Richard Clark wrote:
"In a reality of 359 other possible phase angles, how does a transmitter
happen to always be in-phase to any reflection?"

Connect any generator to any resistor, and current in the resistor is
in-phase with the applied voltage. The Zo of the common transmission
line is a reasonably good resistance. At radio frequencies, Zo is
independent of frequency.

The current in the incident wave is always in-phase with the voltage
applied to a transmission line. The current in the reflected wave is
always 180-degrees out-of-phase with the reflected voltage. It makes no
difference which was inverted by reflection, the volts or tha amps, one,
and only one, of them was flipped upside down. The transmission line can
and does handle the reflected wave.

Standing waves display interference between incident and reflected waves
whiich ideally have in-phase and out-of-phase constituents.

The fact that the Bird wattmeter works is evidence that the theory is
correct at least until a better theory replaces existing theory.

Best regards, Richard Harrison, KB5WZI

On Tue, 7 Sep 2004 09:40:36 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
"In a reality of 359 other possible phase angles, how does a transmitter
happen to always be in-phase to any reflection?"

Connect any generator to any resistor, and current in the resistor is
in-phase with the applied voltage.

Hi Richard,

In a world of mismatches, how does it happen that the transmitter
always sees an in-phase, resistive load? If this were the best of all
worlds, tuner manufacturers would be out of business, salesmen would
starve and Reggie would have nothing to write about.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
In a world of mismatches, how does it happen that the transmitter always
sees an in-phase, resistive load?"

It doesn`t. You can put a capacitor directly across its output
terminals, and the transmitter will energize the capacitor. But, a
transmission line is not a capacitor unless it is a short open circuit,
or the equivalent. A transmission line is a distributed network of
inductance and capacitance. This network transfers emergy in bucket
brigade fashion. The "brigade" presents a resistive impedance to both
the incident wave and to the reflected wave. Zo is an enforcer.

Best regards, Richard Harrison, KB5WZI

On Tue, 7 Sep 2004 10:48:16 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
In a world of mismatches, how does it happen that the transmitter always
sees an in-phase, resistive load?"

It doesn`t. You can put a capacitor directly across its output
terminals, and the transmitter will energize the capacitor.

Hi Richard,

And so it would seem that a too-short antenna would act likewise.
This, then, returns us to my original question for this, a very simple
example that has a currency that is frequent enough in our experience:
how does a transmitter happen to always be "in-phase" to any
reflection?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
how does a transmitter happen to always be "in-phase" to any
reflection?

It doesn't. The reflected voltage can obviously be 90 degrees
out of phase with the forward voltage in which case, the
interference term equals zero, and the superposed voltage is
SQRT(Vf^2+Vr^2), i.e. greater than Vf. (The argument reminds
me of Gary Coffman's one-dimensional "spitting up the fire
hose" argument.)
--
73, Cecil http://www.qsl.net/w5dxp


-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Richard Harrison wrote:
Standing waves display interference between incident and reflected waves
whiich ideally have in-phase and out-of-phase constituents.

In the following, the interference term equals zero.

XMTR--------1/8WL 50 ohm lossless coax------291.5 ohm load

Forward power = 100w, reflected power = 50w.

Vf = 70.7v at zero degrees, If = 1.4a at zero degrees

Vr = 50v at 90 degrees, Ir = 1.0a at -90 degrees

Superposing:

Vtot = SQRT(Vf^2+Vr^2) = 86.6v over-voltage condition

Itot = SQRT(If^2+Ir^2) = 1.72a over-current condition

The phase angle between Vtot and Itot is about 70.4 degrees,
i.e. the source sees a highly reactive impedance.

Vtot is NOT in phase with Vf, Itot is NOT in phase with If,
Vtot is NOT in phase with Itot. Vf is NOT in phase with Vr.
No two voltages are in phase. No two currents are in phase.
--
73, Cecil http://www.qsl.net/w5dxp


-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----