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I am trying to make sense of the filter coefficients.

The author claims that they are running a 3rd order lowpass Butterworth filter and were willing to give their design. They are not the most communicative type and it is typically weeks until I am able to get any sort of answer from them.

The coefficients are as follows:

1.22293 1.394312 1.994313
-0.3339159 -0.6572268 -0.9943293

They also provided the transfer function:

$$ H_k(z) = \frac{b_{k0}+b_{k1}z^{-1}+b_{k2}z^{-2}}{1-a_{k1}z^{-1}-a_{k2}z^{-2}},\quad k = 1, 2, 3. $$

Whey I plug in these coefficients into the FDA toolbox in MATLAB, I get the following frequency response (sampling rate was set to $20833\textrm{ Hz}$):

Frequency Responce

What is the correct way to interpret these numbers? I am somewhat out of my element when it comes to the Digital Filtering and could use some help.

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    $\begingroup$ The transfer-function in the equation is only 2nd order... Did they label the coefficients? $\endgroup$ – Arnfinn Nov 30 '16 at 22:31
  • $\begingroup$ Could you provide a source for the work you are talking about? $\endgroup$ – Laurent Duval Nov 30 '16 at 23:34
  • $\begingroup$ @Laurent Duval The coefficients and the TF expression are all that is provided to us. We are trying to troubleshoot their system that processes signals generated by our equipment and the author/customer is very secretive regarding their implementation details. $\endgroup$ – udushu Dec 1 '16 at 1:12
  • $\begingroup$ @Arnfinn the coefficients were not labeled. Just two rows of numbers in the exact same order as shown in the post. $\endgroup$ – udushu Dec 1 '16 at 1:14
  • $\begingroup$ @udushu Well, I guess you can try various permutations of the coefficients -- trying them out as numerator and denominator and flipping them left-right -- to see if you at the very least can get a stable filter... $\endgroup$ – Arnfinn Dec 1 '16 at 2:08
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Sorry, but that does not make any sense

  1. A 3rd order BW lowpass can be described most simply with a single number: the cutoff off frequency
  2. It's the not numerator a denominator since this would have 4 numbers and the numerator would have the shape of x*[1 3 3 1] and the denominator should start with 4.
  3. It's not the poles and zeros since all the zeros are at -1 and one pole is conjugate complex
  4. It sure as heck isn't state space either
  5. The transfer function is a boiler plate text book copy and also mostly wrong for a third order BW (which has one biqaud and one single real pole section).

You are not missing anything obvious and having you guessing blindly doesn't serve any useful purpose. You have to get back to the client and tell them that you need proper specification or you can't get the work done.

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  • $\begingroup$ I figured that much. I am writing them a strongly worded email about the whole situation. Thanks! $\endgroup$ – udushu Dec 1 '16 at 18:30

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