On 6/29/2015 3:47 PM, Wayne wrote:


"John S" wrote in message ...

On 6/29/2015 10:48 AM, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
metal roof. The antenna is fed with about 25 feet of RG-8, and there is
a tuner at the transmit end.

You use a 16ft vertical as a lead-in? For what and how is that done?

Grammatically, the description of the vertical is a lead in for the
question, not an actual antenna lead.


What are the dimensions of the metal roof?

Somewhat irrelevant to my question. But it's about 20 by 35 feet.
I'm not looking for an analysis of the existing antenna.


While I'm pretty happy with the antenna, I'd like to simplify the
matching.

To what matching do you refer? You don't want to use the tuner, or is
there some other stuff you have not mentioned?

I want the tuner matching to be less awkward on some bands.
I'm willing to live with the existing high SWRs on the upper bands.


Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
vertical? ( I assume the "4" side is on the antenna side.)

You wrote that you were interested in a 16ft vertical. Now it is a
43ft vertical?

Please disregard all about the 16 ft vertical. I'm asking about a 43 ft
vertical 1:4 unun.


I'd expect a better coax to antenna match when the antenna feedpoint is
a high Z (example, at 30 meters), but I'd also expect a worse coax to
antenna match when the feedpoint is a low Z (example, at 10 meters).

Is that the way it works, or is there other magic involved?

All this depends on your answers to the above questions.

So, lets begin again, with no distractions.

What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical.

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?

In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:


"John S" wrote in message ...

On 6/29/2015 10:48 AM, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
metal roof. The antenna is fed with about 25 feet of RG-8, and there is
a tuner at the transmit end.

You use a 16ft vertical as a lead-in? For what and how is that done?

Grammatically, the description of the vertical is a lead in for the
question, not an actual antenna lead.


What are the dimensions of the metal roof?

Somewhat irrelevant to my question. But it's about 20 by 35 feet.
I'm not looking for an analysis of the existing antenna.


While I'm pretty happy with the antenna, I'd like to simplify the
matching.

To what matching do you refer? You don't want to use the tuner, or is
there some other stuff you have not mentioned?

I want the tuner matching to be less awkward on some bands.
I'm willing to live with the existing high SWRs on the upper bands.


Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
vertical? ( I assume the "4" side is on the antenna side.)

You wrote that you were interested in a 16ft vertical. Now it is a
43ft vertical?

Please disregard all about the 16 ft vertical. I'm asking about a 43 ft
vertical 1:4 unun.


I'd expect a better coax to antenna match when the antenna feedpoint is
a high Z (example, at 30 meters), but I'd also expect a worse coax to
antenna match when the feedpoint is a low Z (example, at 10 meters).

Is that the way it works, or is there other magic involved?

All this depends on your answers to the above questions.

So, lets begin again, with no distractions.

What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical.

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?
--
Ian

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


"John S" wrote in message ...

On 6/29/2015 10:48 AM, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a
flat
metal roof. The antenna is fed with about 25 feet of RG-8, and
there is
a tuner at the transmit end.

You use a 16ft vertical as a lead-in? For what and how is that done?

Grammatically, the description of the vertical is a lead in for the
question, not an actual antenna lead.


What are the dimensions of the metal roof?

Somewhat irrelevant to my question. But it's about 20 by 35 feet.
I'm not looking for an analysis of the existing antenna.


While I'm pretty happy with the antenna, I'd like to simplify the
matching.

To what matching do you refer? You don't want to use the tuner, or is
there some other stuff you have not mentioned?

I want the tuner matching to be less awkward on some bands.
I'm willing to live with the existing high SWRs on the upper bands.


Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
vertical? ( I assume the "4" side is on the antenna side.)

You wrote that you were interested in a 16ft vertical. Now it is a
43ft vertical?

Please disregard all about the 16 ft vertical. I'm asking about a 43 ft
vertical 1:4 unun.


I'd expect a better coax to antenna match when the antenna feedpoint is
a high Z (example, at 30 meters), but I'd also expect a worse coax to
antenna match when the feedpoint is a low Z (example, at 10 meters).

Is that the way it works, or is there other magic involved?

All this depends on your answers to the above questions.

So, lets begin again, with no distractions.

What is the purpose (or benefit) of using a 1:4 unun on a 43 ft
vertical.

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.

Does this help?

"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)

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).

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?

--

Rick

"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.

# 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.

In message , Wayne
writes





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.

A length of coax, which has (say) at least 10dB loss at the frequency of
interest, can indeed make a superb SWR dummy load!
--
Ian

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.

Why will the reflections not have losses? Every load that is not an
infinite impedance will absorb some of the signal that would be
reflected. That applies to the transmitter output as well as the
antenna, no?

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?

--

Rick

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?

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?