Complex Z0 - Power : A Proof
[Please use fixed-space font]

It seems that
a proof has been carried out
for the validation
of the so-called
Principle of Conservation of Energy
"P.C.E"
in any point of a Transmission Line
with a Complex characteristic impedance,
which is terminated at any passive load.

But,
I have to make an appeal
for some patience...

[1]
---
Let
the Characteristic Impedance

Zo = Ro + j Xo

and the Gamma

g = a + j b

[2]
---
At the Terminal Load

Zt = Rt + j Xt

a "Reflection Coefficient"
is defined by

pt = (Zt - Zo)/(Zt + Zo)
pt = |pt|.Exp(j tt)

On the load
the Average Power
"Wt"
is expressed as[*]

Wt = k.T

where k 0 is a constant
and

T = 1 - |pt|^2 + j (Xo/Ro)(pt - pt*)

[3]
---
With direct substitution of
Zt and Zo in the last one
and after
_enough_but_straightforward_
algebraic manipulation,
which it is impossible to be reproduced here,
we have for the
Average Power at the Terminal Load

[Wt = 0] = [T = 0] = [Rt = 0]

This is an important partial result
which proves that
P.C.E. is valid
on any passive terminal load
of a Transmission Line with complex Zo.

[4]
---
Now,
at any other point on the Transmission Line
at a distance d from the Terminal Load,
a "Reflection Coefficient" is introduced from

pd = pt.Exp(-2.g.d)[*]

and the following are defined

Zd = (1 + pd)/( 1 - pd)
pd = (Zd - Zo)/(Zd + Zo)
pd = |pd|.Exp(j td)

and the relation

|pd| = |pt|.Exp(-2.a.d)

holds between the two
Reflection Coefficients

[5]
---
As in above [2],
the Average Power at any point at distance d is

Wd = k.D

where

D = 1 - |pd|^2 + j (Xo/Ro)(pd - pd*) =
D = 1 - Exp(-4.a.d).|pt|^2 - 2.(Xo/Ro).Exp(-2.a.d).|pt|.Sin(td)

Obviously,
P.C.E. holds,
if and only if

D = 0

or equivalently

2.(Xo/Ro).Exp(-2.a.d).|pt|.Sin(td) = 1 - Exp(-4.a.d).|pt|^2

[6]
---
But, we have

0 Exp(-4.a.d) = 1

and

0 = |pt|^2

therefore

Exp(-4.a.d)|pt|^2 = |pt|^2 =
-|pt|^2 = -Exp(-4.a.d)|pt|^2

and finally

1 - |pt|^2 = 1 - Exp(-4.a.d).|pt|^2

Hence,
for P.C.E validation,
it is sufficient to prove

2(Xo/Ro).Exp(-2.a.d).|pt|.Sin(td) = 1 - |pt|^2

But, in addition, we have
for the maximum value of the left part

0 Exp(-2.a.d) = 1
Sin(td) = 1
(Xo/Ro) = |Xo|/Ro

and therefore the left part
has the relation,
to its maximum value

2.(Xo/Ro).Exp(-2.a.d).|pt|.Sin(td) = 2.(|Xo|/Ro).|pt|

[7]
---
Consequently,
it is sufficient to prove

2.(|Xo|/Ro).|pt| = 1 - |pt|^2

or

|pt|^2 + 2.(|X0|/R0).|pt| -1 = 0

(Note:
(Probably,
(this is possible
(with a direct substitution,
(as in [3]

This is a 2nd degree polynomial in |pt|
with roots:

|pt|- = -|Xo|/Ro - Sqrt(|Xo|^2/Ro^2 + 1) 0
|pt|+ = -|Xo|/Ro + Sqrt(|Xo|^2/Ro^2 + 1) 0

After that,
the P.C.E. is valid at any point
if and only if
the "Reflection Coefficient"
at the terminal load is
lower than or equal to the positive root

0 = |pt| = |pt|+

But

|Xo|/Ro = Tan(|to|)

where tO is the argument of Zo, so

|pt|+ = -Tan(|to|) + Sqrt(Tan(|to|)^2 + 1)

After some trigonometric manipulation
the identity
can be proved

-Tan(|t0|) + Sqrt(Tan(|to|)^2 + 1) =
= Sqrt[(1 + Sin(|to|)/(1 - Sin(|to|)]

Therefore,
the P.C.E. is valid,
at any point,
if and only if
the "Reflection Coefficient"
at the Terminal Load is

|pt| = Sqrt[(1 + Sin(|to|)/(1 - Sin(|to|)]

But this is a relation,
which has been already proved
in the Thread

"Complex Z0 [Corrected]"

QED
I hope,
but,
as any other time
d;^)
your comments are welcomed!

Sincerely,

pez
SV7BAX
[*]
---
e.g. Chipman, 1968, p.138

The Missing Step
[Please use fixed-space font]

For any patient reader,
the Missing Step
of the validation of the
Principle of Conservation of Energy
at the Terminal Load,
is now possible to be reproduced here...

[1]
---
Given a Transmission Line with

Zo = Ro + j Xo
Yo = 1/Zo

the following are defined

Vi = V1.Exp[-g.l]
Vr = V2.Exp[+g.l]
V = Vi + Vr

p = Vr/Vi
V = Vi.(1 + p)

Ii = I1.Exp[-g.l]
Ir = I2.Exp[+g.l]
I = Ii + Ir

I1 = Yo.V1
I2 =-Yo.V2
Ii = Yo.Vi
Ir =-Yo.Vr
I = Yo.[Vi - Vr]
I = Yo.Vi.(1 - p)

Z = V/I
Z = Zo.(1 + p)/(1 - p)
p =(Z - Zo)/(Z + Zo)

[2]
---
At any point
the Complex Power is given by

P = V.I* = Vi.(1 + p).Yo*.Vi*.(1 - p*)

where

Yo*= Yo.Yo*/Yo = Zo/|Zo|^2

therefore

P =(|Vi/Zo|^2).Zo.(1 + p).(1 - p*)
P =(|Vi/Zo|^2).Zo.(1 - p* + p -|p|^2)
P =(|Vi/Zo|^2).(Ro + j Xo).[(1 -|p|^2) + (p - p*)]
P =(|Vi/Zo|^2).Ro.[1 + j (Xo/Ro)].[(1 -|p|^2) + (p - p*)]

and since

Vi = V1.Exp[-2.a.l].Exp[-j 2.b.l]
|Vi|^2 = (|V1|^2).Exp[-2.a.l]

we can define

k = (|Vi/Zo|^2).Ro = (|V1/Zo|^2).Ro.Exp[-2.a.l] 0

The Complex Power becomes

P = k.[1 + j (Xo/Ro)].[(1 -|p|^2) + (p - p*)]

where

p - p* = j 2 Im{p}

therefore

P = k.[1 + j (Xo/Ro)].[(1 -|p|^2) + j 2 Im{p}]
P = k.{[(1 -|p|^2) - 2.(Xo/Ro).Im{p}] + j [(Xo/Ro).(1 -|p|^2) + 2 Im{p}]

But

P = PR + j PI
PR = k.{[(1 -|p|^2) - 2.(Xo/Ro).Im{p}]

and

-2.Im{p} = j (p - p*)

Finally,
at any point,
the Average Power is

PR = k.[(1 -|p|^2) + j (Xo/Ro).(p - p*)]

[3]
---
At any point at a distance l from the input

0 = l = L
p =(V2/V1).Exp[2.g.l]

At the Terminal Load

l = L
pt =(V2/V1).Exp[2.g.L]
Z = Zt
pt =(Zt - Zo)/(Zt + Zo)

hence

Wt = PR(p=pt)
Wt = k.T
T = 1 -|pt|^2 + j (Xo/Ro).(pt - pt*)

But we have

|pt|^2 =(|Zt - Zo|^2)/(|Zt + Zo|^2)

pt - pt* = (Zt - Zo)/(Zt + Zo) - (Zt* - Zo*)/(Zt* + Zo*) =
= [(Zt - Zo).(Zt* + Zo*) - (Zt + Zo).(Zt* - Zo*)]/(|Zt + Zo|^2)

Therefore

T.(|Zt + Zo|^2).Ro =
=(|Zt + Zo|^2).Ro - (|Zt - Zo|^2).Ro +
j Xo.[(Zt - Zo).(Zt* + Zo*) - (Zt + Zo).(Zt* - Zo*)]

= Ro.[(Rt + Ro)^2 + (Xt + Xo)^2 - (Rt - Ro)^2 - (Xt - Xo)^2] +
+ j Xo.[Zt.Zt* + Zt.Zo* - Zo.Zt* - Zo.Zo* - Zt.Zt* + Zt.Zo* - Zo.Zt* + Zo.Zo*) =

= Ro.[Rt^2 + Ro^2 + 2.Rt.Ro + Xt^2 + Xo^2 + 2.Xt.Xo -
-Rt^2 - Ro^2 + 2.Rt.Ro - Xt^2 - Xo^2 + 2.Xt.Xo]
+j 2.Xo.(Zt.Zo* - Zo.Zt*)

= 4.Ro.[Rt.Ro + Xt.Xo] + j 2.Xo.(Zt.Zo* - Zo.Zt*)

But

j 2.Xo.(Zt.Zo* - Zo.Zt*) =
j 2.Xo.[(Rt + j Xt).(Ro - j Xo) - (Ro + j Xo)(Rt - j Xt)} =
j 2.Xo.[Rt.Ro - j Rt.Xo + j Xt.Ro + Xt.Xo - Ro.Rt + j Ro.Xt -j Xo.Rt - Xo.Xt]=
j 2.Xo.[ j {-Rt.Xo + Xt.Ro + Ro.Xt - Xo.Rt}]=
-2.Xo.[-Rt.Xo + Xt.Ro + Ro.Xt - Xo.Rt]

Hence

T.(|Zt + Zo|^2).Ro =
4.Ro^2.Rt + 4.Ro.Xt.Xo + 2.Rt.Xo^2 - 2.Xo.Xt.Ro - 2.Xo.Ro.Xt + 2.Xo^2.Rt =
4.Ro^2.Rt + 4.Xo^2.Rt =
4.(|Zo|^2).Rt

Finally we take

T.(|Zt + Zo|^2).Ro = 4.(|Zo|^2).Rt

from which

[Wt = 0 ] = [T = 0] = [Rt = 0]

QED, I hope.

And as usually d;^)
"Comments are welcomed"...

Sincerely,

pez
SV7BAX

On Thu, 11 Sep 2003 09:13:21 +0300, "pez" wrote:

Unfortunately enough
my proof on this subject is
_Wrong_ and rejected.

Obviously,
the relation

-Tan(|t0|) + Sqrt(Tan(|to|)^2 + 1) =
= Sqrt[(1 + Sin(|to|)/(1 - Sin(|to|)]

is *not* an identity
but an equation one,
which has probably
only the zero as solution
in the [-45 , 45] degrees
interval of Zo Argument.

For example,
put to = 45 degrees
to get the result

-1 = 1

But further,
it is not just this.

The whole matter of the validation of
the Principle of Conservation of Energy
_at_any_point_
of a
Uniform Transmission Line with Complex Z0,
has to be put under investigation,
starting,
once again,
from the beginning.

I am terribly sorry for any inconvenience.

Sincerely,

pez
SV7BAX


Hello,

Identity or equation. This is a matter of importance to you alone as
others would have moved right past it without close attention (this
includes me too). The significance of it is not explained which is
more important (part of knowing what to discard from writing). In
other words, if it is a deep layer in the logic, then including it
adds nothing of insight. If it is pivotal, important, then the rest
of the material should be clipped away.

As for "Conservation of Energy" this should be a test at the end, not
a goal. Too many correspondents start out proving this "law" and
offering nothing notable along the way. Stasis offers a sure proof of
the "law" and is very boring. We endure too many static sermons
already ;-)

I have to admit that in my other response (other correspondence) to
your notice of rejection; that I mistook that to be rejection by
others. However, I also offered that the merit of that rejection
marks the critic. You, again, show a very commendable trait that
associates you with consistent logic.

Develop the math for Web publication, and argue it here by reference.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote -

Develop the math for Web publication, and argue it here by reference.

=================================

Why do you persist in inventing non-existent problems?

Dear Mr. Richard Clark,

I am most grateful for your useful comments,
especially
for those expressions which amplify my poor vocabulary.

Sincerely yours,

pez
SV7BAX