On Apr 1, 9:34 am, Cecil Moore wrote:
Keith Dysart wrote:
The impedance
encountered by the reflected wave at the generator is the
same as the generator's source impedance.

No, the generator's source impedance is *NOT* the
impedance encountered by the reflected wave. Please
reference w2du's article again.

http://www.w2du.com/r3ch19a.pdf

Forget about the conjugate match and concentrate on the
non-dissipative source resistance being different from
what you are calling the generator's source impedance.
An *active* source creates a source impedance looking back
into the source that is *different* from what you are
calling the generator impedance.

It would appear that you are confusing the possible complexities
of a class C power amplifier with the simplicity of the generator
in the experiment I proposed. A 2 Amp current source
in parallel with a 450 Ohm resistor does not, as far as I can
tell, have a 'non-dissipative source resistance'. It has a
dissipating source resistor. This is not a particularly efficient
implementation, but is certainly a possible one. Now that I
have clarified that there is a dissipating source resistor will
this allow you to use superposition to solve the problem?

Just for your convenience, a reminder of the problem:
- generator with 450 Ohm source resistance drives
- a line with 450 Ohm characteristic impedance
- terminated by a 75 Ohm load
- the generator is set such that it would output
450 Volts into a 450 Ohm load

Question:
- Will there be ghosts?
If the answer is yes...
- What is the magnitude of the first re-reflection?
Ancillary question:
- What 'forward power' will a directional wattmeter in the
450 Ohm line indicate?
- What 'reverse power' will a directional wattmeter in the
450 Ohm line indicate?

If necessary for answering the question:
- The line can be assumed to be 31 wavelengths long.
- The generator can be assumed to be a 2 Amp current source
in parallel with a 450 Ohm resistor.

....Keith