YechielA
Junior Member level 1
Hello,
According to FCC the E field at 3 meters should not exceed 500uV/m.
So let us calculate the total radiateted power assuming isotropic radiator.
[1] Pt=(|E|^2/ηo)*(4*Π*D^2) WHERE E IS EQUEL 500uV/m and D=3 meters.
on the other hand, according to FRIIS equation, the free space path loss is given by: (λ/4*Π*D)^2 so that,
[2] Pt= (|E|^2/ηo)*[1/(λ/4*Π*D)^2]=(|E|^2/ηo)*(4*Π*D/λ)^2.
But both equation contradict and give different results.
Any idea ???
According to FCC the E field at 3 meters should not exceed 500uV/m.
So let us calculate the total radiateted power assuming isotropic radiator.
[1] Pt=(|E|^2/ηo)*(4*Π*D^2) WHERE E IS EQUEL 500uV/m and D=3 meters.
on the other hand, according to FRIIS equation, the free space path loss is given by: (λ/4*Π*D)^2 so that,
[2] Pt= (|E|^2/ηo)*[1/(λ/4*Π*D)^2]=(|E|^2/ηo)*(4*Π*D/λ)^2.
But both equation contradict and give different results.
Any idea ???