I`m building a QFH NOAA weathersat antenna and wish to use low loss t/v coax
which is
72ohm; but is more manageable and less lossy at VHF than the 50ohm RG58
specified
over a 100ft run!.... ( RF pre-amps aint cheap! ).......

What are the pros and cons... can i use CT100 72/75ohm..... i believe i can
but, will the impedence affect
the 4 turn choke ????...

Thanks.

Lee......de G6ZSG.....

On Mon, 14 Aug 2006 07:44:10 GMT, "Lee"
wrote:

I`m building a QFH NOAA weathersat antenna and wish to use low loss t/v coax
which is
72ohm; but is more manageable and less lossy at VHF than the 50ohm RG58
specified
over a 100ft run!.... ( RF pre-amps aint cheap! ).......

What are the pros and cons... can i use CT100 72/75ohm..... i believe i can
but, will the impedence affect
the 4 turn choke ????...

Thanks.

Lee......de G6ZSG.....

It will work perfectly, because the TV coax is good HF quality.

Why do you use a 4 turn choke?

w.

"Helmut Wabnig" *_.-_- wrote in
message ...
On Mon, 14 Aug 2006 07:44:10 GMT, "Lee"
wrote:


It will work perfectly, because the TV coax is good HF quality.

Why do you use a 4 turn choke?

w.

Sorry, typo!.........Balun!!

Thanks, thought so.

Lee.....de G6ZSG......

On Mon, 14 Aug 2006 07:44:10 GMT, "Lee" wrote:

I`m building a QFH NOAA weathersat antenna and wish to use low loss t/v coax
which is
72ohm; but is more manageable and less lossy at VHF than the 50ohm RG58
specified
over a 100ft run!.... ( RF pre-amps aint cheap! ).......

What are the pros and cons... can i use CT100 72/75ohm..... i believe i can
but, will the impedence affect
the 4 turn choke ????...

Thanks.

Lee......de G6ZSG.....

Hi Lee,

Consider this: If your QFH has a 50-ohm terminal impedance, the mismatch is only
1.44: 1. Therefore, the loss due to the mismatch is 0,14 dB, insignifiant--use
the 72-ohm line and forget the miniscule mismatch.

Even if the mismatch was 2:1, the reflection loss is only 0.51 dB.

Concerning the 4-turn choke, nothing happens to the matching operation, because
nothing inside the coax changes due to the coiling of the coax.

Walt, W2DU

Please leave me out of this. I'm mentally handicapped!

I don't know, never did know, how to use an old fashioned, mid-20th
century Smith Chart.
----
Reg.





"Walter Maxwell" wrote in message
...
On Mon, 14 Aug 2006 07:44:10 GMT, "Lee"
wrote:

I`m building a QFH NOAA weathersat antenna and wish to use low loss
t/v coax
which is
72ohm; but is more manageable and less lossy at VHF than the 50ohm
RG58
specified
over a 100ft run!.... ( RF pre-amps aint cheap! ).......

What are the pros and cons... can i use CT100 72/75ohm..... i
believe i can
but, will the impedence affect
the 4 turn choke ????...

Thanks.

Lee......de G6ZSG.....

Hi Lee,

Consider this: If your QFH has a 50-ohm terminal impedance, the
mismatch is only
1.44: 1. Therefore, the loss due to the mismatch is 0,14 dB,
insignifiant--use
the 72-ohm line and forget the miniscule mismatch.

Even if the mismatch was 2:1, the reflection loss is only 0.51 dB.

Concerning the 4-turn choke, nothing happens to the matching
operation, because
nothing inside the coax changes due to the coiling of the coax.

Walt, W2DU

Reg Edwards wrote:
I don't know, never did know, how to use an old fashioned, mid-20th
century Smith Chart.

Reg, I'm curious how you would solve this stub problem
without a Smith Chart.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Stub sections are lossless. Z01 = 600 ohms and is 45
degrees long. Z02 = 50 ohms and is 45 degrees long.
What is the impedance looking into the stub from the
source?
--
73, Cecil http://www.qsl.net/w5dxp

On Mon, 14 Aug 2006 14:58:32 GMT, Cecil Moore
wrote:

Reg Edwards wrote:
I don't know, never did know, how to use an old fashioned, mid-20th
century Smith Chart.

Reg, I'm curious how you would solve this stub problem
without a Smith Chart.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Stub sections are lossless. Z01 = 600 ohms and is 45
degrees long. Z02 = 50 ohms and is 45 degrees long.
What is the impedance looking into the stub from the
source?

I missed the significance of this problem Cecil.

Is it principally a theoretical (being lossless) problem that a Smith
chart can solve, or does it have some other significance?

Whilst a Smith chart is great for visualising transmission line
problems, a great way for visually mapping impedance over a range of
frequencies, it isn't the most practical way to solve practical
problems when we have access to the computing power commonly available
to designers today.

Owen

PS: I think the problem you have given can be solved with simple trig:
find the reactance of the Z02 section using one trig term, find the
length of Z01 that would deliver that reactance using one trig term,
add that length and the actual length of Z01 section, find the
reactance of the Z01 section using one trig term. I could do that in a
flash with a scientific hand calculator while you were sharpening your
pencil.

It is a trivial problem either way, and can only ever be an
approximation of a practical problem.
--

Owen Duffy wrote:
Cecil Moore wrote:
Reg, I'm curious how you would solve this stub problem
without a Smith Chart.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Stub sections are lossless. Z01 = 600 ohms and is 45
degrees long. Z02 = 50 ohms and is 45 degrees long.
What is the impedance looking into the stub from the
source?

I missed the significance of this problem Cecil.

Is it principally a theoretical (being lossless) problem that a Smith
chart can solve, or does it have some other significance?

It's just a mental exercise with a hidden significance. This
is the type of problem that I would solve with a Smith Chart.

How about a solution? What impedance does the source see? The
physical length of the stub is 90 degrees. What is the electrical
length of the stub in degrees?
--
73, Cecil http://www.qsl.net/w5dxp

On Mon, 14 Aug 2006 23:04:30 GMT, Owen Duffy wrote:



PS: I think the problem you have given can be solved with simple trig:
find the reactance of the Z02 section using one trig term,

Z=-j50*cot(45)=-j50

find the
length of Z01 that would deliver that reactance using one trig term,

l=acot(50/600)=85.2

add that length and the actual length of Z01 section, find the

Z01'=85.2+45=130.2

reactance of the Z01 section using one trig term. I could do that in a

X=-j600*cot(130.2)=j507.7

flash with a scientific hand calculator while you were sharpening your
pencil.

It is a trivial problem either way, and can only ever be an
approximation of a practical problem.

But you wouldn't get that accuracy from the Smith chart.

Owen
--

On Mon, 14 Aug 2006 23:04:30 GMT, Owen Duffy wrote:

On Mon, 14 Aug 2006 14:58:32 GMT, Cecil Moore
wrote:

Reg Edwards wrote:
I don't know, never did know, how to use an old fashioned, mid-20th
century Smith Chart.

Reg, I'm curious how you would solve this stub problem

I missed the significance of this problem Cecil.

Hi Owen,

Forgive the slur from the world of fixed sports, the significance is a
cross-Atlantic, Australian tag team troll.

73's
Richard Clark, KB7QHC