On Jan 1, 4:16*pm, Cecil Moore wrote:
Keith Dysart wrote:
The Norton or Thévenin equivalent circuits seem *capable of positive
reflection coefficients. *

Either can be positive, negative, or zero depending
on the value of the output impedance compared to Z0.

Would you please quote a reference that addresses
the subject of reflection coefficients from Thevenin
or Norton equivalent sources?

To the best of my knowledge, there is absolutely no
requirement that a Thevenin or Norton equivalent
circuit exhibit the same reflection coefficient
at the circuit it replaces.

It happens without effort because the output impedance
of the Norton/Thevenin equivalent circuit is the same
as the circuit it replaces, otherwise it is not an
equivalent.

...Keith

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
As you well know, the convention is to apply a negative
sign to positive energy flowing in the opposite direction
from the "forward" energy which is arbitrarily assigned
a plus sign.

Let's see even one reference that mentions explicitly the concept of
applying a negative sign to positive energy. Not power, not voltage,
not current, not waves, but energy.

If you keep feigning ignorance like that Gene, you are
going to lose all respect. If the Poynting vector has
a negative sign, as used by Ramo & Whinnery, that sign
is an indication of the *direction of energy flow*,
see quote below.

From Ramo & Whinnery:

The Poynting vector is "the vector giving *direction* and
magnitude of *energy flow*". When Ramo & Whinnery hang a
sign on a Poynting vector in a transmission line, it is
an indication of the direction of energy flow.

For pure standing waves,
"The average [NET] value of Poynting vector is zero
at every cross-sectional plane; this emphasizes the
fact that on the average as much energy is carried
away by the reflected wave as is brought by the
incident wave."

What? Reflected waves "carrying" energy? Shame on
Ramo & Whinnery for contradicting the rraa gurus.

It is impossible to satisfy you, Gene. When I quote
reference after reference about reflected power, you
say power doesn't reflect. When I change it to reflected
energy, you ask for a reference.


Cecil,

Still up to your tricks? I ask for reference on a scalar quantity, and
you respond with some stuff about vectors.

If you want to continue to misinterpret the experts and believe that
power, energy, or whatever flows in opposite directions at a single
point at the same time, go right ahead. I suppose such beliefs expounded
on RRAA are quite harmless in the grand scheme of world affairs.

For future reference, however, just remember: Fields first, then power
or energy. That's the way superposition really works.

(By the way, I am not the one who made the point about power vs. energy.
That must have been someone else.)

73,
Gene
W4SZ

On Jan 1, 4:57*pm, Cecil Moore wrote:
Keith Dysart wrote:
Could you better describe how you determine that the source has a Z0
equal to the line Z0? *I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.

This will do it.

We have a black box. We use a 12vdc battery and a current
meter to measure the impedance of the black box. The
current meter reads zero when we connect the 12vdc battery
to the black box terminals. What is the impedance inside
the black box since test V/I = infinity?

Recall that impedance is the slope and to obtain
the slope you need to do two measurements. Only if
the entity is completely passive can you make only
one measurement because in that case the line passes
through the origin which is the implicit other
measurement.

With two measurements you will obtain the correct
impedance.

....Keith

On Jan 1, 6:00*pm, Roger wrote:
Roy Lewallen wrote:
Roger wrote:

Could you better describe how you determine that the source has a Z0
equal to the line Z0? *I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.

Probably the simplest way is to put the entire source circuitry into a
black box. Measure the terminal voltage with the box terminals open
circuited, and the current with the terminals short circuited. The ratio
of these is the source impedance. If you replace the box with a Thevenin
or Norton equivalent, this will be the value of the equivalent circuit's
impedance component (a resistor for most of our examples).

If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance; if it consists
of a perfect current source in parallel with a resistance, the source Z
will be the resistance. You can readily see that the open circuit V
divided by the short circuit I of these two simple circuits equals the
value of the resistance.

The power output of the Thévenin equivalent circuit follows the load.

Sorry, I don't understand this. Can you express it as an equation?

There seems to be some confusion as to the terms "Thévenin equivalent
* circuit", "ideal voltage source", and how impedance follows these
sources.

Two sources we all have access to are these links:

Voltage source:
*http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

I don't disagree with anything I read there.

But you may not quite have the concept of impedance
correct.

The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.
When there is a source present, this
line does not pass through the origin. Only for
passive components does this line pass through
the origin in which case it becomes V/I.

Because the voltage to current plot for an ideal
voltage source is horizontal, the slope is 0 and hence
so is the impedance. For an ideal current source
the slope is vertical and the impedance is infinite.

Hoping this helps clarify....

...Keith

On Jan 1, 9:03*pm, Cecil Moore wrote:
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
*and I think very revealing. *Yes, if we use part of the model, we must
use it all the way. *To do otherwise would be error, or worse.

Roy and Keith don't seem to realize that the zero source
impedance for the ideal voltage source is only when the
source is turned off for purposes of superposition. They
conveniently avoid turning the source voltage on to complete
the other half of the superposition process. When the
source signal and the reflected wave are superposed at
the series source resistor, where the energy goes becomes
obvious. Total destructive interference in the source
results in total constructive interference toward the load.
See below.

You have been a supporter of this theory for a long time.

Yes, I have. I am a supporter of the principles and laws of
physics. Others believe they can violate the principle of
conservation of energy anytime they choose because the
principle of conservation of energy cannot be violated -
go figure.

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) * or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

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

Can you make this all work for a pulse, or a step
function?

How do you compute
Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
for a pulse or a step?

Or is your approach limited to sinusoids?

...Keith

"Cecil Moore" wrote in message
...
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.

clip some

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.


Well, I would like to think I could understand it, but maybe something is
wrong like Roy suggests. I think it is all falling into place, but our
readers are not all in agreement.

Could I ask a couple of questions to make sure I am understanding your
preconditions?

Is this a Thevenin source? If so, what is the internal resistor set to in
terms of Z0?

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource


For condition number 2 below, is this a Thevenin equivalent resistance?

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

OK. A few reactions.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless the
cos(A) is negative. Something seems wrong here, probably my understanding.

2. The two voltages should be equal, therfore the power delivered by each
to the source resister would be equal.

3. The power delivered to the source resistor will arrive a different times
due to the phase difference between the two waves.

4. If the reflected power returns at the same time as the delivered power
(to the source resistor), no power will flow. This because the resistor in
a Thevenin is a series resistor. Equal voltages will be applied to each side
of the resitor. The voltage difference will be zero.

5. If the reflected power returns 180 degrees out of phase with the applied
voltage (to the source resistor), the voltage across the resistor will
double with each cycle, resulting in an ever increasing current.(and power)
into the reflected wave.


For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

This is correct for this condition. The problem with the equation comes
when cos(90) which can easily happen.

I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the
angle between positive peaks of the two waves. This angle will rotate twice
as fast as the signal frequency due to the relative velocity between the
waves..

Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

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

Light, especially solar energy, is a collection of waves of all magnitudes
and frequencies. We should see only about 1/2 of the power that actually
arrives due to superposition. The reflection from a surface should create
standing waves in LOCAL space that will contain double the power of open
space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be
correct for a collection of waves, but incorrect for the example.

I tried to read your article a couple of weeks ago, but I found myself not
understanding. I have learned a lot since then thanks to you, Roy, and
others. I will try to read it again soon.

So what do you think Cecil?

73, Roger, W7WkB

Correction. Please note the change below. I apologize for the error.

Roger Sparks wrote:
"Cecil Moore" wrote in message
...
Roger wrote:
The principles of superposition are mathematically usable, not too hard,
and I think very revealing. Yes, if we use part of the model, we must
use it all the way. To do otherwise would be error, or worse.

clip some

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.


Well, I would like to think I could understand it, but maybe something is
wrong like Roy suggests. I think it is all falling into place, but our
readers are not all in agreement.

Could I ask a couple of questions to make sure I am understanding your
preconditions?

Is this a Thevenin source? If so, what is the internal resistor set to in
terms of Z0?

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource


For condition number 2 below, is this a Thevenin equivalent resistance?

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

OK. A few reactions.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless the
cos(A) is negative. Something seems wrong here, probably my understanding.

2. The two voltages should be equal, therfore the power delivered by each
to the source resister would be equal.

3. The power delivered to the source resistor will arrive a different times
due to the phase difference between the two waves.

4. If the reflected power returns at the same time as the delivered power
(to the source resistor), no power will flow. This because the resistor in
a Thevenin is a series resistor. Equal voltages will be applied to each side
of the resitor. The voltage difference will be zero.

5. If the reflected power returns 180 degrees out of phase with the applied
voltage (to the source resistor), the voltage across the resistor will
double with each cycle, resulting in an ever increasing current.(and power)
into the reflected wave.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

This is correct for this condition. The problem with the equation comes
when cos(90) which can easily happen.

I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is the
angle between positive peaks of the two waves. This angle will rotate twice
as fast as the signal frequency due to the relative velocity between the
waves..
No, this angle is fixed by system design. If the system changes, this
angle will rotate twice as fast as the angle change of a single wave.
The equation should be OK.
Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

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

Light, especially solar energy, is a collection of waves of all magnitudes
and frequencies. We should see only about 1/2 of the power that actually
arrives due to superposition. The reflection from a surface should create
standing waves in LOCAL space that will contain double the power of open
space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be
correct for a collection of waves, but incorrect for the example.

I tried to read your article a couple of weeks ago, but I found myself not
understanding. I have learned a lot since then thanks to you, Roy, and
others. I will try to read it again soon.

So what do you think Cecil?

73, Roger, W7WkB

Keith Dysart wrote:
On Jan 1, 6:00 pm, Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.
Probably the simplest way is to put the entire source circuitry into a
black box. Measure the terminal voltage with the box terminals open
circuited, and the current with the terminals short circuited. The ratio
of these is the source impedance. If you replace the box with a Thevenin
or Norton equivalent, this will be the value of the equivalent circuit's
impedance component (a resistor for most of our examples).
If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance; if it consists
of a perfect current source in parallel with a resistance, the source Z
will be the resistance. You can readily see that the open circuit V
divided by the short circuit I of these two simple circuits equals the
value of the resistance.
The power output of the Thévenin equivalent circuit follows the load.
Sorry, I don't understand this. Can you express it as an equation?
There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.

Two sources we all have access to are these links:

Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

I don't disagree with anything I read there.

But you may not quite have the concept of impedance
correct.

The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.
When there is a source present, this
line does not pass through the origin. Only for
passive components does this line pass through
the origin in which case it becomes V/I.

Because the voltage to current plot for an ideal
voltage source is horizontal, the slope is 0 and hence
so is the impedance. For an ideal current source
the slope is vertical and the impedance is infinite.

Hoping this helps clarify....

...Keith

Yes, I can understand that there is no change in voltage no matter what
the current load is, so there can be no resistance or reactive component
in the source. The ideal voltage link also said that the ideal source
could maintain voltage no matter what current was applied. Presumably a
negative current through the source would result in the same voltage as
a positive current, which is logical if the source has zero impedance.

During the transition from negative current passing through the source
to positive current passing through the source, the current at some time
must be zero. How is the impedance of the perfect source defined at
this zero current point? How is the impedance of the attached system
defined?

The resistor in the Thevenin/Norton equivalent source is selected with
some criteria in mind. What I would like to do is to design a Thevenin
source to provide 1v across a 50 ohm transmission line INFINATELY
long, ignoring ohmic resistance. I would like the source resistor to
absorb as much power as the line, so that if power ever returns under
reflected wave conditions, it can all be absorbed by the resistor. I
think the size of such a resistor will be 50 ohms.

Thanks for the clarification.

73, Roger, W7WKB

Second analysis: +0.5 input reflection coefficient

I got enough email responses to my request to make me believe it might
be worthwhile to put together another analysis.

The analysis I did previously is of limited use because of the total
lack of any loss in the system. It has also caused conceptual
difficulties because of the perfect source's -1 reflection coefficient.
So I'll do another analysis with a more realistic source.

The following analysis will illustrate the startup and progress to
steady state of a lossless, one wavelength, open ended 50 ohm
transmission line as before. But connected to the input end of the line
will be a perfect voltage source in series with a 150 ohm resistance.
This is not a Thevenin equivalent circuit, because it isn't meant to be
an equivalent of anything; it's simply a perfect source and a
resistance, both of which are common idealized linear circuit models.
The only difference between this setup and the one I analyzed earlier is
the addition of the source resistance.

I'm going to make a slight change in notation and call the initial
forward and reverse waves vf1 and vr1 respectively, instead of just vf
and vr as before. I apologize if this causes any confusion.

When we initially connect or turn on the voltage source, the
source/resistance combination sees Z0 looking into the line. So the
voltage at the line input is vs(t) * Z0 / (Z0 + Rs), where vs(t) is the
voltage of the perfect voltage source and Rs is the 150 ohm source
resistance, until the reflection of the original forward wave returns.
So I'm going to specify that

vs(t) = 4 * sin(wt)

so that the initial voltage at the line input is simply sin(wt). At any
point that the forward wave has reached, then,

vf1(t, x) = sin(wt - x)

The open far end of the line has a reflection coefficient of +1, so as
before the returning wave is:

vr1(t, x) = sin(wt + x)

(The change from -x to +x is due to the reversal of direction of
propagation. You'll see this with every reflection.) At any time between
when vf1 reflects from the output end (t = 2*pi/w) and when it reaches
the input end of the line (t = 4*pi/w), the total voltage at any point
the returning wave has reached is

vtot = vf1(t, x) + vr1(t, x) = sin(wt - x) + sin(wt + x) [Eq. 1]

When the returning wave vr1 reaches the input end of the line, it
re-reflects to form a new forward wave vf2. But because of the added 150
ohm resistance, the source reflection coefficient is +0.5 instead of the
-1 of the previous analysis, so

vf2(t, x) = .5 * sin(wt - x)

Just after the reflection, the voltage at the line input will be

vf1(t, 0) + vr1(t, 0) + vf2(t, 0) = sin(wt) + sin(wt) + .5 * sin(wt)
= 2.5 * sin(wt)

So we'll see a 2.5 peak volt sine wave at the input from the time the
first returning wave reaches the source until the next one does, or from
t = 4*pi/w to t = 8*pi/w. And anywhere on the line which vf2 has
reached, we'll see

vtot(t, x) = 1.5 * sin(wt - x) + sin(wt + x)

When vf2 hits the far end and reflects, it creates the second reflected wave

vr2(t, x) = .5 * sin(wt + x)

So now at any point where vr2 has reached we'll have

vtot(t, x) = 1.5 * sin(wt - x) + 1.5 * sin(wt + x) [Eq. 2]

There's something worth pointing out here. Look at the similarity
between Eq. 1, which was the total voltage with only one forward and one
reflected wave, and Eq. 2, which is the total with two forward and two
reflected waves. They're exactly the same except in amplitude -- Eq. 2
vtot is 1.5 times Eq. 1 vtot. Also notice that the equation for vf2 is
exactly the same as for vf1 except a constant, and likewise vr2 and vr1.
Every time a new forward wave is created by reflection from the input
end of the line, a new reflected wave is created by the reflection from
the far end. The reflection coefficient at the far end doesn't change,
so the relationship between each forward wave and the corresponding
reflected wave is the same as for any other forward wave and its
reflection. So the ratio of the sum of all forward waves to the sum of
all reflected waves is the same as for any single forward wave and its
reflection. The relationship between forward and reverse waves
determines the SWR, so this is why the source reflection coefficient
doesn't play any role in determining the line SWR. No matter what it is,
it creates a new pair of waves having the same ratio as every other pair.

Let's look at just one more pair of waves.

vf3 = .5 * .5 * sin(wt - x) = 0.25 * sin(wt - x)
vr3 = .5 * .5 * sin(wt + x) = 0.25 * sin(wt + x)

From the time vf3 is created until vr3 returns to the input, or from t
= 8*pi/w to t = 10*pi/w, at the input end of the line we'll see

v(t, 0) = vf1(t, 0) + vr1(t, 0) + vf2(t, 0) + vr2(t, 0) + vf3(t, 0) =
3.25 * sin(wt)

For the previous round trip period the sine wave amplitude was 2.5 volts
and for this round trip period it's 3.25 volts. Where is this going to
end after an infinite number of reflections?

Well, it had better end at 4 volts, the voltage of the perfect source!
At steady state, the impedance looking into the line is infinite, so the
current from the source is zero. (A comparable analysis of the current
waves would show convergence to itot = 0 at the input.) So there's no
drop across the 150 ohm resistor and the line input voltage equals the
voltage of the source.

Let's see if we can show this convergence mathematically. The trick is
to take advantage of a simple formula for the sum of an infinite
geometric series.

If F = a + ra + r^2*a + r^3*a, . . . an infinite number of terms

then F = a / (1 - r) if |r| 1. It's a very useful formula. I learned
it in high school algebra, but see that it's in at least one my calculus
and analytic geometry texts. Try it out, if you want, with a pocket
calculator or spreadsheet.

Going back and looking at the analysis, we had

vtot(t, x) = vf1(t, x) + vr1(t, x)

for the first set of waves. The next set was exactly 1/2 the value of
the first set, due to the +0.5 source reflection coefficient. The next
set was 1/2 that value, and so forth. So the first term of the series
(a) is sin(wt - x) + sin(wt - x) and the ratio of terms is 0.5, so

vtot(t, x)(steady state) = (sin(wt - x) + sin(wt + x)) / (1 - 0.5) =
2 * sin(wt - x) + sin(wt + x)

Using the same trig identity as in the earlier analysis,

vtot(t, x)(steady state) = 4 * sin(wt) * cos(x)

This is the correct steady state solution. You can see that it includes
the standing wave envelope cos(x), as it must.

Because superposition applies, we can add up the infinite number of
forward and reverse waves any way we want, and the total will always be
the same. So let's separately add the forward waves and reverse waves.

vf3/vf2 = vf2/vf1 = 0.5, as it is for the ratio of any two successive
forward waves
vr3/vr2 = vr2/vr1 = 0.5, as it is for the ratio of any two successive
reverse waves

So

vf(t, x)(total) = sin(wt - x) / (1 - 0.5) = 2 * sin(wt - x)

and

vr(t, x)(total) = sin(wt + x) / (1 - 0.5) = 2 * sin(wt + x)

These can be used for steady state analysis, as though there is only one
forward and one reverse wave, and the result will be exactly the same as
if the system was analyzed for each of the infinite number of waves
individually. This is almost universally done; the run-up from the
initial state is usually only of academic interest.

If we add the total forward and reverse waves to get the total voltage,
the result is exactly the same as it was when we summed the pairs of
waves to get the total.

Again I ran a SPICE simulation of the circuit, using a 5 wavelength line
for clarity.

http://eznec.com/images/TL2_input.gif is the voltage at the line input.
As predicted, it starts at 1 volt peak, remains there until the first
reflected wave returns to the input end (at t = 10 sec.), then jumps to
2.5 volts. After another round trip, it goes to 3.25.
http://eznec.com/images/TL2_1_sec.gif shows the voltage one wavelength
(1 second) from the source. Here you can see that the voltage is zero
until the initial forward wave arrives at t = 1 sec. From then until the
reflected wave arrives at t = 9 sec., it's one volt peak. From then (t =
9) until the reflected wave re-reflects from the source and returns (t =
11), we have vf1 + vr1 = sin(wt - x) + sin(wt + x). x is 2*pi radians or
360 degrees, so the voltage is simply 2 * sin(wt). And that's what the
plot shows - a sine wave of 2 volts amplitude. Then vf2 gets added in at
t = 11 sec, for a total of sin(wt - x) + sin(wt + x) + 0.5 * sin(wt -
x), or at the observation x point, 2.5 * sin(wt). At t = 19 sec., vr2
arrives and adds another 0.5 * sin(wt) to the total, raising the
amplitude to 3 volts peak. At t = 21 sec., vf3 arrives, adding another
0.25 * sin(wt) for a total amplitude of 3.25 volts. And so forth. As
with the previous analysis, SPICE shows exactly what the analysis predicts.

http://eznec.com/images/TL2_5_sec.gif is the voltage at the open end of
the line. For the first 5 seconds, the voltage is zero because the
initial wave hasn't arrived. At t = 5 sec., the voltage at the end
becomes vf1 + vr1 = 2 * sin(wt). At 15 sec., it becomes vf1 + vr1 + vf2
+ vr2 = 3 * sin(wt). And so forth, just as predicted by the analysis.

Although I've used a line of the convenient length of one wavelength (or
5 wavelengths for the SPICE run), an open circuited end, and a resistive
source, none of these are required. Exactly the same kind of analysis
can be done with complex loads of any values at the line input and
output, and with any line length. But in the general case, returning
waves don't add directly in phase or out of phase with forward waves,
and a wave undergoes some phase shift other than 0 or 180 degrees upon
reflection. So phase angles have to be included in the descriptions of
all voltages. In the general case it's much easier to revert to phasor
notation, but otherwise the analytical process is identical and the
results just as good.

So far I haven't seen any analysis using alternative theories, ideas of
how sources work, or using power waves, which also correctly predict the
voltage at all times and in steady state.

Because there's so much interest in power, I'll calculate the power and
energy at the line input. But I'll put it in a separate posting.

Roy Lewallen, W7EL

Here's an analysis of the input power and energy of the system similar
to the one described in my "Second analysis: +0.5 input reflection
coefficient" Please refer to that posting when reading this. I'm going
to make just one change to the system in that analysis: I'll use the 5
wavelength line of the SPICE analysis. This has absolutely no effect on
any of the values shown in the analysis except where I gave periods of
time when various equations were valid (e.g. t = 2*pi/w) -- those will
be five times greater. The effect of this change is to allow use of the
SPICE output as a visual aid. When speaking of the SPICE plot, I'll be
referring to http://eznec.com/images/TL2_input.gif, which is the voltage
at the input end of the line.

For this analysis I'm going to let f = 1 Hz, so w = 2*pi Hz.

From the voltage analysis and the SPICE plot, the initial voltage at
the input of the line is sin(wt). So the voltage across the input
resistor is 3 * sin(wt) (+ toward the source), and the current flowing
into the line is (3 * sin(wt)) / 150 = 20 * sin(wt) mA. The average
power being delivered to the line is Vin(rms) * Iin(rms) (since the
voltage and current are in phase) = (0.7071 v. * 14.14 mA) = 10 mW.
Since the line initially presents an impedance of Z0, this should also
be Vin(rms)^2 / Z0 or Iin(rms)^2 / Z0. Let's see: Vin(rms) = 0.7071, so
Vin(rms)^2 / Z0 = 0.5 / 50 = 10 mW; Iin(rms) = 0.01414 A, so Iin(rms)^2
* Z0 = 0.0002 * 50 W = 10 mW. Ok.

The input voltage remains unchanged until the first reflected wave
returns at t = 10 sec (remember, we're using a 5 wavelength line and
frequency of 1 Hz). So during that time we put 10 mW * 10 sec = 0.1
joule of energy into the line. Note that I could have calculated the
instantaneous power for each instant, then integrated it over the first
10 seconds to get the total energy. The result would be identical.

During that first 10 seconds, the resistor is dissipating an average of
Vr(rms) * Ir(rms) = 2.121 * 14.14 = 30 mW. The source is producing
Vs(rms) * Is(rms) = 2.828 * 14.14 = 40 mW. So here's the story so far:

For the first 10 seconds (one round trip):
The source produces 40 mW, for a total of 400 mj (millijoules)
The resistance dissipates 30 mW, for a total of 300 mj
The line gets 10 mW, for a total of 100 mj

All the power and energy is totally accounted for, so far. The sum of
power and energy delivered to the resistor and line equals the power and
energy produced by the source.

As you can see from the SPICE output or the voltage analysis, the input
voltage jumps to 2.5 volts at t = 10 seconds and stays there for the
next 10 seconds. During that time, the voltage across the resistor is (4
- 2.5) * sin(wt), so the current drops to 1.5 sin(wt) / 150 = 10 sin(wt)
mA which has an RMS value of about 7.071 mA. Using the same methods as
before,

For the next 10 seconds:
The source produces 20 mW, for a total of 200 mj
The resistance dissipates 7.50 mW, for a total of 75 mj
The line gets 12.5 mW, for a total of 125 mj

We've again accounted for all the power and energy, with no need to
invoke any kind of power waves.

Next the input voltage goes to 3.25 volts, so the current drops to 5 *
sin(wt) mA and

For the next 10 seconds,
The source produces 10 mW, for a total of 100 mj
The resistance dissipates 1.875 mW, for a total of 18.75 mj
The line gets 8.125 mW, for a total of 81.25 mj

There's something interesting about this energy flow. Notice that the
amount of power produced by the source decreases by a factor of two each
cycle, and the amount of power dissipated by the resistor decreases by a
factor of four each cycle. These are to be expected, since the line
input voltage is increasing each time. But look at the power supplied to
the line -- it actually increases during the second cycle relative to
the first, then drops back. What's happening? Well, the line has an
apparent impedance of Vin/Iin at any given time. Let's see what it is
(the peak values have the same ratio as the RMS values, so I'll use those):

For t = 0 to 10, Vin/Iin = 1/0.020 = 50 ohms
For t = 10 to 20, Vin/Iin = 2.5/0.010 = 250 ohms
For t = 20 to 30, Vin/Iin = 3.75/0.005 = 750 ohms

The input impedance will, of course, approach an infinite value as time
goes on, Vin approaches Vs, and Iin approaches zero. But during the time
interval of 10 to 20 seconds, the apparent line impedance provided a
better "impedance match" for the 150 ohm source than at other times,
which increased the line power input during that time.

Only because we have no energy leaving the line can be calculate a total
energy delivered by the source and dissipated by the resistance over an
arbitrarily long time period. That is, we can find the total energy
delivered if the line were connected and the source left on forever. We
can use the same formula for summing an infinite series as used in the
voltage analysis to find that:

The source produces a total of 400 / (1 - 0.5) = 800 mj
The resistor dissipates a total of 300 / (1 - 0.25) = 400 mj

from which we see that a total of 400 mj has been supplied to the line.
If we were to quickly replace the source and resistor with a 50 ohm
resistor across the line and no source, what should happen? Well, we
have forward and reverse waves of 2 volts peak traveling on the line, so
the end voltage should immediately drop to 2 * sin(wt) as the reverse
traveling wave exits with no further reflections from the input end of
the line, and the forward wave moves toward the far end. So the 50 ohm
resistor would have an RMS voltage of 1.414 volts across it, and a
dissipation of (1.414)^2 / 50 = 40 mW. This constant dissipation should
continue until the line is completely discharged, which will take one
round trip time or 10 sec. The total energy removed from the line and
dissipated during this time is 40 mW * 10 sec = 400 mj, which is exactly
the amount we put into the line during the charging process. All the
energy produced by the source is accounted for. I think I can set this
maneuver up with SPICE, and I'll do so if anyone who might be skeptical
would be convinced by the SPICE output.

A caution is in order. This analysis was greatly simplified by the lack
of any energy storing devices other than the transmission line. That is,
there was no reactance at either end of the line. An equally accurate
analysis could be done with reactances present, but calculation of
energy from power would have to account for the temporary energy storage
in the reactive components. Also, if a resistance-containing load is
connected to the output, it will also dissipate power and energy, so
that would also have to be accounted for. The bottom line is that care
should be taken in applying this method to more complex circuits without
suitable modification.

I've completely accounted for the power and energy leaving the voltage
source, being dissipated in the resistor, and entering the line, at all
times from startup to steady state, and done it quantitatively with
numerical results. And I did this without any mention of propagating
waves of power or energy. It was done very simply using Ohm's law and
elementary power relationships, with the only variable being the line
input voltage calculated from the voltage wave analysis. I offer a
challenge to those who embrace the notion of traveling waves of average
power or other alternative theories to do the same using mathematics and
premises consistent with the alternative theory. Until such an analysis
is produced, I remain unconvinced that any such theory is valid. *Any*
analysis has to produce results consistent with the law of the
conservation of energy, as this one has.

Roy Lewallen, W7EL