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# Re: Need Docs/Phasing Harness

• Subject: Re: [amsat-bb] Need Docs/Phasing Harness
• From: "John P. Toscano" <tosca005@xxxxxxxxxx>
• Date: Sun, 09 Mar 2003 18:52:22 -0600

```Jeff Breitner wrote:

> I'm interested in knowing if anyone has design plans for a phasing
> harness, or if I forwarded a pic of this beast its manufacturer could
> be identified.

Hi, Jeff.  Ordinarily, I would send this only to you, but since this or
similar questions have come up a few times lately, I figured I'd send
it to all.  Those of you who already know all this can feel free to hit
the DELETE button.  Although the reply is long, I hope that its size is
not considered excessive given that it is all ASCII text.

First of all, you have to decide what polarization(s) you want.  If you
are content with horizontal and vertical, you don't need anything
special, just two feedlines to a coaxial relay to pick which set of
elements are connected to the radio.  Some folks also use two feedlines,
one to their SSB radio (horizontal elements) and one to their FM radio
(vertical elements), no relays at all.  But since you asked here, I
assume that you want to at least switch polarities to a single radio,
or obtain circular polarity, or both.

Secondly, you need to determine the amount of offset (along the
direction of the boom) there is between one set of elements and the
other.  I would say "horizontal" vs. "vertical" elements, but there is
a neat way of getting Horizontal, Vertical, Right Circular, and Left
Circular polarity (described below) with fewer parts if you set the
elements at 45 degrees from the horizontal or vertical.  The two most
common cases are:
(a) zero offset, i.e. the "vertical" elements are at exactly
the same positions as the "horizontal" elements, touching one
another or forming a + instead of separate - and | elements.

(b) one quarter-wavelength offset, i.e. one set of elements is
a quarter-wavelength closer to the end of the boom than the
other set, so none of them are touching each other, only the
boom.  At 145 MHz, a quarter-wavelength is about 51.7 cm, or
20.34 inches.

Of course, there is a third, somewhat pathological case, (c)
some offset other than 1/4 wavelength.  Uggh.

To obtain circular polarity, what you need is to feed the two sets of
antenna elements so that they radiate waves that are 1/4 wavelength out
of phase.

That means that if the elements are already 1/4 wavelength apart, you
just feed them in phase and the rest is done for you automatically.

That also means that if the elements have zero offset, you feed them so
that one set of elements has and extra 1/4 wavelength of feedline to get
them 1/4 wavelength out of phase.

In either case, you have one more problem to deal with, namely that two
50 ohm antenas connected in parallel yield a 25 ohm antenna system,
which will be poorly matched to your 50 ohm transceiver.  So you ALSO
need impedance transformation to keep everything as happy as possible.

There are two ways to do this that come to mind.  One is to use
impedance transformers at each antenna that step up the impedance from
50 ohms to 100 ohms, then connect these two "100 ohm antennas" to a Tee
connector in parallel, giving you a 50 ohm antenna system.  The other
is to connect the two antennas to a Tee, then use a single impedance
transformer to step up the 25 ohm result to 50 ohms.

Now, what are these "impedance transformers" that I'm talking about?
Oddly enough, they are things that most hams can build successfully,
and they don't look anything like the electronic component that most of
us think of when we hear the word "transformer", i.e. an iron core with
many turns of insulated wire wrapped around it.

A simple impedance transformer is a quarter-wavelength long piece of
transmission line with a specific impedance, which is different than
the impedance at each of its opposite ends.  In the case of stepping up
a 50 ohm antenna to 100 ohms, so that two of them can be put in
parallel, you need a transmission line of 70.7 ohms that is 1/4
wavelength long.  As it turns out, a piece of 75 ohm coax is close
enough to the ideal impedance to do an excellent job.  Just remember
that you need 1/4 wavelength MULTIPLIED BY THE VELOCITY FACTOR OF THE
TRANSMISSION LINE.  So if the coax you use has a velocity factor of
90%, then you need 90% of 20.34 inches or only 18.3 inches.

Schematically, this arrangement would be built something like the
following.  The diagram will make sense only with a fixed-width font
like Courier due to the use of crude ASCII graphics:

1/4 WL           1/4 WL            1/4 WL
50 ohm           75 ohm    Tee     75 ohm
Antenna 1 ================o================o================ Antenna 2
|
|
Feedline

The single 1/4 wavelength section of 50 ohm cable keeps the impedance at
50 ohms, but performs the 1/4 wavelength phasing function.

The two 1/4 wavelength sections of 75 ohm cable step up each 50 ohm
antenna to 100 ohms, so that when they are paralleled at the Tee, to
go to the feedline, they become 50 ohms again.

If your antennas are already spaced 1/4 wavelength apart, you simply
omit the 1/4 wavelength section of 50 ohm cable, and everything else
is wired up the same.

If your antennas are spaced something OTHER than 1/4 wavelength apart,
you need to change the length of the 50 ohm section so that they will
become 1/4 wavelength out of phase.  So, say that instead of 20.34
inches apart, they are 10 inches apart.  You need an additional
(20.34 - 10 inches) x (velocity factor of 50 ohm cable)
length of 50 ohm coax to do it.  Note that the velocity factor of air
is assumed to be so close to 1.0 that it is being ignored.  But the
velocity factor of coaxial cable is not 1.0, and it is a function of
the exact brand of cable as well as the style of cable, i.e. not all
"RG-8" is the same as all other "RG-8", etc.

The other way to do it is to use a device called a "two-port power
divider".  This is a handy device that combines the functions of the
Tee and the impedance transformer all in one.  One advantage of the
power divider is that you are spared the trauma of using a cheap Tee
only to find that it does a horrible job of passing your signal, as
there are some frankly worthless Tee's out there.  And there is little
need to limit yourself to impedance transformations that can be managed
with the readily-available coax impedances that are out there (just
try to find something other than 50, 75, or 95 ohms sometime!)  The
power divider uses one metal tube inside of another metal tube, so it
is basically an air-dialectric feedline (VF = 1.0, which makes the
calculations simple and reliable), and by varying the diameters of the
two tubes, a wide variety of impedances can be generated.  This ease of
designing any wanted impedance is real important when the devices are
used for their more common purpose, namely stacking 2, 3, 4, 6, 8, or
more VHF/UHF antennas into big arrays for extreme weak-signal work like
EME operation.

Here is a web site that shows you what they look like, and even how you
can build them fairly easily:
http://home.teleport.com/~oldaker/power_dividers.htm

Anyway, with a two-port power divider, the diagram becomes something
like the following:

1/4 WL          2-port
50 ohm      power divider

Antenna 1 =================)o( Antenna 2
|
|
Feedline

Now, you may find it quite inconvenient to attach one side of the
2-port power divider right to the coax connector of the antenna, so
a more practical example might be:

1/4 WL          "X" length       "X" length
50 ohm            50 ohm           50 ohm
Antenna 1 ================o================)o(================ Antenna 2
|
|  \_ 2 port power divider
|
Feedline

The value of "X" above is irrelevant, you only need two pieces of 50 ohm
coax that are exactly the same length ("X") to keep everything in phase.
Also, if you are thinking ahead, you can see that it is somewhat silly
to have two pieces of coax, one 1/4 WL long, and one of length "X", that
are joined by a barrell connector or whatever.  The real solution is to
do something like this:

1/4 WL plus  "X" length          "X" length
of 50 ohm coax                 50 ohm
Antenna 1 ================================)o(================ Antenna 2
|
|  \_ 2 port power divider
|
Feedline

As before, if the two antennas are not designed with zero offset, then
you modify the 1/4 WL section length (or the difference in length
between the two cables) to get the two antennas 1/4 WL out of phase.

As for switchable polarization, I think that the best description of
this that I have seen is somewhere on this web site:
http://ham.te.hik.se/homepage/sm5bsz/

Unfortunately, that server seems to be unavailable as I type this
reply right now, but hopefully it will be back in operation when you
go there to look.

The idea described there is to use four coaxial relays, and four
phasing lines of 1/4, 1/4, 1/2, and 3/4 WL in a particular arrangement
between the relays, and to mount the antennas in an X pattern instead
of an + pattern.  In this manner, you can select horizontal, vertical,
right circular, or left circular polarization at any time, based on
the combination of relays that are energized/un-energized.  The basic
idea is that in addition to phasing the antennas at +1/4 WL or -1/4 WL
to get the two circular polarities, you can also phase then at 0 or
+1/2 WL to get a linear polarity signal at a 45 degree angle to the
elements.  That's the reason for the X setup, so that the two linear
polarities come out | or - for terrestrial FM, terrestrial SSB/CW,
respectively, plus their utility for satellites when vertical or
horizontal happen to be the best polarity at the moment.  Since this
switching arrangement always has the two antennas in parallel, you
will also need either a 2-port power divider to your feedline, or two
75 ohm 1/4 WL impedance transformers from the phase switch box to a
Tee, to keep the impedances correct.

The wiring setup goes something like this:

RELAY                RELAY
1A                  1B

O ====[ 1/4 WL ]==== O
Antenna 1 ===== O                    O ====O   2-port power divider
O ====[ 1/2 WL ]==== O     |  /
| /
|--------------O=== feedline
|
O ====[ 1/4 WL ]==== O     |
Antenna 2 ===== O                    O ====O
O ====[ 3/4 WL ]==== O

RELAY                RELAY
2A                    2B

Imagine all relays to be lined up in the same orientation, i.e.:

O -- normally closed port
O -- common port
O -- normally open port

If all relays are de-energized, both antennas have a 1/4 WL phasing
line inserted, and so the offset is 1/4 - 1/4 = 0, and we get linear
horizontal polarity (at a 45 degree angle to the elements which are
already at 45 degrees to the horizon, remember the antennas are set
up as X, not + in orientation.)

If relays 2A and 2B are energized, but relays 1A and 1B are not, then
we have a 1/4 WL phasing line inserted in the path of antenna 1, but
a 3/4 WL phasing line inserted in the path of antenna 2, giving us an
offset of 3/4 - 1/4 = 1/2 WL, and we get linear vertical polarity.

The other two combinations of 2A+2B off, 1A+1B ON, versus all relays
ON, gives us 1/2 - 1/4 = +1/4 WL, and 1/2 - 3/4 = -1/4 WL for the two
circular polarities.

In this setup, all phasing lines shown, as well as all feedlines
shown or implied, are 50 ohm.  However, to eliminate the 2-port power
divider, one could use two 75 ohm (1/4 WL x velocity factor) feedline
transformers from the common ports of relays 1B and 2B to a Tee, and
then to a 50-ohm feedline at the center connector of the Tee.

Anyway, I think that about covers it.  Feel free to ask if anything
needs clarification.

73 de W0JT
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