How to connect $\omega$ in time domain to physical frequency?

Question

I'm a noobie to DSP, and recently I'm learning DSP for an urgent assignment. There's a problem that has been confused me for days. I hope anybody could shed some light on it, although it might seem to be silly.

The assignment asks me to filter a signal so that it's in $20-800$Hz. After I spent several days to learn DSP, I finally get some idea about frequency domain, DTFT, FIR, IIR, etc. But I still have no idea about how to proceed. Furthermore, I have a huge confusion about frequency: in the frequency domain, we always talk about the radian $\omega$ which is always less than $\pi$, how can I connect this $\omega$ to the physical frequency? Is it right to use $\omega/2\pi$?

Answer 1

It's easier to keep straight if you consider units.

$\omega$ has units of radians per sample, as you stated.

$f_{Hz}$ has units of cycles per second.

$2\pi$ has units of radians per cycle.

$f_s$ has units of samples per second.

So your conversion equation from $\omega$ to $f_{Hz}$ is:

$$ f_{Hz} = \omega \cdot \frac{f_s}{2\pi} $$

Where the units are:

$$ \frac{cycles}{second} = \frac{radians}{sample} \cdot \frac{\frac{samples}{second}}{\frac{radians}{cycle}} $$

And the reverse is:

$$ \omega = f_{Hz} \cdot \frac{2\pi}{f_s} $$

Where the units are:

$$ \frac{radians}{sample} = \frac{cycles}{second} \cdot \frac{\frac{radians}{cycle}}{\frac{samples}{second}} $$

Hope this helps.

Ced

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Follow up:

When you bring the DFT into the picture, you introduce:

$N$ has units of samples per frame.

$k$, the bin index, has units of cycles per frame.

$$ \omega = k \cdot \frac{2\pi}{N} $$

Where the units are:

$$ \frac{radians}{sample} = \frac{cycles}{frame} \cdot \frac{\frac{radians}{cycle}}{\frac{samples}{frame}} $$

And

$$ f_{Hz} = k \cdot \frac{f_s}{N} $$

Where the units are:

$$ \frac{cycles}{second} = \frac{cycles}{frame} \cdot \frac{\frac{samples}{second}}{\frac{samples}{frame}} $$

Answer 2

There are several "frequency" variables in signal processing, the most common all being related by constant factors. I will use the notation from Digital Signal Processing by Proakis & Manolakis:

• $F$: "Ordinary" frequency in units of Hz.
• $\Omega$: Angular frqeuency, in units of radians/second; $\Omega = 2\pi F$
• $f$: Normalized or "digital frequency". $f = F/F_s$, where $F_s$ is the sample rate in Hz. This is a formally unitless quantity, but sometimes its units are listed as being cycles/sample. Sampling produces aliasing, which means that the spectrum of a sampled signal will be periodic with period 1 cycle/sample. For this reason DSP engineers only look at the spectrum of the signal in the principal period of $-1/2 \leq f < 1/2$.
• $\omega$: Normalized radian frequency. $\omega = \Omega/F_s = 2\pi F/F_s$. Sometimes its units are listed as being radians/sample. Because of aliasing, it is only necessary to study the spectrum of a signal from $-\pi \leq \omega < \pi$ in the digital domain.

Other notions of the concept of "frequency" exist (such as instantaneous frequency, and a notion of an "instantaneous spectrum"), but these are more advanced concepts that you probably don't need to worry about yet.

Answer 3

$\omega$ you are referring to which always lies in the range [-π, π] or [0, 2π], is normalised by the sampling frequency Fs. Therefore, the Nyquist frequency would be π(rad/samples).

In your case, 20Hz (physical frequency), in $\omega$ domain will be (2•π•20)/Fs