# Is the path loss exponent $\alpha$ is always 2 in friis equation model?

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

Then the formula would be:

$$P_r = P_tA_rA_t(\lambda/(4\pi))^2 /d^\alpha$$

where $$d$$ is distance between terminals, $$P_r$$ is received power, $$P_t$$ is transmission power, $$A_r, A_t$$ are the rx, tx antenna gains, respectively, and $$\lambda$$ is wavelength.

It seems that the path loss exponent $$\alpha$$ is not mentioned.

But I haven't seen a case where this $$\alpha$$ isn't 2.

Is this $$\alpha$$ always 2 in the friis equation?

And can the simulation environment be used even if this $$\alpha$$ is not 2?

Path loss exponent is not always equal to 2, only in the free space scenario is this true. In general the received power $$P_r$$ is proportional to $$d^{-\alpha}$$, where $$\alpha$$ 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 $$\alpha$$ is probably close to 2 since the situation is approximately free space condition.
For communication in a building, within the same floor, then $$\alpha$$ 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 $$\alpha$$ could be as high as 4-6 depending on construction of the building, etc.