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Keith Dysart wrote:
Cecil Moore wrote: A schematic shows exactly what is happening. There is no path from SGCL1 to R1. There is no path from SGCL2 to R2. SGCL1---1---2------2---1---SGCL2 \ / \ / 3 3 | | R1 R2 There is nothing in the circuit to cause any reflections. So the power dissipated in R2 comes from SGCL1 and the power in R1 comes from SGCL2. Can not happen after cutting the branches. The inclusion of circulators in the example ensures that it is a distributed network example. Cutting the branches is not a valid action in distributed network examples because technically it is a zero current "point" and not a zero current "branch", i.e. the current is not zero throughout the entire branch. See below. Sorry, the lumped circuit model is known to fail for distributed network problems. That's probably why the distributed network model still survives today but has been discarded and forgotten by many in the rather strange rush to use a shortcut method at all costs. Or are you disuputing the validity of cutting branches with zero current? Of course, it is obviously invalid in distributed network problems. We can add 1/2WL of lossless transmission line to the example to see why it is invalid. 1/2WL 50 ohm SGCL1---1---2--+--lossless line--+--2---1---SGCL2 \ / \ / 3 3 | | R1 R2 Your zero current "branch" is now 1/2WL long and in the center of that zero current "branch", the current is at a maximum value of 0.4 amps for 50 ohm signal generator voltages of 10 volts as in your original example. How can the current in the middle of the line be 0.4 amps when the current at both points '+' is zero? Does that 0.4 amps survive a cut at point '+'? There are no reflections anywhere in the system. Since the voltages are equal for the signal generators, we can only conclude that 0.2 amps of traveling wave current is flowing from SGCL1 to R2 and that 0.2 amps of traveling wave current is flowing from SGCL2 to R1. The two current nodes at the '+' points do NOT indicate that zero current is flowing in the 1/2WL line. They only indicate that the two traveling wave currents are equal in amplitude and opposite in phase at the '+' points. Any cut that disrupts the flow of those traveling wave currents is invalid. -- 73, Cecil http://www.w5dxp.com |
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