An antenna question--43 ft vertical
 

Jeff wrote:

On 03/07/2015 08:29, rickman wrote:
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?


The point can be easily proved with a lossy feeder, the lossier the
better. If your assumption is correct then the power delivered to the
antenna would be the Tx power less the cable loss less the reflected
power at the antenna mismatch, however it is the case that can be
measured and seen on computer simulation that all of the power is
delivered to the antenna except that which is dissipated in the cable
loss (for the multiple reflections). Also it can been seen that with
perfect components in the matching circuit no power is dissipated there.

The 1063 ohms that you refer to is not resistive so with perfect Ls & Cs
no power will be dissipated in it.

In the real world the Cs and Ls in the matching unit will have some loss
associated with them but that is a different story.

Jeff


While conjugate matching is the way to transfer the maximum power from a
voltage (or current) generator to a load, it is not the way power
amplifiers are set up. The transmitter normally does not present a
match to signals from the aerial, hence the re-reflection.


--
Roger Hayter

An antenna question--43 ft vertical
 

In message , Roger Hayter
writes
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?

I guess that until reflections are received back from the far end of the
coax, the transmitter will see the 50 ohms Zo (surge impedance) of the
coax. But once things have settled worn, the transmitter neither knows
nor cares what's at the far end. All it knows is that the load presented
to it isn't what it ought to be. But insert a matcher, and it will be as
happy as Larry. The system will work fine, but will suffer the penalty
of the additional SWR losses on the coax, and those of the matcher.
Provide these are not unacceptable, the benefit is that all the matching
can be done in the comfort of shack.
--
Ian

An antenna question--43 ft vertical
 

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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic involved.
A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at
30 meters. Using a 1:4 transformer at the feed point will reduce that to
253 + J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

An antenna question--43 ft vertical
 

In message , John S
writes
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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic
involved. A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at
30 meters. Using a 1:4 transformer at the feed point will reduce that
to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

A fixed-tuned TX will still need a matcher.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

A fixed-tuned TX will probably be reasonably happy with a direct
connection - although maybe even happier with a series capacitor of -J22
ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

The question is really whether the losses with the 4:1 transformer, plus
those of any matcher at the TX end, exceed those when there is no
transformer (but with higher loss on the coax), plus a matcher. Put
another way, for short feeder lengths, is it better to use the
transformer?
--
Ian

An antenna question--43 ft vertical
 

In message , Ian Jackson
writes


although maybe even happier with a series capacitor of -J22 ohms.

Sorry - somebody obviously swapped the '2' and '3' keys.


--
Ian

An antenna question--43 ft vertical
 

"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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic involved. A
bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms at 30
meters. Using a 1:4 transformer at the feed point will reduce that to 253 +
J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A 1:4
transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you understand
that the answers cannot be exact with the info presented.

I hope this helps.

Thanks John.
Yes, we have strayed from the original question, but I have found the
discussion stimulating.
Perhaps a new thread should be started to address those subjects.

If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
diameter radiator, I get impedances in the same ball park as you list.

If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna would
see as a feedline, if a 4:1 unun had 50 ohm coax on the other side), the SWR
plot becomes interesting.

The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
getting below 2.5:1 around 29 MHz.

Is that a valid approach?

An antenna question--43 ft vertical
 

On 7/3/2015 10:17 AM, Ian Jackson wrote:
In message , John S
writes
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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic
involved. A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms
at 30 meters. Using a 1:4 transformer at the feed point will reduce
that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

A fixed-tuned TX will still need a matcher.


That was not part of the original question(s).



At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

A fixed-tuned TX will probably be reasonably happy with a direct
connection - although maybe even happier with a series capacitor of -J22
ohms.

That was not part of the original question(s).

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

The question is really whether the losses with the 4:1 transformer, plus
those of any matcher at the TX end, exceed those when there is no
transformer (but with higher loss on the coax), plus a matcher. Put
another way, for short feeder lengths, is it better to use the transformer?

That was not the question he asked. Please re-read the OP. I was trying
to address his original question(s) as best as I could. In addition I
also said that there were "several disclaimers I could include" which
may involve your personal concerns. I did not want to muddy the waters.

I think I answered Wayne's question(s), but I will wait to hear from him
to see if that is so.

An antenna question--43 ft vertical
 

On 7/3/2015 10:37 AM, 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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic
involved. A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms
at 30 meters. Using a 1:4 transformer at the feed point will reduce
that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

Thanks John.
Yes, we have strayed from the original question, but I have found the
discussion stimulating.
Perhaps a new thread should be started to address those subjects.

If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
diameter radiator, I get impedances in the same ball park as you list.

If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna
would see as a feedline, if a 4:1 unun had 50 ohm coax on the other
side), the SWR plot becomes interesting.

The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
getting below 2.5:1 around 29 MHz.

Is that a valid approach?

I have not done what you have done, but it sounds correct. I'll try to
verify what you have done when time permits.

I really think you know what you are doing. Don't forget that EZNEC can
use transmission lines, transformers, inductors, capacitors, resistors
and other stuff to help in your analysis. Although the true answers come
from the physical implementation, it is very helpful to use EZNEC to
gain insight into the situation. And, I think you know that as well.

An antenna question--43 ft vertical
 

On 7/3/2015 10:37 AM, 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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic
involved. A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms
at 30 meters. Using a 1:4 transformer at the feed point will reduce
that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

Thanks John.
Yes, we have strayed from the original question, but I have found the
discussion stimulating.


Indeed! So have I.


Perhaps a new thread should be started to address those subjects.


Please start one if you feel compelled.


If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
diameter radiator, I get impedances in the same ball park as you list.


Ha! I used 1.5 inches. I will re-do.


If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna
would see as a feedline, if a 4:1 unun had 50 ohm coax on the other
side), the SWR plot becomes interesting.


I've never done that. I will explore this set-up.


The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
getting below 2.5:1 around 29 MHz.

Are we still considering a 10MHz to 30Mhz frequency sweep?

Is that a valid approach?

You might be ahead of me on this.

An antenna question--43 ft vertical
 

"John S" wrote in message ...

On 7/3/2015 10:37 AM, 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.

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

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.)

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?

I think we strayed off the path to answering your original question.

The short answer is that you are correct and there is no magic
involved. A bit longer answer is:

A 43ft vertical will present a feed impedance of 1010 + J 269.2 ohms
at 30 meters. Using a 1:4 transformer at the feed point will reduce
that to 253 + J 67 ohms. That is a bit closer to your 50 ohm line.

At 10 meters, the antenna will present a 147 + J 133 ohms impedance. A
1:4 transformer will reduce that to 37 + J 33 ohms.

There are several disclaimers I could include, but I think you
understand that the answers cannot be exact with the info presented.

I hope this helps.

Thanks John.
Yes, we have strayed from the original question, but I have found the
discussion stimulating.


Indeed! So have I.


Perhaps a new thread should be started to address those subjects.


Please start one if you feel compelled.


If I use EZNEC to model the 43 footer over perfect ground with a 3 inch
diameter radiator, I get impedances in the same ball park as you list.


Ha! I used 1.5 inches. I will re-do.


If I change the "alt SWR Z0" to 200 ohms (presumably what the antenna
would see as a feedline, if a 4:1 unun had 50 ohm coax on the other
side), the SWR plot becomes interesting.


I've never done that. I will explore this set-up.


The plot has SWRs of about 2.5:1 to 5:1 over most of the range, with SWR
getting below 2.5:1 around 29 MHz.

Are we still considering a 10MHz to 30Mhz frequency sweep?

Well, I have been running the SWR across 4 to 30 MHz, but mainly looking at
10 MHz and above.

As for EZNEC and transmission lines, I have never done that, but plan to
when I can. I don't follow how to do it. In the few cases I wanted the
info for a single frequency, I just used a Smith chart.

This thread has given me a lot to consider in improving my whip setup, but
details of the possibilities would run the thread off in the weeds :)