The short answer is that you'd use the magnitude of rho, not its
complex value, to find the magnitude of Vr in terms of the magnitude
of Vf, and from that, the relative value of the powers.

So I suppose you'll get the same values for rho=+1 and rho=-1, since
the magnitude is +1 in both cases. Beware of how you do the calcs:
rho=+j, rho=-j, and rho=(1+j)/sqrt(2) all should also give you
|rho|=1.

But I'll leave the power calcs to you. Resolving things into "forward
power" and "reflected power" for steady-state excitation really
doesn't do a thing for me. I want to know the load presented to the
source, and the power delivered to the line by the source and to the
load by the line, and perhaps some other things like power dissipation
as a function of distance along the line, but I can't think of any
reason why I'd care about "f.p." or "r.p." Now what happens during
transient situations is a completely different story.

Cheers,
Tom


Jim Kelley wrote in message ...
Hi Tom,

Since rho represents the fraction of forward power that is reflected,
what does a negative value for rho indicate?

Thanks,

Jim AC6XG

Tom Bruhns wrote:

Hi Dave (and lurkers),

Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT
If+Ir. You have to be careful about directions, and be careful to
define things and stick with those definitions. I usually use
Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same
physical direction along the line, and then of course
rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it
should be clear. Anyway, rho=+1 at a point of zero net current,
either way you define the current direction, and rho=-1 at a point of
zero net voltage. No Smith chart, and no Zo, needed to figure that
out.

Cheers,
Tom

Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02...
Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited
... the net current is zero (i.e. at an open circuit). Therefore Ir =
-If, and rho is a -1.

Comment?

Tom Bruhns wrote:

I trust we can all agree that the definition of rho is rho=Vr/Vf, and
I trust we can all agree that on a TEM line, the net voltage
Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit),
then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you
chose to use to find rho at a load from the load impedance, Zload, and
the line characteristic impedance, Zo, better work out right for a
short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)"
does NOT work for that simple case, unless Zo*=Zo. But
"rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through
something similar for an open circuit load. Note that rho=-1 for a
short circuit load, and rho=+1 for an open circuit load, independent
of line impedance.
....

Jim Kelley wrote:
Since rho represents the fraction of forward power that is reflected,
what does a negative value for rho indicate?

rho represents the fraction of forward voltage that is reflected.
|rho|^2 represents the fraction of forward power that is reflected.
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:
Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

rho = +1 means there is a zero degree phase shift in the reflected voltage.

rho = -1 means there is a 180 degree phase shift in the reflected voltage.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Jim Kelley wrote:
Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

rho = +1 means there is a zero degree phase shift in the reflected
voltage.

rho = -1 means there is a 180 degree phase shift in the reflected voltage.

So in other word the sign sometimes indicates phase, and other times
indicates a direction in propagation depending on which hand is waving..
Thanks.

73, Jim AC6XG

Jim Kelley wrote:

"Cecil Moore" wrote:
rho = +1 means there is a zero degree phase shift in the reflected voltage.

rho = -1 means there is a 180 degree phase shift in the reflected voltage.

So in other word the sign sometimes indicates phase, and other times
indicates a direction in propagation depending on which hand is waving..
Thanks.

Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the
voltage reflection coefficient is positive and there is no reflected
voltage phase shift. If ZLZ0, then the voltage reflection coefficient
is negative and there is a 180 degree phase shift in the reflected
voltage. The same holds true for the E-field of reflected light.

The RF reflected current convention differs from the reflected light
H-field convention. Kirchhoff's current convention enters into the sign
of the reflected RF current where no such convention exists for light.
--
73, Cecil, W5DXP

Jim,

Take the sign of rho literally. For instance, if Zl=0, a short circuit,
rho=-1; therefore, at the termination, V=V+ +(- V+) = 0. If Zl= open, then
rho=+1, and V= V+ + V+ =2V+

Tam/WB2TT

"Jim Kelley" wrote in message
...
Ur right, thanks Dave. I meant to say voltage rather than power. Let
me ask the question properly.

Tom,
Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

Thanks and 73,

Jim AC6XG



Dave Shrader wrote:

Tom, rho^2 represents the fraction of forward power that is reflected.

The squaring function produces a positive value.

Rho represents the percentage of voltage or current. Rho^2 is the power
function.

Dave

Jim Kelley wrote:
Hi Tom,

Since rho represents the fraction of forward power that is reflected,
what does a negative value for rho indicate?

Thanks,

Jim AC6XG

Tom Bruhns wrote:

Hi Dave (and lurkers),

Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT
If+Ir. You have to be careful about directions, and be careful to
define things and stick with those definitions. I usually use
Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same
physical direction along the line, and then of course
rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it
should be clear. Anyway, rho=+1 at a point of zero net current,
either way you define the current direction, and rho=-1 at a point of
zero net voltage. No Smith chart, and no Zo, needed to figure that
out.

Cheers,
Tom

Dave Shrader wrote in message
news:iBkcb.568264$o%2.253779@sccrnsc02...

Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited
... the net current is zero (i.e. at an open circuit). Therefore Ir =
-If, and rho is a -1.

Comment?

Tom Bruhns wrote:


I trust we can all agree that the definition of rho is rho=Vr/Vf,
and
I trust we can all agree that on a TEM line, the net voltage
Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit),
then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you
chose to use to find rho at a load from the load impedance, Zload,
and
the line characteristic impedance, Zo, better work out right for a
short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)"
does NOT work for that simple case, unless Zo*=Zo. But
"rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through
something similar for an open circuit load. Note that rho=-1 for a
short circuit load, and rho=+1 for an open circuit load, independent
of line impedance.

Cheers,
Tom

(Tom Bruhns) wrote in message
...


It's even easier than that. Using the _definition_ of reflection
coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net
line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or
zero). That's either a real short or a virtual short.

Cheers,
Tom

"David Robbins" wrote in message
...



lets see...

rho = (Z - Zo)/(Z + Zo)
if rho = -1 then

-1=(Z-Zo)/(Z+Zo)
or
-1*(Z+Zo)=(Z-Zo)
or
-Z - Zo = Z - Zo
or
-Z = Z
the only value i know that satisfies that is zero.
lets substitute it back in to be sure...
-1=(0-(50+j50))/(0+(50+j50))
-1=(-50-j50)/(50+j50)
-1=-1(50+j50)/(50+j50)
-1=-1 qed.

Yes, Cecil. I have considered it, and I agree. It just bothers me when
people forget what the minus sign means, and try using it to make
unrealistic claims.

73, Jim AC6XG

Cecil Moore wrote:

Cecil Moore wrote:
Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the
voltage reflection coefficient is positive and there is no reflected
voltage phase shift. If ZLZ0, then the voltage reflection coefficient
is negative and there is a 180 degree phase shift in the reflected
voltage. The same holds true for the E-field of reflected light.

The above assumes ZL and Z0 to be real numbers. The light index
of refraction is usually a real number.
--
73, Cecil, W5DXP

Cecil Moore wrote:
Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the
voltage reflection coefficient is positive and there is no reflected
voltage phase shift. If ZLZ0, then the voltage reflection coefficient
is negative and there is a 180 degree phase shift in the reflected
voltage. The same holds true for the E-field of reflected light.

The above assumes ZL and Z0 to be real numbers. The light index
of refraction is usually a real number.
--
73, Cecil, W5DXP

Jim Kelley wrote:
Yes, Cecil. I have considered it, and I agree. It just bothers me when
people forget what the minus sign means, and try using it to make
unrealistic claims.

At least for real Z0's, it should be consistent. Wouldn't
a rho of 0.5 at 20 degrees would be the same as a rho of
-0.5 at 200 degrees?
--
73, Cecil, W5DXP

No. The voltage reflection coefficient is the ratio of two voltages, and
has nothing to do with their directions. The reverse voltage wave is
always traveling toward the source. The forward voltage wave is always
traveling toward the load. The angle of the reflection coefficient is
the relative phases of these two at the point the reflection coefficient
is being evaluated. It makes sense to speak of the reflection
coefficient as having a "sign" only in the two special cases of zero and
180 degree angles. Otherwise, it has an angle, not a positive or
negative "sign".

The current reflection coefficient is a little more ambiguous because of
the freedom of defining which direction represents positive flow of Ir.
If Ir is defined to be positive toward the load (the more common
definition), then the current reflection coefficient, Ir/If = -Vr/Vf. If
it's defined to be positive toward the source, then Ir/If = Vr/Vf.

Roy Lewallen, W7EL

Jim Kelley wrote:
"Cecil Moore" wrote in message
...

Jim Kelley wrote:

Since rho represents the fraction of forward voltage that is reflected,
what does a negative value for rho indicate?

rho = +1 means there is a zero degree phase shift in the reflected

voltage.

rho = -1 means there is a 180 degree phase shift in the reflected voltage.


So in other word the sign sometimes indicates phase, and other times
indicates a direction in propagation depending on which hand is waving..
Thanks.

73, Jim AC6XG