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Old September 20th 03, 04:58 PM
Radio913
 
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Default Smith Chart Quiz

Hi,

I've just become aware of the answers to these questions, but i thought i
might see if anyone agrees/disagrees with me.

What is the center of the Smith Chart when characteristic impedance
Zo= 50 + j50?

And what will the complex series impedance be for Rho = -1 in this case?

(Hint: it will NOT be a short!)


Slick (Garvin)
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Old September 20th 03, 06:53 PM
Cecil Moore
 
Posts: n/a
Default

Radio913 wrote:
What is the center of the Smith Chart when characteristic impedance
Zo= 50 + j50?


I suspect that the standard Smith Chart is incapable of handling
a Z0 = 50 + j50. On the standard Smith Chart, would the center of
the SWR spiral be 1 + j for a Z0 = 50 +j50? If so, that would make
for some interesting impedance transformations.
--
73, Cecil http://www.qsl.net/w5dxp



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Old September 22nd 03, 03:04 AM
Tom Bruhns
 
Posts: n/a
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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.

  #4   Report Post  
Old September 24th 03, 06:06 PM
Tom Bruhns
 
Posts: n/a
Default

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.

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Old September 24th 03, 06:35 PM
Dave Shrader
 
Posts: n/a
Default

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.




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Old September 25th 03, 04:25 AM
Tom Bruhns
 
Posts: n/a
Default

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.

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Old September 25th 03, 11:41 PM
Jim Kelley
 
Posts: n/a
Default

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.

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Old September 26th 03, 12:05 AM
Dave Shrader
 
Posts: n/a
Default

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.


  #9   Report Post  
Old September 26th 03, 12:24 AM
Jim Kelley
 
Posts: n/a
Default

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.

  #10   Report Post  
Old September 26th 03, 01:43 AM
Roy Lewallen
 
Posts: n/a
Default

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


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