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Continued from the previous two questions

Q3: When $f_o$ isn't a multiple of $F_s$ , can $F_s$ still map to $2\pi$？

Answer: $F_s$ always map to $2\pi$ under digital (normalized) frequency domain.

Q4: Can $F_s$ be mapped to multiple of $2\pi$ like $4\pi$, $8\pi$？

Answer:

Some programs (such as MATLAB) that design filters with real-valued coefficients use the Nyquist frequency $\frac{f_s}{2}$ as the normalization constant. The resultant normalized frequency has units of half-cycles/sample or equivalently cycles per 2 samples."

Under this circumstances $F_s$ is mapped to $\pi$ (half-cycles)

Q5: Let $f_o = mF_s$. For $j2\pi$ times an integer mn, Since n has already been an integer, what's the meaning for multiplying another integer m.

Answer: The meaning of the integer m is to stretch the x-axis periodically.

Pic1 - fo = mF_s

Pic2 - Fs mapped to 2π

Ran
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$f_o$ is continuous on the frequency axis and can take on any values in Hz from $-\infty$ to $+\infty$, representing the frequency of a signal of interest. For discrete time signals it is common to divide the frequency by the sampling rate to get $f_o/f_s$ resulting in a "normalized frequency" in units of cycles/sample. This normalized frequency will have a unique span from $f= -1/2$ to $f = +1/2$ (called the "first Nyquist zone") due to the periodicity in frequency that the sampling process creates. In this case $f_s$ maps specifically to $f=1$. If we change to units of radians/sample, then the unique frequency span from $-\pi$ to $+\pi$ and $f_s$ will map to $2\pi$ radians/sample. This makes sense as for every sample of the clock, the sampling frequency will have revolved $2\pi$ radians, or one cycle.

For more details on normalized frequency, please refer to this post:

What is normalized frequency

Dan Boschen
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  • Hi @Dan Boschen I had been waiting to earn enough reputation to ask you under your discussion with Musse Redi. Luckily my question draws your attention!

    The unit $f_o$ is cycles/sec, thus the normalized freq $\frac{f_o}{f_s}$ set up a relation to $2\pi$ because of "cycles". But $f_s$ is samples/sec, why must it be tied to $2\pi$? Let me modify your example for discussion. For an analog signal given as $$x(t)=\sin(2\pi F t)$$
    When sampled at a new sampling interval of $T_snew = 1.1T_s$ , After being sampled is given as: $$x(nT_snew)= \sin\left(\frac{2\pi F*1.1}{F_s}n\right)$$

     – Ran Feb 14 '22 at 10:00
  • Where the units of normalized frequency, either $\frac{F}{F_s}$ in cycles/sample or $\frac{2\pi F}{F_s}$ in radians/sample is clearly shown.

    Why does the unique span always keep unchanged but not from $-1.1\pi$ ~ $1.1\pi$? Where does coefficient 1.1 go? Thanks!

     – Ran Feb 14 '22 at 10:01
  • When sampled at a new sampling rate: you changed the sampling rate to $1.1 F_s$. By definition it is normalized by the sampling rate - it is as simple as that. The whole point is to put all the continuous time frequencies on a frequency axis relative to the sampling rate - so whatever you use for the sampling rate becomes $f_s=1$ (cycles per sample) in units of normalized frequency. If you want to use radian frequency then that is $2\pi f$ or $\omega_s= 2\pi$ (radians per sample). – Dan Boschen Feb 14 '22 at 11:01
  • $f_s = 1/T_s$, Right? – Dan Boschen Feb 14 '22 at 11:01
  • Now I understand! I omit the very basic definition "normalized by the sampling rate"! Thanks again, Boschen – Ran Feb 15 '22 at 02:16