Walter Maxwell wrote in
:

....
It appears to me that even with all the successive posts on the
subject of power in the standing wave, you all seem to be missing the
ingredient that proves why there is no useable power in the standing
wave. It is because the current and voltage in the standing wave are
90° out of phase. Multiplying E x I under this condition results in
zero power.

Walt,

I am trying to make sense of this and the first issue is what you mean by
the term "standing wave".

The only meaning that seems possible is that it is the magnitude of the
time alternating voltage or current at some displacement along the
transmission line.

If that is the meaning, then the situation you describe of 90° phase
difference between E and I is rather specific, it can only occur with a
distortionless line AND a load that is (s/c OR o/c OR purely reactive).

Is that the case?

If so, should you have stated the assumptions and how does the case you
discuss help in explanation of general principles?


In addition to another comment above that implies that reflected power
is reactive power, this is not true--reflected power is as real as
forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in
phase in the forward direction, they are traveling 180° out of phase
in the rearward direction. Multiplying voltage and current while 180°
different in phase results in the same power as when they are in
phase.

You seem to be inferring that it is legitimate (in a general sense) to
calculate the power of forward and reflected waves as voltage times
current (eg Vf*If).

Isn't the instantaneous power at a point a function of time, and it is p
(t)=v(t)*i(t) and the expansion of that equals Vf*If-Vr*Ir ONLY when the
other two terms of the expansion cancel, and that is the special case of
a distortionless line.

Are you illustrating general principles with a special case without
stating the underlying assumptions.

Why is it that so many attempts to explain transmission line behaviour,
particularly regarding real and imaginary components of power at a point,
aren't consistent with basic AC circuit theory?

Owen

Owen Duffy wrote:
Walter Maxwell wrote:
It appears to me that even with all the successive posts on the
subject of power in the standing wave, you all seem to be missing the
ingredient that proves why there is no useable power in the standing
wave. It is because the current and voltage in the standing wave are
90° out of phase. Multiplying E x I under this condition results in
zero power.

I am trying to make sense of this and the first issue is what you mean by
the term "standing wave".

Walt is right.
Let's look at an arbitrary example of forward and reflected
voltage and current instantaneous phasors at one point on a
particular line.

Vfor = 100v at 45 degrees, Ifor = 2 amps at 45 degrees

The forward voltage and forward current are in phase.

Vref = 100v at -45 degrees, Iref = 2 amps at 135 degrees

The reflected voltage and reflected current are 180 degrees
out of phase.

Now calculate the total voltage and total current.

Total voltage = 2*100cos(45) = 141.4v at 0 deg

Total current = 2*2sin(45) = 2.83a at 90 deg

For a pure standing wave, the instantaneous voltage is
*always* 90 degrees out of phase with the instantaneous
current. There are no V*I*cos(A) watts in a pure standing
wave. There are only V*I*sin(A) VARS.

However, the VARS in the standing wave require energy
which can be converted to watts by I^2*R losses.
--
73, Cecil http://www.w5dxp.com

Cecil, W5DXP wrote:
However the VARS in the standing wave require energy which can be
converted to watts---."

VARS is an acronym for Volt Amps Reactive.
Apparent power can include real power and VARS. I would think that VARS
all have volts and amps in quadrature (at 90 degrees). If so, power is
VI cos theta. WI cos 90 degrees = VI (0)= 0, thus the power in VARS is
0.

Richard Harrison wrote:
Cecil, W5DXP wrote:
However the VARS in the standing wave require energy which can be
converted to watts---."

VARS is an acronym for Volt Amps Reactive.
Apparent power can include real power and VARS. I would think that VARS
all have volts and amps in quadrature (at 90 degrees). If so, power is
VI cos theta. WI cos 90 degrees = VI (0)= 0, thus the power in VARS is
0.

It takes joules of energy for VARS to exist. Any time one
wants to give up the VARS, they can be converted to watts,
just like the energy stored in a capacitor can be converted
to watts by connecting a resistor.

The VARS stored in the standing waves in a transmission line
can be converted to watts by connecting a load equal to the
Z0 of the line. Of course, the standing waves cease to exist
in the process. One cannot have one's cake and eat it too.
--
73, Cecil http://www.w5dxp.com

Owen Duffy wrote:
"If that is the meaning, then the situation you describe of 90-degree
phase difference between E and I is rather specific, it can only occur
with a distortionless line AND a load that is (s/c OR o/c OR purely
reactive).

No. Walter is exactly right, and don`t drag any distortionless line into
the discussion. That is a device for audio circuits. For RF, you only
need a low-loss (Zo = R) transmission line.

Best regards, Richard Harrison, KB5WZI

(Richard Harrison) wrote in news:2484-476F9B75-
:

....
No. Walter is exactly right, and don`t drag any distortionless line into
the discussion. That is a device for audio circuits. For RF, you only
need a low-loss (Zo = R) transmission line.
....

Richard, a lossless line is a special case of a distortionless line (ie all
lossless lines are distortionless lines, but not vice versa).

Specifying a distortionless line is less limiting than restriction to only
lossless lines.

You are actually proposing a greater restriction than I, and it isn't
necessary.

Owen