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In the friis equation, the reference distance is 1m at the log distance path loss.

Then the formula would be:

Pr=PtArAt(λ/(4π))2/dα

where d is distance between terminals, Pr is received power, Pt is transmission power, Ar,At are the rx, tx antenna gains, respectively, and λ is wavelength.

It seems that the path loss exponent α is not mentioned.

But I haven't seen a case where this α isn't 2.

Is this α always 2 in the friis equation?

And can the simulation environment be used even if this α is not 2?

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Path loss exponent is not always equal to 2, only in the free space scenario is this true. In general the received power Pr is proportional to dα, where α is somewhere around 2-6 depending on the situation. The choice of the path loss exponent should match the scenario you're trying to simulate.

For example, for communication at a high altitude with little obstructions then α is probably close to 2 since the situation is approximately free space condition.

For communication in a building, within the same floor, then α may be around 2-3. This is because walls other obstructions attenuate the signal power further than the free space condition.

Things can get even worse if we again consider communication in a building but now if the transmitter is on the first floor and receiver is on the third floor, then α could be as high as 4-6 depending on construction of the building, etc.

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