Walter Maxwell wrote:
Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

I absolutely agree, Walt, "when an EM wave encounters a short circuit ...".
But when "complete destructive interference" occurs, something else happens.
The E-field goes to zero AND the H-field (B-field) goes to zero at the same
time. If you concentrate on the voltage, it looks like a short. If you
concentrate on the current, it looks like an open. It is both or neither.

Complete destructive interference requires that both fields go to zero
simultaneously and emerge as constructive interference in the opposite
direction obeying the rule that E/H=V/I=Z0. It is an energy reflection
that is also an impedance transformation at an impedance discontinuity.

Let's assume that 100v at zero degrees with a current of 2a at 180 degrees
encounters another wave traveling in the same rearward direction of 100v
at 180 degrees with a current of 2a at zero degrees. These two waves cancel.
The voltage goes to zero AND the current goes to zero. Each wave was associated
with 200 watts. So a total of 400 watts reverses directions. Assuming the
destructive interference occurred in a 50 ohm environment and the resulting
constructive interference occurred in a 300 ohm environment, the reflected
wave would be 346 volts at 1.16 amps. It's pretty simple math.

346*1.16 = 400 watts 346/1.16 = 300 ohms

The above quantities represent the destructive/constructive interference.
These quantities must be added to the other voltages and currents that are
present to obtain the net voltage and net current.
--
73, Cecil http://www.qsl.net/w5dxp



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