On Thu, 05 Nov 2009 22:20:32 -0600, tom wrote:

I do have 2 feet.

But not one of them in Perth.

On Thu, 05 Nov 2009 22:24:44 -0800, Richard Clark
wrote:

On Thu, 05 Nov 2009 22:20:32 -0600, tom wrote:

I do have 2 feet.

But not one of them in Perth.

No relation to anyone you are thinking_of/describing/etc, sorry to
ruin your conspiracy theory.

If you want to try and achieve a match to 50 ohms by moving the
adjacent parasitic elements seriously close to the driven folded
dipole, go for it. (I could dust off trusty Elnec and get a result.)
But I'd be surprised if anyone who gives a rats about the consistency
of the result would go down that path.

I am very familair with how the commercial side-mounted dipoles and
yagis are manufactured here in Australia, and I doubt that the rest of
the world is dramatically different. In three simple words - series
coax transformer. Let's agree that with an SMD you don't have
parasitics to play around with - except for tower spacing (which has
an impact on pattern, and variations are used for that end.) The
Aussie manufacturers use eaxactly the same method on the FD on their
yagis. That is why I suggested the O/P look into that approach.

On Fri, 06 Nov 2009 15:49:46 +0800, who where wrote:

But not one of them in Perth.

No relation to anyone you are thinking_of/describing/etc, sorry to
ruin your conspiracy theory.

Your confirmation here doesn't ruin anything. Art would hug you no
matter how you sign. Those he does have a remarkable need for
retaining anonymity. He would have us believe it's because his
supporters are easily bruised in the jostle. The following comment
would support that:

...gives a rats about the consistency
of the result would go down that path.

which is another but perhaps left-handed confirmation.

73's
Richard Clark, KB7QHC

On Fri, 06 Nov 2009 08:18:49 -0800, Richard Clark
wrote:

On Fri, 06 Nov 2009 15:49:46 +0800, who where wrote:

But not one of them in Perth.

No relation to anyone you are thinking_of/describing/etc, sorry to
ruin your conspiracy theory.

Your confirmation here doesn't ruin anything. Art would hug you no
matter how you sign. Those he does have a remarkable need for
retaining anonymity. He would have us believe it's because his
supporters are easily bruised in the jostle. The following comment
would support that:

...gives a rats about the consistency
of the result would go down that path.

which is another but perhaps left-handed confirmation.

73's
Richard Clark, KB7QHC

Whatever - and whoever Art is. I wonder why people like you carry on
at a personal level towards posters whose views you don't share. And
you seem to need the limelight, posting a name and callsign.

I'm describing how the matching IS done commercially. You can crap on
forever if you wish about how you might do it. Fini.

On Sat, 07 Nov 2009 06:32:25 +0800, who where wrote:

you seem to need the limelight, posting a name and callsign.

Yeah, as a longstanding convention for thousands of posters here, it
is a strange thing about being public and open in this world isn't it?

If you can't put your name to it, then any posting is only vacant
spam. "No one at home" informs us all about quality.

On the other hand, you choosing to be anonymous means you could have
us believe you are writing from a cave on the Afghan/Pakistan border
while waiting for your dialysis treatment to finish. Only Ossama and
vampires avoid the limelight - as you call it.

I'm describing how the matching IS done commercially.

Your painted-into-the-corner explanation has nothing to do with the
correlation between exhibited low feedpoint R and the proximity of
passive elements to what would have ordinarily been a very HiZ folded
element.

Fini.

We shall await your next post as

73's
Richard Clark, KB7QHC

who where wrote:

Whatever - and whoever Art is. I wonder why people like you carry on
at a personal level towards posters whose views you don't share. And
you seem to need the limelight, posting a name and callsign.

I'm describing how the matching IS done commercially. You can crap on
forever if you wish about how you might do it. Fini.

The "ways it's done commercially" depends a lot on the desired result.

A choked line into a 50 ohm DE is an easy to do but not optimal method.

It doesn't give best gain or BW or best F/B or best noise temperature
and never ever gives the best combination of them for weak signal work.

But it IS easy.

And it's not always what the commercial antenna builders use. It's what
you have noticed that they sell. Or you might be pushing how much it's
used just a bit.

tom
K0TAR

On Fri, 06 Nov 2009 20:55:19 -0600, tom wrote:

who where wrote:

Whatever - and whoever Art is. I wonder why people like you carry on
at a personal level towards posters whose views you don't share. And
you seem to need the limelight, posting a name and callsign.

I'm describing how the matching IS done commercially. You can crap on
forever if you wish about how you might do it. Fini.

The "ways it's done commercially" depends a lot on the desired result.

A choked line into a 50 ohm DE is an easy to do but not optimal method.

It doesn't give best gain or BW or best F/B or best noise temperature
and never ever gives the best combination of them for weak signal work.

But it IS easy.

And it's not always what the commercial antenna builders use. It's what
you have noticed that they sell. Or you might be pushing how much it's
used just a bit.

If you re-read what I posted, you will notice I stated "series coax
transformer". An in-line impedance transforming section is totally
different to simply stuffing RF choking on the line.

It is the method the three major manufacturers here in Australia
employ on their SMD's and the driven FD's on their yagis.

On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" wrote:

I'll try to get a better picture of the feedpoint for you.
Is there a way to work the .66 wavelength of 50 ohm cable backwards ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?
(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)
Mike

Here's a link to a picture.
http://i395.photobucket.com/albums/p...MFJCollage.jpg
Mike


Hi Mike,

That is a pretty good rendering given the other pix. Have you any
experience with Smith Charts? Still, and all, you need to know the Z
of at least one point to transform to another.

73's
Richard Clark, KB7QHC

On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" wrote:
I'll try to get a better picture of the feedpoint for you.
Here's a link to a picture.
http://i395.photobucket.com/albums/p...MFJCollage.jpg
Mike

You moved resulting in the one area of interest, near the coax
connector, being difficult to see. Can you try again, this time not
moving? Extra credit for putting a piece of graph paper under the
antenna so I extract dimensions.

The point I was trying to make is that the fairly long and exposed
leads at the connector, are perfectly acceptable for low frequencies
(HF) but are NOT acceptable for microwave work at 2.4GHz. The exposed
wires are inductors and/or radiators. My guess is there's a total of
about 4mm of exposed conductor. With a wavelength of 12.5mm, that's
1/3 of a wavelength. Before hitting the balun (or whatever that's
suppose to be), most of the RF will be radiated by the exposed section
of the coax, not the antenna.

I'm not an expert on baluns, but that thing doesn't look right. The
coax cable forms a balun, but the ferrite cores aren't involved except
to do block any RF coming back along the outside of the coax. My
guess(tm), is that the designer attempted to design the folded dipole
feed for 50 ohms, but discovered that the VSWR was far too high. So,
rather than move the feed impedance up to the more common 200 or 300
ohms, and use a 4:1 balun/xformer, he just shoved a bunch of ferrite
beads around the coax in order to "fix" the VSWR problem. It's not
really fixed or even matched. It just doesn't show any VSWR. The
real VSWR, measured at the feed point, is probably quite high.

Is there a way to work the .66 wavelength of 50 ohm cable backwards ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?

50 ohms. If the source, load, and coax are all 50 ohms, then there's
no transformation. You can use any length of 50 ohm coax and it will
still be 50 ohms in and out. Of course, we're assuming that the
MJF-1800 uses 50 ohm coax, not 75 ohm, which would be another story.

(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)

One must suffer before enlightenment. Let's pretend that it's 75 ohm
coax instead of 50 ohms. Let's also ignore the sloppy exposed
conductors at the RF connector. Let's also assume that we don't
really know the impedance of the folded dipole fed antenna.
Unfortunately, I also have to assume that your 0.66 wavelength doesn't
include the velocity factor for the coax making it closer to 0.75
wavelengths (so I can do this without dragging out the Smith Chart).
Odd multiples of 1/4 wavelength will neatly transform the endpoint
impedances according to:
Zcoax = sqrt (Zin * Zout)
or
Zcoax^2 = Zin * Zout
So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax:
Zout = 112.5 ohms
which is a bit closer to what I would expect to see with a folded
dipole antenna.
http://www.antennex.com/preview/New/quarter.htm
The designer could have also done it with 93 ohm coax, but the photo
doesn't look like RG-62/u. However, if he had, it would transform to
173 ohms, which is quite close to a folded dipole.

Bottom line.
I'm not thrilled with the design or construction of the MFJ-1800.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

"Richard Clark" wrote in message
...
On Sun, 8 Nov 2009 11:25:01 -0600, "amdx" wrote:

I'll try to get a better picture of the feedpoint for you.
Is there a way to work the .66 wavelength of 50 ohm cable backwards
ie.
What impedance would be transformed to 50 ohms with .66 wavelength of
50 ohm coax?
(this assumes the little knowledge I have about impedance transformation
using
coax is correct.)
Mike

Here's a link to a picture.
http://i395.photobucket.com/albums/p...MFJCollage.jpg
Mike


Hi Mike,

That is a pretty good rendering given the other pix. Have you any
experience with Smith Charts? Still, and all, you need to know the Z
of at least one point to transform to another.

73's
Richard Clark, KB7QHC

They should have been better, those are pictures I took a couple of years
ago.
I didn't blowup someone elses pictures.
"you need to know the Z of at least one point to transform to another."
I would be happy with the assumption the the impedance at the N connector
is 50 ohms. But I think I have a misunderstanding because, in use you would
add more 50 ohm coax to run from the N connector to the transmiter. Soo,
that .66
wavelength section on the antenna becomes anything you add to it.
AT this point, I have to think the folded loop is forced down to 50 ohms by
it's
surrounding structures and there is no impedance transformation betwen the
loop
and the N connector.
Mike