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
--

"Cecil Moore" wrote in message
m...
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.


Hi Cecil I sure dont want to get involved with any mobil antenna loading
coil discussions, I admit that I'm not qualified.



| 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.

Yeah, but is isnt a line with a constant Zo


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.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 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 I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not even
smart. But, I sure dont see how anyone can conclude there is a phase shift
at the junction of two transmission lines. There is a shunt capacitive
reactance that results from that abrupt change in dimensions, but I wouldnt
have thought it would be enough to result in 40 degrees of phase shift.
..

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.

I think I agree with the concept you are using. You are giving the
condition where a purely capacitive reactor of 50 ohm impedance is required
to be resonated by introducing a series inductor. A short length of higher
impedance transmission line will sure do that. The higher the line Zo,
the shorter it needs to be to resonate that 50 ohm capacitor.


--
73, Cecil http://www.qsl.net/w5dxp

Jerry

"Owen Duffy" wrote in message
...
On Mon, 14 Aug 2006 14:53:53 GMT, "Lee"
wrote:


There are alot of affordable amplifiers designed for TV that you
could
use
at the base of your QFH. You might consider building your owm
amplifier
to
fit in the base of the QFH.
I wouldnt recomend the use of a pre-amp at the antenna for NOAA
satelite
station. They often cause more problems than they solve.
All Electronics has alot of ferrite tubes that can be used to fit
over
the
coax so you wouldnt need the "4 turn choke".

`4 turn Choke Balun`.....typo....


Jerry

Thanks Jerry, i`ll give it some thought as i`m right under some pmr
towers
which breaks through a little from 150megs pagers and a preamp may worsen
things...

Lee,

It is easy to build a preamp with high gain and low noise figure and
it will exhibit superb performance on a test bench in a shielded room
on a signal generator.

In a real world environment, you are unlikely to realise the full
sensitivity of the receiver due to:
- external noise; and
- intermodulation products generated within your receiver (preamp).

It is harder to build a preamp with low intermodulation distortion,
and one method of reducing the results of that intermodulation
distortion is front end filtering to reduce the level of undesired
signals reaching the non-linear devices.

Front end selectivity costs much more money than a low NF preamp
transistor or gasfet.

Whilst wideband preamps are available at low cost, it is quite likely
that they will actually degrade your receiver performance.

It may even be that adding an external filter will improve your S/N
ratio.

An interesting test to perform is to note the S/N ratio, add a small
attenuator to the receiver input, and again measure the S/N ratio. If
the S/N ratio improves, it is an indicator that you have significant
intermodulation distortion and front end filtering may improve the
sensitivity.

I`ll consider your advice very carefully Owen, Thanks....


I listened last night and could hear NOAA 14 on a hand held scanner
(IC-R20) with a 130mm long rubber duckie off my 2m transceiver. It
wasn't good enough for pictures, but it could be heard... so it
shouldn't take a lot of receiver sensititivity to decode it well.
(BTW, I could not hear the bird using a 200mm whip on the scanner...
to much noise from intermod products).

Yes, i`ve done that also....and the handy was quite strong too but the
scanner front end was awful ..... AOR2002..... Ugh!!!..


I know you asked about coax and you are seeking a low loss coax
situation, coax loss might be less important that adequate receiver
front end filtering so that you can realise most of its potential in
the presence of other strong signals. In the absence of that, coax
loss might actually improve S/N!

Surprisingly, i get quite a good picture using my Yaesu 857D also!!! ( with
attenuator in, of course)....


Owen

PS: I recently performed some tests on the new Icom IC-7000 on 144MHz
to determine the usable sensitivity on a wideband antenna, and
although the specified sensitivity is -126dBm, the sensitivity when
connected to a Diamond D-130 at this location was -96dBm, that is 30dB
poorer than spec, and the main contibution was IMD within the IC-7000.
Putting a 10dB attenuator inline improved the sensitivity by 14dB!
--

I have the Icom IC-E90 which has good audio on the NOAA`s too ... also on
the ducky...

Lee.....G6ZSG....

Owen Duffy wrote:
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.

Thanks very much, Owen. I used the Smith chart to get 85, 130,
and 500 above. That's about as good an accuracy as I ever need.
Also MicroSmith says the impedance value is j507.2 ohms.

The stub, which is 45+45 = 90 degrees physically, is electrically
130 degrees long. There is an ~80 degree shift in the Gamma angle
at the 600--50 ohm impedance discontinuity resulting in ~40 degrees
being added to the electrical length of the stub by the impedance
discontinuity.

Let's say we now want to turn that stub into an electrical 1/4WL
stub. If we made the two sections the same number of degrees,
how many degrees would they occupy?

| X deg | X deg |
source====Z01=========Z02====open 1/4WL stub

where Z01=600 ohms and Z02=50 ohms

I get 16.1 degrees for X. A stub that is physically ~1/12WL
long is 1/4WL resonant, i.e. a 90 degree phase shift from end
to end. Does this remind anyone of a base-loaded mobile antenna?
--
73, Cecil http://www.qsl.net/w5dxp

Jerry Martes wrote:
"Cecil Moore" wrote in message
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.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 degrees
long.

Well, it is physically 90 degrees long because the two
physical pieces are physically 45 degrees each. That's a
given. However, the +j500 result tells us that it is
electrically 130 degrees removed from the open circuit
at the far end.

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 I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not even
smart. But, I sure dont see how anyone can conclude there is a phase shift
at the junction of two transmission lines.

There is an abrupt change in the Gamma angle of the reflection
coefficient at the impedance discontinuity. I can show you why
on a phasor graphic. Simplified, it goes something like this.

Itotal = 21.5*sin(25) = 10*sin(65)

where 21.5 is the phasor amplitude of the current in the 50
ohm section at the junction and 10 is the phasor amplitude
of the current in the 600 ohm section at the junction.

The values must be the same even though the magnitude of Z0,
which controls the amplitude of the current, has changed.
If those values must be equal and the amplitude changes because
the Z0 changed, the only other thing that can change is the
phase angle.
--
73, Cecil http://www.qsl.net/w5dxp

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

"Cecil Moore" wrote in message
. ..
Jerry Martes wrote:
"Cecil Moore" wrote in message
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.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 degrees
long.

Well, it is physically 90 degrees long because the two
physical pieces are physically 45 degrees each. That's a
given. However, the +j500 result tells us that it is
electrically 130 degrees removed from the open circuit
at the far end.

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 I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not
even smart. But, I sure dont see how anyone can conclude there is a
phase shift at the junction of two transmission lines.

There is an abrupt change in the Gamma angle of the reflection
coefficient at the impedance discontinuity. I can show you why
on a phasor graphic. Simplified, it goes something like this.

Itotal = 21.5*sin(25) = 10*sin(65)

where 21.5 is the phasor amplitude of the current in the 50
ohm section at the junction and 10 is the phasor amplitude
of the current in the 600 ohm section at the junction.

The values must be the same even though the magnitude of Z0,
which controls the amplitude of the current, has changed.
If those values must be equal and the amplitude changes because
the Z0 changed, the only other thing that can change is the
phase angle.
--
73, Cecil http://www.qsl.net/w5dxp

Hi Cecil

Thanks for pointing me toward learning about reflection coefficient. I
am really surprised that there is such a large amount of phase shift at that
junction.

Jerry

Jerry Martes wrote:
Thanks for pointing me toward learning about reflection coefficient. I
am really surprised that there is such a large amount of phase shift at that
junction.

W8JI has theorized an even larger phase shift there at the
junction of a loading coil and the stinger of a mobile antenna.
Unfortunately, he attributes 100% of the phase shift below
the stinger to that single junction point, while ignoring
the phase shift provided by the loading coil. This thread
is my attempt at setting the technical record straight and
correcting W8JI's new old wives' tale. This discussion
started years ago on QRZ.com where both sides were wrong
and nobody was 100% technically correct, including myself.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
W8JI has theorized an even larger phase shift there at the
junction of a loading coil and the stinger of a mobile antenna.
Unfortunately, he attributes 100% of the phase shift below
the stinger to that single junction point, while ignoring
the phase shift provided by the loading coil. This thread
is my attempt at setting the technical record straight and
correcting W8JI's new old wives' tale. This discussion
started years ago on QRZ.com where both sides were wrong
and nobody was 100% technically correct, including myself.

There is nothing that verifies Cecil's theory except Cecil's theory.

73 Tom