplitude B of the sine wave that comes out of the filter. If the ratio A/B is 2, you have a 1st order filter. If the ratio A/B is 4, you have a 2nd order filter. If the ratio A/B is 8, you have a 3rd order filter. You see the pattern. This is based on the definition of filter order that I gave in my previous post. I am not sure that this is the correct definition of filter order, but it works for the filters we are discussing here.

            You can also determine the complete frequency response of the filter by running sine waves of various frequencies through the filter, and plot the amplitude of the filter output vs. frequency of the input signal. It is a very useful exercise to do this once, to reassure you that the filter is doing what it is supposed to do. With the same approach, you can also verify the cutoff frequency. A sine wave at the cutoff frequency should be attenuated by a factor 1.41, the square root of two.

            Ton van den Bogert

            Comment


            • #7
              Re: Low pass filter order

              This is an excellent reply and provides me with understandable tools to help solve my problem.

              Thank you very much. I really appreciate your help.

              Regards
              Dan

              Comment


              • #8
                Re: Low pass filter order

                I thought it would be helpful to show the (theoretical) frequency response of the 4th order Butterworth filter and the double second order filter (see below).

                As you can see, they are both 4th order filters because their response at high frequencies is the same, and is 12 dB/octave (1/f^4). The difference is around the corner frequency.

                This is why it is not correct to use the term "4th order Butterworth filter" when you use a double second order filter. It's a 4th order filter, but not a 4th order Butterworth filter.

                I found some comments on the internet that an actual digital implementation (e.g. Matlab functions butter and filtfilt) can deviate from the theoretical response, so it may be a good idea to also derive the theoretical response for the digital discrete-time version. Or determine the response "experimentally" by running sine wave signals through the filter as I indicated earlier.

                Ton van den Bogert

                Matlab code:
                Code:
                w = 0.1:0.01:10;
                G1 = 1./sqrt(1+w.^8);    % amplitude response of 4th order Butterworth filter (see equation in Wikipedia)
                G2 = 1./(1+w.^4);         % square of the response of the 2nd order Butterworth filter
                loglog(w,G1,w,G2);
                xlabel('frequency / corner frequency');
                ylabel('amplitude response')
                legend('4th order Butterworth','double 2nd order Butterworth');
                Plot: BWfilters.jpg
                Last edited by Ton van den Bogert; October 18, 2013, 09:42 AM. Reason: picture did not appear

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