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

Cecil Moore wrote:



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?

KISS

But if you did that you couldn't get this thread to last forever.

tom
K0TAR

Tom Ring wrote:
KISS

Keep It Simple Stupid?

But if you did that you couldn't get this thread to last forever.

I have kept it as simple as possible. Wonder why nobody
has ventured an answer?
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
. ..
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

Hi Cecil

Is it posible that the length of the "stubs" change? I'd have thought the
length of the stub is always the same. 45 degrees should always be 45
degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line
looks like 50 ohms capacitive. That 50 ohms looks like something like 500
ohms inductive as viewed 45 degrees back along a 600 ohm line.
I'd guess your point is that 500 ohms of pure inductive reactance is never
seen 90 degrees back from an open, no matter what the Zo of the line

Jerry

Jerry Martes wrote:
Is it posible that the length of the "stubs" change? I'd have thought the
length of the stub is always the same. 45 degrees should always be 45
degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line
looks like 50 ohms capacitive. That 50 ohms looks like something like 500
ohms inductive as viewed 45 degrees back along a 600 ohm line.
I'd guess your point is that 500 ohms of pure inductive reactance is never
seen 90 degrees back from an open, no matter what the Zo of the line

The point I'm eventually going to make is about loading coils
in mobile antennas but let's stick with the above stub example.

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

Z01 = 600 ohms, Z02 = 50 ohms

If the Z0 were constant and the stub was 90 degrees long,
the source would see zero ohms. Yet in our above example
the stub is physically 90 degrees long and the source sees
+j500 ohms. The above stub is electrically 130 degrees long.
There is a 45 degree delay through the Z01 section of stub.
There is a 45 degree delay through the Z02 section of stub.
There is a 40 degree phase shift at the Z01 to Z02 junction.

If we want to turn the above stub into a functional 1/4WL
open stub such that the source sees zero ohms, we can
remove 40 degrees from the Z02 section. If we make the
Z02 section 5 degrees long, the entire stub will be
electrically 90 degrees long, and 1/4WL resonant.
There will be a 45 degree delay through the Z01 section
There will be a 5 degree delay through the Z02 section
There will be a 40 degree phase shift at the Z01 to Z02
junction.
--
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
--