tuned ground plane

Jim Weir wrote in message
...

(John Tvedte)
shared these priceless pearls of wisdom:

-I have a Velocity - and am planning on using a Comant CI-105
-transponder antenna, and place it out on the wing strake area - as I
-do not want it close to the engine and near my AHRS setup.
-
-As this airplane is a glass airplane, and this antenna requires a
-ground
-plane - I have read about a "tuned" ground plane.
-
-The suggestion was to make a 5.2" disc - @1090Mhz, 1/2 wave is
-5.15229..."
-
-Can someone give me a heads up as to what a tuned ground plane really
-is?
-
-Thanks,
-
-John
A "tuned" ground plane is one that is an odd multiple of a quarter wave at the
operating frequency. Odd -- one, three, five, and so on. In
practicality (with
the possible exception of GPS) the norm is ONE quarter wave, or a quarter-wave
groundplane.

Having said that, you must understand that you cannot make a "perfect" circular
("disk") ground plane for a transponder, as it must operate on two frequencies
simultaneously -- 1030 and 1090 MHz. What is quarter wave for one is not
quarter wave for the other. While the error is slight and relatively
insignificant for transponders, it DOES become significant when you start to
talk about things like the aircraft COM band, where the instantaneous bandwidth
is on the order of 11%.

My chosen way to make a transponder ground plane is to start off with a square
of aluminum using the following calculations:

w = 11810 / f
s = 0.487 w

where "w" is a quarter wave in air at frequency "f"

(A) 'w' is mis-identified. This is a _full_ wavelength in air, not a quarter.
Assumes frequency in mHz, gives result in inches.
only accurate to 4 sig figs -- good enough for most 'practical' work.
Use 11811.02 [ (300 million meter/sec) * (inches/meter) / one million ]
for 'high precision' "in vacuum" and then correct for propagation in air
vs. in vacuum (299702547/299792458), giving 11807.45775338017 grin

(B) the 'mysterious' 0.487 is a combination of two things.
1) we want a 1/4 wavelength _radius_, so the "diameter" will be 1/2 wave.
2) we want the 'minimum' dimension of the octagon to be resonant at
1090 mHz, while we did the 'w' calc at 1060 mHz.
3) guess what 1060/1090, _divided_by_2_ is? grin
(0.48623853211009+, for those without a calculator handy)

Thus, from the center, to the _middle_ of any side is a 1/4 wave at 1090 mHz.

And, the distance from the center of the octagon, to a 'corner' is enough
longer than the distance from the center to the 'middle' of a side (a factor
of 1.0823922002+ [sqrt(1**2+(sqrt(2)-1)**2), if anybody cares) that the 1/4
wave resonant frequency on the 'diagonal/ is 1007 mHz.

Thus, 'somewhere' between the middle of a side, and the corner, the length
will be 'right' for a 1030mHz 1/4 wave.


and "s" is the length of one side of the square of aluminum.

Then punch a hole in the exact center of the groundplane for the antenna.

Then cut the corners of the square to make a regular octagon.

SOMEWHERE along the periphery of that octagon will be an exact quarter wave at
both 1030 and 1090 MHz. IF you select w to be halfway between these two
frequencies (i.e. 1060 MHz.).

So, in the above scenario, w = 11.14" and s = 5.42".


Jim

Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com