wrote:

I recently built a modified vertical dipole
from wire and using a pole and trees as supports.
I had hoped it would work on 2 bands 75 and 40 but it only
works on one 75 where the VSWR is 1.3:1 and the bandwidth is 85 khz
for the 2:1 points.

On 40 meters the VSWR is over 10:1 so I thought I could use a 1/4 coax
matching section 50 ohms but this doesnt solve the problem either as the
VSWR doesnt change much with the extra coax section.

I did cut the section properly as the velocity factor for the coax is .78%
which gave me 26.825 ft.

Not sure what I am doing wrong or assuming - but looks like I could
use some input.

Try a half wavelength of coax.

ac6xg

Jim Kelley wrote:



wrote:

I recently built a modified vertical dipole
from wire and using a pole and trees as supports.
I had hoped it would work on 2 bands 75 and 40 but it only
works on one 75 where the VSWR is 1.3:1 and the bandwidth is 85 khz
for the 2:1 points.

On 40 meters the VSWR is over 10:1 so I thought I could use a 1/4 coax
matching section 50 ohms but this doesnt solve the problem either as the
VSWR doesnt change much with the extra coax section.

I did cut the section properly as the velocity factor for the coax is
.78% which gave me 26.825 ft.

Not sure what I am doing wrong or assuming - but looks like I could
use some input.

Try a half wavelength of coax.

ac6xg

why a half wave?

--
I SPILLED SPOT REMOVER ON MY DOG..............AND NOW HES GONE!!

wrote:
Not sure what I am doing wrong or assuming - but looks like I could
use some input.

If there was any way to get a low SWR on a simple
single-wire coax-fed dipole on 75m and 40m at the
same time, half of the ham population would have one.

A soon-to-be-published "Worldradio" article of mine
describes a fixed length dipole with a fixed feedline
length that works well on 40m and 75m without an antenna
tuner.
--
73, Cecil http://www.w5dxp.com

wrote in
newsoidnRiueftdl13YnZ2dnUVZ_rXinZ2d@hawaiiantel. net:

I recently built a modified vertical dipole
from wire and using a pole and trees as supports.
I had hoped it would work on 2 bands 75 and 40 but it only
works on one 75 where the VSWR is 1.3:1 and the bandwidth is 85 khz
for the 2:1 points.

On 40 meters the VSWR is over 10:1 so I thought I could use a 1/4 coax
matching section 50 ohms but this doesnt solve the problem either as
the VSWR doesnt change much with the extra coax section.

The feedpoint impedance of a full wave dipole is likely to be in the
region of 4000 ohms.

A quarter wave transformer needs to be of Zo (50*4000)^0.5 or 447 ohms.
You will not find a practical coax line with such Zo.

If you for instance used a quarter wave of RG58C/U for a transformer, the
input impedance would be 2 ohms (VSWR=25) (taking into account the 5+dB
of loss in the transformer.

Both the length and the Zo of a quarter wave transformer are important to
its operation.

Does this help to explain the results you observed?

Owen


I did cut the section properly as the velocity factor for the coax is
.78% which gave me 26.825 ft.

Not sure what I am doing wrong or assuming - but looks like I could
use some input.

Owen Duffy wrote:

wrote in
newsoidnRiueftdl13YnZ2dnUVZ_rXinZ2d@hawaiiantel. net:

I recently built a modified vertical dipole
from wire and using a pole and trees as supports.
I had hoped it would work on 2 bands 75 and 40 but it only
works on one 75 where the VSWR is 1.3:1 and the bandwidth is 85 khz
for the 2:1 points.

On 40 meters the VSWR is over 10:1 so I thought I could use a 1/4 coax
matching section 50 ohms but this doesnt solve the problem either as
the VSWR doesnt change much with the extra coax section.

The feedpoint impedance of a full wave dipole is likely to be in the
region of 4000 ohms.

its half wave



A quarter wave transformer needs to be of Zo (50*4000)^0.5 or 447 ohms.
You will not find a practical coax line with such Zo.

If you for instance used a quarter wave of RG58C/U for a transformer, the
input impedance would be 2 ohms (VSWR=25) (taking into account the 5+dB
of loss in the transformer.

Both the length and the Zo of a quarter wave transformer are important to
its operation.

Does this help to explain the results you observed?

maybe I should have asked does my coax tune my antenna???????


Owen


I did cut the section properly as the velocity factor for the coax is
.78% which gave me 26.825 ft.

Not sure what I am doing wrong or assuming - but looks like I could
use some input.

--
I SPILLED SPOT REMOVER ON MY DOG..............AND NOW HES GONE!!

wrote:
maybe I should have asked does my coax tune my antenna???????

When the load impedance is not 50 ohms and the
target impedance is 50 ohms, 50 ohm coax cannot
tune your antenna system to a 50 ohm match. In fact,
using 50 ohm coax with an SWR not equal to 1:1 is
one way of guaranteeing that the transmitter will
*never* see 50 ohms without some help from a
tuner or other matching method. (Extensive losses
will also bring the SWR closer to 1:1).

If the target impedance is 50 ohms, only a
characteristic impedance different from 50 ohms
can transform the non-50 ohm impedance to 50
ohms. And that can only be done when the SWR is
not 1:1. Note that in Owen's 1/4WL matching
equation, only if the load and source are both 50
ohms does the equation result in Z0 = 50 ohms.
Such a system is called a "flat" system with a
trivial default SWR of 1:1.

An autotuner at the feedpoint will solve the
problem or a parallel feedline with a Z0 of
300-600 ohms will solve the problem. I have
solved the problem at my QTH with a 148 foot
dipole and 90 feet of 450 ohm ladder-line.
That antenna system is simultaneously resonant
at my QTH on both 7.15 MHz and 3.8 MHz at the
same time. And it is also close to resonance on
17m.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

wrote:
maybe I should have asked does my coax tune my antenna???????

When the load impedance is not 50 ohms and the
target impedance is 50 ohms, 50 ohm coax cannot
tune your antenna system to a 50 ohm match. In fact,
using 50 ohm coax with an SWR not equal to 1:1 is
one way of guaranteeing that the transmitter will
*never* see 50 ohms without some help from a
tuner or other matching method. (Extensive losses
will also bring the SWR closer to 1:1).

If the target impedance is 50 ohms, only a
characteristic impedance different from 50 ohms
can transform the non-50 ohm impedance to 50
ohms. And that can only be done when the SWR is
not 1:1. Note that in Owen's 1/4WL matching
equation, only if the load and source are both 50
ohms does the equation result in Z0 = 50 ohms.
Such a system is called a "flat" system with a
trivial default SWR of 1:1.

An autotuner at the feedpoint will solve the
problem or a parallel feedline with a Z0 of
300-600 ohms will solve the problem. I have
solved the problem at my QTH with a 148 foot
dipole and 90 feet of 450 ohm ladder-line.
That antenna system is simultaneously resonant
at my QTH on both 7.15 MHz and 3.8 MHz at the
same time. And it is also close to resonance on
17m.

can this be done with one feedline?

--
I SPILLED SPOT REMOVER ON MY DOG..............AND NOW HES GONE!!

wrote:
I have
solved the problem at my QTH with a 148 foot
dipole and 90 feet of 450 ohm ladder-line.
That antenna system is simultaneously resonant
at my QTH on both 7.15 MHz and 3.8 MHz at the
same time. And it is also close to resonance on
17m.

can this be done with one feedline?

Yes, I have done it with one feedline and written
an article for "Worldradio" about it.
--
73, Cecil http://www.w5dxp.com

wrote in
:


its half wave


I would not have guessed that from the meagre information you supplied.

Owen