It seems to me that entirely too much information is being put out on
VSWR loss, line attenuation, and "lost" power. The messages below have
been edited to remove the senders name's. They are merely examples, and
by far not the worst. Read them, decide what you think about each
statement, and then at the end, read page 4-10 of Maxwell's
"Reflections"
> Others have posted about power not making it to the antenna are
> reflected back and forth and eventually make it to the antenna and are >
> radiated. This is so absurd that I had to respond to that one.
>From different message
>Yes, it's true. sometimes it makes a difference. Sometimes it's >significant.
>And there is a reference to go to to determine what it >is!!
> Unfortunately, the reference you site below, is too a completely
> different "it" (loss mechanism)! See on below.
>Per the ARRL Handbook, 1994, Chapter 16 (Transmission Lines), page 16-14
> through
>16-15, the following is a summary of Fig 26, a graph showing on the x
> ? axis: % relative power delivered to the antenna; and y axis: VSWR at >
> input end
> > (i.e., at
>the transmitter) of the transmission line, and a family of curves
>depicting the >transmission line loss at the frequency of interest if > the
>load is matched to the line (i.e., the manufacturer's spec dB/ft times # of
>feet used):
> > >For 1:1 VSWR (at xmtr) and lossless transmission line, % power > >
> > >delivered to the antenna = 100%.
>For 1:1 VSWR (at xmtr) and 1 dB loss in transmission line, per mfg., % > power
>delivered to antenna = 80%
>For 1:1 VSWR (at xmtr) and 2 dB loss in transmission line, per mfg., % >power
>delivered to antenna = 63% (eyeball interpolation)
>For 2:1 VSWR (at xmtr) and 1 dB loss in transmission line, per mfg., %
> power = 73% (eyeball again)
>The "lost" power is dissipated in I Squared R losses (H E A T) in the
> copper (or )other metal) in the transmission line.
The following is quoted word for word. ( I hope )
Now we'll consider the effects of line attenuation A ( alpha ) on
the increased loss of power delivered to a load when the load is
mismatched to the transmission line impedance. We'll assume a 3:1 SWR,
where p=.5 ( rho added )and where 100 watts supplied by the transmitter
will yield 133.33 watts of forward power because forward power is
1/(1-p^2) times the supplied power, so 1/.75=1.3333 and 1.3333 times 100
watts equals 133.33 watts. If we initially neglect the losses 100 watts
will also be absorbed in the load, because reflected power p^2=.25 or
25% of the forward power, leaving 75% absorbed the load (1-p^2 ) or
25% of the forward power, and 75% of 133.33 is exactly 100.
Consider a typical realistic case. In 175 feet of RG-8 at 4 MHz,
the matched line attenuation is .5 db. If the load were perfectly
matched to the line ( 1.0 SWR ), the 100 watts delivered by the
transmitter would be attenuated to 89.13 during travel to the load. But
with a 3:1 mismatched load the (emphasis ) ADDITIONAL one-way line
attenuation ( because of the SWR ) is .288 db. Hence, the forward power
at the conjuatematch point would then be 124.78 watts ( .288 dB below
133.33 watts ), and 111.21 watts of power (.5 dB below 124.78 watts )
reach the load. Of that power, 27.8 watts ( 25% )are reflected, leaving
the 83.41 watts to be absorbed. Of the 27.8 watts reflected, 24.78 watts
arrive back at the load to join the 100 watts of source power to
develop the 124.78 watts of forward power. The 5.72 watts difference
between the power absorbed in the (emphasis) MATCHED and the 3:1
(emphasis) MISMATCHED load ( .288 dB) is insignificant. It actually
amounts to less than 1/12 of an S unit. Information of calculating these
values is presented in Chapter 6. ( and no, I'm not going to type Chap.
6 ed ) These values are typical of data obtained during actual routine
measurements in a professional laboratory. ( read last statement again
ed ) They provide additional evidence that reflected power is real,
not fictitious. If it were fictitious power, no more that 66.85 watts
( 75 % of 89.13 watts ) would be available to the 3:1 mismatch. But,
the 83.41 watts actually absorbed in the mismatched load is 93.58% of
the amount absorbed in the matched load, the loss of 6.42% being
completely accounted for in line attenuation alone.
End of quote
Comments: So much for the ARRL graph. It only used a 2:1 mismatch.
So much for reflected power being lost as heat.
I asked that you pay special attention to the statement about "routine
tests is a PROFESSIONAL lab ) Not the mention they were conducted by
a professional of his reputation. And again, please don't associate
the messages at the beginning to any individual...They are not intended
as such.
There will be a short quiz later.
73
Ed
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