|To:||Tom Rauch <firstname.lastname@example.org>, <email@example.com>,"Al Williams" <firstname.lastname@example.org>|
|Subject:||Re: [TowerTalk] dump power?|
|From:||Jim Lux <email@example.com>|
|Date:||Wed, 17 Nov 2004 10:46:16 -0800|
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.
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.
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