On Feb 8, 9:15*am, Cecil Moore wrote:
Keith Dysart wrote:
When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

Nope, Vf doesn't have to change for there to be a
reflection. Looking at what happens to the power
dissipation in the source resistor, Rs, when the
reflected wave arrives, we see that the energy being
supplied by the source drops by 99.85%. Most of
the energy in the forward wave is supplied by the
reflected wave, starting at the time the reflected
wave arrives and causes destructive interference.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

To be technically correct, reflected energy is redistributed
back toward the load during the process of destructive
interference.

The conditions before the reflected wave arrives and
after the reflected wave arrives are extremely different.

Before the reflected wave arrives, the source resistor,
Rs, is dissipating 25 watts.

After the reflected wave arrives, the source resistor,
Rs, is dissipating 0.03845 watts.

Clearly, something drastic has happened and that is:
99.85% of the forward energy originally supplied by the
source has been replaced with reflected energy being
redistributed back toward the load.

Because of destructive interference, reflected energy
*never* flows through the source resistor, Rs, and is
instead redistributed back toward the load.

Nothing else is possible since the source is supplying
only 1.9608 watts during steady-state. 92.3% of the forward
power is not being supplied by the source during steady-state.

The energy incident upon a point must equal the energy
exiting the point. The energy incident upon the
generator terminals is 1.92234 joules/sec from the
source and 23.0777 joules/sec from the reflected wave.
The energy exiting that point is 25 joules/sec. The
reflected wave energy obviously reverses direction and
joins the forward wave and that is what we call a
"reflection".

Are you sure? I thought a reflection was something that
occurred at an impedance discontinuity and the magnitude
of the voltage reflection was defined by

Vr = Vincident * ReflectionCoefficient
= Vincident * (Z2-Z1)/(Z2+Z1)

and that the reflected voltage then added to any wave
already travelling in that direction.

But you are claiming that power can be reflected when
voltage is not. I have never encountered this claim before.

Are you sure?

...Keith

Keith Dysart wrote:
Are you sure? I thought a reflection was something that
occurred at an impedance discontinuity and the magnitude
of the voltage reflection was defined by

Vr = Vincident * ReflectionCoefficient
= Vincident * (Z2-Z1)/(Z2+Z1)

That's true for normal reflections which involve only
one wave. Wave cancellation is a different kind of
energy reflection involving two waves. The energy flow is
canceled in one direction and therefore flows in the
other direction. In optics, it is known as a redistribution
of energy in directions that allow constructive interference.
In a transmission line, there are only two possible directions
so any redistribution of energy due to destructive interference
can be considered to be a reflection in the opposite direction,
the only direction available to constructive interference.

and that the reflected voltage then added to any wave
already travelling in that direction.

True for a single wave reflection. For two interacting
waves, the voltage in one of the waves can simply replace
the voltage in the other wave. In our example, the reflected
voltage simply replaces the source voltage component.

But you are claiming that power can be reflected when
voltage is not. I have never encountered this claim before.

When the reflected wave arrives, it cancels most of the
existing forward wave from the source. The reflected
voltage exactly equals the canceled source voltage in
our example because the source resistance is the same
as the Z0 of the line.

Are you sure?

It is obvious that reflected energy never flows in the
source resistor so it must go in the only other direction
possible. Yes, I am sure. The conservation of energy
principle will allow nothing else.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Are you sure?

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm 23.08w--Pref |
| 48.08w / 1.92w
Vs 45 degrees \ Rl
100 cos(wt) 50 ohm line / 1 ohm
50w \ load
| |
+--------------+----------------------+
gnd

As a matter of interest, let's return to the 45 degree
line, the one to which I objected previously.

The forward power and reflected power are the same as
in the previous example but now the dissipation in Rs
is equal to the sum of the forward power and reflected
power. (It would be nice if all cases were as straight
forward as this special case.)

This is indeed the one special case where everything you
have been saying is true. There is no redistribution of
energy because there is no interference. There is no
interference because at Vg, Vfor and Vref are 90 degrees
out of phase.

These concepts are covered in my magazine article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com