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.

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.

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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