Reflexión electromagnética/Índice/Anexos y demostraciones

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Anexos Demostración Ecuación 3.19 ${\epsilon }_i={\epsilon }_0{\epsilon }_r$ $\mu ={\mu }_0{\mu }_r$ $z<0={\epsilon }_1,G_1,{\mu }_1$ $z>0={\epsilon }_2,G_2,{\mu }_2$ ${\epsilon }_i=H_i$ $H_i=B_i\sin (90-{\theta }_i)$ $B_i=\omega \sqrt{{\mu }_i{\epsilon }_i}$ ${\epsilon }_r=H_r$ $H_r=B_r\sin (90-{\theta }_r)$ $B_r=\omega \sqrt{{\mu }_r{\epsilon }_r}$ ${\epsilon }_t=H_t$ $H_t=B_t\sin (90-{\theta }_t)$ $B_t=\omega \sqrt{{\mu }_t{\epsilon }_t}$ ${\epsilon }_i+{\epsilon }_r={\epsilon }_t$ ${\epsilon }_i={\epsilon }_r+{\epsilon }_t$ ${\epsilon }_r={\epsilon }_i-{\epsilon }_t$ ${\epsilon }_t={\epsilon }_i-{\epsilon }_r$ ${\epsilon }_i={\epsilon }_i-{\epsilon }_t+{\epsilon }_i-{\epsilon }_r$ ${\epsilon }_r+{\epsilon }_t={\epsilon }_i-{\epsilon }_t+{\epsilon }_i-{\epsilon }_r$ $B_r\sin ({\theta }_r)+B_t\sin ({\theta }_t)=B_i\sin ({\theta }_i)-B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_r\sin ({\theta }_r)$ $\theta =-B_r\sin ({\theta }_r)+B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_r\sin ({\theta }_r)$ ${\theta }_r=B_i{\theta }_r={\theta }_i$ $B_t\sin ({\theta }_t)-B_i\sin ({\theta }_i)=-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)$ $\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)-\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)=-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)$ $\omega (\sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)-\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i))=-2\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-(\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i))$ ${\eta }_1={\mu }_i{\epsilon }_i={\mu }_r{\epsilon }_r{{\eta }_2=\mu }_t{\epsilon }_t$ $\omega ({\eta }_2\sin ({\theta }_t)-{\eta }_1\sin ({\theta }_i))=-\omega B_i\sin ({\theta }_i)+\omega B_r\sin ({\theta }_r)-\omega ({\eta }_2\sin ({\theta }_t)+{\eta }_1\sin ({\theta }_i))$ $\frac{\omega ({\eta }_2\sin ({\theta }_t)-{\eta }_1\sin ({\theta }_i))}{({\eta }_2\sin ({\theta }_t)-{\eta }_1\sin ({\theta }_i))}=\frac{(\omega +\omega B_r\sin ({\theta }_r))}{\omega B_i\sin ({\theta }_i)}$ ${\mathrm{\Gamma }}_{||}=\frac{E_r}{E_i}=\frac{{\eta }_{2}sin{\theta }_t-{\eta }_{1}sin{\theta }_i}{{\eta }_{2}sin{\theta }_t+{\eta }_{1}sin{\theta }_i}$ 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 $sin\left({\theta }_B\right)=\sqrt{\frac{{\epsilon }_1}{{\epsilon }_1+{\epsilon }_2}}$ Ec. 3.27 ${\mathrm{\Gamma }}_{\parallel }=\frac{{\eta }_2\sin {\theta }_t-{\eta }_1\sin {\theta }_i}{{\eta }_2\sin {\theta }_t+{\eta }_1\sin {\theta }_i}$Ec. 3.19 $0=\frac{{\eta }_2\sin {\theta }_t-{\eta }_1\sin {\theta }_i}{{\eta }_2\sin {\theta }_t+{\eta }_1\sin {\theta }_i}$ $0={\eta }_2\sin {\theta }_t-{\eta }_1\sin {\theta }_i$ $\sqrt{\frac{{\mu }_2}{{\epsilon }_2}}\sin {\theta }_t=\sqrt{\frac{{\mu }_1}{{\epsilon }_1}}\sin {\theta }_i$ $\sqrt{{\mu }_2{\epsilon }_1}\sin {\theta }_t=\sqrt{{\mu }_1{\epsilon }_2}\sin {\theta }_i$ ${\mu }_2{\epsilon }_1{\sin }^2{\theta }_t={\mu }_1{\epsilon }_2{\sin }^2{\theta }_i$ $\frac{{\mu }_1{\epsilon }_2}{{\mu }_2{\epsilon }_1}{\sin }^2{\theta }_i={\sin }^2{\theta }_t$ $\frac{{\mu }_1{\epsilon }_2}{{\mu }_2{\epsilon }_1}{\sin }^2{\theta }_i=(1-{\cos }^2{\theta }_t)$ $\frac{{\mu }_1{\epsilon }_2}{{\mu }_2{\epsilon }_1}{\sin }^2{\theta }_i=1-\frac{{\mu }_1{\epsilon }_1}{{\mu }_2{\epsilon }_2}{\cos }^2{\theta }_i$ $(\frac{{\mu }_1{\epsilon }_2}{{\mu }_2{\epsilon }_1}-\frac{{\mu }_1{\epsilon }_1}{{\mu }_2{\epsilon }_2}){\sin }^2{\theta }_i=1-\frac{{\mu }_1{\epsilon }_1}{{\mu }_2{\epsilon }_2}$ $\frac{{\mu }_1}{{\mu }_2}(\frac{{\epsilon }_2}{{\epsilon }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}){\sin }^2{\theta }_i=1-\frac{{\mu }_1{\epsilon }_1}{{\mu }_2{\epsilon }_2}$ ${\sin }^2{\theta }_i=\frac{\frac{{\mu }_2}{{\mu }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}}{\frac{{\epsilon }_2}{{\epsilon }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}}$ $\sin {\theta }_i=\sqrt{\frac{\frac{{\mu }_2}{{\mu }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}}{\frac{{\epsilon }_2}{{\epsilon }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}}}$ Para el angulo de brewster se puede considerar que: ${\mu }_2={\mu }_1$ $\sin {\theta }_i=\sqrt{\frac{1-\frac{{\epsilon }_1}{{\epsilon }_2}}{\frac{{\epsilon }_2}{{\epsilon }_1}-\frac{{\epsilon }_1}{{\epsilon }_2}}}$ $\sin {\theta }_i=\sqrt{\frac{\frac{{\epsilon }_2-{\epsilon }_1}{{\epsilon }_2}}{\frac{{\epsilon }_2^2-{\epsilon }_1^2}{{\epsilon }_1{\epsilon }_2}}}$ $\sin {\theta }_i=\sqrt{\frac{{\epsilon }_1({\epsilon }_2-{\epsilon }_1)}{{\epsilon }_2^2-{\epsilon }_1^2}}$ $\sin {\theta }_i=\sqrt{\frac{{\epsilon }_1({\epsilon }_2-{\epsilon }_1)}{({\epsilon }_2-{\epsilon }_1)({\epsilon }_2+{\epsilon }_1)}}$ En este punto se demostró de donde sale la ecuacion 3.27 cuando ${\mu }_2={\mu }_1$ $sin\left({\theta }_B\right)=\sqrt{\frac{{\epsilon }_1}{{\epsilon }_1+{\epsilon }_2}}$ 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. $sin\left({\theta }_B\right)=\frac{\sqrt{{\epsilon }_r-1}}{\sqrt{{\epsilon }_r^2-1}}$Ec. 3.28 ${\mathrm{\Gamma }}_{\parallel }=\frac{-{\epsilon }_r\sin {\theta }_i+\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}}{{\epsilon }_r\sin {\theta }_i+\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}}$Ec. 3.24 Polarización vertical $0=\frac{-{\epsilon }_r\sin {\theta }_i+\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}}{{\epsilon }_r\sin {\theta }_i+\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}}$ $0=-{\epsilon }_r\sin {\theta }_i+\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}$ ${\epsilon }_r\sin {\theta }_i=\sqrt{{\epsilon }_r-{\cos }^2{\theta }_i}$ ${({\epsilon }_r\sin {\theta }_i)}^2={\epsilon }_r-{\cos }^2{\theta }_i$ ${\epsilon }_r^2\sin {\theta }_i^2={\epsilon }_r-{\cos }^2{\theta }_i$ ${\epsilon }_r^2\sin {\theta }_i^2={\epsilon }_r+{\sin }^2{\theta }_i-1$ ${\epsilon }_r^2\sin {\theta }_i^2-{\sin }^2{\theta }_i={\epsilon }_r-1$ $({\epsilon }_r^2-1){\sin }^2{\theta }_i={\epsilon }_r-1$ ${\sin }^2{\theta }_i=\frac{{\epsilon }_r-1}{{\epsilon }_r^2-1}$ En este punto se demostró de donde sale la ecuacion 3.28 par el uso en espacio libre. $\sin {\theta }_i=\sqrt{\frac{{\epsilon }_r-1}{{\epsilon }_r^2-1}}$ 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. ${\epsilon }_i={\epsilon }_0+{\epsilon }_r$ $z<0={\epsilon }_1,G_1,{\mu }_1$ $z>0={\epsilon }_2,G_2,{\mu }_2$ ${\epsilon }_i=H_i$ $H_i=B_i\sin ({\theta }_i)$ $B_i=\omega \sqrt{{\mu }_i{\epsilon }_i}$ ${\epsilon }_r=H_r$ $H_r=B_r\sin ({\theta }_r)$ $B_r=\omega \sqrt{{\mu }_r{\epsilon }_r}$ ${\epsilon }_t=H_t$ $H_t=B_t\sin ({\theta }_t)$ $B_t=\omega \sqrt{{\mu }_t{\epsilon }_t}$ $E_i+-E_r=E_t$ $E_i=E_r+E_t$ $E_r=E_i-E_t$ $E_t=E_i-E_r$ $E_i=E_i-E_t+E_i-E_r$ $E_r+E_t=E_i-E_t+E_i-E_r$ $B_r\sin ({\theta }_r)+B_t\sin ({\theta }_t)=B_i\sin ({\theta }_i)-B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_r\sin ({\theta }_r)$ $0=-B_r\sin ({\theta }_r)+B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_t\sin ({\theta }_t)+B_i\sin ({\theta }_i)-B_r\sin ({\theta }_r)$ ${\theta }_r={\theta }_i$ $B_t\sin ({\theta }_t)-B_i\sin ({\theta }_i)=-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)$ $\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)-\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)=-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-\omega \sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i)-\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)$ $\omega (\sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)-\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i))=-2\omega \sqrt{{\mu }_r{\epsilon }_r}\sin ({\theta }_r)-\omega (\sqrt{{\mu }_t{\epsilon }_t}\sin ({\theta }_t)+\omega \sqrt{{\mu }_i{\epsilon }_i}\sin ({\theta }_i))$ ${\eta }_1=\sqrt{{\mu }_i{\epsilon }_i}$ ${\eta }_2=\sqrt{{\mu }_t{\epsilon }_t}$ $\omega ({\eta }_2\sin ({\theta }_t)-{\eta }_1\sin ({\theta }_i))=-{\omega }_i\sin ({\theta }_i)+{\omega }_r\sin ({\theta }_r)-\omega ({\eta }_2\sin ({\theta }_t)+{\eta }_1\sin ({\theta }_i))$ $\omega \sin {\theta }_i({\eta }_2\sin {\theta }_t-{\eta }_1\sin {\theta }_i)/{\eta }_1\sin {\theta }_i+\sin {\theta }_t=\omega +\omega B_r\sin {\theta }_r$ ${\eta }_2\sin {\theta }_t-{\eta }_1\sin {\theta }_i)/({\eta }_1\sin {\theta }_i+{\eta }_2\sin {\theta }_t)=B_r\sin {\theta }_r/B_i\sin {\theta }_i$ $E_i=B_i\sin {\theta }_i$ $E_r=B_r\sin {\theta }_r$ Realizando los cambios correspondientes obtendremos la siguiente ecuacion: ${\mathrm{\Gamma }}_I=\frac{{\epsilon }_r}{{\epsilon }_i}=\frac{{\eta }_2\sin ({\theta }_t)-{\eta }_1\sin ({\theta }_i))}{{\eta }_2\sin ({\theta }_t)+{\eta }_1\sin ({\theta }_i))}$ Demostración ecuación 3.24 ${\eta }_1=\sqrt{\frac{\mu }{{ϵ}_0}}*{\eta }_2=\sqrt{{\frac{\mu }{ϵ}}_2}$ $\sqrt{{ϵ}_0{\mu }_0}\sin 90-{\theta }_i=\sqrt{{ϵ}_2{\mu }_0}sin90-{\theta }_t$ $\sqrt{\frac{{\mu }_0{ϵ}_0}{{\mu }_0{ϵ}_2}}\sin (90-{\theta }_i)=\sin (90-{\theta }_t)$ $\sqrt{\frac{{ϵ}_0}{{ϵ}_2}}\sin (90-{\theta }_i)=\sin (90-{\theta }_t)$ Entidades trigonométricas $\sin (90-{\theta }_i)=\cos {\theta }_i$ ${\cos }^2{\theta }_i=1-{\sin }^2(90-{\theta }_i)$ $(\sqrt{\frac{{ϵ}_0}{{ϵ}_2}}*\cos {\theta }_i=\cos {\theta }_t{)}^2$ $\frac{{ϵ}_0}{{ϵ}_2}{\cos }^2{\theta }_i={\cos }^2{\theta }_t$ $\frac{{ϵ}_0}{{ϵ}_2}{\cos }^2{\theta }_i=1-{\sin }^2({\theta }_t)$ ${\sin }^2({\theta }_t)=1-\frac{{ϵ}_0}{{ϵ}_2}{\cos }^2{\theta }_i$ $(\frac{{ϵ}_2}{{ϵ}_0}){\sin }^2({\theta }_t)=1-\frac{{ϵ}_0}{{ϵ}_2}{\cos }^2{\theta }_i(\frac{{ϵ}_2}{{ϵ}_0})$ Permitividad del vacío: ${ϵ}_r=\frac{{ϵ}_2}{{ϵ}_0}$ ${ϵ}_r{\sin }^2{\theta }_t={ϵ}_r-{\cos }^2{\theta }_i$ $\sqrt{({ϵ}_r{\sin }^2{\theta }_t={ϵ}_r-{\cos }^2{\theta }_i)}$ $\sqrt{{ϵ}_r}\sin {\theta }_t=\sqrt{{ϵ}_r-{\cos }^2{\theta }_i}$ ${\mathrm{\Gamma }}_{||}=\frac{{\eta }_{2}sin{\theta }_t-{\eta }_{1}sin{\theta }_i}{{\eta }_{2}sin{\theta }_t+{\eta }_{1}sin{\theta }_i}$ ${\mathrm{\Gamma }}_{||}=\frac{\sqrt{{ϵ}_r-{\cos }^2{\theta }_i}-{ϵ}_r\sin {\theta }_i}{\sqrt{{ϵ}_r-{\cos }^2{\theta }_i}+{ϵ}_r\sin {\theta }_i}$ ${\mathrm{\Gamma }}_{||}=\frac{-{ϵ}_r\sin {\theta }_i+\sqrt{{ϵ}_r-{\cos }^2{\theta }_i}}{\sqrt{{ϵ}_r-{\cos }^2{\theta }_i}+{ϵ}_r\sin {\theta }_i}$ 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) )