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