# What is z equal to in z-transform?

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?

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.

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.