An antenna question--43 ft vertical
On 7/2/2015 7:37 PM, John S wrote:
On 7/2/2015 11:18 AM, 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?
No. As I said, one watt is applied at the source end. This is condition
defined by the example and has nothing to do with source matching. The
32% loss is due to transmission line loss. The mismatch at the load end
causes the high SWR which increases the line loss due to high current at
some point in the line as well as increased voltage at other point(s).
The impedance of the 1063+J0 load is transformed to 54+J192 ohms at the
source. However, at an electrical quarter wave away from the antenna,
the impedance is about 2.8+j0 ohms. So that point is a relative hot spot
in the line.
Correction: 0.5+j0 at 1/4 wavelength away from the antenna.
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)
With a lossless transmission line and one watt applied to one end the
other end will have one watt available. The only place the power can go
is into the antenna. To put one watt into 1063 ohms will require .0306
amps and 32.6V at the feed point.
Does this make sense?
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