On Dec 28, 1:02*am, Cecil Moore wrote:
Keith Dysart wrote:

* Cecil Moore wrote:

* * * * * * * * * *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.

Using the distributed network approach, you have added
an infinite number of branches, and now there are two
branches which are appropriate places to make the cut.

The point is that in the above example, there are absolutely
no reflections. When you cut the line you cause reflections
where none existed before.

This is what makes the example so fascinating.

Before the cut, there is a distribution of voltage, current
and power on the line. The functions representing these can
be written as

V(x,t)
I(x,t)
P(x,t)

After the cut(s), the voltage, current and power distributions
are exactly the same: V(x,t), I(x,t), P(x,t).

The cuts changed nothing about the conditions in the circuit.

And yet the claim is made that before the cuts there are no
reflections and after the cut there are a bunch. And yet the
conditions in the circuit are EXACTLY the same.
But befonone, after:bunch. But conditions are exactly
the same.

It is obviously invalid to completely
change the operation of the circuit in that manner.

Just cutting a branch. Completely legal.

Is the operation of the circuit completely different?
Perhaps it is reasonable to view a virtual open as
producing a reflection. You have to work this out for
yourself. But as you do so, keep front and center the
fact that the voltage, current and power distributions
are identical before and after the cut(s). The before
and after circuits are behaving identically.

If you were merely provided with V(x,t), I(x,t) and
P(x,t) you could not determine whether there were cuts
or not.

We could
be sending data from SG1 to R2 and from SG2 to R1. Those data
streams stop when you cut the line.

The specification of the conditions in the experiment mean
that there can be no data stream. If there were data, the
current would not be always 0.

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 '+'?

Absolutely, if the line is lossless. Cut both "+" and the
current in middle of the line still remains.

We both know that is a physical impossibility. Sometimes you
are just forced to accept reality and get on with it.

It is completely possible for a lossline line.
It is no different than a connected capacitor and inductor
which will ring for ever given the appropriate initial
conditions.

...Keith