I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

While your math is correct, your application in my opinion is incorrect.


"H. Adam Stevens, NQ5H" wrote in message
...

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE" wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the bandwidth
where it drops no more than about 90% or so, you can operate confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal strength
in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

73
H.
NQ5H

Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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Fred W4JLE wrote:
Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1

!.7:1???? I must admit, that's a new one on me. :-)

My statements were general, and apply to any SWR.
--
73, Cecil http://www.qsl.net/w5dxp

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One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

See other post.

H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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It's called db, Cecil.


"Cecil Moore" wrote in message
...
H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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H. Adam Stevens, NQ5H wrote:

"Cecil Moore" wrote:
The log of the ratio of two SWRs doesn't seem to have much
meaning.

It's called db, Cecil.

The IEEE Dictionary says the ratio of power, voltage, and
current can be expressed in dB. It specifically states
that dB can only be related to power ratios or to parameters
that are proportional to the square root of power ratios.

SWR1 = [SQRT(Pfor1)+SQRT(Pref1)]/[SQRT(Pfor1)-SQRT(Pref1)]

SWR2 = [SQRT(Pfor2)+SQRT(Pref2)]/[SQRT(Pfor2)-SQRT(Pref2)]

The ratio of two SWRs will not reduce to a power ratio or
to the square root of a power ratio.
--
73, Cecil http://www.qsl.net/w5dxp

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