"W5DXP" wrote in message
...
Tarmo Tammaru wrote:
There was no actual power generated by the source.

How do you know? Was it 100% efficient and consumed no
power?
--
73, Cecil http://www.qsl.net/w5dxp

The output rose to the open circuit value. Actually, I had thought of
building a very efficient class C amplifier and calibrating the output power
in terms of DC collector current, but think it is a waste of time.

Tam/WB2TT



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Tarmo Tammaru wrote:
The output rose to the open circuit value.

Assuming you are talking about voltage, for a Norton equivalent,
that's a very lossy situation. What was your source?
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Tarmo Tammaru wrote:
The output rose to the open circuit value.

Assuming you are talking about voltage, for a Norton equivalent,
that's a very lossy situation. What was your source?
--
73, Cecil http://www.qsl.net/w5dxp
The source was an MFJ269, and as I mentioned, the external series resistor
was 75 Ohms. I should try it with something like 1K. Coax was ~100 feet of
9913. My scope is only good to 100 MHz; so, I have to keep the frequency
down.

BTW, I have met Ed Norton.

Tam/WB2TT

"Tarmo Tammaru" wrote in message ...
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes to -Z0*/Z0,
which, for Zo having phase angle b equates to -1 at angle(-2b).

Agreed. This doesn't prove anything, does it?

How does this allow for the sum of V+ and V- to be 0? That is what you have
across a short.


This is the correct answer, you just interpret it incorrectly.

The reflection coefficient is defined as the reflected voltage
divided by the incident voltage. Sure, the voltage is zero right at
the short, but there is a reflected voltage wave that moves back
towards the generator.

When you have a short, the phase is flipped 180 degrees, which is
exactly what the -1 means. Notice if you had a Zl=infinity, that the
RC would be +1, which would be full reflections INPHASE with the
generator, or in phase with the incident voltage wave.

It's very simple stuff, but many people don't understand this.



.........................................
P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)]

Remember, z is distance from the load.

Tam/WB2TT



ok, you've done a nice job of copying the text you sent me.

As I recall, you said you were not familiar with these diagrams and did not
understand them. What do you want me to do? derive it in a different way?


I want you to tell me the significance of the fact that the
normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr
because Zo is complex in the general case. And how this may make this
example incorrect for this discussion.


Slick

"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes
to -Z0*/Z0,
which, for Zo having phase angle b equates to -1 at angle(-2b).

Agreed. This doesn't prove anything, does it?

How does this allow for the sum of V+ and V- to be 0? That is what you
have
across a short.


This is the correct answer, you just interpret it incorrectly.

The reflection coefficient is defined as the reflected voltage
divided by the incident voltage. Sure, the voltage is zero right at
the short, but there is a reflected voltage wave that moves back
towards the generator.

When you have a short, the phase is flipped 180 degrees, which is
exactly what the -1 means. Notice if you had a Zl=infinity, that the
RC would be +1, which would be full reflections INPHASE with the
generator, or in phase with the incident voltage wave.

It's very simple stuff, but many people don't understand this.

That's the whole point. By the conjugate formula RC is *not* -1. It is -1
with a phase angle. I agree it works for an open circuit, since you can
divide both sides by Zl, and anything divided by infinity is 0



.........................................
P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)]

Remember, z is distance from the load.

Tam/WB2TT



ok, you've done a nice job of copying the text you sent me.

As I recall, you said you were not familiar with these diagrams and did
not
understand them. What do you want me to do? derive it in a different
way?


I want you to tell me the significance of the fact that the
normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr

Obviously true


because Zo is complex in the general case. And how this may make this
example incorrect for this discussion.

Slick

I am not sure which example you mean, but A*/A does not have 0 phase angle,
unless the phase angle of A is 0.

PS I scanned in the other stuff. Compressed TIF

Tam

"Tarmo Tammaru" wrote in message ...

I can't scan the whole chapter, and part of the next. You should be able to
get your library to borrow the book for you. Or, look at some other college
textbook under lossy lines. 5.2B is straightforward enough:

Zo=(1/Yo)=SQRT[(R+jwL)/(G+jwC)]=Ro+jXo

Tam/WB2TT


I've looked at the 9 pages you sent me, and I'm not certain what to
tell you.

They say, "the fact that the rho may exceed unity on a dissipative
line does not violate a condition of power conservation (as it would
on a lossless structure)."

On the one hand, I give them creedence for addressing the
concept...it's what i've been saying all along.
However, they don't explain why a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case. This may or may not make their example moot.

And i don't trust their Smith Chart extended out to 1+sqrt(2) for
a dissipative line. Maybe for an active network, but not a passive
one.

Also, they go from equation 5.12 to 5.13 without showing us how
they got there. Perhaps they just copied it out of another book and
used the formula incorrectly.

Certainly text books can disagree, just as the "experts" often do.

Maybe this book is hard to find because it's out of print because
nobody trusted it.


Slick

"Dr. Slick" wrote in message
om...
However, they don't explain why a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

They make a pont of the fact that they are *not* violating the concept of
conservation of energy

They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case. This may or may not make their example moot.

I don't see the problem. 100 /_30 degrees divided by 2/_5 degrees is 50/_15
degrees. Different phase angle. By general case they mean not the lossless
case.

And i don't trust their Smith Chart extended out to 1+sqrt(2) for
a dissipative line. Maybe for an active network, but not a passive
one.

No idea. Never had to extend a Smith chart

Also, they go from equation 5.12 to 5.13 without showing us how
they got there.

They use the identity e**jx = cos x + jsin x

Perhaps they just copied it out of another book and
used the formula incorrectly.

I don't think a technical textbook is supposed to introduce new concepts
unless they are labeled as such

Certainly text books can disagree, just as the "experts" often do.

Maybe this book is hard to find because it's out of print because
nobody trusted it.

Slick

As I said out front. The book is copyrighted 1960. There is a certain life
to these things.

Tam

"Tarmo Tammaru" wrote in message ...

How does this allow for the sum of V+ and V- to be 0? That is what you
have
across a short.


This is the correct answer, you just interpret it incorrectly.

The reflection coefficient is defined as the reflected voltage
divided by the incident voltage. Sure, the voltage is zero right at
the short, but there is a reflected voltage wave that moves back
towards the generator.

When you have a short, the phase is flipped 180 degrees, which is
exactly what the -1 means. Notice if you had a Zl=infinity, that the
RC would be +1, which would be full reflections INPHASE with the
generator, or in phase with the incident voltage wave.

It's very simple stuff, but many people don't understand this.

That's the whole point. By the conjugate formula RC is *not* -1. It is -1
with a phase angle. I agree it works for an open circuit, since you can
divide both sides by Zl, and anything divided by infinity is 0


It was clear that you misunderstood what the -1 meant.

Ok, -1 with a phase angle. I believe it, if the Zo is complex.
The point is that the ratio of the reflected to incident voltage is
one, and has nothing to with the DC voltage measured at the short
itself.

-1 with a phase angle would simply be 180 degrees phase shift,
plus or minus and additional angle.




I want you to tell me the significance of the fact that the
normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr
because Zo is complex in the general case. And how the example you
sent me may make be incorrect for this discussion.


Slick

"Tarmo Tammaru" wrote in message ...
"Dr. Slick" wrote in message
om...
However, they don't explain why a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

They make a pont of the fact that they are *not* violating the concept of
conservation of energy


But they never explain WHY a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

I don't trust their claims on this.

If you get more power reflected than you send into a passive
network, you are getting energy from nowhere, and are thus violating
conservation of energy.




They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case. This may or may not make their example moot.

I don't see the problem. 100 /_30 degrees divided by 2/_5 degrees is 50/_15
degrees. Different phase angle. By general case they mean not the lossless
case.


I believe you mean 50 @ 25 degrees.




And i don't trust their Smith Chart extended out to 1+sqrt(2) for
a dissipative line. Maybe for an active network, but not a passive
one.

No idea. Never had to extend a Smith chart


Do some research, and you will never see an "extended Smith
Chart" for a passive network. Oh, certainly for a active device, for
stability circles and such, but passive networks can never have a rho
greater than 1.




Also, they go from equation 5.12 to 5.13 without showing us how
they got there.

They use the identity e**jx = cos x + jsin x



Yes? And? How did they get the Zo=(Zn-1)/(Zn+1) from this?




As I said out front. The book is copyrighted 1960. There is a certain life
to these things.

Tam


But it seems to be out of print, perhaps with good reason...


Slick

Opps!

I meant, how did they derive Reflection Coefficient=(Zn-1)/Zn+1)?

They certainly don't show us! This would be key to answering our questions.


Slick