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


Consequently, as a function of $f$, $H(e^{j2\pi f/f_s})$ is periodic with the sampling frequency $f_s$.

  • $\begingroup$ 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? $\endgroup$ – knzy Jan 27 at 15:25
  • $\begingroup$ @knzy: As a function of $\omega$, the DTFT is always $2\pi$-periodic. As a function of $f$, its period equals the sampling frequency. $\endgroup$ – Matt L. Jan 27 at 15:56
  • $\begingroup$ 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 $\omega_s$? $\endgroup$ – knzy Jan 27 at 16:00
  • $\begingroup$ @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. $\endgroup$ – Matt L. Jan 27 at 17:05

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