Recovering DTFT from Z-transform The relationship between the Z-transform and DTFT can be expressed like:
$$ H(e^{j \omega}) = H(z)|_{z = e^{j \omega}}$$
Graphically, evaluating the Z-transform on the unit circle is shown as sweeping $\omega$ as $e^{j \omega}$ is plotted, represented as a phasor rotating around the unit circle.
My question is, isn't there a problem in that the argument of an exponential is supposed to be unitless? Any complex number (on the unit circle) can be represented as $e^{j \phi}$ where $\phi$ is the number's phase and is unitless (radians). But $\omega$ has units of frequency, so what's happening here? Is there implicitly a time unit of 1?
I've seen some texts cancel the units by normalizing the frequency, so instead of following $e^{j \omega}$ you follow $e^{j 2\pi (f / f_s)}$, where $f_s$ is a sampling frequency. How would assuming an arbitrary $f_s$ affect the calculated frequency response?
The $\omega$ in the frequency response of a discrete-time system $H(e^{j\omega})$ is indeed unitless. The frequency response $H(e^{j\omega})$ is periodic with period $2\pi$. If the discrete signal is obtained via sampling with sampling rate $f_s$, then the relation between $\omega$, the actual frequency $f$, and the sampling frequency $f_s$ is
$$\omega=2\pi\frac{f}{f_s}\tag{1}$$
Consequently, as a function of $f$, $H(e^{j2\pi f/f_s})$ is periodic with the sampling frequency $f_s$.