1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $f_s$.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.