On 6/25/2011 12:41 PM, Cecil Moore wrote:
On Jun 25, 12:02 pm, John wrote:
If the reflections toward the source is eliminated, how is it that it
appears to be 50 ohms at that point rather than 291.4 ohms?

You answered your own question - if reflections toward the source are
eliminated in a Z0=50 ohm environment, the apparent (virtual)
impedance cannot be anything except 50 ohms

You said "Since a virtual impedance is result of the superposition of a
forward
wave and a reflected wave, a virtual impedance cannot re-reflect the
reflected wave, i.e. one cannot re-reflect the reflected wave while at
the same time the reflected wave is being used to generate an
impedance."

But, it does. First, it causes the 50 ohms line (looking into the 291.4
ohms line to see a match due to the reflection. Second, the
re-reflection from that discontinuity is half of what maintains the
circulating energy on the line. The other half is the discontinuity of
the non-virtual load.