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Cecil, W5DXP wrote:
"What reflects the other half of the energy?" The previous example I gave was a 25-ohm resistor-load on a 50-ohm line. Change the load to 100 ohms. Now the load cannot accept all the current carried by the incident wave. Lenz`s law says the falling current generates a rising voltage in an attempt to maintain the current. The load-generated voltage is in the same phase as the incident voltage so their sum is greater. Increased voltage across the load reverses phase and direction of the line-current at the too-high load resistance. Thus, direction of the reflection is opposite that of the incident wave. If the load is too small or too large for Zo, some of the incident energy is reflected by the load. The two processes are analogous. When the load value is too small, there is a reversal in the phase of the voltage without change in the phase of the current (1955 Terman page 92). When the load value is too large, there is a reversal in the phase of the current without change in the phase of the voltage (1955 Terman page 89). Those are the necessary and sufficient conditions to reverse the direction of some of the energy in an incident wave on a transmission line. For a complete reversal, a short or an open is required. Best regards, Richard Harrison, KB5WZI |
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