Keith Dysart wrote:
Cecil Moore wrote:
I didn't refuse to explore the example. You told me
to "read no further" so I did just that.

There was an 'if' clause. YOU made the choice to avoid
the opportunity for learning.

It was an obvious joke, Keith. I just analyzed your
analysis and pointed out that your choice was a special
case that made it OK to superpose powers. Shame on you.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
The power provided by the source is equal
to the power dissipated in the source resistor and the
power dissipated in the load resistor.

49.999 = 31.249 + 18.75

So all the energy is accounted for, as expected.

I'm surprised you don't see your own contradiction.

If, as you say, the reflected wave is not reflected by
the source, then the reflected wave flows through the
source resistor to ground and is dissipated.

If the reflected wave is dissipated in the source
resistor, it cannot join the forward wave.

The forward wave is 25 joules/sec. The source is
supplying 18.75 joules/sec. If the reflected wave
is dissipated in the source, where is the other
6.25 joules/sec coming from? Is it mere coincidence
that the reflected wave is 6.25 joules/sec????

Hint: You cannot eat your reflected wave and have it too.

If the forward power is greater than the source power,
the reflected wave is joining the forward wave, i.e.
being reflected by the source.

Here is an example of the reflected wave flowing through
the circulator resistor and being dissipated.

Source---1---2----45 deg 50 ohm feedline---150 ohm load
\ /
|
50 ohms

But in this example Psource = Pforward in order to satisfy
the conservation of energy principle which your example
does not.
--
73, Cecil http://www.w5dxp.com

On Mon, 4 Feb 2008 12:32:39 -0800 (PST)
Keith Dysart wrote:

On Jan 28, 11:30 am, Cecil Moore wrote:
Keith Dysart wrote:
Unfortunately, this is quite wrong. And I continue to
be surprised that you argue that there is a reflection
where there is not an impedance discontinuity.

Since an absence of reflections violates the conservation
of energy principle, there is something wrong with your
assertion and your earlier example was proved to contain
a contradiction.

Psource = Pfor - Pref = Pload

Pfor = Psource + Pref = Pload + Pref

Those equations are true only if reflected energy
does not flow back into the source. I suspect that,
contrary to your assertions, the actual real-world
source presents an infinite impedance to reflected
waves.

Your suspicion is quite incorrect. Real-world generators
which present well defined impedances to the reflected
wave are common.

And your assertion that a lack of reflections will
result in a violation of the conservation of energy
principle is also incorrect.

The equations you present above hold just as well
when there is no reflection at the source.

It might do to analyze a simple example. To simplify
analysis, this example does use ideal components:
there is an ideal 50 ohm transmission line (R=G=0,
uniform 50 ohm impedance along its length), two
ideal resistors (zero tempco, no inductance or
capacitance, just resistance), and an ideal voltage
source (desired voltage is produced regardless of
load impedance, though the external components mean
this source does not need to provide or sink infinite
current).

If you don't think any understanding can be gained
from examples with ideal components, please read
no further.

This simple example has a voltage source in series
with a 50 ohm resistor, driving 45 degrees of 50
ohm line connected to a 150 ohm load.

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs 45 degrees \ Rl
100 cos(wt) 50 ohm line / 150 ohm
| \ load
| |
+--------------+----------------------+
gnd

The voltage source in series with the 50 ohm
resistor forms a generator.

Vs(t) = 100 cos(wt)

or, in phasor notation

Vs = 100 /_ 0d

where d is degrees or 2*pi/360 when converting to radians
is appropriate.

When the source is turned on, a wave travels down the
line. At the generator terminal (Vg), the forward voltage
is:
Vf.g(t) = 50 cos(wt)
Vf.g = 50 /_ 0d
since the voltage divided evenly between the source
50 ohm resistor and 50 ohm impedance of the line.

At the load, the voltage reflection coefficient is

RCvl = (150-50)/(150+50)
= 0.5

The forward voltage at the load is

Vf.l(t) = 50 cos(wt - 45d)
Vf.l = 50 /_ -45d

The reflected voltage at the load is

Vr.l(t) = 0.5 * 50 cos(wt - 45d)
= 25 cos(wt - 45d)
Vr.l = 25 /_ -45d

The voltage at the load is

Vl = Vf.l + Vr.l
= 50 /_ -45d + 25 /_ -45d
= 75 /_ -45d
Vl(t) = 75 cos(wt - 45d)

At the generator, the voltage reflection coefficient is

RCvg = (50-50)/(50+50)
= 0

so there is no reflection and the system has settled
after one round trip.

The voltage at the generator is, therefore,

Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-45d -45d)
= 50 /_ 0d + 25 /_ -90d
= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

Whoa! You started with 50 volts applied to the forward line, then froze the wave by finding the time phase of the wave. Mathematically, you want to find the voltage at the generator which is the sum of the terms Vf.g and Vr.g. The sum of those terms is 50 + 0 = 50, not 55.902 /_ -26.565d.


Currents can be similar computed...

If.g(t) = 1 cos(wt)
If.g = 1 /_ 0d

RCil = -(150-50)/(150+50)
= -0.5

If.l(t) = 1 cos(wt - 45d)
If.l = 1 /_ -45d

Ir.l(t) = -0.5 * 1 cos(wt - 45d)
= -0.5 cos(wt - 45d)
Ir.l = -0.5 /_ -45d

Il = If.l + Ir.l
= 1 /_ -45d - 0.5 /_ -45d
= 0.5 /_ -45d
Il(t) = 0.5 cos(wt - 45d)

And just to check, from V = I*R
R = V / I
= Vl(t) / Il(t)
= (75 cos(wt-45d)) / (0.5 cos(wt-45d))
= 150
as expected.

RCig = -(50-50)/(50+50)
= 0

Ig = If.g + Ir.g
= 1 /_ 0d - 0.5 /_ (-45d -45d)
= 1 /_ 0d -0.5 /_ -90d
= 1.118 /_ 26.565d
Ig(t) = 1.118 cos(wt + 26.565d)

The same error here.
Ig = 1 /_ 0d - 0.5 /_ -90d
= 1 - 0
= 1

Now let us see if all the energy flows balance properly.

The power applied to the load resistor is

Pl(t) = Vl(t) * Il(t)
= 75 cos(wt - 45d) * 0.5 cos(wt - 45d)

Recalling the relation
cos(a) * cos(b) = 0.5(cos(a+b)+cos(a-b))

Pl(t) = 75 * 0.5 * 0.5 (cos(2wt-90d)+cos(0))
= 18.75 cos(2wt-90d) + 18.75 cos(0)
= 18.75 cos(2wt-90d) + 18.75

Since the average of cos is 0, the average power is

Pl.avg = 18.75

To confirm, let us compute using

Pavg = Vrms * Irms * cos(theta)

Pl.avg = (75 * 0.707) * (0.5 * 0.707) * cos(-45d-(-45d))
= 18.75 * 1
= 18.75

Agreement.

Now at the generator end

Pg(t) = Vg(t) * Ig(t)
= 55.902 cos(wt - 26.565d) * 1.118 cos(wt + 26.565d)
= 55.902 * 1.118 * 0.5 (cos(2wt)+cos(-53.130))
= 31.249 cos(2wt) + 31.249 * 0.6
= 31.249 cos(2wt) + 18.75
Pg.avg = 18.75

No. Based on the way I read your CORRECTED sums, the power at the generator Pg should be 50 * 1 = 50.


Good news, the average power into the line is equal to
the average power delivered to the load, as required to
achieve conservation of energy.

Now let us check Pf and Pr

Pf.avg = Vfrms^2 / Zo
= (50*.707)^2 / 50
= 25

Pr.avg = Vrrms^2 / Zo
= (25*.707)^2 / 50
= 6.25

Pnet.avg = Pf.avg - Pr.avg
= 25 - 6.25
= 18.75

Again, as expected, the average energy flow in the line
is equal to the average power into the line and the average
power out of the line.

So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.

It is instructive to continue the cross-checking.
Let us validate the input impedance of the line.

Using a Smith chart, and going towards the source from
a 150 ohm load on a 50 ohm line yields

Zin.smith = 30 - j40
= 50 /_ -53.130

Dividing the voltage by the current at the input to the
line...

Zin = Vg / Ig
= 55.902 /_ -26.565d / 1.118 /_ 26.565d
= 50.002 /_ -53.130
= 30.001 -j 40.001

yields the same result as the Smith chart.

Lastly let's compute the power provided by the source...

Ps(t) = Vs(t) * Is(t)
= 100 cos(wt) * 1.118 cos(wt+26.565)
= 111.8 * 0.5 (cos(2wt+26.565) + cos(-26.565))
= 55.9 cos(2wt + 26.565) + 55.9 * 0.894
= 55.9 cos(2wt + 26.565) + 49.999

Ps.avg = 49.999

and that dissipated in the source resistor

Prs(t) = Vrs(t) * Irs(t)
= (Vs(t)-Vg(t)) * Ig(t)
= (100 cos(wt) - 55.902 cos(wt - 26.565d)) * 1.118 cos(wt +
26.565d)
= 55.902 cos(wt + 26.565d) * 1.118 cos(wt + 26.565d)
= 55.902 * 1.118 * 0.5 (cos(2wt + 53.130) + cos(0))
= 31.249 cos(2wt + 53.130d) + 31.249

Prs.avg = 31.249

The power provided by the source is equal
to the power dissipated in the source resistor and the
power dissipated in the load resistor.

49.999 = 31.249 + 18.75

So all the energy is accounted for, as expected.

When the source impedance is the same as the line
impedance, there are no reflections at the source
and all the energy is properly accounted. There is
no violation of the conservation of energy principle.

...Keith

PS. It is worthwhile to note that there are a large
number (infinite) of source voltage and source
resistor combinations that will produce exactly
the same final conditions on the line. But if
the source resistor is not 50 ohms, then there
will be reflections at the source and it will
take infinite time for the line to settle to
its final state.

With the circuit and conditions presented, the reflected voltage and current return out of phase in time with the source voltage and current. A standing wave condition of 3:1 would always exist throughout the system.
--
73, Roger, W7WKB

On Feb 5, 5:59*am, Cecil Moore wrote:
Keith Dysart wrote:
This simple example has a voltage source in series
with a 50 ohm resistor, driving 45 degrees of 50
ohm line connected to a 150 ohm load.

* * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * *+----/\/\/-----+----------------------+
* * *| * *50 ohm * * * * * * * * * * * * * |
* * *| * * * * * * * * * * * * * * * * * * /
* * Vs * * * * * * * * 45 degrees * * * * *\ * Rl
*100 cos(wt) * * * * * 50 ohm line * * * * / 150 ohm
* * *| * * * * * * * * * * * * * * * * * * \ *load
* * *| * * * * * * * * * * * * * * * * * * |
* * *+--------------+----------------------+
* * gnd

The voltage source in series with the 50 ohm
resistor forms a generator.

Vs(t) = 100 cos(wt)

So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.

Keith, this special case is covered in my Worldradio
energy analysis article at:

http://www.w5dxp.com/energy.htm

I agree with your special case analysis. The reason that
it is a special case is that at the generator terminals:

(Vf)^2 + (Vr)^2 = (Vf+Vr)^2

i.e. you have chosen the one special case where the
superposition of powers yields a valid result.

Quoting my article with some special characters replaced by
ASCII characters:

"If A = 90 deg, then cos(A) = 0, and there is no
destructive/constructive interference between V1 and V2. There
is also no destructive/constructive interference between V3 and
V4. Any potential destructive/constructive interference between
any two voltages is eliminated because A = 90 deg, i.e. the
voltages are superposed orthogonal to each other (almost as if
they were not coherent)."

At the generator terminals, you have chosen the one phase
angle between Vf and Vr that will result in zero interference.
Whether by accident or on purpose is unclear. The phase angle
between Vf and Vr at the generator terminals is 90 degrees
resulting in zero interference. Your analysis is completely
correct for this special case. Now try it for any length of
feedline between 0 deg and 180 deg besides 45 deg and 135 deg.
90 degrees would be very easy to calculate.

I choose 45 degrees because it was not 90, the standard value
used which leaves all sorts of misleading artifacts when doing
analysis. I noticed that a few artifacts also exist at 45 but
I had hoped they would not mislead. I see I was wrong.

Below, I have redone the analysis with a 35 degree line.

The analysis is based on no reflection occurring at the
generator, and the energy flows still balance as expected.

The analysis for a 35 degree line follow.

--------

This simple example has a voltage source in series
with a 50 ohm resistor, driving 35 degrees of 50
ohm line connected to a 150 ohm load.

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs 35 degrees \ Rl
100 cos(wt) 50 ohm line / 150 ohm
| \ load
| |
+--------------+----------------------+
gnd

The voltage source in series with the 50 ohm
resistor forms a generator.

Vs(t) = 100 cos(wt)

or, in phasor notation

Vs = 100 /_ 0d

where d is degrees or 2*pi/360 when converting to radians
is appropriate.

When the source is turned on, a wave travels down the
line. At the generator terminal (Vg), the forward voltage
is:
Vf.g(t) = 50 cos(wt)
Vf.g = 50 /_ 0d
since the voltage divided evenly between the source
50 ohm resistor and 50 ohm impedance of the line.

At the load, the voltage reflection coefficient is

RCvl = (150-50)/(150+50)
= 0.5

The forward voltage at the load is

Vf.l(t) = 50 cos(wt - 35d)
Vf.l = 50 /_ 35d

The reflected voltage at the load is

Vr.l(t) = 0.5 * 50 cos(wt - 35d)
= 25 cos(wt - 35d)
Vr.l = 25 /_ -35d

The voltage at the load is

Vl = Vf.l + Vr.l
= 50 /_ -35d + 25 /_ -35d
= 75 /_ -35d
Vl(t) = 75 cos(wt - 35d)

At the generator, the voltage reflection coefficient is

RCvg = (50-50)/(50+50)
= 0

so there is no reflection and the system has settled
after one round trip.

The voltage at the generator is, therefore,

Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-35d -35d)
= 50 /_ 0d + 25 /_ -70d
= 63.087640 /_ -21.862219d
Vg(t) = 63.087640 cos(wt - 21.86221d)

Currents can be similar computed...

If.g(t) = 1 cos(wt)
If.g = 1 /_ 0d

RCil = -(150-50)/(150+50)
= -0.5

If.l(t) = 1 cos(wt - 35d)
If.l = 1 /_ -35d

Ir.l(t) = -0.5 * 1 cos(wt - 35d)
= -0.5 cos(wt - 35d)
Ir.l = -0.5 /_ -35d

Il = If.l + Ir.l
= 1 /_ -35d - 0.5 /_ -35d
= 0.5 /_ -35d
Il(t) = 0.5 cos(wt - 35d)

And just to check, from V = I*R
R = V / I
= Vl(t) / Il(t)
= (75 cos(wt-35d)) / (0.5 cos(wt-35d))
= 150
as expected.

RCig = -(50-50)/(50+50)
= 0

Ig = If.g + Ir.g
= 1 /_ 0d - 0.5 /_ (-35d -35d)
= 1 /_ 0d - 0.5 /_ -70d
= 0.952880 /_ 29.543247d
Ig(t) = 0.952880 cos(wt + 29.54324d)

Now let us see if all the energy flows balance properly.

The power applied to the load resistor is

Pl(t) = Vl(t) * Il(t)
= 75 cos(wt - 35d) * 0.5 cos(wt - 35d)

Recalling the relation
cos(a) * cos(b) = 0.5(cos(a+b)+cos(a-b))

Pl(t) = 75 * 0.5 * 0.5 (cos(2wt-70d)+cos(0))
= 18.75 cos(2wt-70d) + 18.75 cos(0)
= 18.75 cos(2wt-70d) + 18.75

Since the average of cos is 0, the average power is

Pl.avg = 18.75

To confirm, let us compute using

Pavg = Vrms * Irms * cos(theta)

Pl.avg = (75 * 0.707) * (0.5 * 0.707) * cos(-35d-(-35d))
= 18.75 * 1
= 18.75

Agreement.

Now at the generator end

Pg(t) = Vg(t) * Ig(t)
= 63.087640 cos(wt - 21.862212d) * 0.952880 cos(wt + 29.543247d)
= 63.087640 * 0.952880 * 0.5 (cos(2wt+7.681035)+cos(-51.405466))
= 30.057475 cos(2wt) + 30.057475 * 0.623805
= 30.057475 cos(2wt) + 18.750004
Pg.avg = 18.75

Good news, the average power into the line is equal to
the average power delivered to the load, as required to
achieve conservation of energy.

Now let us check Pf and Pr

Pf.avg = Vfrms^2 / Zo
= (50*.707)^2 / 50
= 25

Pr.avg = Vrrms^2 / Zo
= (25*.707)^2 / 50
= 6.25

Pnet.avg = Pf.avg - Pr.avg
= 25 - 6.25
= 18.75

Again, as expected, the average energy flow in the line
is equal to the average power into the line and the average
power out of the line.

So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.

It is instructive to continue the cross-checking.
Let us validate the input impedance of the line.

Using a Smith chart
(http://education.tm.agilent.com/index.cgi?CONTENT_ID=5),
and going towards the source from a 150 ohm load on a 50 ohm
line yields

Zin.smith = 41.28 - j51.73
= 66.18 /_ -51.41

Dividing the voltage by the current at the input to the
line...

Zin = Vg / Ig
= 63.087640 /_ -21.862219d / 0.952880 /_ 29.543247d
= 66.207329 /_ -51.405466
= 41.300466 -j51.746324

yields the same result as the Smith chart.

Lastly let's compute the power provided by the source...

Ps(t) = Vs(t) * Is(t)
= 100 cos(wt) * 0.952880 cos(wt+29.543247d)
= 95.288000 * 0.5 (cos(2wt+29.543247d) + cos(-29.543247d))
= 47.644000 cos(2wt + 29.543247) + 47.644000 * 0.869984
= 47.644000 cos(2wt + 29.543247) + 41.449507

Ps.avg = 41.449507

and that dissipated in the source resistor

Prs(t) = Vrs(t) * Irs(t)
= (Vs(t)-Vg(t)) * Ig(t)
= (100 cos(wt) - 63.087640 cos(wt - 21.862219d)) * 0.95288
cos(wt + 29.543247d)
= 47.643989 cos(wt + 29.543247d) * 0.952880 cos(wt +
29.543247d)
= 47.643989 * 0.952880 * 0.5 (cos(2wt + 59.086480d) + cos(0))
= 22.699502 cos(2wt + 59.086480d) + 22.699502

Prs.avg = 22.699502

The power provided by the source is equal
to the power dissipated in the source resistor and the
power dissipated in the load resistor.

41.44950 = 22.699502 + 18.75

So all the energy is accounted for, as expected.

When the source impedance is the same as the line
impedance, there are no reflections at the source
and all the energy is properly accounted. There is
no violation of the conservation of energy principle.

...Keith

PS. It is worthwhile to note that there are a large
number (infinite) of source voltage and source
resistor combinations that will produce exactly
the same final conditions on the line. But if
the source resistor is not 50 ohms, then there
will be reflections at the source and it will
take infinite time for the line to settle to
its final state.

On Feb 5, 8:53*am, Roger Sparks wrote:
On Mon, 4 Feb 2008 12:32:39 -0800 (PST)
Keith Dysart wrote:
[snip]
The voltage at the generator is, therefore,

Vg = Vf.g + Vr.g
* *= 50 /_ 0d + 25 /_ (-45d -45d)
* *= 50 /_ 0d + 25 /_ -90d
* *= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

Whoa! *You started with 50 volts applied to the forward line, then froze the wave by finding the time phase of the wave. *Mathematically, you want to find the voltage at the generator which is the sum of the terms Vf.g and Vr.g. *The sum of those terms is 50 + 0 = 50, not *55.902 /_ -26.565d..

Vr.g is the reflection of Vf from the load so it is
a delayed version of Vr.l, the delay being another
45 degrees.

The reflected voltage at the load is
25 /_ -45d

Another 45 degrees makes it
25 /_ -90d
by the time it gets back to the generator terminal.

So the total voltage at the generator is
Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-45d -45d)
= 50 /_ 0d + 25 /_ -90d
= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

[snip]

Similarly for Ig and Ig(t) and Pg.avg and Pg(t).

[snip]

With the circuit and conditions presented, the reflected voltage and current return out of phase in time with the source voltage and current. *A standing wave condition of 3:1 would always exist throughout the system.

Agreed, though it might be slightly more correct to say
that there is a 3:1 VSWR on the transmission line.

...Keith

Keith Dysart wrote:
So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.

Again your analysis violates the conservation of energy
principle. The only way to balance your energy equation
is for your source to supply 25 joules/sec.

IF THE REFLECTED WAVE IS NOT REFLECTED FROM THE SOURCE,
IT FLOWS THROUGH THE SOURCE RESISTOR AND IS DISSIPATED
IN THE SOURCE RESISTOR. THAT REFLECTED WAVE ENERGY IS
THEREFORE NOT AVAILABLE TO THE FORWARD WAVE.

You cannot eat your reflected wave and have it too.

The standard equation is:

Psource = Pfor - Pref = Pload

But you have taken away the reflected energy and
dissipated it in the source resistor. So your new
equation becomes:

Psource = Pfor + ????????
18.75w = 25w + ________

You say the source power is 18.75 joules/sec and the
forward power is 25 joules/sec. If none of the reflected
energy is available to the forward wave, where did the
extra 6.25 joules/sec come from? Is it sheer coincidence
that the reflected wave is associated with 6.25 joules/sec
that are now missing from the above equation?

Here is an example of the reflected wave flowing through
a circulator resistor and being dissipated.

Source---1---2----45 deg 50 ohm feedline---150 ohm load
25w \ / 18.75w
|
50 ohms
6.25w

But in this example Psource = Pfor in order to satisfy
the conservation of energy principle which your example
does not.
--
73, Cecil http://www.w5dxp.com

On Feb 5, 7:40*am, Cecil Moore wrote:
Keith Dysart wrote:
The power provided by the source is equal
to the power dissipated in the source resistor and the
power dissipated in the load resistor.

49.999 = 31.249 + 18.75

So all the energy is accounted for, as expected.

I'm surprised you don't see your own contradiction.

If, as you say, the reflected wave is not reflected by
the source, then the reflected wave flows through the
source resistor to ground and is dissipated.

If the reflected wave is dissipated in the source
resistor, it cannot join the forward wave.

The forward wave is 25 joules/sec. The source is
supplying 18.75 joules/sec. If the reflected wave
is dissipated in the source, where is the other
6.25 joules/sec coming from? Is it mere coincidence
that the reflected wave is 6.25 joules/sec????

Hint: You cannot eat your reflected wave and have it too.

If the forward power is greater than the source power,
the reflected wave is joining the forward wave, i.e.
being reflected by the source.

I wondered if this bit of numerology would lead to
confusion, and apparently it has.

Consider what happens when the source is first turned on.
Immediately
Pf.avg = 25 W

Sometime later the forward wave reaches the load and a
reflected wave is returned
Pr.avg = 6.25 W

Later, this wave arrives back at the generator.

How does this reflected wave of 6.25 W affect the already
present forward wave of 25 W when it arrives back at the
generator. It affects it not one iota. The forward power
remains the same, the forward voltage remains the same
and the forward current remains the same. This is the
meaning of no reflection.

The one thing that changes is the net average power which
decreases to 18.75 W.

I would suggest that you start with the circuit I proposed
and analyze it with the technique of your choice.
What is the final Vf, Vr, Pf.avg, Pr.avg, Pg.avg and Pl.avg?
Are your answers different than mine?
If so, we can explore which approach is in error.
What happens at the generator when the first reflection
arrives back at the generator?
Does Vf and Pf.avg change? Or remain the same?

When Pf.avg remains the same, how can it be claimed that
the reflected wave is re-reflected?

...Keith

On Tue, 5 Feb 2008 07:36:05 -0800 (PST)
Keith Dysart wrote:

On Feb 5, 8:53*am, Roger Sparks wrote:
On Mon, 4 Feb 2008 12:32:39 -0800 (PST)
Keith Dysart wrote:
[snip]
The voltage at the generator is, therefore,

Vg = Vf.g + Vr.g
* *= 50 /_ 0d + 25 /_ (-45d -45d)
* *= 50 /_ 0d + 25 /_ -90d
* *= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

Whoa! *You started with 50 volts applied to the forward line, then froze the wave by finding the time phase of the wave. *Mathematically, you want to find the voltage at the generator which is the sum of the terms Vf.g and Vr.g. *The sum of those terms is 50 + 0 = 50, not *55.902 /_ -26.565d.

Vr.g is the reflection of Vf from the load so it is
a delayed version of Vr.l, the delay being another
45 degrees.

The reflected voltage at the load is
25 /_ -45d

Another 45 degrees makes it
25 /_ -90d
by the time it gets back to the generator terminal.

So the total voltage at the generator is
Vg = Vf.g + Vr.g
= 50 /_ 0d + 25 /_ (-45d -45d)
= 50 /_ 0d + 25 /_ -90d
= 55.902 /_ -26.565d
Vg(t) = 55.902 cos(wt - 26.565d)

OK, 55.902 /_ -26.565d is another way of expressing 50. I should have realized that identity. Sorry to have troubled you.


[snip]

Similarly for Ig and Ig(t) and Pg.avg and Pg(t).

[snip]

With the circuit and conditions presented, the reflected voltage and current return out of phase in time with the source voltage and current. *A standing wave condition of 3:1 would always exist throughout the system.

Agreed, though it might be slightly more correct to say
that there is a 3:1 VSWR on the transmission line.

...Keith
--
73, Roger, W7WKB

Keith Dysart wrote:
How does this reflected wave of 6.25 W affect the already
present forward wave of 25 W when it arrives back at the
generator. It affects it not one iota. The forward power
remains the same, the forward voltage remains the same
and the forward current remains the same. This is the
meaning of no reflection.

Maybe a more drastic example will help.

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| /
Vs 90 degrees \ Rl
100 cos(wt) 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

Current through the source: Is = 0.0392 cos(wt).
The source is supplying 1.96 watts. That is all the
power available to the entire system.

You say the forward power is 25 watts and that none
of the reflected power is reflected at the source.

How can there be 25 watts of forward power when
the source is supplying only 1.96 watts and none
of the reflected energy is joining the forward
wave?
--
73, Cecil http://www.w5dxp.com

On Feb 5, 10:44*am, Cecil Moore wrote:
Keith Dysart wrote:
So, contrary to Cecil's assertion, an analysis based on
'no reflections at the source', has not resulted in any
violation of conservation of energy. This is good.

Again your analysis violates the conservation of energy
principle. The only way to balance your energy equation
is for your source to supply 25 joules/sec.

IF THE REFLECTED WAVE IS NOT REFLECTED FROM THE SOURCE,
IT FLOWS THROUGH THE SOURCE RESISTOR AND IS DISSIPATED
IN THE SOURCE RESISTOR. THAT REFLECTED WAVE ENERGY IS
THEREFORE NOT AVAILABLE TO THE FORWARD WAVE.

Not sure that shouting helps. I do not find any unbalances
with the energy. See below.

You cannot eat your reflected wave and have it too.

The standard equation is:

Psource = Pfor - Pref = Pload

Yes. Indeed. And just changing to use my terminology...

Pgenerator = Pfor - Pref = Pload

For the example at hand...

Pgenerator = 18.75
Pfor - Pref = 18.75
Pload = 18.75

And the source provides 50 W of which 31.25 is dissipated
in the source resistor and 18.75 is delivered to the
line (Pgenerator, above).

All seems well with world and no energy is left unaccounted.

But you have taken away the reflected energy and
dissipated it in the source resistor. So your new
equation becomes:

Psource = Pfor + ????????
18.75w *= 25w *+ ________

As your original equation above

Psource = Pfor - Pref
18.75 = 25 - 6.25

I do not see any issues.

When the source was first turned on, 25 J/s flowed from
the generator into the line.
Pfor = 25 W

After the reflected wave makes it back to the generator,
25 - 6.25 - 18.75 J/s are flowing in the line.
Pfor - Pref = 18.75 W

And the generator output has also reduced from 25 W to
18.75 W, so all is still in balance.

I am having difficulty determining where you think there
is a violation of the conservation of energy principle.

...Keith

PS Do not be fooled by the numerology where 25 + 6.25
gives 31.25. For the 35 degree line, the corresponding
numbers are
Ps = 41.449507
Prs = 22.699502
Pg = 18.75
Pf = 25
Pr = 6.25
Pl = 18.75

Pg = Pf - Pr = Pl
Ps = Prs + Pg
All as expected. No missing energy.