# Z-transform - Digital Signal Processing

#### Anthony Quah

Joined Dec 1, 2007
80
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

Anthony

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#### Mark44

Joined Nov 26, 2007
628
Anthony,
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.

#### AlexK

Joined May 23, 2007
34
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) .

Mark44,

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

Hope that helps.

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