[Lowfer] Active Whips

[JD]

>>>As for JD's situation, he makes a good point. JD is actually using a 40 
>>>foot long "active whip" to receive. I am using an 8 foot long "active 
>>>whip". When you get down to wavelengths that span 2000 meters, what's the 
>>>difference? Answer: pretty much no difference.

And maybe even less difference than that! :-)

* Alert::: slightly long and detailed material follows, but hopefully not 
too overcondensed or too boring *

I forget what length the support for your whip ended up being, Doug, but 
I've heard of some folks up as high as 30-35 feet.  For convenience of 
comparison, let's just pick a 32 foot support for this analysis.  If a 
person then mounts an 8 foot antenna up there, does that make it an 8 foot 
active whip?  Nope, it's a 40 foot active whip, too.

This is because, in the most literal sense, there is no such thing as a 
monopole.  Every antenna, just as every amplifier input, has TWO terminals 
in reality, or no complete circuit exists.    In my setup, the buffer 
amplifier is actually receiving signal from a dipole...one leg of which is a 
relatively skinny 40 foot steel mast sticking up into the air, and the other 
leg of which is a big ball of damp dirt and rock 8000 miles in diameter.

The voltage developed across those two terminals by a passing 
electromagnetic wave of the same polarization as the electrically short 
mast, is then very nearly equal (provided my circuit is high enough 
impedance, both reactively and resistively speaking) to the field strength 
in volts/meter times the effective height of the antenna in meters.  If 
there is not some form of loading to change the linearly tapered current 
distribution in the electrically short element, that effective height is 
_half_ the physical height.  So, an EM wave of 1 microvolt/meter intensity, 
having its electric field in the plane of my mast as it happens to pass by, 
will generate a voltage of:

(1 uV/m * 12.192 m ) / 2 = 6.1 uV/m approx

across my amplifier input terminals.  Not too shabby, assuming QRM and QRN 
aren't generating even more voltage there at the time.

Now, our presumed 8 foot whip...if its amplifier has a connection via the 
support pole, a separate wire, or the coax shield to ground in any way, then 
IT TOO is receiving its input from a dipole...one leg consisting of the 8 
foot whip, and the other consisting of 32 feet of conductor with an 8000 
mile ball of damp dirt at the other end.  How can that be the same as each 
other antenna, despite the difference in appearance?

Let's look a little closer at what's actually happening in the first case, 
my own "40 ft whip."  There's one point in particular to remember.  The 
electric field of the passing wave ends just inside the surface of the 
sorta-conducting big ball of dirt.  So, the total span of the EM wave's 
electric field being intercepted is 12.192 meters long--from the tip of the 
mast to the surface!   But then, as already mentioned, because of the linear 
taper of current distribution in the non-ball-of-dirt leg (you can't 
separate the magnetic effects from the electric in a propagating EM wave), 
the _effective_ length of field being intercepted is only half that, or ~6.1 
meters.  If you followed what I said about the E field terminating just 
inside the surface of the dirtball, you see can why the fiction of the 
electric "monopole" is useful in some aspects of antenna engineering;  the 
dirtball, despite its size, contributes almost nothing to the received 
terminal voltage and hence does not appear in equations involving vertical 
antenna effective height.  But despite that, the circuit is still a dipole 
electrically, because the current induced in the non-dirtball leg still has 
to flow through the dirtball too..

OK, now back to the "8 foot" whip.  The antenna input port of the amplifier 
has an 8-foot conductor on one terminal...and through whatever path, has a 
32 foot connection to the dirtball on the other terminal.  Once again, our 
EM wave passes, and once again the electric field terminates at the 
dirtball.  So, the intercepted electric field component spans from the tip 
of the 8 foot whip to the amp, and then from there, 32 more feet to the 
earth side of the conductor, where the field ends.  And once again, viewed 
across the whole length of the dipole, the current tapers linearly from tip 
to dirtball.  The total voltage appearing across the input terminals of the 
amplifier for a 1 uV/m signal is therefore the contribution of the 8 foot 
leg plus the contribution of the 32 foot-plus-dirtball leg, or:

( 1 uV/m * 2.438 m ) / 2 + ( 1 uV/m *  9.754 m ) / 2 =
( ( 1 uV/m * 2.438 m ) + ( 1 uV/m *  9.754 m ) ) / 2 =
( 1 uV/m * ( 2.438 + 9.754 ) m) / 2 =
(1 uV/m * 12.192 m ) / 2 = 6.1 uV/m approx

When all the associative and distributive mumbojumbo has been done, a 
typical 8 foot whip at 32 feet is not "merely" an 8 foot whip, but is just 
the same as a 40 foot whip at ground level, so far as its ability to 
intercept signal.  An 8 ft whip at 25 feet is equivalent to a 33-footer, 8 
feet at 20 ft is like 28 ft ground mounted, and so on.

Could you make an 8 foot whip act like only 8 feet while still way up in the 
air, or even a 2 inch whip act like 2 inches?  With heroic measures, yes, 
but why would you want to? :)

John