Lanzamiento de martillo/Español/Texto completo

Introducción

El presente trabajo es una compilación la sección 3 del libro Rappaport usado en la clase de Radio enlaces esta sección habla sobre el método de propagación llamado reflexión. en la implementación de radio enlaces, básicamente es mecanismo de propagación del cual se debe tener en cuenta distintos fenómenos como las perdidas, los desvanecimientos, la propagación por trayectos múltiples, Cálculos que afectan los radioenlaces. Para que nuestros enlaces de radio estén correctamente sintonizados teniendo en cuenta estos fenómenos que podrían afectarla.

Métodos de Propagación

La reflexión, la difracción y la dispersión son los tres mecanismos básicos de propagación que influyen en la propagación en un sistema de comunicaciones móviles. Estos mecanismos se explican brevemente en esta sección, y los modelos de propagación que describen estos mecanismos se analizan posteriormente en este capítulo. La potencia recibida (o su recíproca, la pérdida de trayectoria) es generalmente el parámetro más importante predicho por los modelos de propagación a gran escala basados en la física de la reflexión, la dispersión y la difracción. El desvanecimiento a pequeña escala y la propagación por trayectos múltiples (discutidos en el capítulo 4) también pueden describirse mediante la física de estos tres mecanismos básicos de propagación.

Reflexión

Ocurre cuando una onda electromagnética que se propaga incide sobre un objeto que tiene dimensiones muy grandes en comparación con la longitud de onda de la onda que se propaga. Los reflejos se producen desde la superficie de la tierra y desde edificios y paredes.

Ejemplo de Reflexión

Mobile Radio Propagation: Large-Scale Path Loss

The Three Basic Propagation Mechanisms Reflection, diffraction, and scattering are the three basic propagation mech-anisms which impact propagation in a mobile communication system. These mechanisms are briefly explained in this section, and propagation models which describe these mechanisms are discussed subsequently in this chapter. Received power (or its reciprocal, path loss) is generally the most important parameter predicted by large-scale propagation models based on the physics of reflection, scattering, and diffraction. Small-scale fading and multipath propagation (dis-cussed in Chapter 4) may also be described by the physics of these three basic propagation mechanisms. Reflection occurs when a propagating electromagnetic wave impinges upon an object which has very large dimensions when compared to the wavelength of the propagating wave. Reflections occur from the surface of the earth and from buildings and walls. Diffraction occurs when the radio path between the transmitter and receiver is obstructed by a surface that has sharp irregularities (edges). The sec-ondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle, giving rise to a bending of waves around the obstacle, even when a line-of-sight path does not exist between transmitter and receiver. At high frequencies, diffraction, like reflection, depends on the geome-try of the object, as well as the amplitude, phase, and polarization of the incident wave at the point of diffraction. Scattering occurs when the medium through which the wave travels con-sists of objects with dimensions that are small compared to the wavelength, and where the number of obstacles per unit volume is large. Scattered waves are pro-duced by rough surfaces, small objects, or by other irregularities in the channel. In practice, foliage, street signs, and lamp posts induce scattering in a mobile communications system.

Reflection

When a radio wave propagating in one medium impinges upon another medium having different electrical properties, the wave is reflected and partially transmitted. If the plane wave is incident on a perfect dielectric, part of the energy is transmitted into the second medium and part of the energy is reflected back into the first medium, and there is no loss of energy in absorption.

Reflection Example

Reflexión de los Dieléctricos

En la reflexión dieléctrica se muestra una onda incide con un ángulo $\theta _{i}$, con el plano limite entre dos medios dieléctricos frontera, donde parte de la energía se refleja de regreso al primer medio en un ángulo $\theta _{r}$ y otras transmite con $\theta _{t}$.

Planos de incidencia

Plano incidente: se define como el plano en el que se encuentran la onda incidente, reflejada y transmitida

En la figura 3.4 se puede observar que, el campo eléctrico se encuentra polarizado de forma paralela al plano incídete en la figura 3.4.a, y la polarización del campo eléctrico es perpendicular al plano incidente en la 3.4.b.

Geometría para calcular los coeficientes de reflexión entre dos dieléctricos

Los parámetros $\epsilon _{1}$, $\mu _{1}$, $\sigma _{1}$ y $\epsilon _{2}$, $\mu _{2}$, $\sigma _{2}$ representan la permitividad, la permeabilidad y la conductividad respectivamente.

La constante dieléctrica de un dieléctrico perfecto esta relacionada con un valor relativo $\epsilon _{r}$ y $\epsilon _{0}$ tal que $\epsilon$ $=$ $\epsilon _{r}*\epsilon _{0}$ es la permitividad del vacío siendo $8.85*10^{-12}$ ${\frac {F}{m}}$.

Reflection from Dielectrics

Figure 3.4 shows an electromagnetic wave incident at an angle $\theta _{i}$, with the plane of the boundary between two dielectric media. As shown in the figure, part of the energy is reflected back to the first media at an angle $\theta _{r}$, and part of the energy is transmitted (refracted) into the second media at an angle $\theta _{t}$,. The nature of reflection varies with the direction of polarization of the E-field. The behavior for arbitrary directions of polarization can be studied by considering the two distinct cases shown in Figure 3.4. The plane of incidenceis defined as the plane containing the incident, reflected, and transmitted rays [Ram65]. In Figure 3.4a, the E-field polarization is parallel with the plane of incidence (that is, the E-field has a vertical polarization, or normal component, with respect to the reflecting surface) and in Figure 3.4b, the E-field polarization is perpendicular to the plane of incidence (that is, the incident E-field is pointing out of the page towards the reader, and is perpendicular to the page and parallel to the reflecting surface).
In Figure 3.4, the subscripts i, r, t refer to the incident, reflected, and transmitted fields, respectively. Parameters $\epsilon _{1}$, $\mu _{1}$, $\sigma _{1}$ y $\epsilon _{2}$, $\mu _{2}$, $\sigma _{2}$, represent the permittivity, permeability, and conductance of the two media, respectively. Often, the dielectric constant of a perfect (lossless) dielectric is related to a relative value of permittivity, $\epsilon _{r}$, such that $\epsilon =\epsilon _{0}*\epsilon _{r}$ where $\epsilon _{0}$ is a constant given by $8.85x10^{-12}$ ${\frac {F}{m}}$. If a dielectric material is lossy, it will absorb power and may be described by a complex dielectric constant given by

$\epsilon =\epsilon _{0}*\epsilon _{r}-j\epsilon _{'}$

where,

$\epsilon _{'}={\frac {\sigma }{2*\pi *f}}$

and $\sigma$ is the conductivity of the material measured in Siemens/meter. The terms $\epsilon _{r}$ and $\sigma$ are generally insensitive to operating frequency when the material is a good conductor ($f<\sigma )/(\epsilon _{0}\epsilon _{r}$) ). For lossy dielectrics, $\epsilon _{r}$ and $\epsilon _{r}$ are generally constant with frequency, but $\sigma$ may be sensitive to the operating frequency, as shown in Table 3.1. Electrical properties of a wide range of materials were characterized over a large frequency range by Von Hipple [Von54]

geometry for calculating the reflection coefficients between two dielectrics.

El mejor dieléctrico se considera el vacío donde:

$\alpha =0$

$\beta =\omega {\sqrt {\mu _{0}\epsilon _{0}}}$

De esta mera se observar la constante de propagación, que se divide en una parte real y otra imaginaria:

$\gamma =\alpha +j\beta$

En el caso de que un material dieléctrico tiene perdidas de tal manera que absorbe energía, se puede representa por una constante dieléctrica compleja:

$\epsilon =\epsilon _{0}*\epsilon _{r}-j\epsilon _{'}$

donde,

$\epsilon _{'}={\frac {\sigma }{2*\pi *f}}$

Los termino de la $\epsilon _{r}$ y $\sigma$ normalmente no son afectados por variaciones en la frecuencia de la onda cuando el material es un buen conductor $(f<{\frac {\sigma }{(\epsilon _{r}*\epsilon {0})}})$

Cuando hay perdidas las permitividades se puede encontrar constante con la frecuencia, pero la conductividad si puede llegar a variar con la frecuencia de operación (se puede observar en la tabla 3.1).

Tabla 3.1

En la tabla se aprecia el comportamiento de la permitividad y la conductividad cuando es afectada por una frecuencia vemos que la frecuencia en algunos materiales puede afectar bien sea su conductividad o su permitividad relativa, para materiales superconductores la conductividad no es afectada por la variación de frecuencia.

Material	Relative Permittivity	Conductividad	Frecuencia (MHz)
Suelo Pobre	4	0.001	100
Tierra Tipica	15	0.005	100
Tierra Buena	25	0.02	100
Agua de Mar	81	5.0	100
Ladrillo	4.44	0.001	4000
Caliza	7.51	0.028	4000
Vidrio, Corning 707	4	0.00000018	1
Vidrio, Corning 707	4	0.000027	100
Vidrio, Corning 707	4	0.005	10000

Figure 3.4 Parámetros de material en varias frecuencias
$\Gamma _{||}={\frac {E_{r}}{E_{i}}}={\frac {\eta _{2}sin\theta _{t}-\eta _{1}sin\theta _{i}}{\eta _{2}sin\theta _{t}+\eta _{1}sin\theta _{i}}}$

$\Gamma _{L}={\frac {E_{r}}{E_{i}}}={\frac {\eta _{_{2}}sin\theta _{i}-\eta _{1}sin\theta _{t}}{\eta _{2}sin\theta {i}+\eta _{1}sin\theta _{t}}}$ Figure 3.4 Material Parameters at various Frequencies

Because of superposition, only two orthogonal polarizations need be considered to solve general reflection problems. The reflection coefficients for the two cases of parallel and perpendicular E-field polarization at the boundary of two dielectrics are given by

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

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

Ecuaciones de polarización

Debido a la superposición, solo se necesitan dos polarizaciones ortogonales para resolver problemas de reflexión general. Los coeficientes de reflexión para los dos casos de polarización paralela y perpendicular del campo Eléctrico en el límite de dos dieléctricos esta dada por:

Cuando $\eta _{i}$ es la impedancia intrínseca del medio i (i=1,2) y es dada por ${\sqrt {\frac {\mu _{i}}{\epsilon _{i}}}}$ la relación entre el campo eléctrico y magnético para una onda plana uniforme en el medio particular. La velocidad de una onda electromagnética es dada por ${\frac {1}{\sqrt {\mu \epsilon }}}$ y las condiciones límite en la superficie de incidencia obedecen a la Ley Snells:

${\sqrt {\mu _{1}\epsilon _{1}}}\sin(90-\theta _{i})={\sqrt {\mu _{2}\epsilon _{2}}}\sin(90-\theta _{t})$ (3.21)

$\theta _{i}=\theta _{t}$ (3.22),

$E_{i}=E_{r}+E_{t}$

$E_{r}=\Gamma E_{i}$ (3.23.a)

$E_{i}=E_{r}+E_{t}$

$E_{i}-E_{i}\Gamma =E_{t}$

$E_{i}(1-\Gamma )=E_{t}$

$E_{t}=(1+\Gamma )E_{i}$ (3.23.b).

Donde $\Gamma$ es $\Gamma _{||}$ o $\Gamma _{1}$, dependiendo de la polarización. Para el caso en el que el primer medio es el espacio libre y $\mu _{1}=\mu _{2}$, los coeficientes de reflexión para los dos casos de polarización vertical y horizontal se pueden simplificar a: $\Gamma _{||}={\frac {-\epsilon _{r}\sin \theta _{i}+{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}{\epsilon _{r}\sin \theta _{i}+{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}}$ y $\Gamma _{1}={\frac {\sin \theta _{i}-{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}{\sin \theta _{i}+{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}}$

$\mu$=Permeabilidad.
$\Gamma$= Coeficiente de reflexión.
$\Gamma _{||}$= Polarización Vertical.
$\Gamma _{1}$= Polarización horizontal.
$\epsilon _{r}$= Permitividad reflejada.
$\theta _{r}$ = Angulo incidente.
Los subíndices $i,r,t$ se refieren a los campos incidente, reflejado y trasmitido.

Figura 3.6

Se analiza el coeficiente de reflexión de Fresnel cuando se hace una polarización vertical usando dos materiales con diferente índice de permitividad.

Magnitud de los coeficientes de reflección en función del ángulo de incidencia para $\epsilon _{r}=4$, $\epsilon _{r}=12$, utilizando

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

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

where is the intrinsic impedance of the i th medium (i = 1, 2), and is given by ${\sqrt {\frac {\mu _{1}}{\epsilon {1}}}}$ , the ratio of electric to magnetic field for a uniform plane wave in the par-ticular medium. The velocity of an electromagnetic wave is given by ${\frac {1}{\sqrt {\mu \epsilon }}}$, and the boundary conditions at the surface of incidence obey Snell’s Law which, referring to Figure 3.4, is given by

${\sqrt {\mu _{1}\epsilon _{1}}}sin{90-\theta _{i}}={\sqrt {\mu _{2}\epsilon _{2}}}sin{90-\theta _{t}}$

The boundary conditions from Maxwell’s equations are used to derive equations (3.19) and (3.20) as well as equations (3.22), (3.23.a), and (3.23.b).

$\theta _{i}=\theta _{r}$

and,

$E_{r}=\Gamma *E_{i}$

$E_{t}={1+\Gamma }*E_{i}$

where $\Gamma$ is either $\Gamma _{||}$ or $\Gamma _{LL}$, depending on polarization. For the case when the first medium is free space and $\mu _{1}$ $=$ $\mu _{2}$, the reflection coefficients for the two cases of vertical and horizontal polarization can be simplified to

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

and $\Gamma _{1}={\frac {\sin \theta _{i}-{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}{\sin \theta _{i}+{\sqrt {\epsilon _{r}-\cos ^{2}\theta _{i}}}}}$

Figure 3.6

Polarización paralela (Campo Eléctrico en plano de incidencia)

[fig:2]

la geometría en la figura 3.4.

Polarización paralela (Campo Eléctrico en plano de incidencia)

Polarizacion perpendicular (Campo Eléctrico no en plano de incidencia)

Se analiza el coeficiente de Reflexión de Fresnel con una polarización horizontal se puede observar que el ángulo de incidencia para el cual la reflexión es del 100% es únicamente para 0° y luego disminuye progresivamente en ambos casos hasta que la transmisión sea muy alta con un ángulo de incidencia de 90°.

Ejemplo 3.4

Demostrar que el medio 1 es el espacio libre y el medio 2 es un dieléctrico, tanto $\Gamma _{||}$ como $\Gamma _{LL}$ se acercan a 1 cuando $\theta _{i}$ se acerca a 0° independientemente de $\epsilon _{r}$.

Solución Sustituir $\theta _{i}$ $=$ 0° en la ecuación — $\Gamma _{||}={\frac {-\epsilon _{r}sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}{\epsilon _{r}sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}}$

$\Gamma _{||}={\frac {\sqrt {\epsilon _{r}-1}}{\sqrt {\epsilon _{r}-1}}}$$=1$

Sustituir $\theta _{i}$ $=$ 0° en la ecuación — $\Gamma _{L}={\frac {sin0-{\sqrt {\epsilon _{r}-cos^{2}0}}}{sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}}$ $\Gamma _{L}={\frac {-{\sqrt {\epsilon _{r}-1}}}{\sqrt {\epsilon _{r}-1}}}=-1$

Perpendicular Polarization (Electric Field not in plane of incidence)

Magnitude of reflection coefficients as a function of angle of incidence for $\epsilon _{r}$ $=$ 4, $\epsilon _{r}$ $=$ 12, using geometry in figure 3.4.

Example 3.4 Demonstrate that if medium 1 is free space and medium 2 is a dielectric, both $\Gamma _{||}$ and $\Gamma _{LL}$ approach 1. as$\theta _{i}$, approaches 0° regardless of $\epsilon _{r}$.

Solution to Example 3.4

Substituting $\theta _{i}$ $=$ 0° in equation (3.24)

$\Gamma _{||}={\frac {-\epsilon _{r}sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}{\epsilon _{r}sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}}$

$\Gamma _{||}={\frac {\sqrt {\epsilon _{r}-1}}{\sqrt {\epsilon _{r}-1}}}$ = 1</math>

Substituting $\theta _{i}$ $=$ 0° in equation (3.25) $\Gamma _{L}={\frac {sin0-{\sqrt {\epsilon _{r}-cos^{2}0}}}{sin0+{\sqrt {\epsilon _{r}-cos^{2}0}}}}$ $\Gamma _{L}={\frac {-{\sqrt {\epsilon _{r}-1}}}{\sqrt {\epsilon _{r}-1}}}$ $=-1$

Este ejemplo nos ilustra que la tierra puede también ser tomada como reflector ideal debido al ángulo de incidencia de la onda el cual en este caso tiende a tener un ángulo muy pequeño o casi 0° esto hace que la relación de onda estacionaria tienda a ser 1 es decir, que se refleje la onda incidente en la tierra.

Ángulo de Brewster

El ángulo de Brewster es el ángulo en el que se produce la reflexión en el medio de origen. Ocurre cuando el ángulo de incidencia $\theta _{B}$ es tal que el coeficiente de reflexión $\Gamma$ es igual a cero. El ángulo de Brewster viene dado por el valor de $\theta _{B}$ que satisface

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

Para el caso en el que el primer medio es espacio libre y el segundo medio tiene un $\varepsilon _{r}$ de permitividad relativa, la ecuación se puede expresar

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

Ejemplo 3.5
Calcular el ángulo de Brewster para una onda que incide en el suelo con una permitividad de epsilon = 4

$sin\left(\theta _{i}\right)={\frac {\sqrt {\left(4\right)-1}}{\sqrt {\left(4\right)^{2}-1}}}={\sqrt {\frac {3}{15}}}={\sqrt {\frac {1}{5}}}$
$\theta _{i}=sin^{-1}{\sqrt {\frac {1}{5}}}=26.56^{\circ }$

3.5.2 Brewster Angle

The Brewster angle is the angle at which no reflection occurs in the medium of origin. It occurs when the incident angle $\theta _{B}$ is such that the reflection coefficient $\Gamma _{||}$ is equal to zero (see Figure 3.6). The Brewster angle is given by the value of $\theta _{B}$ which satisfies

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

For the case when the first medium is free space and the second medium has a relative permittivity $\epsilon _{r}$, equation (3.27) can be expressed as

$sin\left(\theta _{B}\right)={\frac {\sqrt {\varepsilon _{r}-1}}{\sqrt {\varepsilon _{r}^{2}-1}}}$
Note that the Brewster angle occurs only for vertical (i.e. parallel) polarization.
Example 3.5 Calculate the Brewster angle for a wave impinging on ground having a permitivity of $\epsilon _{r}=4$.

Solution Example 3.5 The Brewster angle can be found by substituting the values for $\epsilon _{r}$ in equation (3.28)

$sin\left(\theta _{i}\right)={\frac {\sqrt {\left(4\right)-1}}{\sqrt {\left(4\right)^{2}-1}}}={\sqrt {\frac {3}{15}}}={\sqrt {\frac {1}{5}}}$
$\theta _{i}=sin^{-1}{\sqrt {\frac {1}{5}}}=26.56^{\circ }$

Anexos

Demostración Ecuación 3.19

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

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
$sin\left(\theta _{B}\right)={\sqrt {\frac {\varepsilon _{1}}{\varepsilon _{1}+\varepsilon _{2}}}}$ Ec. 3.27

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

Polarización vertical

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

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

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

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

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

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

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

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

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

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

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

Demostración de la ecuación 3.20

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

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

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

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

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

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

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

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

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

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

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

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

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

Demostración ecuación 3.24

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

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

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

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

Entidades trigonométricas

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

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

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

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

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

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

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

Permitividad del vacío:

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

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

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

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

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

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

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

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):
    print('ingrese Efectividad relativa:') 
    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):\\

    print ("en grados es:", math.radians(T))

    n=(math.sin(math.radians(T))- math.sqrt(Er-(math.cos(math.radians(T))**2))) 
    d=(math.sin(math.radians(T))+ math.sqrt(Er-(math.cos(math.radians(T))**2))) 
     divi=n/d
    posi=abs(divi)
    print("Gama perpendicular es:",posi)
(
import keyword;
keyword.iskeyword('pass')
import keyword
import math 

def Gamma(Er,T):
    print ("en grados es:", math.radians(T))
    n=(-Er*math.sin(math.radians(T))+ math.sqrt(Er-(math.cos(math.radians(T))**2))) 
    d=(Er*math.sin(math.radians(T))+ math.sqrt(Er-(math.cos(math.radians(T))**2
    divi=n/d
    posi=abs(divi)
    print("Gama paralelo es:",posi)
    )

• Estructura elaborada por los estudiantes Héctor Javier Vega Lozano, Esther Alexandra Ramos Arias, Ferney Genaro Vazques. Estudiantes de octavo semestre de Ingeniería Electrónica en la Universidad de Cundinamarca.