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I guess my question is as follows: if I find FSPL (there will be no buildings or anything blocking/diffracting), can I figure out how far the signal can be transmitted if the receiver gain, loss and sensitivity as well as the transmission power and gain/loss is known.

All I really know is the frequency of the RF (around 730 MHz), the TX Power (6dBm), the antenna gain (-21 dB), RX Sensitivity.

Since I cannot really find FSPL (free-space path loss) because you need the distance transmitted for that, I am kind of at a loss (no pun intended) as to how to find the possible transmitted distance.

So, is there a way to figure out (based on a receiver's loss/gain and sensitivity) what minimum value (I guess dBm) would be needed for a signal to be decipherable? (I have heard that 3 dB or dBm or what ever measures this is a good assumption...)

Also, I know RX power = TX power + gains - losses and FSPL is a loss (which has distance built into the formula). Since **FSPL** (dB) = 20log(**distance**) + 20log(**frequency**) + 32.45 (for kilometers) that means that distance = 10^(**FSPL** (db) - 20log(**frequency**) - 32.45)/20. The problem is that there is no way to find the FSPL value in the second equation because in order to find that, you need the distance which is what we are trying to find in the first place!! So, you would be able to solve the initial equation (with RX power = TX power + gains - losses) equal to distance, but you would have no way of finding th FSPL value.

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Thanks for your help, Phil! It turns out we cannot even estimate the distance because the TX power outputted can not truly be found.

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there will be no buildings or anything blocking/diffracting

Really? What about the ground? Unless you are in space, you will need to incorporate some kind of path model into your calculations. The two-ray ground reflection model is one simple option.

The problem is that there is no way to find the FSPL value in the second equation because in order to find that, you need the distance which is what we are trying to find in the first place!

Yes there is. Use algebra. You can take your path loss (based on the Friis transmission equation) and solve for $d$:

$$ \begin{align} G_\text{FSPL} &= 20 \log_{10}(d_\text{km}) + 20log_{10}(f_\text{MHz}) + 32.45 \\ \log_{10}(d_\text{km}) &= (G_\text{FSPL} - 20log_{10}(f_\text{MHz}) - 32.45) / 20 \\ d_\text{km} &= 10^{ (G_\text{FSPL} - 20log_{10}(f_\text{MHz}) - 32.45) / 20} \end{align} $$

Now, you merely need to calculate a link budget to determine how much you can afford in path loss. Replace $G_\text{FSPL}$ with this number and solve the equation, and you now have the distance beyond which your path losses are too much.

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  • $\begingroup$ Yes, the ground will be there. :) Also, I'm working on a link budget, but there are just so many unknowns... thank you for the response, though. The two-ray ground reflection model is useful, but were you trying to help or just confuse me more! :P Just kidding; I guess I'll have to put some elbow grease in... $\endgroup$ – dylnmc Jul 15 '14 at 16:17
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    $\begingroup$ @dylnmc You can certainly simply the path model if you don't need a very accurate prediction. Just be sure you understand the consequences. If you have plenty of extra margin, then you can even skip the planning entirely, set it up, measure the SNR, then (optionally) reduce power or increase data rate until you are making the most efficient use of the channel, whatever that means for your particular application. $\endgroup$ – Phil Frost - W8II Jul 16 '14 at 14:27

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