Richard Harrison wrote:
Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached
to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load.

Cecil wrote:
"Destructive interference on the XMTR side of "+" (where 50-ohm cable
meets the 300-ohm Q-matching section) eliminates the reflections."

It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z)
is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The
numbers are right, so in my opinion, the Q-section converts the
1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the
transmitter`s load. The 50-ohms presented by the Q-section to the cable
should result in a match and thus there should be no reflected energy in
the 50-ohm cable to cause interference. The reflections should all be in
the 300-ohm Q-section.

Exactly, thanks to destructive interference. In S-parameter terms:

b1 = a1*s11 + a2*s12

where b1 is the normalized reflected voltage back toward the source
and is equal to zero because complete destructive interference between
a1*s11 and a2*s12 causes their phasor sum to equal zero.

Those two voltage components cancel to zero toward the source. That
destructive interference is necessary and sufficient for a matched
system. The result is constructive interference toward the load.
--
73, Cecil http://www.qsl.net/w5dxp


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