Re: Back to fundamentals

Owen Duffy wrote: 

An alternative is to calculate the power collected by a lossless,
matched receiver as Pr=S*A.

In this case, S=0.3**2/(120*pi)

Kraus derives A (the effective apperture) for a short dipole to be
3/8/pi*wavelength**2.

This gives the power collected by the receiver as 6.4mW. If the
antenna and receiver were disected by the ground plane, wouldn't there
be 3.2mW developed in each half of the receiver load?

I found another source (Ramo et al) which directly gives the ratio of power in the load of a matched receiving antenna to the power applied to a transmitting antenna, in terms of the effective apertures of the antennas. This doesn't require the intermediate step of calculating field strength. The equation is:

  Wr/Wt = (Aer * Aet) / (lambda^2 * r^2)

where:
Wr, Wt are received and transmitted power respectively
Aer, Aet are the receiving and transmitting antenna effective apertures
lambda = wavelength
r = distance between the antennas

Note that effective aperture, like other measures of an antenna pattern, is a function of the direction from the antenna. So this equation is correct regardless of antenna orientation as long as Aer and Aet are correctly calculated.

Letting

  K = 3/(8 * pi) ~ 0.1194

we can write the equation cited by Owen for effective aperture of a short dipole in its most favored direction (broadside) in free space as

  Ae = K * lambda^2

(This is the effective aperture of an infinitesimally short dipole. However, it changes very little with length when the dipole is electrically short. The effective aperture broadside to a half wave dipole is only 10% greater.)

Then we get that

Wr = Wt * (K^2 * lambda^4) / (lambda^2 * r^2)
   = Wt * K^2 * lambda^2 / r^2

For Reg's example, lambda = 15 meters and r = 1 km. Since this is a free space analysis involving dipoles so far, I'll apply 2 kW (Wt) to the transmitting dipole, resulting in Wr = 6.412 mW.

Now we can split the model exactly in half with a ground plane. On the top of the ground plane, the transmitting antenna has exactly half the applied power, or 1 kW, which is what we had in Reg's example. Half the load power is in the upper plane also, so we have 3.206 mW for the load power in Reg's example setup. This is very close to the 3.234 mW result from the EZNEC model.

The antenna's effective height (that is, the ratio of induced voltage to field strength) has been at issue. As Reg pointed out, ~ 3 mW at the load requires an effective height of 0.5 meter for the 1 meter high antenna. (I incorrectly gave it as 1 m in an earlier posting.) I did find an explicit equation for effective height for a vertical over perfect ground, in King, Mimno, and Wing (Dover edition, p. 165). This also confirms that the correct effective height is 0.5 m for the 1 m electrically short vertical antenna over ground.

I'm satisfied that we have the answer to Reg's question. It's been an educational process for me -- thanks for posing it.

One final note, regarding the NEC applied plane wave. My earlier statement that the resulting field is twice the plane wave source magnitude when a ground plane is present is true only when the plane wave is applied over perfect ground at exactly grazing incidence (zenith angle = 90 deg.). If applied from other angles the resulting field strength will be different. If you apply a vertically polarized wave over a ground plane, I believe the resulting field strength will look like the pattern from a vertical radiator over a perfect ground plane -- strongest when applied at the horizon, decreasing when applied at higher angles, and dropping to zero if applied from directly overhead. I haven't confirmed this, but believe it's necessary in order to get a receiving pattern that's the same as the transmitting pattern. So use it with caution when a ground plane is present, and don't casually make assumptions about the resulting field.

Roy Lewallen, W7EL
.

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