# How to use Shannon Equation

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

And the received power would be

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

(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)

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.

• Possibly better link, that's not behind a paywall: researchgate.net/profile/Ziye-Jia/publication/… Commented Nov 26, 2022 at 0:27
• Thanks, I have updated the link in my post with it. Commented Nov 26, 2022 at 1:33

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 $$c_{iu}^t = B_{iu}^t \log_2 \left(1 + \frac{P_{i^t}^{tr} G_{iu}}{\sigma^2}\right) = B_{iu}^t \log_2 \left(1 + \frac{P_{i^t}^{tr} \gamma_0}{l_{iu}^2(t)}\right).$$
• 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. Commented Nov 26, 2022 at 0:35
• Hi @TimWescott , thanks for your check and exposition. Just want to make sure , did I derive the formula correctly ? $$R = B \cdot log_2(1+\frac{P_tG_tG_rc^2}{(4\pi df)^2 \cdot Noise})$$ Commented Nov 26, 2022 at 1:45