[Lowfer] Active Whips

[Douglas D. Williams]

Great explanation John. So, when we raise the height of our "active whip"
antennas, say, an additional ten feet by adding another ten foot long
section of support pole, and then see an improvement in signal levels, it
isn't necessarily because the "whip" part of the antenna is higher, but
because the support pole (or coax shield if the support pole is non
conductive) is actually part of the antenna, so we have effectively
increased the length of the antenna by using a longer support pole?

I wonder why some "active antenna" manufacturers caution users that long
support poles (i.e. placing the antenna much higher than 20 feet above
ground) are not necessary? Could it be that they are concerned that too
much antenna length (since the  support pole and/or coax shield is actually
contributing to the signal level at the receiver)  could overload the
amplifier and cause IMD?

73, Doug KB4OER





On Thu, Jan 24, 2013 at 3:52 AM, JD <listread at lwca.org> wrote:

> 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
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