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 ?



1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.