On 7/2/2015 3:52 PM, Wayne wrote:


"rickman" wrote in message ...

On 7/2/2015 12:18 PM, Wayne wrote:


"John S" wrote in message ...

On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S
writes
On 6/29/2015 3:47 PM, Wayne wrote:

snipped to shorten

Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will
be very high impedance at that frequency and a 1:4 unun will
theoretically bring that impedance down closer to the feed line
impedance.

Does this help?

It was been pointed out to me that the figures for feeder loss with an
imperfect SWR are only correct when the length is fairly long (at least
an electrical wavelength?). How much loss does 25' of RG-8 really have
at 12MHz, when there's a halfwave hanging on the far end?

# A *resonant* half wave at 12MHz is about 36.7 feet long and it presents
# an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end.
The
# current at the antenna end is 0.0245A while one watt is applied at the
# source end. This means that the power applied to the antenna is about
# 0.687W. So, about 68% of the applied power reaches the antenna.

# So, about 32% of the power is lost in the RG-8 for this example.

I'm just trying to understand this, so let me ask a question about your
example.

Isn't the 32% lost a function of not having a conjugate match maximum
power transfer?
If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline,
wouldn't maximum power be transferred?
(Even with a SWR of about 21:1)

# Transferred where? The match at the transmitter output only matches the
# output to the line. There are still reflections from the mismatch at
# the antenna. These reflections result in extra losses in the line as
# well as power delivered back into the transmitter output stage
# (especially with a perfect impedance match).

Well, I put a few (unrealistic) qualifiers into my question: a
transmitter with a a 1063 ohm output (not 50), and a lossless RG-8.

Thus, the back and forth reflections would not have attenuation.
And the transmitter and load are conjugately matched for maximum power
transfer.

Your quoting style is very confusing. If you use with a space at the
front of lines you are quoting it will show up the same as everyone
else's quotes.

Why will the reflections not have losses? Every load that is not an
infinite impedance will absorb some of the signal that would be
reflected. That applies to the transmitter output as well as the
antenna, no?

A matched impedance does not mean no losses. It means the maximum
transfer of power. These are not at all the same thing.


# But I don't see anyone taking wavelength vs. feed line length into
# account. If the wavelength is long compared to the feed line I believe
# a lot of the "bad" stuff goes away. But then I am used to the digital
# transmission line where we aren't really concerned with delivering
# power, rather keeping a clean waveform of our (relatively) square waves.
# So I guess a short feed line doesn't solve the SWR problems... or does
# it?

The attenuation at a given high SWR depends upon the the matched
feedline loss, as reflections encounter that loss with every forward or
backward trip.
Thus feedline length/attenuation should be considered.

As a young man I was given a problem of solving poor antenna performance
on an aircraft band fixed station antenna. The SWR at the transmitter
was close to 1:1, but the antenna didn't work well.
I climbed up on the tower and found that the coax had never been
connected to the antenna. That was with about 400 feet of coax at 120 MHz.

So how was the SWR 1:1?

--

Rick