Roger wrote:
Keith Dysart wrote:
On Dec 30, 5:30 pm, Roger wrote:


I don't recall any examples using perfect CURRENT sources. I think a
perfect current source would supply a signal that could respond to
changing impedances correctly. It should solve the dilemma caused by
the rise in voltage which occurs when when a traveling wave doubles
voltage upon encountering an open circuit, or reversing at the source.

What do you think?

A perfect current source has an output impedance of
infinity, just like an open circuit. The reflection
coefficient is 1.

Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

...Keith

This disagrees with Roy, who assigns a -1 reflection coefficient when
reflecting from a perfect voltage source.

It appears you're confusing perfect voltage and current sources.

A perfect voltage source has a zero impedance, so if it's connected to a
transmission line with no series resistance, it presents a reflection
coefficient of -1. A perfect current source has an infinite impedance,
so if it's connected to a transmission line with no parallel resistance,
the reflection coefficient is +1, as Keith says.

The Norton or Thévenin equivalent circuits seem capable of positive
reflection coefficients. That is all that I am looking for.

Reflection coefficients are complex numbers, so they can't properly be
described as "positive" or "negative" except in the special cases of +1
and -1. In all other cases, the can only be described by their magnitude
and angle, or real and imaginary component. Under normal circumstances,
reflection coefficients can have any magnitude from zero to one, and any
angle. A Thevein equivalent, like the circuit it's replacing, can
present any possible reflection coefficient. For example, a Thevenin
equivalent circuit having an impedance of 19 - j172 (that is, the
equivalent consists of a perfect voltage source in series with a 19 ohm
resistance and a capacitance of 172 ohms reactance) will present a
reflection coefficient of about 0.8 - j0.5 to a 50 ohm transmission
line. This is also true of a Norton equivalent consisting of a perfect
current source shunted by an impedance of 19 - j172 ohms.

Of course, a voltage source and series impedance not acting as an
equivalent for any other circuit can also be used to drive the line, and
the source impedance will equal the impedance that's in series with the
perfect source. Likewise a current source with a parallel impedance.

. . .

I must have missed something, because I can't understand why there is an
insistence that a negative reflection coefficient must exist at the
source for the 1/2 or 1 wavelength long transmission line fed at one end.

No one has insisted on that at all. As I've said, any reflection
coefficient can exist at the source. It depends solely on the impedances
of the transmission line and the source. Both Keith and I have given the
equation describing the simple relationship, and you can find it also in
many references. It was -1 for my example only because I used a perfect
voltage source with no series impedance.

Roy Lewallen, W7EL