Derive as equações de \alpha e \beta abaixo a partir das equações \gamma=\alpha+j \beta, \quad \gamma^{2}= j \omega \mu(\sigma+j \omega \epsilon)
\begin{aligned} & \alpha=\omega \sqrt{\frac{\mu \varepsilon}{2}\left[\sqrt{1+\left[\frac{\sigma}{\omega \varepsilon}\right]^{2}}-1\right]} \\\ & \beta=\omega \sqrt{\frac{\mu \varepsilon}{2}\left[\sqrt{1+\left[\frac{\sigma}{\omega \varepsilon}\right]^{2}}+1\right]} \end{aligned}

Passo 1

O problema trata da constante de propagação num meio condutor, dada por
\gamma=\alpha+j\beta,
e satisfaz
\gamma^{2}=j\omega\mu\bigl(\sigma+j\omega\varepsilon\bigr).
Queremos isolar as partes real \alpha e imaginária \beta.

Passo 2

Escrevendo \gamma=\alpha+j\beta e elevando ao quadrado, temos
(\alpha+j\beta)^{2}=\alpha^{2}-\beta^{2}+j\,2\alpha\beta.
Por outro lado, expandindo o lado direito,
\gamma^{2} = j\omega\mu\,\sigma + j^{2}\,\omega\mu\,\omega\varepsilon = -\omega^{2}\mu\varepsilon + j\,\omega\mu\,\sigma.
Igualando parte real e parte imaginária:
\begin{cases} \alpha^{2}-\beta^{2}=-\omega^{2}\mu\varepsilon,\\\\ 2\,\alpha\beta=\omega\mu\,\sigma. \end{cases}

Passo 3

Para achar \alpha^{2}+\beta^{2}, usamos
(\alpha^{2}+\beta^{2})^{2} =(\alpha^{2}-\beta^{2})^{2}+(2\alpha\beta)^{2} =(\omega^{2}\mu\varepsilon)^{2}+(\omega\mu\,\sigma)^{2} =\omega^{2}\mu^{2}\bigl(\sigma^{2}+\omega^{2}\varepsilon^{2}\bigr),
logo
\alpha^{2}+\beta^{2} =\omega\mu\sqrt{\sigma^{2}+\omega^{2}\varepsilon^{2}}.
Agora
\alpha^{2} =\frac{(\alpha^{2}+\beta^{2})+(\alpha^{2}-\beta^{2})}{2} =\frac{\omega\mu\sqrt{\sigma^{2}+\omega^{2}\varepsilon^{2}} -\omega^{2}\mu\varepsilon}{2},
\beta^{2} =\frac{(\alpha^{2}+\beta^{2})-(\alpha^{2}-\beta^{2})}{2} =\frac{\omega\mu\sqrt{\sigma^{2}+\omega^{2}\varepsilon^{2}} +\omega^{2}\mu\varepsilon}{2}.

Passo 4

Fazendo a raiz e escrevendo \sqrt{\sigma^{2}+\omega^{2}\varepsilon^{2}} =\omega\varepsilon\sqrt{1+\bigl(\tfrac{\sigma}{\omega\varepsilon}\bigr)^{2}},
obtém-se
\alpha =\omega\sqrt{\frac{\mu\varepsilon}{2} \Bigl(\sqrt{1+\bigl(\tfrac{\sigma}{\omega\varepsilon}\bigr)^{2}}-1\Bigr)}, \quad \beta =\omega\sqrt{\frac{\mu\varepsilon}{2} \Bigl(\sqrt{1+\bigl(\tfrac{\sigma}{\omega\varepsilon}\bigr)^{2}}+1\Bigr)}.

Resposta

\alpha=\omega\sqrt{\frac{\mu\varepsilon}{2} \Bigl[\sqrt{1+\bigl(\tfrac{\sigma}{\omega\varepsilon}\bigr)^{2}}-1\Bigr]}, \quad \beta=\omega\sqrt{\frac{\mu\varepsilon}{2} \Bigl[\sqrt{1+\bigl(\tfrac{\sigma}{\omega\varepsilon}\bigr)^{2}}+1\Bigr]}.

Exercícios relacionados