"Dr. Slick" wrote:

wrote in message ...

And yes, |rho| can be greater than unity for a passive load.

...Keith

Absolute Rubbish.. Could you produce a passive circuit that
will reflect a greater voltage than what you feed it? I'd
LOVE to see that...

Several examples have been presented, but rather than accepting
them, you changed the definition of rho. Perhaps you could build
one of these circuits to determine if modifying the definition
of rho was appropriate.

The ratio Pref/Pfwd is directly related to the ratio [rho].

Pref/Pfwd = [rho]**2 Absolute value brackets are a must!

Consider that after the absolute value brackets, the phase information
is gone. But since we are going to a ratio of average (RMS)
values OR peak values of power, it doesn't matter.

In other words, if you use V**2/R, the "V" can be either peak or
RMS, it doesn't matter, because it is a ratio. And of course, the "R"
doesn't matter either. And of course, the phase information is gone
with
the absolute value brackets.

If you agree that the Pref/Pfwd ratio cannot be greater than 1

Which I haven't since Pref and Pfwd are just computed numbers and
the result for some circuits is that Pref/Pfwd is greater than 1.
Of course, Pnet is not equal to Pfwd-Pref in these circumstances
so there is no violation of basic physics. It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).

for a passive network, then neither can the [Vref/Vfwd]= rho be
greater
than 1 either.

....Keith

wrote:
It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).

Funny how they can heat resistors both at the load end and at the source
end when the source is equipped with a circulator+load.

Incidentally, I've come up with a proof that there are no reflections
at a voltage null in a homogeneous transmission line. Consider a 50
ohm lossless feedline. At a voltage null -

rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

wrote:
It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).

Funny how they can heat resistors both at the load end and at the source
end when the source is equipped with a circulator+load.

Not surprising. The energy comes from the source.

Incidentally, I've come up with a proof that there are no reflections
at a voltage null in a homogeneous transmission line. Consider a 50
ohm lossless feedline. At a voltage null -

rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction

That's what you get when you use the surge impedance and compute
surge rho. Try steady state impedance for steady state rho.

rho = (0-50)/(0+50) = -1 as expected

....Keith

wrote:
That's what you get when you use the surge impedance and compute
surge rho. Try steady state impedance for steady state rho.

rho = (0-50)/(0+50) = -1 as expected

Heh, heh, the steady state impedances are V/I ratios which are results
incapable of causing anything. Your logic is a closed loop. The V/I ratios
cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios.
Can you name any other result in reality that causes itself? Hint: Only
physical impedance discontinuities can cause reflections. Image impedances
are incapable of causing anything since they are an end result.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

wrote:
That's what you get when you use the surge impedance and compute
surge rho. Try steady state impedance for steady state rho.

rho = (0-50)/(0+50) = -1 as expected

Heh, heh, the steady state impedances are V/I ratios which are results
incapable of causing anything. Your logic is a closed loop. The V/I ratios
cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios.
Can you name any other result in reality that causes itself? Hint: Only
physical impedance discontinuities can cause reflections. Image impedances
are incapable of causing anything since they are an end result.

Was it not only a few days ago that there was agreement that

If it is acceptable to claim that sometimes there is no reflection
at an impedance discontinuity, then it must also be acceptable to
claim that sometimes there is a reflection where there is no
discontinuity.

?

....Keith

Cecil:

[snip]
rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction
--
73, Cecil http://www.qsl.net/w5dxp
[snip]

Hey... that's my line.

That's Mother Nature's reflection coefficient!

At every infinitesimal length along a transmission line rho = (Zo - Zo)/(Zo
+ Zo) = 0.

It's not nice to fool Mother Nature!

--
Peter K1PO
Indialantic By-the-Sea, FL.

wrote in message ...

And yes, |rho| can be greater than unity for a passive load.

...Keith

Absolute Rubbish.. Could you produce a passive circuit that
will reflect a greater voltage than what you feed it? I'd
LOVE to see that...

Several examples have been presented, but rather than accepting
them, you changed the definition of rho. Perhaps you could build
one of these circuits to determine if modifying the definition
of rho was appropriate.



The definition of Rho has been set for "God-knows-how-long!"

Way before you or I were born, i would imagine!

Rho is the magnitude of Gamma, which is the complex voltage
reflection coefficient.

BUT, some people write Gamma is Rho, so just in case, it's wise
to say Power RC = [rho]**2 , with absolute value brackets.

Ok, show us a circuit, I'm waiting to see.



The ratio Pref/Pfwd is directly related to the ratio [rho].

Pref/Pfwd = [rho]**2 Absolute value brackets are a must!

Consider that after the absolute value brackets, the phase information
is gone. But since we are going to a ratio of average (RMS)
values OR peak values of power, it doesn't matter.

In other words, if you use V**2/R, the "V" can be either peak or
RMS, it doesn't matter, because it is a ratio. And of course, the "R"
doesn't matter either. And of course, the phase information is gone
with
the absolute value brackets.

If you agree that the Pref/Pfwd ratio cannot be greater than 1

Which I haven't since Pref and Pfwd are just computed numbers and
the result for some circuits is that Pref/Pfwd is greater than 1.
Of course, Pnet is not equal to Pfwd-Pref in these circumstances
so there is no violation of basic physics. It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).


ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


Slick

Dr. Slick wrote:
The definition of Rho has been set for "God-knows-how-long!"

Actually, 'rho' has contradictory definitions. (Z2-Z1)/(Z2+Z1)
is not always the same value as Sqrt(Pref/Pfwd) because of
interference energy.
--
73, Cecil http://www.qsl.net/w5dxp



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Slick:

[snip]
I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


Slick
[snip]

It sure will be an interesting event for your Daiwa when the transmitter
hooked to the other end is keyed!

--
Peter K1PO
Indialantic By-the-Sea, FL.

"Dr. Slick" wrote:
ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.

I would never claim there is a passive circuit which violated
the rules of conservation of energy but if you define
A = Vf**2/Z0
and
B = Vr**2/Z0
then there are circuits for which B is greater than A and if you
accept the often held view that Pfwd = A and Prev = B then YOU
are also claiming the Pref is greater than Pfwd. I claim that A
and B don't, in reality, have much to do with power at all.

On the other hand Vr Vf has been demonstrated by several
examples of which Roy's August 20 post

http://groups.google.ca/groups?dq=&hl=en&lr=&ie=UTF-8
&threadm=vl88o381tks0cc%40corp.supernews.com
&prev=/groups%3Fdq%3D%26num%3D25%26hl%3Den%26lr%3D%26ie%3 DUTF-8
%26group%3Drec.radio.amateur.antenna%26start%3D25

is just one example.

This example had a line with Z0 of 68-j39 ohms connected to a load
with impedance 10+j50 ohms.

While this example demonstrated lossy lines, for this analysis
we can simplify the line to its Thevenin equivalent:
- an ideal voltage source
- producing a 10 kHz sinusoid
- at 52.68 V
- with a source impedance of 68-j39 ohms
connected to a load of 10+j50 ohms.

The incident voltage is 26.34 V.

Using polar notation...
- source voltage: 52.68/_ 0.0 [52.68+j0.0] V
- source impedance: 78.4/_ -29.8 [68-j39] ohms
- load impedance: 51.0/_ 8.7 [10+j50] ohms

Total circuit impedance is source + load impedance:
78.4/_ -29.8 + 51.0/_ 8.7 = 78.8/_ 8.03 [78+j11]
Circuit current (voltage/impedance)
52.68/_ 0.0 / 78.8/_ 8.03 = 0.669/_ -8.03 [0.662-j0.0934]
Voltage at load (current * impedance)
0.669/_ -8.03 * 51.0/_ 8.7 = 34.1/_ 70.7 [11.3+j32.2]
which agrees with Roy's.
Reflected voltage (load voltage - incident)
34.1/_ 70.7 - 26.34/_ 0.0 = 35.5/_ 115.1 [-15.0+j32.2]
which also agrees.

So reflected voltage is greater than incident voltage
which leads to rho being greater than unity.

Now about that Daiwa....

Directional wattemeters compute the Vf and Vr using
Vf = (V + I*Z0) /2
Vr = (V - I*Z0) /2

Your Daiwa is probably calibrated for Z0 = 50 ohms, but let's
assume we can recalibrate for Z0 = 78.4/_ -29.8 [68-j39] ohms.

Then it will obtain
Vf = (34.1/_ 70.7 + 0.669/_ -8.03 * 78.4/_ -29.8) /2
= 26.34/_ 0.0 [26.34+j0.0]
Vr = (34.1/_ 70.7 - 0.669/_ -8.03 * 78.4/_ -29.8) /2
= 35.5/_ 115.1 [-15.0+j32.2]
as expected. Please note that your Daiwa DOES think that
Vr is greater than Vf.

Assuming your Daiwa works like most directional wattmeters
it will feed these voltages (appropriately scaled) to a
meter which will move linearly in response to the voltage.

So if your Daiwa had a linear scale it would have no
difficulty showing Vf and Vr (except that it would need
to be adjusted for the different Z0) and it would show
a greater Vr than Vf (i.e. rho 1).

But displaying power (even though meaningless in this case)
is somewhat more difficult. Your Daiwa likely computes power
by having non-linear markings on the meter representing
V**2/Z0. This works fine for real Z0, but will not do for
complex Z0. For this, you need more sophosticated computation
than is possible with just a non-linear scale, so the
power indicated by your Daiwa will be quite incorrect.

But it does get Vf and Vr correct (assuming it is adjusted for
the different Z0).

....Keith