how to calculated distance from tx power attenuation

[arr_baobao]

let say i have a 2 device one side is tx and the other side is rx
both device is 3m away.
i start to tx at 10dB and check the packet loss at rx.
i keep on attenuate the power in tx until the packet loss is too high.
from the attenuated power can i calculate what is the range of the device that will still perform reasonable( low packet loss).
say now at -25dBm tx power i saw the packet loss is too high.
what is the max range of the device. how do i calculate.
i know when we double the distance we will get -6dB free space loss. can i use this formula since my tx power is dbm not db.

hope you guys understand what i want to ask =)

 

[FvM]

can i use this formula since my tx power is dbm not db.

dBm is absolute power quantity, dB is relative. 10 dBm - (-25) dBm = 35 dB attenuation. Corresponds to about factor 50 distance in free field. 3 m may be still near field, depending on the frequency, so it's a bit inaccurate but order of magnitude should be correct.

 

[vfone]

For example, in free-space if you transmit a 2.529 GHz signal with TX power of +10dBm, at 3m (which is about 50dB path loss) you may get about -40dBm at RX input (considering 0dB gain for RX and TX antennas, and no other losses).
From -40dBm @ 3m if you start to attenuate the TX conducted signal in 6dB steps theoretically you will get:
-46dBm @ 6m
-52dBm @ 12m
-58dBm @ 24m
-64dBm @ 48m

 

[volker@muehlhaus]

You might also look at free space path loss calculation:

https://www.pasternack.com/t-calculator-fspl.aspx

 

[arr_baobao]

For example, in free-space if you transmit a 2.529 GHz signal with TX power of +10dBm, at 3m (which is about 50dB path loss) you may get about -40dBm at RX input (considering 0dB gain for RX and TX antennas, and no other losses).
From -40dBm @ 3m if you start to attenuate the TX conducted signal in 6dB steps theoretically you will get:
-46dBm @ 6m
-52dBm @ 12m
-58dBm @ 24m
-64dBm @ 48m

thanks. so much clearer now.

btw can i use the same on tx side?
if i tx 20dB, i attenuate 6dB device still working mean distance double. then i continue to attennuate another 6dB
packet loss still acceptable, i double the distance from previous distance.

can this work? without knowing the rssi in receiver side?

 

[volker@muehlhaus]

Yes, your understanding is correct.