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I just start learning how to calculate Shannon capacity and trying to understand the relationship between power and path loss.

I would like to make sure if I use the equation correctly.

The path loss can be express as

enter image description here

And the received power would be

enter image description here

If I would like to calculate the Shannon capacity , it should be like

enter image description here

(where B is bandwidth, c is speed of light, d is distance between transmitter & receiver, f is central frequency)

However, I found a paper that calculate the formula without square of the term in received power. (I mark the term with red lines)

enter image description here

Can anyone explain this to me , please ?

Am I doing the calculation incorrectly or is there an error in the formula in the paper ?

Thanks.

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From this link, which is not behind a paywall, the author's own (1) says $$G_{iu} = \delta_0/l^2_{iu}(t).$$

Then their (2) says $$c_{iu}^t = B_{iu}^t \log_2 \left(1 + \frac{P_{i^t}^{tr} G_{iu}}{\sigma^2}\right) = B_{iu}^t \log_2 \left(1 + \frac{P_{i^t}^{tr} \gamma_0}{l_{iu}^2(t)}\right).$$

Then their (3) apparently drops the exponent from the distance.

So -- their paper is in error. Any far-field communication in this 3D universe of ours is going to have power dropping off as distance squared.

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  • $\begingroup$ You may want to send an email to the authors asking them for clarity -- chances are they just forgot the exponent when they went from $l_{iu}^2$ to $l_{us}^t$, because they'd put a $t$ in the superscript. Even if it is just a clerical error and any computations they did are correct, they should still appreciate knowing there's a misprint. $\endgroup$
    – TimWescott
    Commented Nov 26, 2022 at 0:35
  • $\begingroup$ Hi @TimWescott , thanks for your check and exposition. Just want to make sure , did I derive the formula correctly ? $$ R = B \cdot log_2(1+\frac{P_tG_tG_rc^2}{(4\pi df)^2 \cdot Noise}) $$ $\endgroup$
    – Henry Lu
    Commented Nov 26, 2022 at 1:45
  • $\begingroup$ Unless there's something very odd hidden in their notation -- yes. $\endgroup$
    – TimWescott
    Commented Nov 26, 2022 at 3:06
  • $\begingroup$ Thank you , I really appreciated your help. $\endgroup$
    – Henry Lu
    Commented Nov 26, 2022 at 3:21

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