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

Derivation of Reflection Coefficient vs SWR
 

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

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

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

Derivation of Reflection Coefficient vs SWR
 

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

Derivation of Reflection Coefficient vs SWR
 

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

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

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

Psource = Pfor - Pref = Pload

Pfor = Psource + Pref = Pload + Pref

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

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

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

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

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

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

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

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

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

Vs(t) = 100 cos(wt)

or, in phasor notation

Vs = 100 /_ 0d

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

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

At the load, the voltage reflection coefficient is

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

The forward voltage at the load is

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

The reflected voltage at the load is

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

The voltage at the load is

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

At the generator, the voltage reflection coefficient is

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

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

The voltage at the generator is, therefore,

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

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


Currents can be similar computed...

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

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

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

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

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

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

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

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

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

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

The power applied to the load resistor is

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

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

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

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

Pl.avg = 18.75

To confirm, let us compute using

Pavg = Vrms * Irms * cos(theta)

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

Agreement.

Now at the generator end

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

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


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

Now let us check Pf and Pr

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

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

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

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

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

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

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

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

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

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

yields the same result as the Smith chart.

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

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

Ps.avg = 49.999

and that dissipated in the source resistor

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

Prs.avg = 31.249

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

49.999 = 31.249 + 18.75

So all the energy is accounted for, as expected.

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

...Keith

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

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

Derivation of Reflection Coefficient vs SWR
 

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.

Derivation of Reflection Coefficient vs SWR
 

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

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

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

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

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

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

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

[snip]

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

[snip]

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

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

...Keith

Derivation of Reflection Coefficient vs SWR
 

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

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

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

You cannot eat your reflected wave and have it too.

The standard equation is:

Psource = Pfor - Pref = Pload

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

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

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

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

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

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

Derivation of Reflection Coefficient vs SWR
 

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

49.999 = 31.249 + 18.75

So all the energy is accounted for, as expected.

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

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

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

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

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

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

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

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

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

Later, this wave arrives back at the generator.

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

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

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

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

...Keith

Derivation of Reflection Coefficient vs SWR
 

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

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

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

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

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

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

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

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

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


[snip]

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

[snip]

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

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

...Keith
--
73, Roger, W7WKB

Derivation of Reflection Coefficient vs SWR
 

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

Maybe a more drastic example will help.

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

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

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

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

Derivation of Reflection Coefficient vs SWR
 

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

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

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

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

You cannot eat your reflected wave and have it too.

The standard equation is:

Psource = Pfor - Pref = Pload

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

Pgenerator = Pfor - Pref = Pload

For the example at hand...

Pgenerator = 18.75
Pfor - Pref = 18.75
Pload = 18.75

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

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

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

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

As your original equation above

Psource = Pfor - Pref
18.75 = 25 - 6.25

I do not see any issues.

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

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

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

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

...Keith

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

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

Derivation of Reflection Coefficient vs SWR
 

On Feb 5, 2:41*pm, Cecil Moore wrote:
Keith Dysart wrote:
How does this reflected wave of 6.25 W affect the already
present forward wave of 25 W when it arrives back at the
generator. It affects it not one iota. The forward power
remains the same, the forward voltage remains the same
and the forward current remains the same. This is the
meaning of no reflection.

Maybe a more drastic example will help.

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

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

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

How can there be 25 watts of forward power when
the source is supplying only 1.96 watts and none
of the reflected energy is joining the forward
wave?

The answer would be clear to you if you were to
complete the analysis.

Please compute:
Pfor
Pref
Vfor
Vref
at the point Vg, both before and after the
reflected wave makes it back to the point Vg.

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

...Keith

Derivation of Reflection Coefficient vs SWR
 

Keith Dysart wrote:
Cecil Moore wrote:
How can there be 25 watts of forward power when
the source is supplying only 1.96 watts and none
of the reflected energy is joining the forward
wave?

The answer would be clear to you if you were to
complete the analysis.

Please don't ignore the question. How does a 1.96
watt source support a forward power of 25 watts
without reflected energy joining the forward energy?
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

Keith Dysart wrote:
Cecil Moore wrote:

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm ~23w--Pref |
| 0.04w / 1.92w
Vs 90 degrees \ Rl
100 cos(wt) 1.96w 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection. The redistribution of
energy due to interference is a well known and
well understood phenomenon in optical physics but has
been virtually ignored in the field of RF engineering.

It is not a conventional reflection. It is wave
cancellation in action. It is obeying the following
equation:

Ptot = P1 + P2 - 2*SQRT(P1*P2) where
-2*SQRT(P1*P2) is the destructive interference term.

If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

Nope, not true. The phase angle
between the voltages determines which direction the
energy flows. If the above example is changed to
1/2WL instead of 1/4WL, the forward and reflected
voltages and powers remain the same but the interference
at the source resistor changes from destructive to
constructive and the source resistor heats up.
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm ~23w--Pref |
| 0.04w / 1.92w
Vs 90 degrees \ Rl
100 cos(wt) 1.96w 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection. The redistribution of
energy due to interference is a well known and
well understood phenomenon in optical physics but has
been virtually ignored in the field of RF engineering.

It is not a conventional reflection. It is wave
cancellation in action. It is obeying the following
equation:

Ptot = P1 + P2 - 2*SQRT(P1*P2) where
-2*SQRT(P1*P2) is the destructive interference term.

Is this the total applied power or the remaining power (Prem)(which is
really the reflected power), where

Prem = Z0(If - Ir)^2
= Z0((If^2) +(Ir^2) - 2(If*Ir))

I think these (yours and mine) are both the same equation except that I
changed the terms so that we could use it on a transmission line. Z0 is
transmission line impedance, If is forward current, Ir is reflected current

Just to clarify,
2*SQRT(P1*P2)
= 2*SQRT((Z*I1^2)*(Z*I2^2)
= 2*Z(I1*I2)

The equation should apply at all locations on the transmission line.


If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

Nope, not true. The phase angle
between the voltages determines which direction the
energy flows. If the above example is changed to
1/2WL instead of 1/4WL, the forward and reflected
voltages and powers remain the same but the interference
at the source resistor changes from destructive to
constructive and the source resistor heats up.

73, Roger, W7WKB

Derivation of Reflection Coefficient vs SWR
 

Roger Sparks wrote:
The equation should apply at all locations on the transmission line.

What you say is true but there is a caveat. Remember
that I have previously said that interference can exist
without wave cancellation. As long as Z0 remains constant
up and down a transmission line, there is no forward and
reflected wave interaction. They pass each other like
"ships in the night".

The resistor is a physical object. When the forward and
reflected waves superpose at the *resistor*, there is
an obvious interaction, i.e., a permanent interference.
One only has to observe the power dissipated in the resistor
to ascertain that it is not the same as the forward power
plus the reflected power unless the E-fields of the two
waves are 90 degrees apart, a condition for which zero
interference exists.

What is interesting is a procedure for determining the
current through Rs using only powers. We have to be
pretty accurate with the powers to do that. We know
the forward power is 25w. The power reflection coefficient
at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes
the reflected power from the load equal to 23.077 watts.
When the forward wave and reflected wave superpose at
the 50 ohm source resistor, they are 180 degrees out of
phase which makes the interference term minus and
therefore destructive.

25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w

That's the dissipation in the 50 ohm source resistor.
So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a

Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt)

And that is, indeed, the current through the 50 ohm
source resistor, Rs, using only an energy analysis.

All energy, except 0.03845w, is "redistributed", i.e.
reflected back toward the load. This is all explained
in my energy analysis article at:

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

Derivation of Reflection Coefficient vs SWR
 

"Cecil Moore" wrote in message
. net...
Roger Sparks wrote:
The equation should apply at all locations on the transmission line.

What you say is true but there is a caveat. Remember
that I have previously said that interference can exist
without wave cancellation. As long as Z0 remains constant
up and down a transmission line, there is no forward and
reflected wave interaction. They pass each other like
"ships in the night".

The resistor is a physical object. When the forward and
reflected waves superpose at the *resistor*, there is
an obvious interaction, i.e., a permanent interference.
One only has to observe the power dissipated in the resistor
to ascertain that it is not the same as the forward power
plus the reflected power unless the E-fields of the two
waves are 90 degrees apart, a condition for which zero
interference exists.

What is interesting is a procedure for determining the
current through Rs using only powers. We have to be
pretty accurate with the powers to do that. We know
the forward power is 25w. The power reflection coefficient
at the load is [(50-1)/(50+1)]^2 = 0.92311. That makes
the reflected power from the load equal to 23.077 watts.
When the forward wave and reflected wave superpose at
the 50 ohm source resistor, they are 180 degrees out of
phase which makes the interference term minus and
therefore destructive.

25w + 23.077w - 2*SQRT(25w*23.077w) = 0.03845w

That's the dissipation in the 50 ohm source resistor.
So the RMS current is SQRT(0.03845w/50ohm)= 0.02773a

Ig(t) = 1.414(0.02773)cos(wt) = 0.039216 cos(wt)

And that is, indeed, the current through the 50 ohm
source resistor, Rs, using only an energy analysis.

All energy, except 0.03845w, is "redistributed", i.e.
reflected back toward the load. This is all explained
in my energy analysis article at:

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


I disagree. If the current through the 50 ohm source resistor is 0.039216,
then the power in the 50 ohm source resistor cannot be 0.03845w. Namely
because 0.039216 * 0.039216 * 50 = 0.076895w (eye-squared-are, matey).
Either your power is wrong or your current is wrong. My own calculations
show that your current is correct. Therefore, your energy analysis appears
to be in error.

John

Derivation of Reflection Coefficient vs SWR
 

John KD5YI wrote:
I disagree. If the current through the 50 ohm source resistor is 0.039216,
then the power in the 50 ohm source resistor cannot be 0.03845w. Namely
because 0.039216 * 0.039216 * 50 = 0.076895w (eye-squared-are, matey).
Either your power is wrong or your current is wrong. My own calculations
show that your current is correct. Therefore, your energy analysis appears
to be in error.

You seem to have forgotten that since 0.039216 is the maximum
value of current you need to divide the 0.076895w by two.

0.076895w/2 = 0.03845 watts

If we used the RMS value of the current, i.e. 0.039216/1.414
equals 0.027734a, then 50(0.027734)^2 = 0.03845 watts and
all is well.
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

On Feb 6, 8:02*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:

* * * * * * Rs * * * Vg * Pfor--25w * * * * Vl
* * * *+----/\/\/-----+----------------------+
* * * *| * *50 ohm * * * ~23w--Pref * * * * |
* * * *| * *0.04w * * * * * * * * * * * * * */ *1.92w
* * * Vs * * * * * * * * 90 degrees * * * * *\ * Rl
* *100 cos(wt) 1.96w * * 50 ohm line * * * * / *1 ohm
* * * *| * * * * * * * * * * * * * * * * * * \ *load
* * * *| * * * * * * * * * * * * * * * * * * |
* * * *+--------------+----------------------+
* * * gnd
When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection.

When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

You may find the Excel spreadsheet on this page
http://keith.dysart.googlepages.com/...oad,reflection
interesting.

It computes the various results for the circuit we
have been discussing. Input Vs, Rs, Rl and line length
and it will compute the various voltages, currents and
powers.

After entering the circuit parameters, click "First".
The circuit conditions at the end of the first round
trip will be computed. Click "Next" to have any
re-reflection included into the forward conditions
for the next round trip.

When Rs is 50 ohms (i.e. it matches the line impedance),
the circuit conditions settle to their final state
immediately, as would be expected when there is no
reflection at the generator. Change Rs to a value other
than 50 and it takes many clicks of "Next" before the
conditions settle. At each click, Vf changes as some of
Vr is re-reflected and added to Vf.

...Keith

Derivation of Reflection Coefficient vs SWR
 

"Cecil Moore" wrote in message
...
John KD5YI wrote:
I disagree. If the current through the 50 ohm source resistor is
0.039216, then the power in the 50 ohm source resistor cannot be
0.03845w. Namely because 0.039216 * 0.039216 * 50 = 0.076895w
(eye-squared-are, matey). Either your power is wrong or your current is
wrong. My own calculations show that your current is correct. Therefore,
your energy analysis appears to be in error.

You seem to have forgotten that since 0.039216 is the maximum
value of current you need to divide the 0.076895w by two.


Yes, I forgot to use the RMS value of the source.

John

Derivation of Reflection Coefficient vs SWR
 

Keith Dysart wrote:
When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

Nope, Vf doesn't have to change for there to be a
reflection. Looking at what happens to the power
dissipation in the source resistor, Rs, when the
reflected wave arrives, we see that the energy being
supplied by the source drops by 99.85%. Most of
the energy in the forward wave is supplied by the
reflected wave, starting at the time the reflected
wave arrives and causes destructive interference.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

To be technically correct, reflected energy is redistributed
back toward the load during the process of destructive
interference.

The conditions before the reflected wave arrives and
after the reflected wave arrives are extremely different.

Before the reflected wave arrives, the source resistor,
Rs, is dissipating 25 watts.

After the reflected wave arrives, the source resistor,
Rs, is dissipating 0.03845 watts.

Clearly, something drastic has happened and that is:
99.85% of the forward energy originally supplied by the
source has been replaced with reflected energy being
redistributed back toward the load.

Because of destructive interference, reflected energy
*never* flows through the source resistor, Rs, and is
instead redistributed back toward the load.

Nothing else is possible since the source is supplying
only 1.9608 watts during steady-state. 92.3% of the forward
power is not being supplied by the source during steady-state.

The energy incident upon a point must equal the energy
exiting the point. The energy incident upon the
generator terminals is 1.92234 joules/sec from the
source and 23.0777 joules/sec from the reflected wave.
The energy exiting that point is 25 joules/sec. The
reflected wave energy obviously reverses direction and
joins the forward wave and that is what we call a
"reflection".
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

On Feb 8, 9:15*am, Cecil Moore wrote:
Keith Dysart wrote:
When we analyze this circuit we find that there is
no voltage re-reflection when the wave gets back
to the generator. This is clear because Vf does not
change, which it would have to do if any of Vr was
re-reflected.

Nope, Vf doesn't have to change for there to be a
reflection. Looking at what happens to the power
dissipation in the source resistor, Rs, when the
reflected wave arrives, we see that the energy being
supplied by the source drops by 99.85%. Most of
the energy in the forward wave is supplied by the
reflected wave, starting at the time the reflected
wave arrives and causes destructive interference.

And yet it appears that you are claiming that power
is reflected at the generator. How can power be
reflected if voltage is not?

To be technically correct, reflected energy is redistributed
back toward the load during the process of destructive
interference.

The conditions before the reflected wave arrives and
after the reflected wave arrives are extremely different.

Before the reflected wave arrives, the source resistor,
Rs, is dissipating 25 watts.

After the reflected wave arrives, the source resistor,
Rs, is dissipating 0.03845 watts.

Clearly, something drastic has happened and that is:
99.85% of the forward energy originally supplied by the
source has been replaced with reflected energy being
redistributed back toward the load.

Because of destructive interference, reflected energy
*never* flows through the source resistor, Rs, and is
instead redistributed back toward the load.

Nothing else is possible since the source is supplying
only 1.9608 watts during steady-state. 92.3% of the forward
power is not being supplied by the source during steady-state.

The energy incident upon a point must equal the energy
exiting the point. The energy incident upon the
generator terminals is 1.92234 joules/sec from the
source and 23.0777 joules/sec from the reflected wave.
The energy exiting that point is 25 joules/sec. The
reflected wave energy obviously reverses direction and
joins the forward wave and that is what we call a
"reflection".

Are you sure? I thought a reflection was something that
occurred at an impedance discontinuity and the magnitude
of the voltage reflection was defined by

Vr = Vincident * ReflectionCoefficient
= Vincident * (Z2-Z1)/(Z2+Z1)

and that the reflected voltage then added to any wave
already travelling in that direction.

But you are claiming that power can be reflected when
voltage is not. I have never encountered this claim before.

Are you sure?

...Keith

Derivation of Reflection Coefficient vs SWR
 

Keith Dysart wrote:
Are you sure? I thought a reflection was something that
occurred at an impedance discontinuity and the magnitude
of the voltage reflection was defined by

Vr = Vincident * ReflectionCoefficient
= Vincident * (Z2-Z1)/(Z2+Z1)

That's true for normal reflections which involve only
one wave. Wave cancellation is a different kind of
energy reflection involving two waves. The energy flow is
canceled in one direction and therefore flows in the
other direction. In optics, it is known as a redistribution
of energy in directions that allow constructive interference.
In a transmission line, there are only two possible directions
so any redistribution of energy due to destructive interference
can be considered to be a reflection in the opposite direction,
the only direction available to constructive interference.

and that the reflected voltage then added to any wave
already travelling in that direction.

True for a single wave reflection. For two interacting
waves, the voltage in one of the waves can simply replace
the voltage in the other wave. In our example, the reflected
voltage simply replaces the source voltage component.

But you are claiming that power can be reflected when
voltage is not. I have never encountered this claim before.

When the reflected wave arrives, it cancels most of the
existing forward wave from the source. The reflected
voltage exactly equals the canceled source voltage in
our example because the source resistance is the same
as the Z0 of the line.

Are you sure?

It is obvious that reflected energy never flows in the
source resistor so it must go in the only other direction
possible. Yes, I am sure. The conservation of energy
principle will allow nothing else.
--
73, Cecil http://www.w5dxp.com

Derivation of Reflection Coefficient vs SWR
 

Keith Dysart wrote:
Are you sure?

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm 23.08w--Pref |
| 48.08w / 1.92w
Vs 45 degrees \ Rl
100 cos(wt) 50 ohm line / 1 ohm
50w \ load
| |
+--------------+----------------------+
gnd

As a matter of interest, let's return to the 45 degree
line, the one to which I objected previously.

The forward power and reflected power are the same as
in the previous example but now the dissipation in Rs
is equal to the sum of the forward power and reflected
power. (It would be nice if all cases were as straight
forward as this special case.)

This is indeed the one special case where everything you
have been saying is true. There is no redistribution of
energy because there is no interference. There is no
interference because at Vg, Vfor and Vref are 90 degrees
out of phase.

These concepts are covered in my magazine article at:

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