Z-transform - Digital Signal Processing

Discussion in 'Homework Help' started by Anthony Quah, Jun 12, 2008.

  1. Anthony Quah

    Thread Starter Active Member

    Dec 1, 2007
    hi guy,

    I am study Digital Signal Processing, I need some help to explain how to get from the s -> z transform...it didnt have math work include so i little confuse..i already attach the material

    z-transform.jpg is my question
    tutorial pdf is my full question and answer...

    may god help me on this...exam coming soon....Thanks

  2. Mark44

    Well-Known Member

    Nov 26, 2007
    There are a few things I don't understand from the thumbnail you attached. At the beginning you have:

    1. Hc(s) = H'c(s) evaluated at s = s/omega_sub_i.

    This doesn't make any sense to me. Is this saying that the function H is equal to its own derivative? That does happen for one function, but it doesn't happen generally.

    2. How can you evaluate the derivative at s = s/omega_sub_i? Am I misreading your writing where s looks like the numeral 5?

    3. In Step 3 in the thumbnail, you are evaluating Hc(s) at s = 2/T * (1 - z^(-1)) / (1 + z^(-1)). In the next line, T has vanished. Is T = 1?

    OK, those are my questions.

    To get from Hc(s) = 0.2 / (s^6 + ... + 1.02s + 0.2),

    it looks like they replaced s in the expression above with 2 * (1 - z^(-1)) / (1 + z^(-1)). That substitution would lead, I think, to about a page of some ugly algebra, but shouldn't be more complicated than that.
  3. AlexK

    Active Member

    May 23, 2007
    You could just substitute s with the bilinear transformation in the given laplace transform but that, as said, would probably lead to very lengthy algebra.

    Perhaps a better way will be to do the substitution in the general form of the Butterworth filter (see attachment) .


    H'c(s) is not the derivative, but a LP to LP frequency transformation of the butterworth filter.

    Hope that helps.
    Last edited: Jun 12, 2008