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I would like to know how to apply a low pass filter (Butterworth) to a digital signal. So, I have some values of the signal, let's say $S(t)$. Those values are equally spaced in time. I have read the Wikipedia's article, but I do not understand how to apply the transfer function $H$. I assume that I have to use a Z-transformation and apply it on $S$ but I am not certain to know how to do that. By the way, the part Digital implementation of this article is not really detailed.

Can someone help me to understand those things ? I would really appreciate any answers :)

P.S. I know there are functions in Python (e.g. scipy) that handle that issue, but I would like to know the mathematics behind it.

Thanks.

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What you need to do is find a digital filter $H(z)$ that behaves similar to the $H(s)$ of your Butterworth filter. That is, you need to read about IIR implementation.

There are several methods described in signal processing books about this subject. I recommend you to check for example "Essentials of Digital Signal Processing" from Lathi and Green.

Two of the most common methods are the "Impulse invariance" and the "Bilinear transform".

  • The impulse invariance requires you to find the impulse response $h(t)$ of your $H(s)$. After doing this, you sample it replacing $t$ by $nT$, to finally Z-transform your $h[n]$ into $H(z)$. Bare in mind that this method only works for band-limited filters (lowpass and bandpass) because of aliasing issues. You should determine an appropriate small enough T to avoid this.

  • The bilinear transform, on the other hand, consists on applying the following mapping of your variable $s$ into $z$:

    $s=\frac{2}{T}(\frac{z-1}{z+1})$

    This method has a drawback as well, which is the frequency warping, you can read about it here https://en.wikipedia.org/wiki/Bilinear_transform#Frequency_warping, but is in general easy to counter by pre-warping your original $H(s)$.

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  • $\begingroup$ Great answer, thanks Axel Mancino. $\endgroup$ – kakarotto Sep 11 at 9:34
  • $\begingroup$ Hmm I have another little question to ask. I do not know $T$, I mean I only have $S(t)$ for $t \in [1,2,...,m]$. So, is it possible to know $T$ ? $\endgroup$ – kakarotto Sep 11 at 12:19
  • $\begingroup$ In both methods described the smaller the T, the better result you will obtain. For the first method, T will determine the sampling frequency and thus the repetitions of your original spectrum. Choose a big sampling frequency for smaller spectral superposition (this will depend on the order of your filter and on how much alias you tolerate). For the second method, non-linearity in the frequency mapping from analog to digital (warping) is a tangent function, which is almost lineal near the origin, so lowering T helps since the digital filter singularities get closer to zero. $\endgroup$ – Axel Mancino Sep 12 at 20:44
  • $\begingroup$ Thanks for explaining this. I did some researches and I found that $T$ is set so that the cut-off frequency of $H(s)$ is the same as $H(z)$. I will try to find $T$ with $N=3$ (order of my filter) which makes the cut-off frequency $\omega_c$ identical in both cases. It would be easy as I already know the formula of Butterworth filters. I would be glad to share you my findings. $\endgroup$ – kakarotto Sep 13 at 11:46
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In order to filter a discrete-time signal with a linear and time-invariant (LTI) filter, you need to implement a difference equation:

$$y[n]=-a_1y[n-1]-a_2y[n-2]-\ldots -a_Ny[n-N]+\\+b_0x[n]+b_1x[n-1]+\ldots+b_Nx[n-N]\tag{1}$$

where $x[n]$ is the input sequence, $y[n]$ is the output sequence, and $a_i$ and $b_i$ are the filter coefficients, which determine the properties of the filter (e.g., low pass or high pass, Butterworth or Chebyshev characteristic, etc.). The number of past values of $x[n]$ and $y[n]$ in $(1)$ is determined by the filter order $N$. The higher the order, the more degrees of freedom does the filter have, and the better it can approximate a desired response. In practice, very high orders are undesirable because of numerical problems.

The difference equation $(1)$ corresponds to a filter transfer function

$$H(z)=\frac{b_0+b_1z^{-1}+\ldots+b_Nz^{-N}}{1+a_1z^{-1}+\ldots+a_Nz^{-N}}\tag{2}$$

Some information on how to obtain the filter coefficients $a_i$ and $b_i$ is given in Axel Mancino's answer.

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  • $\begingroup$ Thanks a lot for your answer, I am starting to understand things. Hope people will vote up your answer (I have no enough reputations to do so). $\endgroup$ – kakarotto Sep 11 at 9:36
  • $\begingroup$ Hmm I have another little question to ask. I do not know $T$, I mean I only have $S(t)$ for $t\in[1,2,...,m]$. So, is it possible to know $T$ ? $\endgroup$ – kakarotto Sep 11 at 13:56

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