Richard Harrison wrote:
"It is generally assumed that current distribution on an infinitesimally
thin antenna---is sinusoidal, and that the phase is constant over
1/,2-WL interval, changing abruptly by 180-degrees between intervals."

I just realized that the IRE reference I gave in another thread is for
a l/a=75 example, i.e. not for an l/a=infinity example.

But we can deduce the answer from what Kraus says on page 187. "A
sinusoidal current distribution may be regarded as the standing wave
produced by two uniform (unattenuated) traveling waves of equal
amplitude moving in opposite directions along the antenna."

And Yes, for a lossless unterminated transmission line, the current
distribution of the standing waves is sinusoidal. But not so for
a transmission line with losses. See "Transmission Lines & Networks",
by Johnson, Fig 4.11 or "Transmission Lines" by Chipman, Fig. 8-10.

We know a radiating dipole has "losses" due to radiation. Therefore,
the current distribution will be more like the two above graphs than
a pure sinusoid. For a real-world current distribution on a real
world dipole, an attenuation factor must be included. That makes
the reflected current less than the forward current which moves
the phase angle away from what it would be on a lossless transmission
line (or on an antenna that didn't radiate).

On a lossless transmission line, the forward current may be 1 at 45 degrees
while the reflected current is 1 at -45 degrees. Thus the net current would
be 1.414 at zero degrees. But on a real-world antenna, the forward current
may be 1 at 45 degrees while the reflected current is 0.85 at -45 degrees.
which would be approximately 1.3 at 4.4 degrees.
--
73, Cecil http://www.qsl.net/w5dxp



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