relationship between gain and electric field

[kae_jolie]

what is the relationship between gain and electric field?

My answer is:

radiation intensity (U) = ½ Re(E x H*) • r²

Directivity (D) = U/Uave

Gain (G) = e D

Therefore, G = e (U/Uave) = e [(½ Re(E x H*) • r²)/(Uave)]

Therefore, electric field (E) is directly proportional to Gain (G), so as E increases Gain will increase and vice versa.

Is this a correct statement? If not, please explain.

Thanks.

 

[WimRFP]

Gain is related to power flux densities (radiation intensity), so if the gain doubles (given same net input power), the radiation intensity doubles (for the direction where the gain doubled). This means both E-field and H- field increases with 1.414 (sqrt(2) ).

So E is proportional with sqrt(G) G = gain, not in dB notation.

 

[noor_]

that is true ,not in db notation

 

[kae_jolie]

Thanks, WimRFP. You are basically saying that G = E^2

Could you tell me which book I can find this relationship in? I need to read this in more detail.

Thanks.

 

[WimRFP]

Hello,

G = Uaut/Uiso, measured at same distance and same input power.

U = power flux density of far field [W/m^2] , aut = antenna under test, iso = isotropic radiator with 100% efficiency (so no heat loss).

U = E*H, Zfreespace (Zfs) = 377 Ohm = E/H, so U = E^2/Zfs. Hence

G = Eaut^2/Eiso^2 = (Eaut/Eiso)^2