On Apr 19, 12:27 pm, Cecil Moore wrote:
Richard Harrison wrote:
How much energy will continue to enter the stub? Practically none
because there is no difference of potential between the source and the
reflection.

In a lossless stub, zero energy will continue to enter the stub.
In fact, in a lossless stub, the source can theoretically be
completely disconnected and everything remains the same
including 100% re-reflection of reflected waves by the new
open circuit.

I've always liked this example. By extension, on a line multiple
quarter wavelengths long, you can disconnect the line at ANY
of the voltage maxima and see exactly the same result as you
describe above. Are those travelling waves really crossing the
voltage maxima? Or are they being reflected? Or are they just
reflected at the maxima that actually occur at a discontinuity?
But there is no discontinuity at the source, so why would they
be reflected there? But if they are reflected at the
non-discontinuity at the source, then they should be reflected
at the non-discontinuities in the line as well. But wait, those
non-discontinuities are virtual opens and shorts, so maybe
they reflect at virtual ones as well. But no, it has been agreed
that reflections only happen at physical discontinuities.

And what if the stub is not a quarter (or multiple) wavelength
long? Then where does the reflection at the source occur?
At the last maximum or minimum along the line? Inside the
generator? It works for any length of line. Somehow the
generator knows exactly what reactance to supply?

It is definitely best to recognize that when the source impedance
is equal to the line impedance, there is no reflection at the
source. And it does not matter if the source impedance is
achieved with a circulator, a resistor, a multiplicity of resistors,
feedback, or what have you; there is no reflection if the
impedance is the same as the line. No discontinuity, no
reflection. A simple rule. Works at a load. Works at a
generator. Works along the line. Always works.

....Keith