Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

Since any reference to voltage on an antenna seems to be verboten, I
have avoided any such reference. Otherwise, here is a quote from Kraus:

"It is generally assumed that the current distribution of an infinitesimally
thin antenna is sinusoidal, and that the phase is *constant* over a 1/2WL
interval, changing abruptly by 180 degrees between intervals."

Don't you just love the phrase, "It is generally assumed ..."? He doesn't
say the current is sinusoidal. He doesn't say the phase is constant over
a 1/2WL interval.

For that general assumption to be true, the reflected current would have
to equal the forward current on a standing-wave antenna. But we know it
doesn't. However, this implies that the reflected current arriving back
at the feedpoint is not extremely/severely attenuated. Here's a cute
ballpark analysis sure to drive the gurus crazy.

Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward
voltage to forward current is 600 ohms. The ratio of reflected voltage to
reflected current is 600 ohms. Assume the feedpoint current is one amp and
the feedpoint voltage is 50 volts. Assume the forward current and the reflected
current are in phase at the feedpoint. Assume the forward voltage and reflected
voltage are 180 degrees out of phase at the feedpoint. This is enough information
to solve for the ratio of forward current to reflected current at the feedpoint.
Assuming the net current is one amp at the feedpoint, I get 0.542 amps for
the forward current and 0.458 amps for the reflected current, i.e. the reflected
current is 85% of the value of the forward current. Remember, that is a ballpark
estimate.

That means the current only decreases by 15% in its round trip to the
end of the antenna and back. Same for the voltage. Oops, I mentioned
voltage - sorry. But please note that the power loss is a lot higher
than 15% since both the current and voltage are reduced by the same 15%.

The argument seems to occur due to the ignoring of the component waves
on a standing wave antenna. Such is the steady-state model seduction attended
by its sacred cows. Who wants to join me in a beer bust and barbecue?
--
73, Cecil http://www.qsl.net/w5dxp



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