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Re: Ground wave transmission, was G-line



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