I am just starting to learn discrete filters and I could use some help. I understand continuous signals and filters.

I am trying to understand the math behind discrete filtering. For example in the s domain, a simple low pass filter can be recognized as $$\dfrac{1}{\tau s+1}$$ How does sampling rate affect a discrete low pass filter?

  • 1
    $\begingroup$ The cut-off frequency of a discrete-time low pass filter is specified relative to the sampling frequency. So by changing the sampling frequency, you also change its cut-off frequency. Is that what you're asking? $\endgroup$
    – Matt L.
    May 22, 2015 at 8:36
  • $\begingroup$ I guess what I am trying to understand where if the equation above was brought to the z-domain, where would the sampling time have a role? $\endgroup$
    – Adam
    May 22, 2015 at 12:59

1 Answer 1


IF we take the sampling period as T.First I will convert the filter to time domain impulse response as $$ h(s)=\frac{1}{\tau s+1}\\h(s)=\frac{1/\tau}{s+\frac{1}{\tau}} $$ this in time domain is $$ h(t)=\frac{e^{-t/\tau}}{\tau} $$ IF the above signal is sampled at time T the it can be written as $$ h(Tn)=\frac{e^{\frac{-Tn}{\tau}}}{\tau}\\h(n)=\frac{e^{\frac{-Tn}{\tau}}}{\tau} $$ if Z transform taken for the impulse response.Then the equation is $$ H(z)=\frac{1/\tau}{1-e^{\frac{T}{\tau}}z^{-1}} $$ Here we can see the sampling time coming in denominator which could put the pole inside or outside the unit circle

  • $\begingroup$ I understand the first part, up to h(n). Could you explain how you got H(z) from h(n)? $\endgroup$
    – Adam
    May 22, 2015 at 17:43
  • $\begingroup$ is T/\tau in H(z) supposed to be negative? $\endgroup$
    – Adam
    May 22, 2015 at 18:14
  • $\begingroup$ since h(n) is a geometric sequence we can find standard H(z) for h(n) as $$ \frac{1}{1-a^{-1}z^{-1}}$$ here a is $$ e^{T/\tau}$$ and T\/tau should be negative $\endgroup$
    – Vinith
    May 23, 2015 at 2:42

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