# 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 $$f_s$$.

• Thanks a lot for the response- to clarify, this applies for cases of $x_d(n)=x_c(n/f_s)$, in which case $x_d(n)$'s DTFT will be $f_s$-periodic, correct? And does this mean that the DTFT will not necessarily be $2 \pi$ periodic anymore, or will it be periodic in $2 \pi$ anyway, it just might not be the fundamental period anymore?
– knzy
Jan 27 '21 at 15:25
• @knzy: As a function of $\omega$, the DTFT is always $2\pi$-periodic. As a function of $f$, its period equals the sampling frequency. Jan 27 '21 at 15:56
• But couldn't you re-express the DTFT as $H(e^{j2\pi \omega / \omega_s})$ (just converting frequencies in Hz to rad/s), which implies that the DTFT as a function of $\omega$ will be periodic in $\omega_s$?
– knzy
Jan 27 '21 at 16:00
• @knzy: Well, now you just use $\omega$ for what I called $f$. In discrete time it's common to use $\omega$ as normalized frequency in radians, just the way I used it in my answer. Jan 27 '21 at 17:05