Roy:

[snip]
b = v - Ri = Zi - Ri = (Z - R)i Volts.

First of all, you're speaking of a circuit with a source impedance R and
load impedance Z, rather than a terminated transmission line. Forward
and reflected wave terminology is widely used in S parameter analysis,
which also uses this model, so I'll be glad to follow along to see if
and how S parameter terminology differs from the transmission line
terminology we've been discussing so far. Please correct me where my
assumptions diverge from yours.

Your "classical definition" of b isn't one familiar to me. v + Ri would
of course be the source voltage (which I'll call Vs). So v - Ri is Vs -
2*Ri. Where does this come from and what does it mean?
[snip]

Yes it should be familiar to you because it is the most common definition
and one you seem to agree with. I presume that you are not used to seeing
the use
of the symbols "a" and "b" for those quantities. The use of "a" and "b" is
widely
used in Scattering Formalism and is less confusing to many than using
subscripts.
In your terminology above the "Vs" symbol is nothing more than the incident
voltage
usually given the symbol "a" in the Scattering Formalism, or the symbol V
with a "+"
sign subscript in many developments. Often you will find authors use a V
with a plus
sign "+" as a subscript to indicate the "a" voltage and a V with a minus
sign "-" subscript
to idicate the "b" voltage. Personally I find the use of math symbols "-"
and "+" or other
subscripts to variables to be confusing, I much prefer the use of "a" and
"b" for forward
or incident and reflected voltages.

Simply put, if a generator with "open circuit" voltage "a" and "internal
impedance" R
is driving a load Z [Z could be a transmission line driving point impedance,
for instance
Z would be the characteristic or surge impedance Zo of a transmission line
if
the generator was driving a semi-infinite line.] then v is the voltage drop
across Z
and I is the current through Z, and so...

a = v + Ri = Zi - Ri = (Z - R)i

is simply the [usual] forward voltage or incident voltage applied by the
generator to the
to the load Z, which may be a lumped element load or if you prefer to talk
about transmission
lines, Z can be just the driving point impedance of a transmsision line,
whatever you wish.

Then its'just a simple application of Ohms Law tosee that b = v - Ri = Zi -
Ri = (Z - R)i is
the [usual] reflected voltage. b is just the difference between the voltage
across Z which is
calculated as Zi and the voltage that would be across Z if Z was actually
equal to R. i.e. the
reflected voltage b is just the voltage that would exist across Z if there
was an "image match"
between Z and R. [If Z is the Zo of a semi-infinite transmsision line you
could call this a Zo match].

Taking the ratio of "b" to "a" just yeilds the [usual] reflection
coefficient as
b/a = (Z - R)i/(Z + R)i = (Z - R)/(Z + R). A well known result. Simple?

[snip]
From your equation, and given source voltage Vs, i = Vs/(R+Z).
Therefore, your "classical definition" of reflected voltage b is, in
terms of Vs, Vs*((Z-R)/(Z+R)).

and the incident voltage a would be the Thevinins equivalent voltage
across
the sum of Z and R, i.e.

a = (Z + R)i
[snip]

Yep you got it all right!

[snip]
Since i = Vs(Z+R), you're saying that a = the source voltage Vs (from
your two equations). So what you're calling the "incident voltage" is
simply the source voltage Vs.
[snip]

Yes, mathematically "a" = "Vs", what else would it be? Nothing mysterious
about
that. The incident voltage is always simply the open circuit voltage of the
source. In
words a is not the source voltage because the source is a Thevinin
equivalent made up
of the ideal voltage generator Vs = a behind the "internal" source impedance
R. A better
way to describe Vs = a in words would be the incident voltage a is the "open
circuit source
voltage".

[snip]
Let's do a consistency check. The voltage at the load should be a + b =
Vs + Vs*((Z-R)/(Z+R)) = Vs*2Z/(Z+R). Inspection of the circuit as I
understand it shows that the voltage at the load should be half this
value. So, we already diverge. Which is true:
[snip]

No, the voltage at the load is not (a + b) rather it is [the quite obvious
by Ohms Law] v = Zi.

and the sum of the incident and reflected voltage is simply

a + b = (v + Ri) + (v - Ri) = 2v = 2Zi

Now if there is an "image match" and the "unknown" Z is actually equal to R,
i.e. let Z = R
in all of the above, then...

a = Vs

b = 0

a + b = 2Ri

and i = Vs/2R = a/2R.

[snip]
1. I've goofed up my algebra (a definite possibility)
[snip]

Only a little :-)

[snip]
2. I've misinterpreted your circuit, or
[snip]

No you have it correct!

[snip]
3. The voltage at the load is not equal to the sum of the forward and
reflected voltages a and b, as you use the terms "forward voltage" and
"reflected voltage". If v isn't equal to a + b, then what is the
relationship between v, a, and b, and what are the physical meanings of
the forward and reflected voltages?
[snip]

I showed those relationships above. There is nothing new here... these are
the [most] widely accepted definitions of incident and reflected voltages.

[snip]
I'd like to continue with the remainder of the analysis, but can't
proceed until this problem is cleared up.

Roy Lewallen, W7EL
[snip]

OK, let's carry on.

--
Peter K1PO
Indialantic By-the-Sea, FL.