Reflexión electromagnética/Índice/Anexos y demostraciones

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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 (&quot;en grados es:&quot;, 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(&quot;Gama perpendicular es:&quot;,posi) ( import keyword; keyword.iskeyword('pass') import keyword import math def Gamma(Er,T): print (&quot;en grados es:&quot;, 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(&quot;Gama paralelo es:&quot;,posi) )