Can the Z-transform be used for both continuous and discrete functions?

[milesyoung]

Hi,

I'm currently studying some introductory discrete control theory and I've run into a problem with the Z-transform, although I could pose the same question regarding the Laplace transform. I know I'm completely off with this question but I've just stared myself blind to it. Here goes:

I have some continuous function of time e(t) and I sample it to get the discrete function e(kT) where k = 0,1,2,... and T is the sample time. Thus:

$$E(z) = \mathcal{Z}\left\{e(kT)\right\} = e(0T)z^{-0} + e(1T)z^{-1} + e(2T)z^{-2} + \ldots$$

Now, the discrete function e(k) would give me the following instead:

$$E(z) = \mathcal{Z}\left\{e(k)\right\} = e(0)z^{-0} + e(1)z^{-1} + e(2)z^{-2} + \ldots$$

My question is then: I seem to be able to assign the same symbol E(z) to both series according to the definition of the Z-transform, but the two series are not generally equal. This is probably a semi-retarded question, but what am I missing here?

 

[AlephZero]

The sampled function is bandwidth-limited by the sampling rate and the Nyquist criterion.
A continuous function is not bandwidth limited in general.

Depending how far you have progressed through your course, you might not have met this issue yet, but will definitely meet it later on. It is one of the key issues using digital signal processing in practice, for any application.

 

[milesyoung]

I think my question more relates to some basic notation.

How can

$$E(z) = \mathcal{Z}\left\{e(k)\right\}$$

and

$$E(z) = \mathcal{Z}\left\{e(kT)\right\}$$

Wouldn't that imply that

$$\mathcal{Z}\left\{e(k)\right\} = \mathcal{Z}\left\{e(kT)\right\}$$

But this is clearly not true in general.