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.