On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.

--

Rick

On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


I also used Z = 20 -j130 ohms for the antenna.

The cable to ATU match has no effect on SWR nor does it change the
impedance seen at the transmitter end of the cable.

Does your online calculator include cable loss and velocity factor? I
used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for
the metric numbers, but that's what my chart uses.

On 9/30/2015 9:37 PM, John S wrote:
On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


I also used Z = 20 -j130 ohms for the antenna.

The cable to ATU match has no effect on SWR nor does it change the
impedance seen at the transmitter end of the cable.

Does your online calculator include cable loss and velocity factor? I
used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for
the metric numbers, but that's what my chart uses.

How do any of those things enter into the VSWR calculation? VSWR is
defined solely by the match of the cable and antenna impedances.

I don't know much about the calculator I used, some random online thing.

http://chemandy.com/calculators/retu...calculator.htm

--

Rick

On 9/30/2015 8:49 PM, rickman wrote:
On 9/30/2015 9:37 PM, John S wrote:
On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but
was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint
approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that
leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in
the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


I also used Z = 20 -j130 ohms for the antenna.

The cable to ATU match has no effect on SWR nor does it change the
impedance seen at the transmitter end of the cable.

Does your online calculator include cable loss and velocity factor? I
used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for
the metric numbers, but that's what my chart uses.

How do any of those things enter into the VSWR calculation? VSWR is
defined solely by the match of the cable and antenna impedances.

You are correct except that the SWR is higher at the antenna end than at
the opposite end due to cable attenuation although it might be
insignificant for short cables and low frequencies.

The velocity factor has no direct effect except it makes the physical
length longer than the electrical length which will increase the loss
over that which would be estimated if it is not included.

I don't know much about the calculator I used, some random online thing.

http://chemandy.com/calculators/retu...calculator.htm


I encourage you to get a Smith charting program and take some time to
get familiar with it. Very enlightening. I have one I recommend, but I
think I posted that one before when you were present.

http://www.fritz.dellsperger.net/

On 10/1/2015 4:24 AM, Jeff wrote:

Does your online calculator include cable loss and velocity factor? I
used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for
the metric numbers, but that's what my chart uses.

How do any of those things enter into the VSWR calculation? VSWR is
defined solely by the match of the cable and antenna impedances.

I don't know much about the calculator I used, some random online thing.

http://chemandy.com/calculators/retu...calculator.htm


Yes they do have an effect.

Cable loss will improve the VSWR as the cable gets longer and the loss
increases so the VSWR reduces.

You are referring to the VSWR at the ATU rather than at the antenna?
John already mentioned that.


The velocity factor determines how long the cable appears electrically
compared to the physical length.

The calculator that you linked to takes no account of the cable length
and what the mismatch, or vswr, at the load looks like at the other end
of a length of cable.

If the cable impedance matched the system impedance, 50 ohms in this
case, and it were totally lossless then the VSWR at the tx end of the
cable would be the same as at the antenna end, BUT the phase of the
mismatch would change, ie your -j120 could end up as being +j120 or
anywhere in between depending on cable length.

Now if the feeder is an impedance other than 50 ohms it act as a
transformer and both the real and imaginary parts of the impedance seen
at the Tx end will change. Depending on length and what impedance the
feeder is it may or may not improve things.

Jeff




--

Rick

On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


Wow! Is this an interesting discussion, or what?

"John S" wrote in message ...

On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


# Wow! Is this an interesting discussion, or what?

Well, I think it is The antenna system has a lossy coax between the
antenna feed and the ATU.

My understanding is that the SWR on the line is what it is, regardless of
the ATU settings.

In fact, I use an RF ammeter between the ATU and the antenna and tune for
maximum smoke. When I look at the SWR reading on the transmitter panel, it
seems to more or less coincide with max RF current.

I'll have to drop out of the discussion for a week or so....going off the
grid on vacation.

On 9/30/2015 8:58 PM, Wayne wrote:


"John S" wrote in message ...

On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote:


"Ian Jackson" wrote in message
...

In message , rickman
writes
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff
writes
On 29/09/2015 14:31, Jerry Stuckle wrote:
On 9/29/2015 4:40 AM, Jeff wrote:


So let's get back to the original question. Was it ever really
answered? I think it was made slightly more complicated by the fact
that the antenna feedpoint impedance was not purely resistive, but was
actually around 20-j130 (at 14 MHz),

Was there any advantage in having the coax paralleled (both for 20
ohms resistive, and for 20-j130)?

Expanding on the original question.... Antenna feedpoint approximately
20-j130
The ATU drives the antenna through about 15 feet of coax.
Assuming that the ATU provides a +j130 conjugate match, does that leave
the coax with a SWR of 50/20= 2.5:1?
If so, then I will not bother with considering 2 parallel coax.

The ATU will have no effect on the SWR in the 15ft of coax. The SWR is
determined entirely by the mismatch between the coax and your antenna.
The ATU simply transforms the impedance it sees to give the transmitter
the load it needs (50 + j0).

I will plead ignorance of ATUs, but wouldn't they also be designed to
match the impedance seen at the cable?


Using a Smith chart, I estimate that your load (antenna) will be
transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what
your ATU will see for a load. The chart also shows about 17:1 SWR in the
coax and there is nothing I can do at the transmitter to change it.

If I now change the coax to Zo of 25 ohms, the impedance at the
transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes
about 40:1.

It seems to me that you would only be hurting yourself.

Are you calculating the SWR based on the cable to ATU match or the cable
to antenna match? Using an online calculator I get 20:1 VSWR with a 50
ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your
numbers? I used Z = 20 -j130 ohms for the antenna.


# Wow! Is this an interesting discussion, or what?

Well, I think it is The antenna system has a lossy coax between the
antenna feed and the ATU.

In Wayne's case it is not a great loss until the SWR gets great.

My understanding is that the SWR on the line is what it is, regardless
of the ATU settings.

Yes, that is so.

In fact, I use an RF ammeter between the ATU and the antenna and tune
for maximum smoke. When I look at the SWR reading on the transmitter
panel, it seems to more or less coincide with max RF current.

That's good.

I'll have to drop out of the discussion for a week or so....going off
the grid on vacation.

I wish you a happy and rewarding vacation. Cheers.