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# Anexos

## Demostración Ecuación 3.19

${\displaystyle \varepsilon _{i}=\varepsilon _{0}\varepsilon _{r}}$
${\displaystyle \mu =\mu _{0}\mu _{r}}$
${\displaystyle z<0=\varepsilon _{1},G_{1},\mu _{1}}$
${\displaystyle z>0=\varepsilon _{2},G_{2},\mu _{2}}$
${\displaystyle \varepsilon _{i}=H_{i}}$
${\displaystyle H_{i}=B_{i}\ \sin {(90\ -\ \theta _{i})}}$
${\displaystyle B_{i}=\omega {\sqrt {\mu _{i}\varepsilon _{i}}}}$
${\displaystyle \varepsilon _{r}=H_{r}}$
${\displaystyle H_{r}=B_{r}\ \sin {(90\ -\ \theta _{r})}}$
${\displaystyle B_{r}=\omega {\sqrt {\mu _{r}\varepsilon _{r}}}}$
${\displaystyle \varepsilon _{t}=H_{t}}$
${\displaystyle H_{t}=B_{t}\ \sin {(90\ -\ \theta _{t})}}$
${\displaystyle B_{t}=\omega {\sqrt {\mu _{t}\varepsilon _{t}}}}$
${\displaystyle \varepsilon _{i}+\varepsilon _{r}=\varepsilon _{t}}$
${\displaystyle \varepsilon _{i}=\varepsilon _{r}+\varepsilon _{t}}$
${\displaystyle \varepsilon _{r}=\varepsilon _{i}\ -\ \varepsilon _{t}}$
${\displaystyle \varepsilon _{t}=\varepsilon _{i}\ -\ \varepsilon _{r}}$
${\displaystyle \varepsilon _{i}=\varepsilon _{i}\ -\ \varepsilon _{t}+\varepsilon _{i}\ -\ \varepsilon _{r}}$
${\displaystyle \varepsilon _{r}\ +\ \varepsilon _{t}=\varepsilon _{i}\ -\ \varepsilon _{t}+\varepsilon _{i}\ -\ \varepsilon _{r}}$
${\displaystyle B_{r}\sin {(\theta _{r})}+B_{t}\sin {(\theta _{t})}=B_{i}\sin {(\theta _{i})}-B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{r}\sin {(\theta _{r})}}$
${\displaystyle \theta =-B_{r}\sin {(\theta _{r})}+B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{t}\sin {(\theta _{t})}+B_{i}\sin {(\theta _{i})}-B_{r}\sin {(\theta _{r})}}$
${\displaystyle \theta _{r}=\ B_{i}\theta _{r}\ =\ \theta _{i}}$
${\displaystyle B_{t}\sin {(\theta _{t})}-B_{i}\sin {(\theta _{i})}=-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}}$
${\displaystyle \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}-\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}=-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})}-\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}}$
${\displaystyle \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}-\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})})=-2\omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin {(\theta _{r})}-(\omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin {(\theta _{t})}+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin {(\theta _{i})})}$
${\displaystyle \eta _{1}=\mu _{i}\varepsilon _{i}=\mu _{r}\varepsilon _{r}{\eta _{2}=\mu }_{t}\varepsilon _{t}}$
${\displaystyle \omega (\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})=-\omega B_{i}\sin {(\theta _{i})}+\omega B_{r}\sin {(\theta _{r})}-\omega (\eta _{2}\sin {(\theta _{t})}+\eta _{1}\sin {(\theta _{i})})}$
${\displaystyle {\frac {\omega (\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})}{(\eta _{2}\sin {(\theta _{t})}-\eta _{1}\sin {(\theta _{i})})}}={\frac {(\omega +\omega B_{r}\sin {(\theta _{r})})}{\omega B_{i}\sin {(\theta _{i})}}}}$
${\displaystyle \Gamma _{||}={\frac {E_{r}}{E_{i}}}={\frac {\eta _{2}sin\theta _{t}-\eta _{1}sin\theta _{i}}{\eta _{2}sin\theta _{t}+\eta _{1}sin\theta _{i}}}}$

## Demostración del ángulo de Brewster

Es importante mencionar que se demostrará en distintos espacios se probará:

En primer lugar, es cuando la permitividad sea distinta en ambos medios. para demostrar se usan las ecuaciones 3.27 y 3.19 que relacionan la premitividad distinta
${\displaystyle sin\left(\theta _{B}\right)={\sqrt {\frac {\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}}}}$ Ec. 3.27

${\displaystyle \Gamma _{\parallel }={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}\sin \theta _{i}}}}$Ec. 3.19

${\displaystyle 0={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}\sin \theta _{i}}}}$

${\displaystyle 0=\eta _{2}\sin \theta _{t}-\eta _{1}\sin \theta _{i}}$

${\displaystyle {\sqrt {\frac {\mu _{2}}{\varepsilon _{2}}}}\sin \theta _{t}={\sqrt {\frac {\mu _{1}}{\varepsilon _{1}}}}\sin \theta _{i}}$

${\displaystyle {\sqrt {\mu _{2}\varepsilon _{1}}}\sin \theta _{t}={\sqrt {\mu _{1}\varepsilon _{2}}}\sin \theta _{i}}$

${\displaystyle \mu _{2}\varepsilon _{1}\sin ^{2}\theta _{t}=\mu _{1}\varepsilon _{2}\sin ^{2}\theta _{i}}$

${\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=\sin ^{2}\theta _{t}}$

${\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=\left(1-\cos ^{2}\theta _{t}\right)}$

${\displaystyle {\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}\cos ^{2}\theta _{i}}$

${\displaystyle \left({\frac {\mu _{1}\varepsilon _{2}}{\mu _{2}\varepsilon _{1}}}-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}\right)\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}}$

${\displaystyle {\frac {\mu _{1}}{\mu _{2}}}\left({\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}\right)\sin ^{2}\theta _{i}=1-{\frac {\mu _{1}\varepsilon _{1}}{\mu _{2}\varepsilon _{2}}}}$

${\displaystyle \sin ^{2}\theta _{i}={\frac {{\frac {\mu _{2}}{\mu _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}$

${\displaystyle \sin \theta _{i}={\sqrt {\frac {{\frac {\mu _{2}}{\mu _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}}$

Para el angulo de brewster se puede considerar que: ${\displaystyle \mu _{2}=\mu _{1}}$

${\displaystyle \sin \theta _{i}={\sqrt {\frac {1-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}{{\frac {\varepsilon _{2}}{\varepsilon _{1}}}-{\frac {\varepsilon _{1}}{\varepsilon _{2}}}}}}}$

${\displaystyle \sin \theta _{i}={\sqrt {\frac {\frac {\varepsilon _{2}-\varepsilon _{1}}{\varepsilon _{2}}}{\frac {\varepsilon _{2}^{2}-\varepsilon _{1}^{2}}{\varepsilon _{1}\varepsilon _{2}}}}}}$

${\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{1}\left(\varepsilon _{2}-\varepsilon _{1}\right)}{\varepsilon _{2}^{2}-\varepsilon _{1}^{2}}}}}$

${\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{1}\left(\varepsilon _{2}-\varepsilon _{1}\right)}{\left(\varepsilon _{2}-\varepsilon _{1}\right)\left(\varepsilon _{2}+\varepsilon _{1}\right)}}}}$

En este punto se demostró de donde sale la ecuacion 3.27 cuando ${\displaystyle \mu _{2}=\mu _{1}}$

${\displaystyle sin\left(\theta _{B}\right)={\sqrt {\frac {\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}}}}$

En segundo lugar, para el caso en el que la onda se propague en el espacio libre se usa la ecuacion 3.28 y la 3.24.

${\displaystyle sin\left(\theta _{B}\right)={\frac {\sqrt {\varepsilon _{r}-1}}{\sqrt {\varepsilon _{r}^{2}-1}}}}$Ec. 3.28

${\displaystyle \Gamma _{\parallel }={\frac {-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}{\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}}}$Ec. 3.24

Polarización vertical

${\displaystyle 0={\frac {-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}{\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}}}$

${\displaystyle 0=-\varepsilon _{r}\sin \theta _{i}+{\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}$

${\displaystyle \varepsilon _{r}\sin \theta _{i}={\sqrt {\varepsilon _{r}-\cos ^{2}\theta _{i}}}}$

${\displaystyle \left(\varepsilon _{r}\sin \theta _{i}\right)^{2}=\varepsilon _{r}-\cos ^{2}\theta _{i}}$

${\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}=\varepsilon _{r}-\cos ^{2}\theta _{i}}$

${\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}=\varepsilon _{r}+\sin ^{2}\theta _{i}-1}$

${\displaystyle \varepsilon _{r}^{2}\sin \theta _{i}^{2}-\sin ^{2}\theta _{i}=\varepsilon _{r}-1}$

${\displaystyle \left(\varepsilon _{r}^{2}-1\right)\sin ^{2}\theta _{i}=\varepsilon _{r}-1}$

${\displaystyle \sin ^{2}\theta _{i}={\frac {\varepsilon _{r}-1}{\varepsilon _{r}^{2}-1}}}$

En este punto se demostró de donde sale la ecuacion 3.28 par el uso en espacio libre.

${\displaystyle \sin \theta _{i}={\sqrt {\frac {\varepsilon _{r}-1}{\varepsilon _{r}^{2}-1}}}}$

## Demostración de la ecuación 3.20

La demostración de la ecuación presentada en el libro Rappaport, la ecuación 3.20

${\displaystyle {\frac {E_{r}}{E_{i}}}={\frac {\eta _{2}\sin \theta _{t}-\eta _{1}*\sin \theta _{i}}{\eta _{2}\sin \theta _{t}+\eta _{1}*\sin \theta _{i}}}}$

Partiendo de la ecuacion presentada anteriormente, empezaremos a hacer el analisis mediante la grafica b de la pagina 452 del libro de Matthew.

${\displaystyle \varepsilon _{i}=\varepsilon _{0}+\varepsilon _{r}}$

${\displaystyle z<0=\varepsilon _{1},G_{1},\mu _{1}}$
${\displaystyle z>0=\varepsilon _{2},G_{2},\mu _{2}}$

${\displaystyle \varepsilon _{i}=H_{i}}$
${\displaystyle H_{i}=B_{i}\ \sin {(\theta _{i})}}$
${\displaystyle B_{i}=\omega {\sqrt {\mu _{i}\varepsilon _{i}}}}$

${\displaystyle \varepsilon _{r}=H_{r}}$
${\displaystyle H_{r}=B_{r}\ \sin {(\theta _{r})}}$
${\displaystyle B_{r}=\omega {\sqrt {\mu _{r}\varepsilon _{r}}}}$

${\displaystyle \varepsilon _{t}=H_{t}}$
${\displaystyle H_{t}=B_{t}\ \sin {(\theta _{t})}}$
${\displaystyle B_{t}=\omega {\sqrt {\mu _{t}\varepsilon _{t}}}}$

${\displaystyle E_{i}+-E_{r}=E_{t}}$
${\displaystyle E_{i}=E_{r}+E_{t}}$
${\displaystyle E_{r}=E_{i}\ -\ E_{t}}$
${\displaystyle E_{t}=E_{i}\ -\ E_{r}}$
${\displaystyle E_{i}=E_{i}\ -\ E_{t}+E_{i}\ -\ E_{r}}$
${\displaystyle E_{r}\ +E_{t}=E_{i}\ -\ E_{t}+E_{i}\ -\ E_{r}}$

${\displaystyle B_{r}\sin(\theta _{r})\ +\ B_{t}\sin(\theta _{t})\ =\ B_{i}\sin(\theta _{i})\ -\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{r}\sin(\theta _{r})\ }$
${\displaystyle 0=-B_{r}\sin(\theta _{r})\ +\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{t}\sin(\theta _{t})\ +\ B_{i}\sin(\theta _{i})\ -\ B_{r}\sin(\theta _{r})}$
${\displaystyle \theta _{r}\ =\ \theta _{i}}$
${\displaystyle B_{t}\sin(\theta _{t})\ -\ B_{i}\sin(\theta _{i})\ =\ -\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\sin(\theta _{r})\ -\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ +\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ -\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ }$
${\displaystyle \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ -\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ =-\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})-\ \omega {\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i})\ -\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})\ }$
${\displaystyle \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})\ -\ \omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i}))\ =-2\ \omega {\sqrt {\mu _{r}\varepsilon _{r}}}\ \sin(\theta _{r})-\ \omega ({\sqrt {\mu _{t}\varepsilon _{t}}}\sin(\theta _{t})+\omega {\sqrt {\mu _{i}\varepsilon _{i}}}\sin(\theta _{i}))\ }$
${\displaystyle \eta _{1}={\sqrt {\mu _{i}\varepsilon _{i}}}}$

${\displaystyle \eta _{2}={\sqrt {\mu _{t}\varepsilon _{t}}}}$
${\displaystyle \omega (\eta _{2}\sin(\theta _{t})-\eta _{1}\sin(\theta _{i}))=-\omega _{i}\ \sin(\theta _{i})+\omega _{r}\sin(\theta _{r})-\ \omega (\eta _{2}\sin(\theta _{t})+\eta _{1}\sin(\theta _{i}))}$
${\displaystyle \omega \sin {\theta _{i}}(\eta _{2}\sin {\theta _{t}}-\eta _{1}\sin {\theta _{i}})\ /\ \eta _{1}\ \sin {\theta _{i}}\ +\sin {\theta _{t}}\ =\ \omega +\omega B_{r}\sin {\theta _{r}}}$

${\displaystyle \eta _{2}\sin {\theta _{t}}-\eta _{1}\sin {\theta _{i}})/(\eta _{1}\sin {\theta _{i}}+\eta _{2}\sin {\theta _{t}})=B_{r}\sin {\theta _{r}}/B_{i}\sin {\theta _{i}}}$

${\displaystyle E_{i}=B_{i}\sin {\theta _{i}}}$
${\displaystyle E_{r}=B_{r}\sin {\theta _{r}}}$
Realizando los cambios correspondientes obtendremos la siguiente ecuacion:
${\displaystyle \mathrm {\Gamma } _{I}={\frac {\varepsilon _{r}}{\varepsilon _{i}}}={\frac {\eta _{2}\sin {(\theta _{t})-\eta _{1}\sin {(\theta _{i}))}}}{\eta _{2}\sin {(\theta _{t})+\eta _{1}\sin {(\theta _{i}))}}}}}$

## Demostración ecuación 3.24

${\displaystyle \eta _{1}={\sqrt {\frac {\mu }{\epsilon _{0}}}}*\eta _{2}={\sqrt {{\frac {\mu }{\epsilon }}_{2}}}}$

${\displaystyle {\sqrt {\epsilon _{0}\mu _{0}}}\sin {90-\theta _{i}}={\sqrt {\epsilon _{2}\mu _{0}}}sin{90-\theta _{t}}}$

${\displaystyle {\sqrt {\frac {\mu _{0}\epsilon _{0}}{\mu _{0}\epsilon _{2}}}}\sin {(90-\theta _{i})}=\sin {(90-\theta _{t})}}$

${\displaystyle {\sqrt {\frac {\epsilon _{0}}{\epsilon _{2}}}}\sin {(90-\theta _{i})}=\sin {(90-\theta _{t})}}$

${\displaystyle \sin {(90-\theta _{i})}=\cos {\theta _{i}}}$

${\displaystyle \cos ^{2}{\theta _{i}}=1-\sin ^{2}{(90-\theta _{i})}}$

${\displaystyle ({\sqrt {\frac {\epsilon _{0}}{\epsilon _{2}}}}*\cos {\theta _{i}}=\cos {\theta _{t}})^{2}}$

${\displaystyle {\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}=\cos ^{2}{\theta _{t}}}$

${\displaystyle {\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}=1-\sin ^{2}{(\theta _{t})}}$

${\displaystyle \sin ^{2}{(\theta _{t})}=1-{\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}}$

${\displaystyle ({\frac {\epsilon _{2}}{\epsilon _{0}}})\sin ^{2}{(\theta _{t})}=1-{\frac {\epsilon _{0}}{\epsilon _{2}}}\cos ^{2}{\theta _{i}}({\frac {\epsilon _{2}}{\epsilon _{0}}})}$

${\displaystyle \epsilon _{r}={\frac {\epsilon _{2}}{\epsilon _{0}}}}$

${\displaystyle \epsilon _{r}\sin ^{2}{\theta _{t}}=\epsilon _{r}-\cos ^{2}{\theta _{i}}}$

${\displaystyle {\sqrt {(\epsilon _{r}\sin ^{2}{\theta _{t}}=\epsilon _{r}-\cos ^{2}{\theta _{i}})}}}$

${\displaystyle {\sqrt {\epsilon _{r}}}\sin {\theta _{t}}={\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}}$

${\displaystyle \Gamma _{||}={\frac {\eta _{2}sin\theta _{t}-\eta _{1}sin\theta _{i}}{\eta _{2}sin\theta _{t}+\eta _{1}sin\theta _{i}}}}$

${\displaystyle \Gamma _{||}={\frac {{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}-\epsilon _{r}\sin {\theta _{i}}}{{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}+\epsilon _{r}\sin {\theta _{i}}}}}$

${\displaystyle \Gamma _{||}={\frac {-\epsilon _{r}\sin {\theta _{i}}+{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}}{{\sqrt {\epsilon _{r}-\cos ^{2}{\theta _{i}}}}+\epsilon _{r}\sin {\theta _{i}}}}}$

Se demuestra la ecuación 3.24

ESTE PROGRAMA REALIZA LA OPERACIÓN DE LA FUNCIÓN GAMA

import keyword;
#keyword.iskeyword('pass')
import keyword
import math
from Lxtx2 import *
print('Coeficiente de reflexión')
print('presione 1 para continuar')
S=int(input())
if (S==1):
Er=int(input())
print('ingrese theta: ')
T=int(input())

#print (Er, T)
Gamma(Er,T)

(
import keyword;
keyword.iskeyword('pass')
import keyword
import math

def Gamma(Er,T):\\

divi=n/d
posi=abs(divi)
print(&quot;Gama perpendicular es:&quot;,posi)
(
import keyword;
keyword.iskeyword('pass')
import keyword
import math

def Gamma(Er,T):
divi=n/d
posi=abs(divi)
print(&quot;Gama paralelo es:&quot;,posi)
)