Derivation of Reflection Coefficient vs SWR
 

On Jan 27, 10:57 am, Roger Sparks wrote:
On Sat, 26 Jan 2008 19:24:22 -0800 (PST)

Keith Dysart wrote:
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?"

Yes, only the model I use substitutes a battery for the signal generator that you are using.

A battery is not the same since it has a very low output impedance.
A battery in series with a 50 ohm resistor would offer a reasonable
match to a 50 ohm transmission line.

The returning wave can recharge the battery, but how does the current stop? Or does it ever stop? From a practical aspect, the current must stop, but I can not explain how except by resistance that absorbs the circulating power.

That is the challenging part to understand when too much emphasis
is placed on the existance of energy being transported with the
"reflected power".

When I was learning this stuff, I did many examples with matched
generators (a battery with a 50 ohm resistor is a good example).
With step functions, it is easy to compute the final state because
you can just treat it as a DC circuit for analysis.

We must keep the limitations of our models in mind.

True, but over limiting is not good either.

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

The reflected wave that does not encounter an impedance at the generator must be on an infinitely long line, and therefore the conditions must have changed between the time of launch and return.

I don't follow the association between generator impedance and
length of line.

For a 50 ohm line, a matched generator has a 50 ohm output
impedance. The returning wave does not encounter an impedance
discontinuity so is not reflected. It disappears into the
the generator.

It makes more sense to think that the reflected voltage of (using your example) 50v meets 50v of forward voltage, and therefore finds an infinite resistance, coming to a stop.

The current does stop (since it can not flow into the
infinite resistance at the end of the line), and a 50 V step
propagates back along the line.

This 50 V reverse propagating step plus the 50 V already on
the line produces a total of 100 V on the line.

If the generator was constructed as a 100 V battery in series
with a 50 ohm resistor, then when the step arrives back at
the generator, there is 100 V on both sides of the source
resistor (battery on one side, line on the other) and the
current through the source resistor becomes zero. No further
current flows into the line.

If that happens, doesn't the same condition occur as soon as the reflected wave is first generated at the line end? The current really stops as soon as the end is reached, with the energy contained in the magnetic field converted to electric field energy visible as voltage. If this happened, the reflected wave could be better described as an electric "jump", similar to a hydraulic jump found in open channel flow of liquids, where kinetic energy is converted to potential energy.

There do seem to be some similarities, though there is likely
trouble if the analogy is carried to far.

This idea of an electric "jump" requires not a reflection occuring without a discontinuity, but a moving wave front that absorbs the traveling wave, bringing it to a stop (in this case).

The current has stoppped, but has the "wave"?

Vf, Vr, If, Ir, Pf, Pr are still computable using the normal
formulae.

....Keith

Derivation of Reflection Coefficient vs SWR
 

On Sat, 2 Feb 2008 13:01:22 -0800 (PST)
Keith Dysart wrote:

On Jan 27, 10:57 am, Roger Sparks wrote:
On Sat, 26 Jan 2008 19:24:22 -0800 (PST)

Keith Dysart wrote:
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?"

Yes, only the model I use substitutes a battery for the signal generator that you are using.

A battery is not the same since it has a very low output impedance.
A battery in series with a 50 ohm resistor would offer a reasonable
match to a 50 ohm transmission line.

The returning wave can recharge the battery, but how does the current stop? Or does it ever stop? From a practical aspect, the current must stop, but I can not explain how except by resistance that absorbs the circulating power.

That is the challenging part to understand when too much emphasis
is placed on the existance of energy being transported with the
"reflected power".

When I was learning this stuff, I did many examples with matched
generators (a battery with a 50 ohm resistor is a good example).
With step functions, it is easy to compute the final state because
you can just treat it as a DC circuit for analysis.

We must keep the limitations of our models in mind.

True, but over limiting is not good either.

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

The reflected wave that does not encounter an impedance at the generator must be on an infinitely long line, and therefore the conditions must have changed between the time of launch and return.

I don't follow the association between generator impedance and
length of line.

For a 50 ohm line, a matched generator has a 50 ohm output
impedance. The returning wave does not encounter an impedance
discontinuity so is not reflected. It disappears into the
the generator.

OK, I think we are saying the same thing. The return wave DISAPPEARS.

It makes more sense to think that the reflected voltage of (using your example) 50v meets 50v of forward voltage, and therefore finds an infinite resistance, coming to a stop.

The current does stop (since it can not flow into the
infinite resistance at the end of the line), and a 50 V step
propagates back along the line.

This 50 V reverse propagating step plus the 50 V already on
the line produces a total of 100 V on the line.

If the generator was constructed as a 100 V battery in series
with a 50 ohm resistor, then when the step arrives back at
the generator, there is 100 V on both sides of the source
resistor (battery on one side, line on the other) and the
current through the source resistor becomes zero. No further
current flows into the line.

If that happens, doesn't the same condition occur as soon as the reflected wave is first generated at the line end? The current really stops as soon as the end is reached, with the energy contained in the magnetic field converted to electric field energy visible as voltage. If this happened, the reflected wave could be better described as an electric "jump", similar to a hydraulic jump found in open channel flow of liquids, where kinetic energy is converted to potential energy.

There do seem to be some similarities, though there is likely
trouble if the analogy is carried to far.

This idea of an electric "jump" requires not a reflection occuring without a discontinuity, but a moving wave front that absorbs the traveling wave, bringing it to a stop (in this case).

The current has stoppped, but has the "wave"?

Vf, Vr, If, Ir, Pf, Pr are still computable using the normal
formulae.

...Keith

Here is a link to a web site discussing reflections and more.

http://www.ivorcatt.co.uk/em.htm

He has some drawings that pertain to this discussion. He shows reflections and crossings, but no one-way transmission lines.
--
73, Roger, W7WKB

Derivation of Reflection Coefficient vs SWR
 

"Roger Sparks" wrote in message
...
On Sat, 2 Feb 2008 13:01:22 -0800 (PST)
Keith Dysart wrote:

On Jan 27, 10:57 am, Roger Sparks wrote:
On Sat, 26 Jan 2008 19:24:22 -0800 (PST)

Keith Dysart wrote:
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?"

Yes, only the model I use substitutes a battery for the signal
generator that you are using.

A battery is not the same since it has a very low output impedance.
A battery in series with a 50 ohm resistor would offer a reasonable
match to a 50 ohm transmission line.

The returning wave can recharge the battery, but how does the current
stop? Or does it ever stop? From a practical aspect, the current
must stop, but I can not explain how except by resistance that absorbs
the circulating power.

That is the challenging part to understand when too much emphasis
is placed on the existance of energy being transported with the
"reflected power".

When I was learning this stuff, I did many examples with matched
generators (a battery with a 50 ohm resistor is a good example).
With step functions, it is easy to compute the final state because
you can just treat it as a DC circuit for analysis.

We must keep the limitations of our models in mind.

True, but over limiting is not good either.

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

The reflected wave that does not encounter an impedance at the
generator must be on an infinitely long line, and therefore the
conditions must have changed between the time of launch and return.

I don't follow the association between generator impedance and
length of line.

For a 50 ohm line, a matched generator has a 50 ohm output
impedance. The returning wave does not encounter an impedance
discontinuity so is not reflected. It disappears into the
the generator.

OK, I think we are saying the same thing. The return wave DISAPPEARS.

It makes more sense to think that the reflected voltage of (using your
example) 50v meets 50v of forward voltage, and therefore finds an
infinite resistance, coming to a stop.

The current does stop (since it can not flow into the
infinite resistance at the end of the line), and a 50 V step
propagates back along the line.

This 50 V reverse propagating step plus the 50 V already on
the line produces a total of 100 V on the line.

If the generator was constructed as a 100 V battery in series
with a 50 ohm resistor, then when the step arrives back at
the generator, there is 100 V on both sides of the source
resistor (battery on one side, line on the other) and the
current through the source resistor becomes zero. No further
current flows into the line.

If that happens, doesn't the same condition occur as soon as the
reflected wave is first generated at the line end? The current really
stops as soon as the end is reached, with the energy contained in the
magnetic field converted to electric field energy visible as voltage.
If this happened, the reflected wave could be better described as an
electric "jump", similar to a hydraulic jump found in open channel flow
of liquids, where kinetic energy is converted to potential energy.

There do seem to be some similarities, though there is likely
trouble if the analogy is carried to far.

This idea of an electric "jump" requires not a reflection occuring
without a discontinuity, but a moving wave front that absorbs the
traveling wave, bringing it to a stop (in this case).

The current has stoppped, but has the "wave"?

Vf, Vr, If, Ir, Pf, Pr are still computable using the normal
formulae.

...Keith

Here is a link to a web site discussing reflections and more.

http://www.ivorcatt.co.uk/em.htm

He has some drawings that pertain to this discussion. He shows
reflections and crossings, but no one-way transmission lines.
--
73, Roger, W7WKB

I just love this quote from that material "...Thus do these clapped-out
radio men..."

Though it's all above my head, I have indeed met some clapped out radio men.

Derivation of Reflection Coefficient vs SWR
 

Roger Sparks wrote:
Here is a link to a web site discussing reflections and more.
http://www.ivorcatt.co.uk/em.htm

Interesting - especially about the inductance
(loading coil) acting like a transmission line.
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

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.

Derivation of Reflection Coefficient vs SWR
 

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

Derivation of Reflection Coefficient vs SWR
 

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

Derivation of Reflection Coefficient vs SWR
 

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

I didn't refuse to explore the example. You told me
to "read no further" so I did just that.

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

Ideal components have led to such outrageous concepts
as zero delays through 75m bugcatcher loading coils.
Do you really believe in zero delays through those
big honking 75m loading coils?
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

On Feb 5, 12:05*am, Cecil Moore wrote:
Keith Dysart wrote:
I observe that you refuse to explore the example that
demonstrates that this is not an issue.

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.

I notice that your own examples are often constructed
with ideal components, so this aversion to ideal
components is not universal. Perhaps only when it
can be conveniently used to avoid situations where
your deeply held beliefs may be threatened?

...Keith

Derivation of Reflection Coefficient vs SWR
 

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