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Originally Posted by NetFox
Thanks, Thorn.
I have another question. From the data I got at distances from 2.5m to 50m, the average received data match the the log-distance path loss model perfectly with a path loss exponent of 1.44. However, the data was meassured outdoor, so I guess this path loss exponent might be too small.
Is it possible that the indications given by the card are just signal strengh but power? So when calculating the power, I need to square the strength, in this way, the path loss exponent will become 2.88, which might make more sense.
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I'm not exactly clear on what you are asking. The signal strength is the received power.
Many a newbe has been annoyed here by me because I've asked them if they've done the math/ It gives you a good idea if something is possible or not. (You can't argue with physics.) However, don't get so hung up on the math that you miss the point of whether something will work or not.
One thing to keep in mind about the log-distance path loss model (or any math model with radios) is that the model is usually derived from ideal lab conditions using tuned test gear to remove all possible variables. The exact power level is known. The exact receive sensitivity is known. The exact loss of the cable and connectors are all known. No interfering RF sources are nearby. Etc, etc.
Out in the real world we have to deal with some (a lot of) unknowns. The antennae aren't perfect. The transmit level may be off slightly, as may the RX sensitivity. The lobes of the antennea on these cards have funny shapes. All of these things mean that doing the math just puts you in a ballpark and is probably never going to work out exactly the same as on paper.
That's why we have fudge factors of 10dBm on signal paths rather than 1 or 2dBm.