The fourier transform of a continuous time signal $x(t)$ is $X(ω)$ where the unit of $ω$ is radians/second. And for a discrete signal $x(n)$, the DTFT is $X(e^{jω})$ where the unit of $ω$ is radians. Clearly the $ω$ for FT and DTFT shouldn't mean the same thing.


What is the difference in interpretation of $ω$ (rads/s) and $ω$ (rads)?

If we interprete $X(ω_i)$ as the ampitude or $|X(ω_i)|^2$ as the energy associated with the frequency component $ω_i$, then how should we interprete $X(e^{jω_i})$ or $|X(e^{jω_i})|^2$?

Thanks in advance.

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    $\begingroup$ In discrete time, the frequencies are in radians per sample. Samples are dimensionless. $\endgroup$ – Juancho Oct 29 '17 at 14:40
  • $\begingroup$ @Juancho what is the physical significance or interpretation of rads/sample? I understand rads/sec means "phase change per second". But, "phase change per sample" doesn't quite ring a bell. (meaning I don't understand) $\endgroup$ – Sadist_Tanmoy Oct 30 '17 at 3:42

The unit rads/sample (or just rads, leaving "/sample" implicit) for a discrete signal means how phase increases between consecutive samples (the exact same way rads/second means how phase increases between consecutive seconds). Some examples with real sinusoids (visualizing a phasor in the complex plane might help, its projection onto the real axis would be the values in these examples):

  1. Signal $x(n) = [...,1,-1,1,-1,1,-1...]$ has frequency $\pi$ rads/sample because it moves forwards (counter-clockwise) $\pi$ rads from sample to sample.
  2. Signal $x(n) = [...,1,0,-1,0,1,0,-1,0...]$ has frequency $\pi/2$ rads/sample because it moves forwards (counter-clockwise) $\pi/2$ rads from sample to sample.
  3. Signal $x(n) = [...,-1,0,1,0,-1,0,1,0...]$ has frequency $-\pi/2$ rads/sample because it moves backwards (clockwise) $\pi/2$ rads from sample to sample.

There is a limited bandwidth in the discrete domain $(-\pi, \pi]$, beyond that, the spectrum repeats itself. Also, $2 \cdot f_{max} < f_{sampling}$. So, $f_{sampling}/2$ rads/second maps to $\pi$ rads/sample (or just rads...).

Notations are just notations, $X(e^{j\omega})$ helps to state the cyclic nature of the DTFT, since $f(\omega) = e^{j\omega}$ has a period of $2\cdot\pi$ (so, if you plug in 34 $\pi$ rads/sample, you know, without seeing the expression of $X(e^{j\omega}$), that it will be the same as plugging in 0 rads/sample). In your question, both expressions mean exactly the same.

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Most people learn (or try) DSP from this book

Oppenheim, and Schaefer, (1999). Discrete-time signal processing. Pearson Education.

and there $\Omega$ is used for continuous time and $\omega$ is for discrete time, which is how they keep their notation consistent. Usually both are continuous, which can lead to some confusion when you consider the DFT and/or Fourier series.

In the Olden Days, Yea DSP was primarily practiced by people who had a background in continuous time signals derived from some physical problem, and notational conventions for $\omega$ were typically $2\pi f$ (in yonder rads/sec).

Most books back then, tended to start with some general Linear Systems theory, and then go to sampling. I think the most recent version of Oppenheim and Schaefer wait until Chapter 4 to cover sampling.

People who taught and wrote about DSP, recognized that including a sample rate wasn't really necessary. My first DSP class was taught by a guy who did adaptive beamforming and he introduced the concept of unity sampling, $f_s=1 \text{ sample/second}$ so all our filter frequency responses were plotted between -1/2 to 1/2 Hz. You just needed to scale everything by your actual sample rate when you got back to your physical problem.

It became clearer over time that there were time series, like Stock Market time series that were inherently discrete in time and a lot of DSP didn't really require reference to time, and things like image processing really didn't either, so to make things more general, $\omega$ as radians seems to have become the most common notation. You can also sell your books to finance majors now.

The notation $X(e^{j\omega})$ is handy when you plug $e^{j\omega}$ into a Z transform (if the unit circle is in the region of convergence), and reinforces the concept that a function or composition of a periodic function $e^{j\omega}$, is periodic.

The notation $X(\omega)$ needs to be interpreted in the context of where it is, could be discrete time, could be continuous time.

So, the answer to your question is: Context.

As an American English speaker, it would be natural for me to assume that this is the complete story but someone who learned from the Russian (or French, or ....) DSP literature would most likely have a different explanation.

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  • $\begingroup$ I really appreciate for the time that you took to answer. Thank you. But I'm afraid the answer didn't quite help. It was abstract. I want specific answer. Rewording my question would be "If $|X(e^{jw_i})|^2$ mean the energy associated with $w_i$, then what is $w_i$? I need to know the physical interpretation. Of course for discrete time signals." I would be grateful if could edit your answer and talk on this question more and less on historical background." Thank you very much. $\endgroup$ – Sadist_Tanmoy Oct 30 '17 at 4:16
  • $\begingroup$ Physical interpretation comes from the physics. If the time series represented a voltage the magnitude squared of the voltage would be a wattage across a load. Energy is Joules, The magnitude squared at a frequency is a number. The frequency is a number. What is the physical interpretation of 5.? $\endgroup$ – user28715 Oct 30 '17 at 4:54
  • $\begingroup$ Well, the frequnce may be a number but it does have physical interpretation. 5 Hz or 10pi rads/sec would mean the voltage signal has a period of 1/5 seconds and it reaches the positive peak 5 times in a second or it changes its phase by 10pi rads in a second and in so many other ways. But, what would 5 rads/sample be interpreted into? It changes its phase by 5 rads per sample? I am a beginner student. So, I am not arguing rather trying clarify my conception. Thank you. $\endgroup$ – Sadist_Tanmoy Oct 30 '17 at 5:09
  • $\begingroup$ just as for discrete-time signals $x[n]$ having Z-Transform as $X(z)$ and Fourier Transform as $X(z) \Bigg|_{z=e^{j \omega}}$ or simply $X(e^{j \omega})$, then for continuous-time signals $x(t)$, they have Laplace Transform of $X(s)$ and Fourier Transform $X(s) \Bigg|_{s=j \Omega}$ or simply $X(j \Omega)$. I stay away from the simpler "$X(\Omega)$" unless the argument is "ordinary" frequency and using a slightly different expression of Fourier Transform: $$ X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2\pi f t}\, dt $$ $\endgroup$ – robert bristow-johnson Oct 30 '17 at 5:50

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