Richard Harrison wrote:
If everything is symmetrical, the self-impedances and the mutual
impedances of the two elements should be equal, producing equal powers
into each element.

Unfortunately, it is not that easy except under special circumstances.
The element with leading phase often has a different feedpoint impedance
than the element with lagging phase. For instance:

Given two 1/4WL monopoles, 1/4WL apart, and fed 90 degrees apart with
one amp each will exhibit a gain of 3 dBi in one direction. However,
the feedpoint of one element is 20-j20 and the feedpoint of the other
element is 50+j20. The feedpoint voltages are obviously not equal so
to equalize the current magnitudes takes some juggling. That' what
Roy's BASIC program does - finds a solution if one exists.

I once had a BASIC program that calculated the mutual impedances given
the feedpoint impedance of one element alone and the feedpoint impedances
of the two elements during operation but I seem to have misplaced it.
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
"Are the patterns in the handbook all equal power division?"

The subscript says:
"The two elements are assumed to be thin and self-resonant, with
equal-amplitude current flowing at the feed-point."

I think the "Antenna Book" authors were familiar with Kraus` Fig. 11-11
on page 290 of the 1950 edition of "Antennas". Kraus` Fig. 11-11 is
similar to the "Antenna Book" Fig 11 on page 8-8 of the 19th edition.
Kraus makes a point of G.H. Brown`s equal power observations.

I am familiar with the negative-resistance tower occasionally found in a
broadcast array. John E. Cunningham says in the "Complete Broadcast
Antenna Handbook":
"In an array of four towers or more, the resistive part of the
driving-point impedance of one or more of the towers often has a
negative value. This means that the tower obtains its energy through the
mutual impedance between it and the other towers of the array. This is a
confusing situation, but if it is carefully thought out, it will cause
no serious problems. We know the following things concerning the
negative tower:

1. The tower must carry a current of the proper magnitude and phase.
2. The direction of this current is 180-degrees out of phase with what
it would be in a tower having a positive base resistance.
3. We need some method of controlling the magnitude and phase of the
tower current.

The simplest, although not the most efficient way of handling the
negative-resistance tower is to terminate it through a matching network
to a resistor, as swhown in Fig.11-15. The energy that the negative
tower actually gets from the other towers is thus dissipated in the
resistor. The magnitude and phase of the current may be controlled by
the parameters of the network. Naturally, this isn`t a very efficient
arranngement, particularly if the negative tower handles a substantial
amount of current.

The preferred way to handle a negative tower is to feed the energy back
to the power divider, where it will be passed back into the feeder
system again. In this way, all of the energy is radiated rather than
some being dissipated in a resistor.

Figure 11-16 shows an arrangement for recovering power from a
negative-resistance tower.----"

Since I can`t do diagrams, I suggest finding a copy of the book. It`s a
good one.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
....
The preferred way to handle a negative tower is to feed the energy back
to the power divider, where it will be passed back into the feeder
system again. In this way, all of the energy is radiated rather than
some being dissipated in a resistor.

This makes sense, but I wonder if this condition can be made to hold
true
over the bandwidth of the transmitted signal. Would this scheme result
in
a system that had such a high Q that it would quickly degrade the
further away from the carrier frequency you got (i.e. mismatch at the
sideband frequencies)?

regards
L

Not sure I understand. Why not use sections of different impedance coax
to raise the effective antenna impedance of each antenna to 100 Ohms,
then use different lengths of 50 Ohm coax from the "Tee" adapter to
these matching sections. The delay difference will cause the pattern to
be steered in some direction. I assume you will be using crossed
dipoles. When the two 100 Ohm antennas are connected together through
the Tee, it will result in a system impedance of 50 Ohms. As long as
you use resonant dipoles, current and voltage will be in phase in its
respective dipole, so the impedance will not change. You are only
providing a phase difference between the two dipoles by changing the
path length the signals travel from the transmitter to each antenna. It
should work. We basically did this when I worked at the U.S. Navy ELF
transmitter site.


Steve Nosko wrote:
"...easy way ..."? To change the phase, yes.... To change the pattern,
Probably not. The impedance changes with the phase relationship. Antennas
feed power to each other. Try searching for "Phased arrays".

acepilot wrote:
As long as
you use resonant dipoles, current and voltage will be in phase in its
respective dipole, so the impedance will not change.

This isn't always true. The feedpoint impedances of two resonant
dipoles when phased together are not usually resistive. For instance
using EZNEC:

A 33 ft. dipole at 66 ft. has a feedpoint impedance of 70+j0 ohms
on 7.298 MHz.

Two 33 ft. dipoles at 66 ft. spaced 33 ft. apart and fed 90 degrees
out of phase have the following feedpoint impedances on 7.298 MHz:

109+j34

29-j33

Please describe how you would achieve equal magnitude currents 90
degrees apart into those dipoles given those feedpoint impedances.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
acepilot wrote:

As long as you use resonant dipoles, current and voltage will be in
phase in its respective dipole, so the impedance will not change.


This isn't always true. The feedpoint impedances of two resonant
dipoles when phased together are not usually resistive. For instance
using EZNEC:

A 33 ft. dipole at 66 ft. has a feedpoint impedance of 70+j0 ohms
on 7.298 MHz.

Two 33 ft. dipoles at 66 ft. spaced 33 ft. apart and fed 90 degrees
out of phase have the following feedpoint impedances on 7.298 MHz:

109+j34

29-j33

Please describe how you would achieve equal magnitude currents 90
degrees apart into those dipoles given those feedpoint impedances.

I just ran this matching problem through Roy's (W7EL) SIMPFEED BASIC
program downloadable from http://www.eznec.com

Using 300 ohm twinlead, the program said that one feedline should be
27 degrees and the other should be 168.5 degrees. For 7.298 MHz and
VF=0.9, that's lengths of 9.1 ft. and 56.78 ft. I plugged them into
EZNEC and it worked great - a phased dipole array with 10.5 dBi gain
in one direction. The above lengths ensure that equal currents flow
in both elements and gives a 50 ohm SWR of about 2:1 at the junction
of the two 300 ohm feedlines.

Incidentally, there were no solutions using 50 or 75 ohm coax.
--
73, Cecil http://www.qsl.net/w5dxp



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Oops, my bust I guess. Damn computer programs! ;) Well, I probably
should have stated that this should be "tried at your own risk". I
guess that my point was that it should be possible to steer a signal by
using crossed dipoles. We do it at ELF frequencies for submarine comms,
so it should work at any other frequency as well...just takes a little
dinkin' around...but hey, that's what autotuners are for ;)

Scott
N0EDV



Cecil Moore wrote:
Cecil Moore wrote:

acepilot wrote:

As long as you use resonant dipoles, current and voltage will be in
phase in its respective dipole, so the impedance will not change.



This isn't always true. The feedpoint impedances of two resonant
dipoles when phased together are not usually resistive. For instance
using EZNEC:

A 33 ft. dipole at 66 ft. has a feedpoint impedance of 70+j0 ohms
on 7.298 MHz.

Two 33 ft. dipoles at 66 ft. spaced 33 ft. apart and fed 90 degrees
out of phase have the following feedpoint impedances on 7.298 MHz:

109+j34

29-j33

Please describe how you would achieve equal magnitude currents 90
degrees apart into those dipoles given those feedpoint impedances.


I just ran this matching problem through Roy's (W7EL) SIMPFEED BASIC
program downloadable from http://www.eznec.com

Using 300 ohm twinlead, the program said that one feedline should be
27 degrees and the other should be 168.5 degrees. For 7.298 MHz and
VF=0.9, that's lengths of 9.1 ft. and 56.78 ft. I plugged them into
EZNEC and it worked great - a phased dipole array with 10.5 dBi gain
in one direction. The above lengths ensure that equal currents flow
in both elements and gives a 50 ohm SWR of about 2:1 at the junction
of the two 300 ohm feedlines.

Incidentally, there were no solutions using 50 or 75 ohm coax.

And one sidenote...

Cecil, I think you were implying that the dipoles you modeled were
parallel to each other, correct? Our ELF antennas were dipoles that
were perpendicular to each other. In theory, there should be minimal
interaction between them because of the nulls off of each end of the
antennas, correct? Somebody else mentioned that the antennas, when
driven, feed power into each other. Placing them at 90 degrees to each
other should minimize interaction, would it not? More stuff to digest!

Anyhow, whoever wants to give it a try, have fun and let us know how
well it works!!

Scott
N0EDV

acepilot wrote:

Oops, my bust I guess. Damn computer programs! ;) Well, I probably
should have stated that this should be "tried at your own risk". I
guess that my point was that it should be possible to steer a signal by
using crossed dipoles. We do it at ELF frequencies for submarine comms,
so it should work at any other frequency as well...just takes a little
dinkin' around...but hey, that's what autotuners are for ;)

Scott
N0EDV



Cecil Moore wrote:

Cecil Moore wrote:

acepilot wrote:

As long as you use resonant dipoles, current and voltage will be in
phase in its respective dipole, so the impedance will not change.




This isn't always true. The feedpoint impedances of two resonant
dipoles when phased together are not usually resistive. For instance
using EZNEC:

A 33 ft. dipole at 66 ft. has a feedpoint impedance of 70+j0 ohms
on 7.298 MHz.

Two 33 ft. dipoles at 66 ft. spaced 33 ft. apart and fed 90 degrees
out of phase have the following feedpoint impedances on 7.298 MHz:

109+j34

29-j33

Please describe how you would achieve equal magnitude currents 90
degrees apart into those dipoles given those feedpoint impedances.



I just ran this matching problem through Roy's (W7EL) SIMPFEED BASIC
program downloadable from http://www.eznec.com

Using 300 ohm twinlead, the program said that one feedline should be
27 degrees and the other should be 168.5 degrees. For 7.298 MHz and
VF=0.9, that's lengths of 9.1 ft. and 56.78 ft. I plugged them into
EZNEC and it worked great - a phased dipole array with 10.5 dBi gain
in one direction. The above lengths ensure that equal currents flow
in both elements and gives a 50 ohm SWR of about 2:1 at the junction
of the two 300 ohm feedlines.

Incidentally, there were no solutions using 50 or 75 ohm coax.

Scott, N0EDV wrote:
"Anyhow, whoever wants to give it a try, have fun and let us know how it
works!!"

I don`t have EZNEC but it seems to be very useful.

Crossed dipoles are often used as quasi omnidirectional antennas.
Broadcasters call it a "turnstile".

If the two dipoles are fed in phase, a figure-eight pattern results.
This is similar to the pattern of a single dipole but at 45-degrees to
the crossed dipoles.

The usual practice is to excite the dipoles 90-degrees out-of-phase.
This produces a nearly circular pattern in the plane of the turnstile.

One could get a choice of patterns with crossed dipoles by feeding them
in-phase, out-of-phase, and in quadrature for an omni pattern. Then
using the dipoles one at a time, two more diirectional patterns are
available.

Best regards, Richard Harrison, KB5WZI

acepilot wrote:
Cecil, I think you were implying that the dipoles you modeled were
parallel to each other, correct? Our ELF antennas were dipoles that
were perpendicular to each other. In theory, there should be minimal
interaction between them because of the nulls off of each end of the
antennas, correct? Somebody else mentioned that the antennas, when
driven, feed power into each other. Placing them at 90 degrees to each
other should minimize interaction, would it not?

Yep, that's true, and a turnstile is an example. But for a phased beam,
one needs maximum interaction. The dipoles in my example are 1/4WL
apart, parallel, and in the same horizontal plane.

Incidentally, one of the disadvantages of Roy's SIMPFEED program is
that one needs to know the mutual coupling impedance between the
elements. For a two-element system, with identical elements, there
is a way to use EZNEC to calculate (estimate) the mutual coupling
impedance, Rm +/- jXm.

For two identical (resonant) elements, the feedpoint impedances reported
by EZNEC will be of the form, (Rs +/- Xm) +/- jRm, where Rs is the
resonant resistance of a single element alone (second element
open-circuited).

For instance, in my earlier example of two 33 ft dipoles, 33 ft apart
at a height of 66 ft, fed 90 degrees apart - the feedpoint impedances
a

109+j34 and 29-j34

That makes Rm = 34 ohms and makes Rs (109+29)/2 = 69 ohms, which
makes Xm = -j39 ohms. Those Rm and Xm values can then be plugged
into Roy's SIMPFEED program to obtain the length of the feedlines.

Note that two phased 20m dipoles work just fine as a beam on 17m.
All it takes is different phasing of the feedlines.
--
73, Cecil http://www.qsl.net/w5dxp



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