Re: Ground wave transmission, was G-line

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>

At 09:21 AM 9/20/2005, you wrote:
> Original poster: Ed Phillips <evp@xxxxxxxxxxx>
>
> " > I understand about the buried antenna but question the image
>  > antenna bit.  That analogy applies to the radiation pattern
>  > of an antenna above a perfect (infinitely conducting) ground
>  > plane but not to an antenna embedded in lossy soil.
>
> I think Robert is right.  After all, the antenna above a ground
> plane is just a special case of an antenna above a boundary
> between two dissimilar materials.  If there's a difference in
> dielectric constant or specific resistance, there'll be partial
> reflection from the boundary, and you can always consider the
> reflected component to originate from an imaginary mirror of the
> antenna."
>
>         The issue is to the reflection coefficient at the boundary.  If there
> really was a wave propagating in the earth (at the normal "buried
> antenna" depths I've seen quoted - a few feet) then there is indeed a
> discontinuity at the boundary between the earth and the air above it,
> but I don't know how to calculate how much power is reflected BACK INTO
> THE EARTH from that discontinuity.

The usual reflection equation will tell you.
For normal incidence on the boundary starting on side 1 and going 
towards side 2

R = [(n1 - n2)/(n1+n2)]^2
T = (n2/n1) * (2*n1/(n1+n2))^2
where n1 and n2 are the respective indices of refraction = sqrt(epsilon)

Since the reflected fraction is a square, it matters not whether n1 
or n2 is greater (it does change the phase of the reflected wave, 
though!.. if n2>n1 then the reflected wave is inverted)

For oblique incidence, it depends on the polarization, and you need 
the Fresnel equations:

define
 thetaI as the angle of incidence (relative to normal)
thetaR is the reflected wave (thetaR = thetaI) (law of reflection)
thetaT is the angle of the transmitted wave  (sin(thetaT)/sin(thetaI) 
= n1/n2: Snell's law)

define alpha = cos(thetaT)/cos(thetaI)   (=1 for normal incidence)
       beta = n2/n1  (or v1/v2, if you like)

For polarization of E field perpendicular to the plane of incidence 
(i.e. V pol in the antenna case)
reflected wave intensity
Er = (1-alpha*beta)/(1+alpha*beta) * Ei
transmitted wave intensity
Et = 2/(1+alpha*beta) * Ei

R = (Er/Ei)^2 = [(1-alpha*beta)/(1+alpha*beta)]^2
T = cos(thetaT)/cos(ThetaI) * v1/v2 * (Et/Ei)^2 = 
alpha*beta*(2/(1+alpha*beta))^2

For E field parallel to the plane of incidence (H pol)

Er = (alpha-beta)/(alpha + beta) * Ei
Et = 2/(alpha + beta) * Ei

R = ((alpha-beta)/(alpha+beta)]^2
T = alpha*beta*(2/(alpha+beta))^2



>  In the case of the "image antenna"
> for an antenna above a [perfectly conducting] earth all of the power is
> reflected at the boundary.  In the case of "real earth" that is by no
> means true and in attempting to calculate the vertical pattern of an
> antenna above earth (or the loss in distant transmission due to
> multipath reflections from the ground) it's necessary to know both the
> real and imaginary parts of the ground conductivity.
>
>         Bottom line is that I still don't think there's much of an "above the
> earth" image antenna for a "typical" underground antenna.
>
> Ed