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.

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.

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.

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.

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.

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.

The voltage reflection coefficent, sometimes designated uppercase gamma,
sometimes lowercase rho, is the ratio of the reflected voltage to
forward voltage. It doesn't represent power at all, and you have to make
some assumptions of questionable validity to try and associate it with a
power.

People who like to speak of "forward power" and "reverse power"
calculate that the ratio of "reverse power" to "forward power" is equal
to the square of the magnitude of the voltage reflection coefficient,
providing that the characteristic impedance of the line is assumed to be
purely real.

The reflection coefficient is a complex number, so it isn't restricted
to "positive" and "negative", but can have any phase angle. Its
magnitude, like the magnitude of any complex number, is a real number,
so is square is always positive. (Complex numbers can be written with a
negative magnitude, but this isn't commonly done. If it were, though,
the square of the magnitude would still be positive.) Incidentally, the
magnitude of the reflection coefficient is also often designated as
lowercase rho, so any question about "rho" is unclear unless you say
whether you're talking about the complex reflection coefficient or just
its magnitude.

The complex reflection coefficient can have a value of exactly -1, or a
magnitude of 1 with a phase angle of 180 degrees. This simply means that
the reflected voltage is equal in magnitude to the forward voltage, and
exactly out of phase with it. The sum of forward and reflected voltage
has to equal the total voltage at any point along the line. If you
terminate a line with a short circuit, the reflection coefficient is -1
and the forward and reverse voltages at the end of the line are equal
and opposite, so they add up to zero. Which is the voltage you have at a
short circuit.

Roy Lewallen, W7EL

Jim Kelley wrote:
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

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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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.

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?

Jim,

I suppose your question has been answered sufficiently (Thanks, Roy,
Cecil and Tam), but I'd like to offer a bit different viewpoint than
is implied by your "fraction that is reflected."

I prefer to think of it not as "a fraction that's reflected" but
rather as a resolution of a particular voltage and current into two
modes. There are two modes of propagation supported by TEM line, one
in each direction along the line. If you excite a line to
steady-state at one frequency, there will be some sinusoidal current
at each point along the line, and some sinusoidal voltage across the
line at each point along its length. (You can have a load at one end
and a source at the other, or two sources each with its own internal
impedance, one at each end, so long as they are on the same
frequency.) That set of voltages and currents can be resolved
mathematically into two components, one corresponding to the mode of
propagation in one direction and one corresponding to the mode in the
other direction. Rho is simply a number representing that resolution
at the point on the line where that rho is measured (or calculated).
It's a complex number because it represents both phase and amplitude.

(Note that our resolution of measured voltage and current into the two
modes generally assumes that we know the line's Zo, and the degree to
which we don't know that will introduce an error in our determination
of rho. But that's a whole 'nuther topic...)

Cheers,
Tom