Re: [TowerTalk] dump power?

[Jim Lux <jimlux@earthlink.net>]

At 11:51 AM 11/17/2004 -0500, Tom Rauch wrote:
> It's also very important to add the ideal current
> distribution in  unidirectional arrays is NOT when power
> distribution is equal. As a matter of fact, optimum
> distribution in an array like this is NEVER when power is
> equal, and it is NEVER when element impedances are equal.
>
> 73, Tom W8JI

Careful when you use absolutes like NEVER... One can spend a bit of time 
arguing what "optimum" might mean, but, considering some sort of forward 
gain metric, in an idealized directive array, with isotropic radiators, 
with  elements separated far enough, equal power distribution (and 
corespondingly, equal current magnitudes) would be the optimum power 
division.  A limiting case in the opposite direction is to put all the 
power into one radiator and none in the rest. There are weightings that can 
change the beamwidth and sidelobe levels (binomial 1:2:1 is one... fatter 
beamwidth, lower sidelobes, or edge weighted Chebyshev, for instance: 
narrower beamwidth, but higher sidelobes).



Equal element currents are sort of like maximally flat Butterworth filters. 
(for the same reason, as it happens... they both have to do with Fourier 
transforms of discrete approximations to a continuous distribution)

-------
Here's an analysis for two elements.  Say you can divide the total power in 
any ratio, a defines the fraction of power going to element 1, (1-a) is the 
power going to the second element.

Considering a direction where the Efield is maximized (on the boresight of 
the pattern), the total E field is the sum of the Efields from each of the 
elements.  The Efield is proportional to the square root of the power fed 
to that element:

Etotal = sqrt(a * Ptotal) + sqrt((1-a)*Ptotal)
a bit of algebra:  Etotal = sqrt(Ptotal) * ( sqrt(a) + sqrt(1-a))

Now, we look at what value of a maximizes this (i.e. the most forward gain)...
One can do it with calculus (find the derivative, set it to zero, etc.) or 
by another approach:
Square the term you want to maximize (since the point at which the square 
is maximized is the point at which the value is maximized)

(sqrt(a) + sqrt(1-a))^2 = (sqrt(a) + sqrt(1-a))*(sqrt(a) + sqrt(1-a))

= sqrt(a)*sqrt(a) + 2 * sqrt(a)*sqrt(1-a) + sqrt(1-a)*sqrt(1-a)
= a + 2 * sqrt(a *(1-a)) + (1-a) = a + (1-a) + 2*sqrt(a*(1-a))
= 1 + 2 * sqrt(a- a^2)
We now want to find the point where the square root term is 
maximized.  THis is the same as finding where (a-a^2) is maximized, which 
is at a=0.5 (or, equal power distribution).

Some discrete examples:
If a =1 or a=0, then the square root operand is 0, so the value is 1
if a = 0.5, then we have 1+2*sqrt(.5 - .25) = 1+2*.5 = 2

You can do a similar analysis for any number of elements..
3 elements: equal power division provides forward gain of 3, 1:2:1 division 
gives forward gain of 2.91

--------------

If you consider element losses, in general,  it makes no difference how the 
currents are divided, if the loss resistance is the same for each element. 
Consider this: each element can be viewed as the sum of a radiation 
resistance (power which gets radiated as an EM wave) and a loss resistance 
(power that gets turned into heat).  The ratio of lost power to radiated 
power is the same, regardless of what the current is.  Say your elements 
are 10% efficient, and you're transmitting 100W.  You can divide your power 
evenly, 50W to each element, and you'll radiate 45W and lose 5W, for a 
total of 90W radiated, 10W lost.  Or, you can put all the power into one 
element, where you'll radiate 90W and lose 10W.


If there is any mutual coupling between elements (which IS the case in most 
practical amateur arrays), equal currents leads to unequal power, since the 
feedpoint impedances are not equal.  From an efficiency standpoint, this 
isn't a big deal (since the loss characeristics of the elements are 
probably similar), although, one would want to wonder about circulating 
reactive currents in the feedlines and the tuning 
networks/hybrid.  Hopefully, though, your elements are far enough apart 
that the mutual component is reasonably small (say, on the order of the 
self impedance?).. that is, farther than about 1/8-1/4 wavelength apart.

This might seem contradictory to the "optimum power division" analysis 
above, but really, it's more a reflection of the fact that the relation 
between the radiation resistance of the elements and the feed point 
impedance isn't so nice and cleanwhen there is mutual coupling.  You might 
be feeding power into a feedpoint, and some of that power is radiated into 
the far field, and some other part of that power winds up in another 
element, and is then radiated into the far field. This is, after all, what 
happens in a Yagi.  The power is being transferred in the reactive near 
fields.  Imagine it as you have a set of idealized resistive radiators, and 
a virtual network of circuits interconnecting those radiators (you can 
consider it as a box with 4 inputs (from your phasing network) and 4 
outputs (to the idealized radiators)).  The currents (and powers) you 
observe on the inputs don't necessarily reflect what the currents (and 
powers) are on the outputs.  Most of the "art" of antenna design, in fact, 
is in creating the desired current distribution on the radiating elements, 
with as simple a network to feed the inputs to the box as possible. In a 
Yagi with a single feedline, you fool with element lengths and spacings to 
change the "inside of the box", and have only one input. In a phased array, 
you get a bit more explicit control, since you have multiple inputs to the box.

_______________________________________________

See: http://www.mscomputer.com  for "Self Supporting Towers", "Wireless Weather Stations", and lot's more.  Call Toll Free, 1-800-333-9041 with any questions and ask for Sherman, W2FLA.

_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk