Owen Duffy wrote:
Why is it necessary to complicate the analysis with tracking multiple re-
reflections, potentially an infinite number of reflections of diminishing
significance, an analysis that converges in the limit on the answer given
by the solution of the source V/I characteristic and (Vf+Vr)/If-Ir) at
the input end of the line (which is the equivalent input impedance). Note
that (Vf+Vr)/If-Ir) at the input end of the line is determined solely by
the tranmission line propagation constant, length, Zo and the far end
load impedance, for avoidance of doubt, source impedance is not
relevant.

The answer is subtle. Consider the following lossless example.

XMTR--x--1WL 450 ohm line--y--1WL 450 ohm line--50 ohm load

The SWR is 9:1 everywhere on the 450 ohm line.
(Vf+Vr)/(If+Ir) = 50 ohms at points x and y. There are reflections
to the left of y but no reflections to the left of x. Why?

The answer is the interference
pattern set up by the 50 ohm environment left of x and the
450 ohm environment to the right of x. Total destructive
interference is occurring to the left of x and total constructive
interference is occurring to the right of x. That cannot be said
of point y yet the V/I ratio is identical to x. The physical rho
at point y is zero. The physical rho at point x is 0.8. That's
the difference. Reflections occur only at physical impedance
discontinuities.

In S-parameter terms at x: b1 = (s11)(a1) + (s12)(a2)

In RF terms at x: Vref1 = rho1(Vfor1) + tau2(Vref2)

The two terms to the right of the equals sign are the voltages
that engage in wave cancellation resulting in a Z0-match at x.

So to answer your question: The 50 ohm virtual impedance at point
y is incapable of causing reflections even though it has an identical
V/I ratio to point x. The physical impedance discontinuity at x is
fully capable of causing reflections along with the ensuing
interference.
--
73, Cecil http://www.w5dxp.com