On Jan 26, 12:15*pm, Roger Sparks wrote:
On Fri, 25 Jan 2008 19:13:31 -0800 (PST)





Keith Dysart wrote:
On Jan 24, 10:33*pm, Roger Sparks wrote:
[snip]
By examining this derivation, the reader can see that power and energy
is reflected when a wave encounters a discontinuity. *The reader can
also see that more power is present on the transmission line than is
delivered to the load.

This is the conventional phraseology for describing the behaviour at
the impedance discontinuity.

Allow me to offer a specific example for which this phraseology is
inappropriate.

Consider a 50 V step function generator with an output impedance of
50 ohms driving a 50 ohm line that is 1 second long terminated in an
open circuit.

Turn on the generator. A 50 V step propagates down the line. The
generator is putting 50 J/s into the line. One second later it
reaches the open end and begins propagating backwards.
After two seconds it reaches the generator. The voltage at the
generator is now 100 V and no current is flowing from the
generator into the line. In the 2 seconds, the generator put
100 joules into the line which is now stored in the line.
The line is at a constant 100 V and the current is zero everywhere.

Computing Pf and Pr will yield 50 W forward and 50 W reflected.
And yet no current is flowing anywhere. The voltage on the line
is completely static.

And yet some will claim that 50 W is flowing forward and 50 W
is flowing backwards.

Does this seem like a reasonable claim for an open circuited
transmission line with constant voltage along its length and
no current anywhere?

I do not find it so.

...Keith

This is a reasonable observation for a static situation where energy is stored on a transmission line. *

If the example contained an ongoing consideration, like "Where does the power move to?", then it would be reasonable to consider that the wave continued to move, simply to avoid the complication of what EXACTLY happens when a wave starts and stops.

So have you thought about "where does the power go?"

When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), where does the
reflected power go?
Does it enter the generator?
Is it dissipated somewhere?

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

...Keith

Keith Dysart wrote:
When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), ...

On the contrary, it is redistributed back toward
the load in the process of destructive interference
and becomes constructive interference associated
with the forward wave. Whether you call that a
reflection or not, the fact that the forward
power equals the source power plus the reflected
power tells us that reflected power being dissipated
in the source would violate the conservation of
energy principle.

where does the reflected power go?

Power doesn't flow so
reflected power doesn't "go" anywhere. It is the
reflected energy that is flowing, i.e. going
somewhere. If the power were flowing, its
dimensions would be joules/sec/sec. Maybe
you would like to try to explain the physical
meaning of those dimensions?

Does it enter the generator?

forward power = source power + reflected power

A component of the forward energy is equal in
magnitude to the reflected energy so no, it
doesn't enter the source.

Is it dissipated somewhere?

Not during steady-state. During steady-state it
is being used for impedance transformation. After
steady-state, it is dissipated either in the
source or in the load as a traveling wave.

Answers to these questions will quickly lead to
doubts about the *reality* of "reflected power".

Reflected power is the Poynting vector associated
with the reflected wave. It exists at a point. It's
average magnitude (indirectly measured by a Bird) is
Re(ExH*)/2. It's direction is the direction of the
flow of reflected energy.

If you don't believe in reflected energy, let your
fingers become the circulator load resistor for
an open-circuit stub being driven by a KW source.
--
73, Cecil http://www.w5dxp.com

On Jan 27, 10:26*am, Cecil Moore wrote:
Keith Dysart wrote:
When the generator is matched to the line so that
the reflected wave does not encounter an impedance
discontinuity when it arrives back at the generator
(and therefore is not reflected), ...

On the contrary, it is redistributed back toward
the load in the process of destructive interference
and becomes constructive interference associated
with the forward wave. Whether you call that a
reflection or not, the fact that the forward
power equals the source power plus the reflected
power tells us that reflected power being dissipated
in the source would violate the conservation of
energy principle.

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.

Some parts of the rest of your post are correct by
coincidence, but since the underlying premise of
reflections where there is no discontinuity is
incorrect, I have snipped it.

But this debate has been had before. You do not want
to understand how the output impedance of a generator
affects a returning signal. I have offerred references
and you have refused to look. I have offerred spice
simulations, and you have refused to look. When the
discussion moves to simpler generators so that the
behaviour can be studied, you will declare them
uninteresting because they do not represent "real
ham transmitters". You will make jokes about 10 cent
resistors, not realizing that is how real test
equipment prevents re-reflection. (How well would
a TDR work, if any substantial amount of the return
was reflected?)

When you decide that you do not want to argue that
reflections occur where there is no impedance
discontinuity, and are willing to study output
immpedance, the learning can begin.

...Keith

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.

When you decide that you do not want to argue that
reflections occur where there is no impedance
discontinuity, ...

You simply fail to recognize the impedance
discontinuity.

Please perform the following experiment to prove
there is no impedance discontinuity and no distortion
and get back to us.

zero ohm
TV RCVR--+--TV source--+--Z0 stub----------open
|
Z0 source impedance resistor
|
GND

The source output goes into the stub. Reflections
occur at the open end of the stub and flow back
through the zero impedance source to be dissipated
in the Z0 source resistor and displayed without
distortion on the TV RCVR. If what you say is true,
it should be duck soup for you to prove.
--
73, Cecil http://www.w5dxp.com

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)

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)

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

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.

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

Please provide an example of an amateur radio transmitter
with a well-defined impedance seen by reflected waves.
If you had done that 30 years ago, the argument would
never have existed in the first place.

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

One cannot have one's source power and eat it too.
If the reflected energy is part of the forward energy,
it cannot also flow into the source.

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

How can there be no reflection at the source when
the source is dissipating zero power?

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

If those ideal components are far removed from being
realizable in the real-world, you have yourself a deal.
Until you can provide an example of a real-world source
approaching a zero real-world impedance, there is no
reason to read any further. You might as well just say,
"God causes everything." and be done with it.
--
73, Cecil http://www.w5dxp.com

On Feb 4, 5:54*pm, Cecil Moore wrote:
Keith Dysart wrote:
Your suspicion is quite incorrect. Real-world generators
which present well defined impedances to the reflected
wave are common.

Please provide an example of an amateur radio transmitter
with a well-defined impedance seen by reflected waves.
If you had done that 30 years ago, the argument would
never have existed in the first place.

Non-sequitor. If the impedance is not well-defined,
then it is impossible to draw any conclusions about
reflections at the generator. So any claim of total
reflection in such an environment would be invalid.

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

One cannot have one's source power and eat it too.
If the reflected energy is part of the forward energy,
it cannot also flow into the source.

I observe that you refuse to explore the example that
demonstrates that this is not an issue.

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

How can there be no reflection at the source when
the source is dissipating zero power?

In the example, the source is not dissipating zero
power. You should consider examining the example.

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

If those ideal components are far removed from being
realizable in the real-world, you have yourself a deal.
Until you can provide an example of a real-world source
approaching a zero real-world impedance, there is no
reason to read any further. You might as well just say,
"God causes everything." and be done with it.

Any excuse to avoid the risk of learning that your
beliefs are incorrect.

Take a chance. Try to find the flaws in the analysis
of this simplest of examples. If there are flaws,
they should not take you long to locate.

...Keith

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

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.

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