On 7/2/2015 11:18 AM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?

No. As I said, one watt is applied at the source end. This is condition
defined by the example and has nothing to do with source matching. The
32% loss is due to transmission line loss. The mismatch at the load end
causes the high SWR which increases the line loss due to high current at
some point in the line as well as increased voltage at other point(s).

The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the
source. However, at an electrical quarter wave away from the antenna,
the impedance is about 2.8+j0 ohms. So that point is a relative hot spot
in the line.

If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

With a lossless transmission line and one watt applied to one end the
other end will have one watt available. The only place the power can go
is into the antenna. To put one watt into 1063 ohms will require .0306
amps and 32.6V at the feed point.

Does this make sense?

"Jerry Stuckle" wrote in message
...
On 7/2/2015 6:08 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...
It's a shame they've dumbed down the exams so much that you don't need
to know anything to hold a ticket any more. Back when I got mine, even
the General Class was tougher than the Extra Class today - and it was
administered by the FCC, with no public question pool or cheat sheets.
You had to actually know something other than just memorizing a few
answers.

And at the time, the Amateur Extra was harder than the First Class
Radiotelephone.


I had my First Phone at the age of 22 back in 1972. Passed the 2nd and
first the same day on the first try.

And back at you on trying to teach someone that will not learn. Seems that
several on here think you are wrong most of the time.

"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

# Transferred where? The match at the transmitter output only matches the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a
transmitter with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

# Your quoting style is very confusing. If you use with a space at the
# front of lines you are quoting it will show up the same as everyone
# else's quotes.

It's a problem with my newsreader not doing the proper job.



#Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case, with
a conjugate match on both ends, wouldn't there be maximum power transmission
regardless of the SWR?

......just a question I'm posing to the group.

With no line losses, and a conjugate match, is the SWR of any consequence?






A matched impedance does not mean no losses. It means the maximum
transfer of power. These are not at all the same thing.


# But I don't see anyone taking wavelength vs. feed line length into
# account. If the wavelength is long compared to the feed line I believe
# a lot of the "bad" stuff goes away. But then I am used to the digital
# transmission line where we aren't really concerned with delivering
# power, rather keeping a clean waveform of our (relatively) square waves.
# So I guess a short feed line doesn't solve the SWR problems... or does
# it?

The attenuation at a given high SWR depends upon the the matched
feedline loss, as reflections encounter that loss with every forward or
backward trip.
Thus feedline length/attenuation should be considered.

As a young man I was given a problem of solving poor antenna performance
on an aircraft band fixed station antenna. The SWR at the transmitter
was close to 1:1, but the antenna didn't work well.
I climbed up on the tower and found that the coax had never been
connected to the antenna. That was with about 400 feet of coax at 120
MHz.



# So how was the SWR 1:1?

It was a long run of coax at 120 MHz. The reflected wave was was attenuated
considerably by the time it returned to the source.

On 7/2/2015 8:53 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it
presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

# Transferred where? The match at the transmitter output only matches
the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a
transmitter with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

# Your quoting style is very confusing. If you use with a space at the
# front of lines you are quoting it will show up the same as everyone
# else's quotes.

It's a problem with my newsreader not doing the proper job.


#Why will the reflections not have losses?

Because the assumption I posed was for a lossless line. In that case,
with a conjugate match on both ends, wouldn't there be maximum power
transmission regardless of the SWR?

You aren't grasping the issue. Losses are *not* only in the
transmission line. When a reflected wave returns to the transmitter
output, it is not reflected 100%. If the output and transmission line
are matched exactly, 50% of the reflected wave reaching the output will
be reflected and 50% will be dissipated in the output stage.

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?

--

Rick

On 7/2/2015 7:37 PM, John S wrote:
On 7/2/2015 11:18 AM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?

No. As I said, one watt is applied at the source end. This is condition
defined by the example and has nothing to do with source matching. The
32% loss is due to transmission line loss. The mismatch at the load end
causes the high SWR which increases the line loss due to high current at
some point in the line as well as increased voltage at other point(s).

The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the
source. However, at an electrical quarter wave away from the antenna,
the impedance is about 2.8+j0 ohms. So that point is a relative hot spot
in the line.

Correction: 0.5+j0 at 1/4 wavelength away from the antenna.

If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

With a lossless transmission line and one watt applied to one end the
other end will have one watt available. The only place the power can go
is into the antenna. To put one watt into 1063 ohms will require .0306
amps and 32.6V at the feed point.

Does this make sense?

On 7/2/2015 8:51 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...
On 7/2/2015 6:08 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message
...
It's a shame they've dumbed down the exams so much that you don't need
to know anything to hold a ticket any more. Back when I got mine, even
the General Class was tougher than the Extra Class today - and it was
administered by the FCC, with no public question pool or cheat sheets.
You had to actually know something other than just memorizing a few
answers.

And at the time, the Amateur Extra was harder than the First Class
Radiotelephone.


I had my First Phone at the age of 22 back in 1972. Passed the 2nd and
first the same day on the first try.

And back at you on trying to teach someone that will not learn. Seems that
several on here think you are wrong most of the time.



I've got you there. First phone in 1970 at age 18 - before I started
college. Amateur I also passed Second and First the same day. My
Amateur Extra came 9 months later, but only because I had to hold a
General for two years before I could take the Extra exam.

Did you ever use your first phone? I was an engineer for one broadcast
station and chief engineer for another. I also repaired everything from
$40 cb sets to multi-million dollar mainframe computers. And even did
some digital design work back on the 70's.

But it's quite obvious from your updates that you have no idea what
you're talking about. You're an appliance operator with no
understanding of what's going on underneath the covers. And yes, I know
there are some idiotic trolls here who don't think I know what I'm
talking about. My EE professors would disagree with them.


--
==================
Remove the "x" from my email address
Jerry Stuckle

==================

On 7/2/2015 7:56 PM, Jeff Liebermann wrote:
On Thu, 02 Jul 2015 14:55:53 -0400, Jerry Stuckle
wrote:

It's easy enough to demonstrate that you're wrong.

1. Setup your favorite HF xmitter and attach a Tee connector to the
antenna connector, your favorite VSWR meter, a length of 50 ohm coax,
and a 50 ohm load.
2. Transmit and convince yourself that the VSWR is 1:1. Make sure
the transmitter is not into ALC.
3. Now, take another 50 ohm dummy load and connect it to the Tee
connector. The transmitter now sees 25 ohms, so the PA stage has half
the normal gain. You may need to increase the drive level to obtain
the same RF power as before.
4. Measure the VSWR again. It should also be 1:1.


Of course it will be. You have 50 ohms on one end. But you're not
measuring what the TRANSMITTER sees. The fact that you may have to
increase drive level indicates the circumstances have changed.

Looking back towards the transmitter, the 50 ohm coax cable sees a 2:1
mismatch of 25 ohms (the alleged 50 ohms from radio in parallel with
another 50 ohms from the extra dummy load). I know this part works
because I've demonstrated it twice to the local non-believers.


Sure. But you don't have power going from one resistor to the
transmitter and the other resister, so your measurement is meaningless -
and you are as full of crap as you normally are.

Stick your VSWR meter between the power source (the transmitter) and
BOTH loads (i.e. before the T). You will see a 2:1 SWR.


Now, we go into uncharted territory and do it again with a 75 ohm coax
and a 75 ohm dummy load at the far end (antenna end) of the coax. Same
procedure.
1. Check the VSWR and it should be 1.5:1.
2. Connect a 50 ohm dummy load to the Tee connector, and measure the
VSWR again. It should still be 1.5:1.

Looking back towards the transmitter, the 75 ohm coax cable sees the
same 2:1 mismatch of 25 ohms. If you want to go further, I think it
can be demonstrated that almost any number of extra dummy loads at the
Tee connector will still produce the same 1.5:1 VSWR.


Again, the dummy load is not producing any power, so adding something to
the other end of the T will have no effect. So once again your "test"
is meaningless and you are full of crap.

Now in this case if you connect the SWR bridge before the T, you will
show a 1.5:1 with one 75 ohm load, and a 1.3:1 SWR with two 75 ohm loads
(37.5 ohms).

SWR measurements are only valid when the VSWR meter is connected between
the power generator (transmitter) and the total load (one or both dummy
loads). Connecting between one leg of the T and the load only shows
VSWR for that leg - but not the entire system.

I'll try it on the bench, but I have other plans for the holiday
weekend. If I find time, and manage to get all the junk off my
workbench, I'll give try it.


Go ahead - continue to mae a fool of yourself. You're real good at it.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had the
math to back me up. The only total reflection I am aware of is an open
circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the reflected
wave by the matching circuit. I believe the example you gave was Z of
1063 -j0. Isn't that a real impedance of 1063 ohms which is equivalent
to a resistor? Resistors dissipate power don't they?

--

Rick

In message , rickman
writes
On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had
the math to back me up. The only total reflection I am aware of is an
open circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the
reflected wave by the matching circuit. I believe the example you gave
was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is
equivalent to a resistor? Resistors dissipate power don't they?

I have to admit that I am, to some extent, confused.

Maybe it helps to look at the situation from the point of view that the
matching circuit doesn't 'know' that there is a reflected wave. All it
sees is the impedance looking into the sending end of the coax - and
this is whatever is on the antenna end, transformed by the length of
coax. The load the matching unit sees could be replaced with the same
physical values of L, C and R, so there IS nowhere for a reflected wave
to exist.

Provided the TX sees a 50 ohm load when looking into the input of the
matcher, there will be no theoretical losses. However, a real-life
matcher WILL have loss, and so will the coax. Also, the coax will have a
loss greater than when it is matched, mainly because of the 'I -squared
R' (literal) hot-spots.
--
Ian

Ian Jackson wrote:

In message , rickman
writes
On 7/3/2015 2:50 AM, Jeff wrote:

Are you suggesting that the conjugate match will reflect back to the
antenna 100% of the original reflected wave from the antenna?


Yes, it must.

For example with an external ATU that provides a conjugate match it is
clearly the case that if a 1:1 VSWR is achieved then no reflected power
reaches the TX. (as shown on an SWR meter between the Tx and ATU.)

I am very certain that this assumption is not correct. I wish I had
the math to back me up. The only total reflection I am aware of is an
open circuit which of course absorbs no power at all.

Here is a point. The VSWR only shows no power being sent back to the
txmt output. That does not mean no power is absorbed from the
reflected wave by the matching circuit. I believe the example you gave
was Z of 1063 -j0. Isn't that a real impedance of 1063 ohms which is
equivalent to a resistor? Resistors dissipate power don't they?

I have to admit that I am, to some extent, confused.

Maybe it helps to look at the situation from the point of view that the
matching circuit doesn't 'know' that there is a reflected wave. All it
sees is the impedance looking into the sending end of the coax - and
this is whatever is on the antenna end, transformed by the length of
coax. The load the matching unit sees could be replaced with the same
physical values of L, C and R, so there IS nowhere for a reflected wave
to exist.

Provided the TX sees a 50 ohm load when looking into the input of the
matcher, there will be no theoretical losses. However, a real-life
matcher WILL have loss, and so will the coax. Also, the coax will have a
loss greater than when it is matched, mainly because of the 'I -squared
R' (literal) hot-spots.

Surely it *is* the reflected wave that mediates the transformation of
the aerial impedance to what is seen at the transmitter end? The
transmitter sees the vector sum of all the waves traversing the
transmission line at that point. Or else how would it "know" what was
happening at the other end?

--
Roger Hayter