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Let's consider the Shannon - Hartley formula for the channel capacity \$C\$:

$$C = B \log_2 (1 + S/N)$$

I have two doubts about it:

1) is it correct for TDMA, FDMA, or CDMA? And why?

2) Can a variation of the dimensions of the radio cells (base transceiver stations) reduce/increase the value of \$S/N\$?

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  • \$\begingroup\$ For Q1, D in those acronyms is for division. So each protocol only divide channel capacity for each user. \$\endgroup\$ – Rokta Sep 25 '19 at 6:38
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1) It's accurate for all the *DMAs. FDMA is pretty straightforward, since B is the width of the channel. TDMA is the same math, but each user only has access to the channel for a fraction of the frame time, so the channel is shared among N users...each one gets a fraction of the total channel capacity. CDMA is similar in concept, but a bit more complex in implementation...same theoretical result, though.

2) Not sure what you mean by "variation of the dimensions of the radio cells," but better power and sensitivity, better antennas and placement, or increase in infrastructure density can definitely affect S/N.

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