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kanbankaren | 1 month ago
Yeah. Even in free space. For example, attenuation at 1 km for 144 MHz (ham VHF band) is about -76 dB while for 2.4 GHz, it is about -100 dB. That 24 dB drop could mean, the signal is below the noise floor of your receiver unless you increase the RF power output which means more battery drain.
For example, BT audio gets cut just moving to the next room despite the RF power of BT transmitters being ~ 5mW( 7 dBm ) and at 10m, the attenuation is -60 dB(just free space loss which is ideal condition), so 53 dBm (7-60) at the receiver is usually sufficient, yet they struggle.
lxgr|1 month ago
> For example, attenuation at 1 km for 144 MHz (ham VHF band) is about -76 dB while for 2.4 GHz, it is about -100 dB.
This is a common misunderstanding of the free-space path loss formula, which is expressed in terms of the idealized isotropic radiator, the length of which is frequency-dependent. In other words, this calculation is assuming a proportionally (much) smaller antenna for the 2.4 GHz case.
With the same antenna size, the path loss is exactly the same. After all, where else should the radiated energy go?
kanbankaren|1 month ago
What do you mean? The size of the dipole or monopole antenna is dependent on the wavelength, so obviously the 2.4 GHz is just a few centimeters and not the same size as a VHF antenna.
> After all, where else should the radiated energy go?
Well, most of RF energy is wasted. There are software that can plot the radiation pattern, but even without knowing the exact pattern, very little RF energy is received at the target.