In some places, it is said that z is equal to:
$$z = e^s \quad where \quad s = \sigma + j \Omega $$
But in some places, it is said that z is equal to:
$$z = e^{sT_s} \quad $$ where Ts is a sampling interval.
Why are there two different definitions?
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
When discussing the z-transform, it is important to note that there are two different frequencies that are innately involved. In many text books, the two frequencies are denoted by $\omega$ (discrete-time) and $\Omega$ (continuous-time). We'll use this convention.
The z-transform uses the "discrete frequency" $\omega$ which is derived from the "analog frequency" $\Omega$ via sampling at a rate $T_s$ as
$$\omega = {\Omega}T_s$$
If we're on the unit circle, which your question implies as the starting condition, then the $\sigma$ term in $s = \sigma + j\Omega$ goes to zero and so
$$s = \sigma + j\Omega \Rightarrow s = j\Omega$$
We can now relate the z-transform as the discrete-time equivalent of the Laplace transform and so
$$z = e^{sT_s} = e^{j{\Omega}T_s} = e^{j\omega}$$
Which is exactly the substitution made between the Fourier transform of a discrete sequence $x[n]$ and it's z-transform:
$$X(e^{j\omega}) = \sum_{-\infty}^{\infty}x[n]e^{-j\omega n}$$
$$X(z) = \sum_{-\infty}^{\infty}x[n]z^{-n}$$
I have never seen $z$ be defined by $z = e^{s}$, which implies the use of the continuous-time frequency and is not how the z-transform is defined.
So the confusion might come down to abuse of notation between how $\omega$ and $\Omega$ are defined in the sources you are referencing.
$\begingroup$
$\endgroup$
In more general sense the z in Z-transform is z=r*e^jw . where r is the radius and w is the frequency. The Fourier transform is evaluated on the unit circle of the z-plane diagram where r=1.
