Where does it go? (mismatched power)
On Jun 18, 10:35*am, Cecil Moore wrote:
On Jun 18, 5:57*am, K1TTT wrote:
you edited too much out of context... it doesn't matter which method
you use to calculate where it goes as long as you know the limitations
of your method.
Exactly! The method that most RF engineers use does not tell how the
energy flows. Witness the posters who are implying that a reflected
wave contains no energy because they cannot figure out where that
energy goes or how it gets there. In his food-for-thought example,
w7el completely ignores the role of superposition/interference in
redistributing the energy, a phenomenon known and understood in the
field of optical physics for at least a century.
--
73, Cecil, w5dxp.com
because, at least when i do circuit design, i NEVER calculate energy,
and only at the end calculate power if i really need to. voltage and
current or E and H are the simpler things to handle and totally
sufficient for all circuit analysis, either lumped or distributed.
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