The formula you found is correct in the sense that because of a 1:3 SWR the
power delivered to a non reactive load by an ideal 50 Ohm generator is 1.25
dB less than it would with a 1:1 SWR.
If the actual load, that one producing the stated SWR, determines an
increase or a decrease in the system efficiency because a generator (that's
real and not ideal) it's a practical consequence that may largely vary but's
finally moderate.
Added loss by 1:3 in a line with low losses (short) is not so important
likewise the possible (moderate) increase in efficiency. Added loss in a
lossy line with 1:3 SWR is much more important than the theoretical 1.25 dB
and in 99% of the cases fully overtakes the amount of the the eventual
effciency increase.
73,
Mauri I4JMY
----- Original Message -----
From: "Tom Rauch" <w8ji@contesting.com>
To: <towertalk@contesting.com>; "Nat Heatwole" <heatwole@clark.net>
Sent: Saturday, May 26, 2001 6:18 AM
Subject: Re: [TowerTalk] dB Loss Equivilents for SWR Values
> Pardon the lecture that follows, but I think you have been misled.
> This really is worth getting into!
>
> > Whoa. Thanks for all of the fast responses. The most popular suggestion
> > appeared to be looking in the ARRL Handbook. I checked out the handbook
> > and with some consideration I figured out the following formula to get
> > equivalent coax db loses from an SWR value:
> >
> > ((SWR - 1) / (SWR + 1)) ^ 2 * 100% = % Lost
> > 10log(100/(100 - % Lost)) = Loss In dB
> >
> > So if your SWR's 3.0/1:
> >
> > ((3 - 1) / (3 + 1)) ^ 2 * 100% = 25% Lost
> > 10log(100/(100 - 25)) = 1.25db Loss
>
> A 3:1 SWR does NOT cause a 25% power loss. It might cause an
> increase in loss, but that increase can be anything. SWR can even
> cause a DECREASE in line loss!
>
> The ONLY case where that formula is correct is the special case
> where the source is a perfect resistive source without any
> shutdown circuits. The source Z must also match the line
> impedance exactly, and the line itself must be lossless.
>
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