Mobile Antenna Matching Question
 

I put a mount on my Dodge pickup, bolted to the bed rail with 4 stainless
steel bolts. I placed a 20M Hamstick on the mount and the lowest I could get
the SWR to drop at my desired frequency was 1.7 to 1. So I placed a small
matching coil at the base of the antenna (also made by the same manufacturer
as the antenna). As I started trying to bring the SWR down I noticed that
anywhere that I touched the coil with the shorting wire would make the SWR
rise. At the very end of the coil - away from the mounting point, the rise
was minimal. As I moved the clip closer to the coil mounting point the SWR
rose.

The same thing happens with my 40M hamstick.

What does this tell me? Am I doing something incorrectly in the matching
process?

Thanks for any and all advice.

WB5CYS
Texas

Well, judging from the description, I'd say you need a cap
instead of the coil. It's sorta telling me you have enough
coil already. Part of your loading coil in the whip is
already acting as a matching coil. I think you probably
need a cap to tune out some reactance. I often use a
simple MFJ random wire tuner as a mobile matcher.
That would dial that down pretty low I would think.
But this is just a guess, being I can't see it...
In my trucks, I have the small tuner, and also a "dollar
special" matching coil. I use the tuner more often I
think...Most of my antenna setups with the glass stick
bases, have a few turns of matching coil wound into
the base. MK

wb5cys wrote:
I put a mount on my Dodge pickup, bolted to the bed rail with 4 stainless
steel bolts. I placed a 20M Hamstick on the mount and the lowest I could get
the SWR to drop at my desired frequency was 1.7 to 1. So I placed a small
matching coil at the base of the antenna (also made by the same manufacturer
as the antenna). As I started trying to bring the SWR down I noticed that
anywhere that I touched the coil with the shorting wire would make the SWR
rise.

Anything you do to a resonant antenna will usually raise the
SWR. For the coil to work, you need to start with an antenna
that is too short. You need to adjust the stinger as well as
adjusting the coil. First adjust the stinger to resonance
without the coil. Then shorten the stinger to raise the SWR
by making the antenna capacitive. Then add the coil. Alternate
between stinger adjustment and coil adjustment until satisfied.
--
73, Cecil http://www.qsl.net/w5dxp


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First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

That being said, the strident need to be 1:1 can be met in several ways. The
antenna will probably be around 35 ohms. giving you the 1.7:1 you observe.
You can wind a toroid (t-106-2) with 14 turns of wire. tap one of the wires
at 11 turns and feed the antenna. A 50pf cap between the antenna and ground
might also prove usefull.

to wind the toroid, start with a black and white wire, wind 14 times through
the hole. Connect the black start end to the coax center. Connect the white
start to the black finish. Connect the white finish to the coax shield, and
vehicle ground. At turn 11 of the black wire, put a tap. This tap goes to
the antenna.

The SWR meter should now read 1:1 and you won't get out one bit better.

The SWR on my base feedlines are 9:1 by design. I run a screwdriver on the
car and once I am below 2:1, the radio is happy and so am I.



"wb5cys" wb5c ...
I put a mount on my Dodge pickup, bolted to the bed rail with 4 stainless
steel bolts. I placed a 20M Hamstick on the mount and the lowest I could
get
the SWR to drop at my desired frequency was 1.7 to 1. So I placed a small
matching coil at the base of the antenna (also made by the same
manufacturer
as the antenna). As I started trying to bring the SWR down I noticed that
anywhere that I touched the coil with the shorting wire would make the SWR
rise. At the very end of the coil - away from the mounting point, the rise
was minimal. As I moved the clip closer to the coil mounting point the SWR
rose.

The same thing happens with my 40M hamstick.

What does this tell me? Am I doing something incorrectly in the matching
process?

Thanks for any and all advice.

WB5CYS
Texas

Fred W4JLE wrote:
First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

That being said, the strident need to be 1:1 can be met in several ways. The
antenna will probably be around 35 ohms. giving you the 1.7:1 you observe. . .

I'm curious -- how did you determine from the SWR that the antenna
impedance would be resistive, and assuming it is, that it would be less
than 50 ohms rather than greater? Perhaps you've measured a Hamstick on
a typical mobile mount?

Roy Lewallen, W7EL

Roy Lewallen wrote:
I'm curious -- how did you determine from the SWR that the antenna
impedance would be resistive, and assuming it is, that it would be less
than 50 ohms rather than greater? Perhaps you've measured a Hamstick on
a typical mobile mount?

It's impossible to bring an impedance greater than 50
ohms down to 50 ohms by installing a shunt reactance
so the resonant feedpoint resistance of any mobile antenna
using a shunt reactance to achieve 50 ohms is, by definition,
less than 50 ohms.

if Rs = 50 and Xs 0 then Rp 50

Assume one can adjust the feedpoint impedance of a 40m
bugcatcher to 25+j25 or 25-j25 by adjusting the length
of the stinger. If one shunts 25+j25 with a -j50 cap,
the feedpoint impedance will be 50+j0 ohms. If one shunts
25-j25 with a +j50 coil, the feedpoint impedance will
be 50+j0 ohms.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil Moore wrote:
It's impossible to bring an impedance greater than 50
ohms down to 50 ohms by installing a shunt reactance ...

My meaning above may not have been 100% clear.
I meant to say: It's impossible to bring an impedance,
having a resistive component greater than 50 ohms, down
to 50 ohms by installing a shunt reactance.
--
73, Cecil http://www.qsl.net/w5dxp

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Only going from the last one I fooled with. I was unable to persuade the
owner he would be better off tuning for field strength than SWR and the
solution I purposed is one of many that will make a SWR meter happy along
with those folks who are cult members of the church of no reflections.

I make no claims that any other antenna in like or different circumstances
will ever exhibit exactly 35 +j0 Ohms. If I gave that impression to the
learned members of this conference, I humbly apologize for once again making
a broad statement with out a list of qualifiers. Mea Culpa :)

The ubiquitous SWR meter, the greatest generator of misinformation since
Baghdad Bob.





"Roy Lewallen" wrote in message
...
Fred W4JLE wrote:
First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

That being said, the strident need to be 1:1 can be met in several ways.
The
antenna will probably be around 35 ohms. giving you the 1.7:1 you
observe. . .

I'm curious -- how did you determine from the SWR that the antenna
impedance would be resistive, and assuming it is, that it would be less
than 50 ohms rather than greater? Perhaps you've measured a Hamstick on
a typical mobile mount?

Roy Lewallen, W7EL

Fred:

I wonder if most arent really aware that SWR is not real in a lot of
cases where the term "swr" is bandied about (big deal, I don't believe
in time either)--I have experienced large feedline radiation on almost
perfect reading swr meters--I accept this as real--and still run with it
(even if I lose 10 watts of 100, I don't sweat it anymore, heck, if
Cecil splits his bottle of wine with me--I could careless about 20 watts
grin)...

.... however, I have noticed that it (the swr meter) slows the need to
keep soldering new mrf4?? mosfets into the finals and linears of this
new age--it is a cheap quick fix--and if transmitter loading is good,
the finals are cool, I am good with it... and some just wanna get on the
air... I will mess with an ant for best received signal--from there on
out it is just crank power till we are at max--if needed--and if that
fails a gentlemans admission of failure... and a gentleman ALWAYS uses
minimum power for effective communication!!! sly grin

.... the true art of dancing fairies on pins, and speaking in shakespeare
to obfuscate the ideas which I express, and beginning to hold myself as
superior in intelligence and mind games--I hold for a later date--which
seems to be approaching far to fast for my liking... check back with me
next week, I may have changed my mind... grin

Warmest regards,
John
"Fred W4JLE" wrote in message
...
Only going from the last one I fooled with. I was unable to persuade
the
owner he would be better off tuning for field strength than SWR and
the
solution I purposed is one of many that will make a SWR meter happy
along
with those folks who are cult members of the church of no reflections.

I make no claims that any other antenna in like or different
circumstances
will ever exhibit exactly 35 +j0 Ohms. If I gave that impression to
the
learned members of this conference, I humbly apologize for once again
making
a broad statement with out a list of qualifiers. Mea Culpa :)

The ubiquitous SWR meter, the greatest generator of misinformation
since
Baghdad Bob.





"Roy Lewallen" wrote in message
...
Fred W4JLE wrote:
First, I would say that 1.7 to 1 is fine, leave it alone. It is
what is
expected.

That being said, the strident need to be 1:1 can be met in several
ways.
The
antenna will probably be around 35 ohms. giving you the 1.7:1 you
observe. . .

I'm curious -- how did you determine from the SWR that the antenna
impedance would be resistive, and assuming it is, that it would be
less
than 50 ohms rather than greater? Perhaps you've measured a Hamstick
on
a typical mobile mount?

Roy Lewallen, W7EL

snip

I'm curious -- how did you determine from the SWR that the antenna
impedance would be resistive, and assuming it is, that it would be
less
than 50 ohms rather than greater? Perhaps you've measured a Hamstick
on
a typical mobile mount?

Roy Lewallen, W7EL





Roy spurred me to type:

I use a screwdriver matched with a simple toroidal autoformer.
I tuned the antenna for a purely resistive load using an analyzer.
Turns out it was 20 - 30 ohms on 40 meters, IIRC.
Then I wound 14 turns of copper wire around a toroid, connecting one end to
the coax center
and the other to coax shield and chassis ground.
Then I tapped the antenna on to the autoformer such that,
looking down the coax, the analyzer saw 50 ohms.
I use this matching configuration for 160 - 20.
I get the best VSWR and field strength of many variations I've tried over
the years.
On the higher bands I just hook up the coax and tune for minimum VSWR at the
transceiver.

73, H.,

H. Adam Stevens
NQ5H

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE" wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the bandwidth
where it drops no more than about 90% or so, you can operate confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the source.
This is not necesssarily the case, but gives us the least signal strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as well.

73
H.
NQ5H

H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the source.
This is not necesssarily the case, but gives us the least signal strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

While your math is correct, your application in my opinion is incorrect.


"H. Adam Stevens, NQ5H" wrote in message
...

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE" wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the bandwidth
where it drops no more than about 90% or so, you can operate confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal strength
in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

73
H.
NQ5H

Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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You have created an all encompassing case from the original SWR of 1.7:1 No
tuner needed. Under these conditions, the amount of reflected energy
radiated approaches Ivory Soap.


"Richard Clark" wrote in message
...
On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

On Sun, 5 Jun 2005 10:02:17 -0400, "Fred W4JLE"
wrote:

You have created an all encompassing case from the original SWR of 1.7:1

Thanx Fred,

No one has ever been able to answer when the Transmitter rereflects
energy, why they need a tuner.

73's
Richard Clark, KB7QHC

Roy
Worst case.
Get it?
H.

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal
strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db
is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

Like I said to Roy.
"Worst case."
Get it?
OF COURSE it isn't absorbed in the source.
H.


"Fred W4JLE" wrote in message
...
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

While your math is correct, your application in my opinion is incorrect.


"H. Adam Stevens, NQ5H" wrote in message
...

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE"
wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what
is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the
bandwidth
where it drops no more than about 90% or so, you can operate
confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal
strength
in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db
is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

73
H.
NQ5H

One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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Ivory Soap. Excellent!
73
H.


"Fred W4JLE" wrote in message
...
You have created an all encompassing case from the original SWR of 1.7:1
No
tuner needed. Under these conditions, the amount of reflected energy
radiated approaches Ivory Soap.


"Richard Clark" wrote in message
...
On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

In TV broadcasting reflections from the antenna back to the transmitter will
be reflected by the transmitter to the antenna and the signal will be
rebroadcast albeit at somewhat less power. Then depending on the length of
transmission line the viewer may see ghosting.
In audio I don't know why and I have run my Collins 30S-1 into ladder line
with a 14 to SWR with no one except me knowing!

--

73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Sun, 5 Jun 2005 10:02:17 -0400, "Fred W4JLE"
wrote:

You have created an all encompassing case from the original SWR of 1.7:1

Thanx Fred,

No one has ever been able to answer when the Transmitter rereflects
energy, why they need a tuner.

73's
Richard Clark, KB7QHC

H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

Fred W4JLE wrote:
Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1

!.7:1???? I must admit, that's a new one on me. :-)

My statements were general, and apply to any SWR.
--
73, Cecil http://www.qsl.net/w5dxp

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H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.

73
H.


"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

See other post.

It's called db, Cecil.


"Cecil Moore" wrote in message
...
H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sun, 05 Jun 2005 21:58:06 GMT, "Henry Kolesnik"
wrote:

In TV broadcasting reflections from the antenna back to the transmitter will
be reflected by the transmitter to the antenna and the signal will be
rebroadcast albeit at somewhat less power.

H. Adam Stevens, NQ5H wrote:

"Cecil Moore" wrote:
The log of the ratio of two SWRs doesn't seem to have much
meaning.

It's called db, Cecil.

The IEEE Dictionary says the ratio of power, voltage, and
current can be expressed in dB. It specifically states
that dB can only be related to power ratios or to parameters
that are proportional to the square root of power ratios.

SWR1 = [SQRT(Pfor1)+SQRT(Pref1)]/[SQRT(Pfor1)-SQRT(Pref1)]

SWR2 = [SQRT(Pfor2)+SQRT(Pref2)]/[SQRT(Pfor2)-SQRT(Pref2)]

The ratio of two SWRs will not reduce to a power ratio or
to the square root of a power ratio.
--
73, Cecil http://www.qsl.net/w5dxp

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H. Adam Stevens, NQ5H wrote:
Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.


Let's see, I put a 100 volt zero impedance source in series with a 50
ohm resistor, connect that to a half wave transmission line terminated
with 150 ohms. The current will be 100/200 = 0.5 amp, the power in the
150 ohm load is 37.5 watts, the power in the 50 ohm source resistor is
12.5 watts. The SWR is 3:1, the forward power is 50 watts, the reverse
power is 12.5 watts. Sure enough, the power in the source resistor
equals the reverse power. Good job. That sure must be the worst case,
all right.

Just to check, I'll change the load resistor to 16.67 ohms. Now the
current is 1.5 amps, the power in the 16.67 ohm load is 37.5 watts, and
the power in the source resistor is 112.5 watts. The SWR is still 3:1,
the forward power is 50 watts just like before, and the reverse power is
12.5 watts just like before. Hm. The reverse power is 12.5 watts, but
the source resistor is now dissipating 112.5 watts. Must be worse than
the worst case.

Well, shoot, maybe the source resistor dissipates all the reverse power
*plus* some more power that comes from somewhere else. So let's try a
200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load
resistor is 32 watts, and the power in the 50 ohm source resistor is 8
watts. The SWR is 4:1, the forward power is 50 watts, and the reverse
power is 18 watts. Oops, the source resistor is only dissipating 8 watts
but the reverse power is 18 watts. Not only isn't it dissipating all the
reverse power, but it isn't even dissipating that extra power that came
from somewhere else when we connected the 16.67 ohm resistor. Wonder
where the other 10 watts of reverse power went?(*)

So using your simple criterion of a zero impedance source and resistor
equal to the transmission line impedance, and by only changing the load
resistance, we've got cases whe

-- The source resistor dissipation equals the reverse power
-- The source resistor dissipation is greater than the reverse power
-- The source resistor dissipation is less than the reverse power

And none of these will explain the loss figure you gave earlier.

Guess I don't know better after all.

Anyone who's interested can find more interesting cases in "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/. And those who aren't
interested, well, you're welcome to believe what you choose. Just don't
look too closely at the evidence.

(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy Lewallen, W7EL

Yes, SWR is a dimensionless quantity. dB is, as far as I know, defined
only for power, voltage and current ratios, as the IEEE Dictionary
implies. Since it's defined differently for power than for voltage or
current (so that an increase or reduction in one quantity represents the
same number of dB increase or decrease in the other), anyone using it
for something else would have to clarify how it would be defined in that
context.

Roy Lewallen, W7EL

Cecil Moore wrote:
H. Adam Stevens, NQ5H wrote:

"Cecil Moore" wrote:
The log of the ratio of two SWRs doesn't seem to have much
meaning.

It's called db, Cecil.


The IEEE Dictionary says the ratio of power, voltage, and
current can be expressed in dB. It specifically states
that dB can only be related to power ratios or to parameters
that are proportional to the square root of power ratios.

SWR1 = [SQRT(Pfor1)+SQRT(Pref1)]/[SQRT(Pfor1)-SQRT(Pref1)]

SWR2 = [SQRT(Pfor2)+SQRT(Pref2)]/[SQRT(Pfor2)-SQRT(Pref2)]

The ratio of two SWRs will not reduce to a power ratio or
to the square root of a power ratio.

On Sun, 05 Jun 2005 21:58:06 GMT, "Henry Kolesnik"
wrote:

In TV broadcasting reflections from the antenna back to the transmitter will
be reflected by the transmitter to the antenna and the signal will be
rebroadcast albeit at somewhat less power.

Hi Hank,

That would pretty much reveal the SWR if we knew, wouldn't it? If
"somewhat less power" was in 1.2:1 ratio, we wouldn't care so much,
but how would the viewer feel about such service?

Then depending on the length of
transmission line the viewer may see ghosting.

I think we, or another correspondent and I have dealt with that at one
time. At the time I believe it was called "fringing," not "ghosts."
The difference being that what were called ghosts at the dawn of the
TV era were separated by fractions of an inch rather than fractions of
a mm. As such, ghosts couldn't have been originated by anything
shorter than mile length transmission lines that were poorly
terminated at both ends. Instead, ghosts were actually transmission
path length differentials in a multipath situation.

In audio I don't know why and I have run my Collins 30S-1 into ladder line
with a 14 to SWR with no one except me knowing!

Well, if this is meant to be analogous to fringing/ghosting, I suppose
its because a microsecond blur at AF is entirely inaudible. Or are we
speaking of SSTV? However, this begs the question, How did you know?
All the Collins equipment I taught at school didn't come with a SWR
meter. It was wholly unnecessary if you performed the standard
tune-up. Matter of fact, back then the only SWR meter I saw was for
Ham gear. The finals' tank performed every function of matching as
any tuner.

However, with the KWT-6, we did use an external tuner, 180-V1
(although I may have this mixed up with another model), for coax
feedlines. This was more for its automatic feature where the
transmitter could be tuned up with a 50 Ohm load, and the automatic
tuner simply did the job of presenting it with the transformed load.

However, returning to the point of a transmitter rereflecting a
reflection; I know the bare KWT-6 into ladder line employs its tank to
protect its final tubes. Without that safeguard, I have seen plates
melt - something no one here wants to call dissipation lest it be
evidence of an internal resistance. The bare tubes with their native
very hi Z would rereflect like nothing else - and this begs the
observation - how could you get original any power out of them, past
the tremendous mismatch? The tuner/final tank comes back into the
equation, and rereflection goes out the window as a property of the
transmitter and returns to the domain of matching.

If anyone wants to constrain the entire crusade of the rereflecting
transmitter to the tube set feeding ladder line - then feel free to do
so. However, I don't think I've ever seen a mobile tube rig feeding
ladder line - no doubt one day I will. We will probably talk about
efficiency. :-)

73's
Richard Clark, KB7QHC

However, I don't think I've ever seen a mobile tube rig feeding
ladder line - no doubt one day I will. We will probably talk about
efficiency. :-)

It was done in 1936.
http://web.wt.net/~nm5k/mobile36.jpg
Cover pix from a 1936 QST...Forgot
what month...
But I still think I prefer coax...
Their "ladder line" looked to be a twisted wire feeder.
The call on that vehicle was W9MSY...
With the short feedline run on a mobile,
even coax is pretty low loss...
I never used an SWR meter when I was a
novice...I had an old viking valiant that would
tune nearly anything...You didn't need a meter...
You just loaded it up to full plate current and went
with it...My TS 830 is like that to a lesser extent..
If it loads within the loading range, it's good nuff...
No point in even putting a meter on it...Adding a
tuner, would just add some loss...MK

Roy Lewallen wrote:
(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

Here's my earlier thought example again.

100w----one second long lossless feedline----load, rho=0.707

SWR = (1+rho)/(1-rho) = 5.828:1
Source is delivering 100 watts (joules/sec)
Forward power is 200 watts (joules/sec)
Reflected power is 100 watts (joules/sec)
Load is absorbing 100 watts (joules/sec)

It can easily be shown that 300 joules of energy have been
generated that have not been delivered to the load, i.e.
those 300 joules of energy are stored in the feedline.
The 300 joules of energy are stored in RF waves which
cannot stand still and necessarily travel at the speed of
light. TV ghosting can be used to prove that the reflected
energy actually makes a round trip to the load and back.
A TDR will indicate the same thing.

Choosing to use a net energy shortcut doesn't negate the
laws of physics.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Roy Lewallen wrote:

(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

Sorry, I misquoted there. Walter Johnson doesn't say anything
about Poynting Vectors. The above is from: "Fields and Waves ..."
by Ramo, Whinnery, and Van Duzer, page 350, where they assert:

Pz-/Pz+ = |rho|^2

The reflected power Poynting Vector divided by the forward
power Poynting Vector equals the power reflection coefficient.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil,

I will presume that your reference to Walter Johnson is with regard to
his book, "Transmission Lines and Networks", published in 1950.

I have been unable to find any mention of Poynting Vectors or Power Flow
Vectors in my copy.

Would you be so kind as to identify the page number(s) describing these
concepts?

73,
Gene
W4SZ


Cecil Moore wrote:

[snip]

Roy, none of my textbook authors think the reflection model
is flawed. Walter Johnson goes so far as to assert that there
is a Poynting (Power Flow Vector) for forward power and a
separate Poynting Vector for reflected power. The sum of those
two Power Flow Vectors is the net Poynting Vector.

[snip]

Gene Fuller wrote:

Cecil,

I will presume that your reference to Walter Johnson is with regard to
his book, "Transmission Lines and Networks", published in 1950.

I have been unable to find any mention of Poynting Vectors or Power Flow
Vectors in my copy.

Would you be so kind as to identify the page number(s) describing these
concepts?

Gene,
My next posting admitted my senility. I was quoting Ramo & Whinnery,
not Walter Johnson. In "Fields and Waves in Communication Electronics",
page 325, an equation is given for Pz+, "The Poynting vector for the
positive traveling wave ...". It continues: "Similarly, the Poynting
vector for the negatively traveling wave is always in the negative 'z'
direction except when it is zero."

On page 350 it gives the ratio of the forward Poynting vector to the
rearward Poynting vector as the power reflection coefficient. Sorry
for my faulty memory.
--
73, Cecil http://www.qsl.net/w5dxp


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