exercicio2.3
\large{\alpha = \omega \sqrt{\frac{\mu\epsilon}{2} \left(\sqrt{1 + \left(\frac{\sigma}{\omega \epsilon}\right)^2} –  1 \right)}}
\large{\beta = \omega \sqrt{\frac{\mu\epsilon}{2} \left(\sqrt{1 + \left(\frac{\sigma}{\omega \epsilon}\right)^2} +  1 \right)}}

Calculando elementos a elementos do fator de atenuação e do fator de fase:

\omega = 2\pi f\omega = 2\pi 20*10^6\boxed{\omega = 12,566 . 10^7  rad/s}\boxed{\large{\frac{\mu.\epsilon}{2} = \frac{4\pi . 10^{-7} * 1,2*8,85 . 10^{-12}}{2}} =  6,673 . 10^{-18}}\boxed{\large{\sqrt{1 + \left(\frac{\sigma}{\omega \epsilon}\right)^2} = \sqrt{1 + \left(\frac{10^{-4}}{12,556.10^7  * 1,2 * 8,85.10^{-12}}\right)^2} = \sqrt{1 + \left(0,075\right)^2}  = 1,00281}}

Portanto voltando as equações do fator de atenuação e do fator de base temos:

\large{\alpha = 12,556.10^7 \sqrt{6,673.10^{-18} \left(1,00281 –  1 \right)}  = 12,556.10^7 \sqrt{6,673.10^{-18} \left(0,00281 \right)}  =  12,556.10^7 * 0,1369 * 10^{-9} }

\boxed{\large{\therefore \alpha = 0,0172  Np/m}}  

\large{\beta = 12,556.10^7 \sqrt{6,673.10^{-18} \left(1,00281 +  1 \right)} =  12,556.10^7 \sqrt{6,673.10^{-18} \left(2,00281 \right)}  =  12,556.10^7 * 3,6558 * 10^{-9}}

\boxed{\large{\therefore \beta = 0,459  rad/m}}

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