"Tarmo Tammaru" wrote in message ...

How does this allow for the sum of V+ and V- to be 0? That is what you
have
across a short.


This is the correct answer, you just interpret it incorrectly.

The reflection coefficient is defined as the reflected voltage
divided by the incident voltage. Sure, the voltage is zero right at
the short, but there is a reflected voltage wave that moves back
towards the generator.

When you have a short, the phase is flipped 180 degrees, which is
exactly what the -1 means. Notice if you had a Zl=infinity, that the
RC would be +1, which would be full reflections INPHASE with the
generator, or in phase with the incident voltage wave.

It's very simple stuff, but many people don't understand this.

That's the whole point. By the conjugate formula RC is *not* -1. It is -1
with a phase angle. I agree it works for an open circuit, since you can
divide both sides by Zl, and anything divided by infinity is 0


It was clear that you misunderstood what the -1 meant.

Ok, -1 with a phase angle. I believe it, if the Zo is complex.
The point is that the ratio of the reflected to incident voltage is
one, and has nothing to with the DC voltage measured at the short
itself.

-1 with a phase angle would simply be 180 degrees phase shift,
plus or minus and additional angle.




I want you to tell me the significance of the fact that the
normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr
because Zo is complex in the general case. And how the example you
sent me may make be incorrect for this discussion.


Slick

Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT

"Tarmo Tammaru" wrote in message ...
Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT


You are correct. But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.


Slick

"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with
V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only
Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT


You are correct. But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.

************************************************** *****

Read this again. The *normal* equation works. It is the ARRL/Besser equation
that does not.

Tam

************************************************** *****


Slick

"Tarmo Tammaru" wrote in message ...

Read this again. The *normal* equation works. It is the ARRL/Besser equation
that does not.

Tam


I can admit that i'm wrong, but only if you give me good
reasons, which is something you have failed to do.

Read THIS again.

But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.


Slick