View Single Post
  #2   Report Post  
Old September 6th 03, 07:34 PM
pez
 
Posts: n/a
Default

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