What is Pfwd and Pref with Complex Zo?
 

On 6 Sep 2003 22:21:19 -0700, (Dr. Slick) wrote:

wrote in message ...

But, going back to the original question, wbat is Pfwd and Prev on a
line
with complex Z0?

...Keith


I'd like to know the answer to this question too!

If we are talking about only DISSIPATED power, do we have
to say P=Vrms**2/Re(Zo)? Taking only the real part of the Zo?
And if phase doesn't mean anything for power, how can we use a
complex Zo in the denominator?

And if the Pfrd originates in Zo, Prev is loaded by Zo,
then even if Zo is complex, can you still not say:

[rho]**2 = power RC? Because the Zo cancels out anyways
(a ratio)?

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Slick

I've been reading a lot concerning this subject on this NG. It appears
to me that a lot of words have been spilt over only an academic issue.
The reason is that the loss in coax the we use at HF, or maybe even at
MF, is so low as to be insignificant. The voltage/current phase angle
becomes significant on the lines we use only at low frequencies in the
audio range. If the discussion is only to provide further education on
the subject, fine, but some readers might get the impression that a
little line attenuation is a worrisome thing.

Walt, W2DU

Tom Bruhns wrote:
"If you want dissipated power in a TEM line, then P = Irms^2*R +
Erms^2*G (at a particular frequency where R and G have fixed values)."

Sure. Total loss is the sum of the series and shunt losses.

Tom also wrote:
"Since Irms and Erms are functions of position along the line, P is a
per-unit-length quantity like R and G and total power is found by
integrating the incremental P over the length of the length of the line
you are interested in."

Yes. Loss is a dB per 100 ft. quantity and loss is cumulative over its
length. But, I wouldn`t worry about the volts and amps produced by SWR.
The forward and reflected waves don`t oscillate in value. Only their
interference pattern does that and it is of no consequence. The forward
and reflected waves are smoothly attenuated by "alpha", the attenuation
constant, which is a function of frequency.

Alpha has the same value for the incident and reflected waves, but
though the dB per 100 ft is the same for a wave traveling in either
direction on the line, the reflected wave is likely much smaller than
the incident wave, and the loss produced by the reflected wave will be
much smaller too. The total loss is the sum of the losses produced by
the incident and reflected waves. As was shown yesterday, the ARRL
Antenna Book has charts to determine the added loss caused by standing
waves (actually caused by the reflected power that produces SWR). These
convenient charts almost eliminate arithmetic in determining additional
loss to be expected given the SWR.

Best regards, Richard Harrison, KB5WZI

Keith wrote:
"Consider a slotted line used to measure voltages for the computation of
VSWR."

O.K. I`m looking at my PRD 250-A. Its slot is about 10 inches long or
about 25.4 cm. The distance between a maximum and a minimum voltage
point in an standing wave pattern is 1/4-wavelength.

The velocity factor is not quite as high as that of free-space. This
tends to shorten the distance between maxima and minima in the slotted
line.

If you could get a good voltage sample anywhere within a 25 cm slot, you
could get one maximum and one minimum in a standing wave pattern at a
frequency where the slot was at least 1/4-wavelength, but you might have
to adjust feedline length to place the pattern in a favorable slotted
line location.

What frequency has a 1/4-wavelength of 25 cm? My calculation says: 300
MHz.

Slotted lines are called trough lines in the U.K. I believe.

SWR is more easily determined with an SWR meter or a wattmeter. These
don`t require a slotted line`s 1/4-wave or more minimum of space for
operation.

The maximum and minimum voltages on a transmission line are not as
significant as the forward and reflected powers because the difference
between these powers is the power delivered to the load and is also the
power supplied by the transmitter as the transmission line has no
storage capacity beyond that required to completely energize the line in
both directions by the traveling wave.

Best regards, Richard Hsarrison, KB5WZI

Tom Bruhns wrote:
"Would you then say that R3 is dissipating---0.50 watts?"

Certainly not. The two 1.5 V cells don`t put any current through each
other because they have no potential difference. The two resistors, one
in series with each cell, each drop 0.5 wolts from 1/2 amp through 1 ohm
in each case. So, there`s one volt across the common 1-ohm resistor, R3.

1 V x 1 A = 1 watt dissipated in R3.

Best regards, Richard Harrison, KB5WZI