Question on SWR
 

I make here reference to the well-known chart, shown in almost all the ARRL
Antenna Handbooks, setting the relationship among the real SWR (i.e. that
measured at antenna), the SWR measured at transmitter and the "line loss".

In the text they explain that the "line loss" to be considered when using the
chart is the line loss under matched conditions (i.e. that given by the line
manufacturer), that is without taking into account the extra loss caused by SWR.

They make the example of a line having a (matched) loss of 1.0dB with an SWR of
4.5 at antenna. The graph shows that the corresponding SWR at transmitter is
3.0.

Unless I am wrong, a simple calculation shows that, in the considered example,
the SWR at transmitter is about 2.3 rather than 3.0.

Here it goes (please note that, for the assumed SWR at antenna of 4.5, the extra
loss caused by SWR is just 1.0dB):

- actual loss on the forward wave: 1.0dB (matched) + 1.0dB (extra by SWR), for a
total of 2.0dB
- return loss corresponding to an SWR at antenna of 4.5: 4.0dB
- actual loss on the reflected wave: 1.0dB (matched) + 1.0dB (extra by SWR), for
a total of 2.0dB
- return loss at transmitter: 8.0dB
- SWR measured at transmitter (corresponding to a return loss of 8.0dB): about
2.3

It is interesting to note that the chart would give an identical result if by
"line loss" they would mean the total line loss (that is also including the
extra loss due to SWR) rather than just the matched line loss (as they state in
the text).

Any comment?

Thanks and 73

Tony I0JX

Question on SWR
 

I realized my error!

In my budget I counted the extra line loss caused by SWR twice:

- first time when I have added 1 dB to the loss of the forward and the return
wave

- second time when I took some power out of the antenna (to account for an SWR
of 4.5) and delivered it back to the transmitter

ARRL is always correct!

Sorry for the useless noise

Tony I0JX

Question on SWR
 

"Antonio Vernucci" wrote in message
...
I realized my error!

In my budget I counted the extra line loss caused by SWR twice:

- first time when I have added 1 dB to the loss of the forward and the
return wave

- second time when I took some power out of the antenna (to account for an
SWR of 4.5) and delivered it back to the transmitter

ARRL is always correct!

All Hail the ARRL!

Question on SWR
 

"Antonio Vernucci" wrote in
:

I realized my error!
....
ARRL is always correct!

The ARRL information on "extra loss due to VSWR" is may be incomplete in
that it may not the assumptions that underly the formula used for the
graphs.

The very concept that SWR necessarily increases loss from the matched line
loss figure is flawed.

Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to
calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms
(both VSWR=10).

Now refer to the ARRL... does it explain the difference?

Owen

Question on SWR
 

Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete in
that it may not the assumptions that underly the formula used for the
graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

Question on SWR
 

"Owen Duffy" wrote in message
...
"Antonio Vernucci" wrote in
:

I realized my error!
...
ARRL is always correct!

The ARRL information on "extra loss due to VSWR" is may be incomplete in
that it may not the assumptions that underly the formula used for the
graphs.

The very concept that SWR necessarily increases loss from the matched line
loss figure is flawed.

Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to
calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms
(both VSWR=10).

Now refer to the ARRL... does it explain the difference?

Owen

yeah, when you use the full complex Z0 and probably the full transmission
line equations it gets a bit more complex. but for amateur use that graph
is close enough. the difference between 5 and 500 ohm loads of .07db or so
for 100m just ain't worth quibbling about for normal amateur hf use. unless
you want to argue it out with cecil.

Question on SWR
 

Dave wrote:
... unless you want to argue it out with cecil.

Sorry, I have no argument with Owen. I do have
an argument with people who say replacing the
RG-213 on a G5RV with 300 ohm twinlead all the
way to the tuner will reduce losses on 80m -
tain't so.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

Question on SWR
 

On Fri, 17 Oct 2008 13:36:04 GMT, "Dave" wrote:

Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to
calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms
(both VSWR=10).

Now refer to the ARRL... does it explain the difference?

Owen

yeah, when you use the full complex Z0 and probably the full transmission
line equations it gets a bit more complex. but for amateur use that graph
is close enough. the difference between 5 and 500 ohm loads of .07db or so
for 100m just ain't worth quibbling about for normal amateur hf use. unless
you want to argue it out with cecil.

To catch Owen's drift, you really need to perform his suggested
exercise.

73's
Richard Clark, KB7QHC

Question on SWR
 

"Dave" wrote in
:

....
yeah, when you use the full complex Z0 and probably the full
transmission line equations it gets a bit more complex. but for
amateur use that graph is close enough. the difference between 5 and
500 ohm loads of .07db or so for 100m just ain't worth quibbling about
for normal amateur hf use. unless you want to argue it out with
cecil.

You either misread my example (it was 1m not 100m) or you labour under
the misapprehension that loss per unit length under mismatched conditions
is constant at all displacements along the cable.

When approximations that hold under some conditions replace the
underlying principles, we dumb down. The formula and graphs for
"additional loss due to VSWR" without statement of the assumptions under
which it is valid are an example... where now, so many people accept the
concept that VSWR necessarily increases loss.

The OP was trying to reconcile calculated losses in a particular
scenario, and one of the contributions to error was the "additional loss
due to VSWR". It is fine with me that understanding doesn't appeal to you
Dave, but you don't need to press that approach on the rest of us.

Owen

(PS: if we take a length of 50 ohm coax sufficiently short that current
distribution is approximately uniform, and consider the losses under
matched conditions and under a 500 ohm load with same load power where
voltage is three times and current is one third, it is intuitive that
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G), that loss will be LESS (than Matched Line Loss)...
approximately one tenth in that case... so why swallow the ROT that high
VSWR necessarily increases loss.)

Question on SWR
 

since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)

A quick question. If most of the the cable loss is due to I^2R, how can one
explain that the foam versions of common coaxial cables show a much lower loss
than versions having solid PE insulation?

For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz, while
RG-213 foam at only 4.5 dB. If G is relatively unimportant with regard to loss,
how can one explain that a change of insulation material yields such a
tremendous change in loss?

Thanks and 73

Tiony I0JX

Question on SWR
 

In message , Antonio
Vernucci writes
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)

A quick question. If most of the the cable loss is due to I^2R, how can
one explain that the foam versions of common coaxial cables show a much
lower loss than versions having solid PE insulation?

For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz,
while RG-213 foam at only 4.5 dB. If G is relatively unimportant with
regard to loss, how can one explain that a change of insulation
material yields such a tremendous change in loss?

Thanks and 73

Tiony I0JX

Lower k dielectric larger diameter inner conductor lower resistance
lower loss.
--
Ian

Question on SWR
 

"Antonio Vernucci" wrote in
:

since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)

A quick question. If most of the the cable loss is due to I^2R, how
can one explain that the foam versions of common coaxial cables show a
much lower loss than versions having solid PE insulation?

If you construct a cable of similar outside dimensions but using a foam
dielectric, it needs a larger diameter inner conductor. That accounts for
the lower loss at lower frequencies (typically below the GHz range.)


For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz,
while RG-213 foam at only 4.5 dB. If G is relatively unimportant with
regard to loss, how can one explain that a change of insulation
material yields such a tremendous change in loss?

See above.

If you use my calculator (link in earlier posting), it gives you the
coefficients of two terms of the loss model, one is due to I^2R and the
other V^2G. You can evaluate them at any given frequency and determine
the contribution that conductor and dielectric losses make at that
frequency for that cable type.

Does that help?

Owen


Thanks and 73

Tiony I0JX

Question on SWR
 

Antonio Vernucci wrote:
For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz,
while RG-213 foam at only 4.5 dB. If G is relatively unimportant with
regard to loss, how can one explain that a change of insulation material
yields such a tremendous change in loss?

Those statements about most loss being due to I^2*R
losses are at *HF* frequencies. 144 MHz is NOT HF.
The difference in RG-213 and RG-213 foam is only 0.2
dB at 10 MHz while the difference between RG-58 and
RG-213 is about 0.7 dB at 10 MHz.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

Question on SWR
 

"Owen Duffy" wrote in message
...
"Dave" wrote in
:

...
yeah, when you use the full complex Z0 and probably the full
transmission line equations it gets a bit more complex. but for
amateur use that graph is close enough. the difference between 5 and
500 ohm loads of .07db or so for 100m just ain't worth quibbling about
for normal amateur hf use. unless you want to argue it out with
cecil.

You either misread my example (it was 1m not 100m) or you labour under
the misapprehension that loss per unit length under mismatched conditions
is constant at all displacements along the cable.

When approximations that hold under some conditions replace the
underlying principles, we dumb down. The formula and graphs for
"additional loss due to VSWR" without statement of the assumptions under
which it is valid are an example... where now, so many people accept the
concept that VSWR necessarily increases loss.

The OP was trying to reconcile calculated losses in a particular
scenario, and one of the contributions to error was the "additional loss
due to VSWR". It is fine with me that understanding doesn't appeal to you
Dave, but you don't need to press that approach on the rest of us.

Owen

(PS: if we take a length of 50 ohm coax sufficiently short that current
distribution is approximately uniform, and consider the losses under
matched conditions and under a 500 ohm load with same load power where
voltage is three times and current is one third, it is intuitive that
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G), that loss will be LESS (than Matched Line Loss)...
approximately one tenth in that case... so why swallow the ROT that high
VSWR necessarily increases loss.)

why should i swallow your rot that shows when you push the limits of the
equations you get results that 'seem' to defy logic. I understand perfectly
well what you are saying, and i do understand the full complex forms of the
transmission line equations. what i am saying is that for most 'normal'
amateur use the graph presented in the arrl book is adequate. if you insist
on pushing computerized calculations to the absurd limits i'll lump you in
with art and his over optimized ball of wire antenna.

Question on SWR
 

Antonio Vernucci wrote:
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)

A quick question. If most of the the cable loss is due to I^2R, how can
one explain that the foam versions of common coaxial cables show a much
lower loss than versions having solid PE insulation?

For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz,
while RG-213 foam at only 4.5 dB. If G is relatively unimportant with
regard to loss, how can one explain that a change of insulation
material yields such a tremendous change in loss?


In reasonably well constructed coax cables, the main source of loss up
to about 1GHz is the I^2R loss in the centre conductor. The inside of
the shield carries an equal (and opposite) current, but the current
density is lower so the I^2R loss there is less important. Dielectric
loss is usually less important still.

In low-loss cables that have the same outside diameter as the classic PE
cables they are replacing, the reduction in loss is almost entirely due
to a larger centre conductor. But that change cannot be made on its own.
In order to maintain a 50 ohm impedance and keep the same outside
diameter too, it is necessary to reduce the dielectric constant of the
insulation material.

In other words, they're using foam or semi-airspaced construction
because they *have* to. Replacing some of the solid PE with gas may make
a small contribution to the lower losses, but nowhere near so much as
the advertisers would have you believe. The main contributor is always
the reduced I^2R loss in a larger centre conductor.



--

73 from Ian GM3SEK

Question on SWR
 

Owen Duffy wrote in
:

....
If you use my calculator (link in earlier posting), it gives you the
coefficients of two terms of the loss model, one is due to I^2R and
the other V^2G. You can evaluate them at any given frequency and
determine the contribution that conductor and dielectric losses make
at that frequency for that cable type.

Lest someone confuses this with an incorrect calculation or estimate:

From TLLC, the matched line loss in dB of LMR400 (a foam coax of similar
OD to RG213) is 3.941e-6*f^0.5+1.031e-11*f. The first term is due to R
and the second due to G.

At 144MHz, the percentage of power lost per meter due to R is (1-10^-
(3.941e-6*f^0.5)/10)*100 is 1.08%. If you do similar for G, the loss is
0.034%, so loss in R is more than 30 times loss in G.

The numbers lead to a better understanding.

Does this make sense? Did I get it correct?

Owen

(BTW for RG213 at 144MHz, the percentage of power lost per meter due to R
is more than 6 times loss in G. Most of the loss advantage of LMR400
comes from reduction of the R loss component per metre from 1.6% to 1.1%
due to the larger diameter inner conductor.)


Owen

Question on SWR
 

Antonio Vernucci wrote:
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)

A quick question. If most of the the cable loss is due to I^2R, how can
one explain that the foam versions of common coaxial cables show a much
lower loss than versions having solid PE insulation?

For instance RG-213 is rated at 8.5dB loss for 100 meters at 144 MHz,
while RG-213 foam at only 4.5 dB. If G is relatively unimportant with
regard to loss, how can one explain that a change of insulation material
yields such a tremendous change in loss?

Thanks and 73

Tiony I0JX

144 MHz isn't HF, which is where the original statement is valid. At
frequencies above around 50 MHz, depending on the dielectric, the
dielectric loss starts to be more significant.

Another trap for the unwary, when comparing coax losses, has to do with
skin effect and the thickness of the copper or silver cladding on the
center conductor. You could have an air insulated coax with silver
plated over stainless steel where the loss is actually greater at low
frequencies than higher, because the skin depth is greater at low
frequencies and the current is flowing mostly in the SS, rather than the
copper. (such coax is used in cryogenic applications, lest one think
it's overly contrived as an example)

Question on SWR
 

In reasonably well constructed coax cables, the main source of loss up
to about 1GHz is the I^2R loss in the centre conductor. The inside of the
shield carries an equal (and opposite) current, but the current density is
lower so the I^2R loss there is less important. Dielectric loss is usually
less important still.

Ian and others,

thanks for your clear explanation, but I still have a doubt that you may kindly
clarify.

The 300-ohm TV flat ribbon specifications show an attenuation generally lower
than that of plain RG-8, despite the conductors of the ribbon are by far thinner
than those of RG-8 (especially than the cable shield).

What am I missing now?

Thanks & 73

Tony I0JX

Question on SWR
 

"Antonio Vernucci" wrote in
:

The 300-ohm TV flat ribbon specifications show an attenuation
generally lower than that of plain RG-8, despite the conductors of the
ribbon are by far thinner than those of RG-8 (especially than the
cable shield).

Under matched line conditions, the 300 ohm line transfers the power at
higher voltage and lower current, one sixth of the current, so even though
the conductors might seem relatively thin (RF R is proportional to diameter
for wide spaced line), I^2R loss is lower.

Owen

Question on SWR
 

Antonio Vernucci wrote:

In reasonably well constructed coax cables, the main source of loss up
to about 1GHz is the I^2R loss in the centre conductor. The inside of
the shield carries an equal (and opposite) current, but the current
density is lower so the I^2R loss there is less important. Dielectric
loss is usually less important still.

Ian and others,

thanks for your clear explanation, but I still have a doubt that you may
kindly clarify.

The 300-ohm TV flat ribbon specifications show an attenuation generally
lower than that of plain RG-8, despite the conductors of the ribbon are
by far thinner than those of RG-8 (especially than the cable shield).

What am I missing now?

300 ohms vs 50 ohms.

Since IR losses dominate at these frequencies, reducing current reduces
loss. The higher Z means more voltage and less current for the same
power. Loss will be 1/36th, assuming all the conductor sizes are the same.

Question on SWR
 

Owen Duffy wrote in
:

....
though the conductors might seem relatively thin (RF R is proportional
to diameter for wide spaced line), I^2R loss is lower.

Of course, that should read "RF R is inversely proportional to diameter for
wide spaced line"

Owen

Question on SWR
 

Jim Lux wrote:

144 MHz isn't HF, which is where the original statement is valid. At
frequencies above around 50 MHz, depending on the dielectric, the
dielectric loss starts to be more significant.


The analyses I've done using published and realistic values for
dielectric (PE and PTFE) and conductor loss indicate that dielectric
loss doesn't become significant until the 1 - 10 GHz region, well above VHF.

Another trap for the unwary, when comparing coax losses, has to do with
skin effect and the thickness of the copper or silver cladding on the
center conductor. You could have an air insulated coax with silver
plated over stainless steel where the loss is actually greater at low
frequencies than higher, because the skin depth is greater at low
frequencies and the current is flowing mostly in the SS, rather than the
copper. (such coax is used in cryogenic applications, lest one think
it's overly contrived as an example)

It's unusual to find a plating thickness that's less than several skin
depths thick except at MF and below, or perhaps the low end of HF. It
does happen, though. I have some 0.1 inch diameter 75 ohm cable which
has a very small center conductor which is made of several strands of
even smaller Copperweld wire (thick copper over steel). Even though the
copper is probably a sizable fraction of the total wire diameter, the
wire is so small in diameter that the copper isn't several skin depths
thick at lower HF. It has very noticeably excessive loss at 7 MHz,
something I discovered the hard way one Field Day.

Roy Lewallen, W7EL

Question on SWR
 

In message , Jim Lux
writes
Antonio Vernucci wrote:
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)
A quick question. If most of the the cable loss is due to I^2R, how
can one explain that the foam versions of common coaxial cables show a
much lower loss than versions having solid PE insulation?
For instance RG-213 is rated at 8.5dB loss for 100 meters at 144
MHz, while RG-213 foam at only 4.5 dB. If G is relatively unimportant
with regard to loss, how can one explain that a change of insulation
material yields such a tremendous change in loss?
Thanks and 73
Tiony I0JX

144 MHz isn't HF, which is where the original statement is valid. At
frequencies above around 50 MHz, depending on the dielectric, the
dielectric loss starts to be more significant.

Another trap for the unwary, when comparing coax losses, has to do with
skin effect and the thickness of the copper or silver cladding on the
center conductor. You could have an air insulated coax with silver
plated over stainless steel where the loss is actually greater at low
frequencies than higher, because the skin depth is greater at low
frequencies and the current is flowing mostly in the SS, rather than
the copper. (such coax is used in cryogenic applications, lest one
think it's overly contrived as an example)

Indeed. If you compare a frequency response of a reel of coax with a
plated centre conductor (say, copper on steel) with one with a solid
inner conductor, the former often has a noticeable kink at around 40 or
50MHz. A long time ago, we found this out at work when trying to find
drop cables and miniature cables which we could be used as simulations
of large-diameter, low-loss CATV trunk cables (for use in the design
lab).

At least one major CATV supplier had lots of very large reels of various
lengths the 'real thing' (no pun intended) in a massive trailer parked
immediately on the other side of lab wall. These were patched through
the wall to the test benches. The engineer could then test and adjust
the flatness of the frequency response of wideband amplifiers against
the length of appropriate cable. Thankfully, this technology has largely
been superseded by optical fibre/fiber equipment!
--
Ian

Question on SWR
 

In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete
in that it may not the assumptions that underly the formula used for
the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.
--
Ian.

Question on SWR
 

I'd never thought of that. I suppose it applies to any situation where the
feeder is electrically short, and the majority of the current is less than it
would be when matched. I presume that the moral is that formulas only really
work when the feeder is electrically long enough for you to be concerned about
what the losses might be.
--

Conversely, for a very short line closed on 5 ohm (instead of 500 ohm), the
extra loss caused by SWR would be higher than that shown on the ARRL graph
(apart from the fact that, when attenuation is so low, the extra attenuation is
generally not of much interest, nor it can be read on the ARRL chart).

Evidently the ARRL chart shows some average between the two cases. On the other
hand they probably had no better way to synthetically illustrate a concept
without giving too many details.

73

Tony I0JX

Question on SWR
 

The 300-ohm TV flat ribbon specifications show an attenuation
generally lower than that of plain RG-8, despite the conductors of the
ribbon are by far thinner than those of RG-8 (especially than the
cable shield).

Under matched line conditions, the 300 ohm line transfers the power at
higher voltage and lower current, one sixth of the current, so even though
the conductors might seem relatively thin (RF R is proportional to diameter
for wide spaced line), I^2R loss is lower.

Thanks, but shouldn't the current ratio be the square root of 6 instead of 6?

73

Tony I0JX

Question on SWR
 

Since IR losses dominate at these frequencies, reducing current reduces loss.
The higher Z means more voltage and less current for the same power. Loss
will be 1/36th, assuming all the conductor sizes are the same.

On another thread I was arguing that, if I am correct, the current ratio should
be the square root of 6, not 6. Accordingly the loss would be 1/6, instead of
1/36.

73

Tony I0JX

Question on SWR
 

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

Owen's explicit method should get it right in all cases. If you select
say 0.125 wavelengths of RG213 in Owen's online calculator, the
predicted loss with a 100 ohm load resistance is *less* than the matched
loss. If you change the load to 25 ohms, the predicted loss is *greater*
than the matched loss.

Both of these results make perfect physical sense because the largest
part of the loss is proportional to the square of the current, which
will be greater with the lower-resistance load. The two different
resistances correctly give different results, yet they both have a VSWR
of 2 (based on the 50-ohm system impedance). This shows that VSWR does
not contain sufficient information to give an explicit single-valued
result.

Owen's program will accept a VSWR input, but it correctly posts a bold
red warning that the result is an estimate. If you let the program
select the worst-case load impedance for the supplied value of VSWR,
you're back on track and it can calculate an explicit result.


Although we're debating fractions of a milliBel here, the debate has
shown how often the terms "VSWR" and "load impedance" are used
interchangeably - which they aren't. It isn't a big mistake here, but it
can be in other applications. For example, a solid-state PA designed
for a 50 ohm load will respond very differently to load *impedances* of
100 or 25 ohms, yet the load *VSWR* is the same in both cases.



--

73 from Ian GM3SEK

Question on SWR
 

Ian Jackson wrote:
I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

I suspect the ARRL additional loss due to SWR charts
are based on 1/2WL increments of feedlines.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

Question on SWR
 

"Antonio Vernucci" wrote in
:

....
Under matched line conditions, the 300 ohm line transfers the power
at higher voltage and lower current, one sixth of the current, so
even though the conductors might seem relatively thin (RF R is
proportional to diameter for wide spaced line), I^2R loss is lower.

Thanks, but shouldn't the current ratio be the square root of 6
instead of 6?

Yes, of course... my mistake.

Owen

Question on SWR
 

Ian White GM3SEK wrote in
:

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

I think the ARRL graph is based on a well known, but apparently not well
understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

On terminology, I prefer to not use the term 'extra loss due to VSWR'
because the name implies to many, that it is always positive. IMHO a
better way to speak of the loss is as line loss under mismatched
conditions... and those conditions are more specific than just a VSWR
figure.

In a lot of cases, the approximation is sufficiently accurate... but you
lose visibility of the error when you assume that the approximation is
ALWAYS sufficiently accurate, a Rule of Thumb or ROT for short.

Owen

Question on SWR
 

Owen Duffy wrote in
:

....
I think the ARRL graph is based on a well known, but apparently not
well understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

The formula is developed by integrating I^2 over an electrical half wave
of line on one side or the other from the observation point, according
to PS. (I don't know the origin of the formula, I am not suggesting that
PS invented it, not that he didn't.)

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave. A requirement for this is that line loss
must be relatively low, that the exponential real term in the
transmission line equations is close to zero.

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

For a line that is many half waves, the formula is fine so long as VSWR
is approximately constant (now a very low loss line). If the line is
longer than many half waves, but not an exact integral number of half
waves, then the error in the partial section will be somewhat diminished
relatively by the loss in the complete half wave sections.

If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

So, in summary, the formula is good for low loss half wave lines, or
even longish random length low loss lines, but not good for short random
length lines or very long lines.

So, why is the formula so popular?

Could it be that it underpins one of the popular myths of ham radio,
that VSWR necessarily increases line loss?

Modern computation tools are better than the 70 year old graphical
method. Publication of the formula without qualification with the
underlying assumptions treats the reader as a dummy.

Owen

Question on SWR
 

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

Question on SWR
 

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

Question on SWR
 

"Antonio Vernucci" wrote in
:

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the
current and voltage maxima and minima and the relative contribution
of R and G to loss. So, the formula may have significant error for
short lines that are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same
at both ends for it to not matter which end is the observation point,
so therefore the first assumption is that VSWR is approximately equal
at both ends of the half wave.
If a practical line is very long, it cannot qualify as having a
constant VSWR (unless it is 1, in which case the formula is
unnecessary), so the formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at
the antenna, and that it DOES take into account that, for a lossy
line, the line portions closer to the transmitter are subjected to a
lower SWR.

Approximations that depend on VSWR as a load metric can:
1. depend on the integral over a half wave of very low loss line then
apply it as a constant loss loss per unit length; or
2. treat the forward and reflected waves as waves independently subject
to a constant loss per unit length.

I explained the sources of error in extending (1) to the general case in
my earlier post.

Case (2) assumes that the attenuation is the result of (vector) addition
of the power that is lost independently from each travelling wave at any
point, whereas the power lost is due to the effect of currents and
voltage resulting from vector addition of the voltages and currents of
the two waves at each point.

In some scenarios, they may be good approximations, but there are also
scenarios where they are poor approximations. (I gave an example in an
earlier posting where they both fail.)

The reality is that on a practical mismatched line, loss per unit length
is not constant with displacement. See my notes on VSWR at varying
displacement on a practical line, see
http://www.vk1od.net/VSWR/displacement.htm .

Look at Fig 9.

Note that the loss vs displacement line does not have a constant slope,
and anything that ignores that is ignoring an aspect of the problem.

Note that in the example, the red line dips below the blue line (meaning
loss under mismatched conditions is LESS than matched line loss at some
lengths). Any method that prevents that result is ignoring an aspect of
the problem.

The loss under mismatch conditions does depend on load impedance, and if
you throw away some of the detail and reduce it to load VSWR, then you
increase the scope for error.

Is there application for the approximations? Certainly, I use them... but
in the knowledge that they are approximations and an awareness of where
they are not good approximations and may not produce an adequate answer
for the current problem.

Your original posting was about reconciling the chart with some examples,
and I noted that the chart itself is a source of significant error in
some scenarios.

BTW, your calculations seem to fall into case (2), and if so, are subject
to the same errors... though they may reconcile well with a chart based
on that approximation.

Owen