Question

Calculate the exact reflection and transmission coefficients, without assuming ${\mu }_1={\mu }_2={\mu }_0$. Confirm that R + T = 1.

Short answer

Answer

The expression for the reflection coefficient is$R={\left(\frac{1-\beta }{1+\beta }\right)}^2$ and transmission coefficient isrole="math" localid="1658739234777" $T=\beta \frac{{\epsilon }_{or}^2}{{\epsilon }_I^2}$ . The proof for role="math" localid="1658739263094" $R+T=1$is shown.

Step-by-step solution

Determine the formulas

Consider the formula for the reflection coefficient as

$$\begin{gathered} R=\frac{I_R}{I_I} \\ =\frac{{\left({\mathbf{\epsilon }}_{\mathbf{0}\mathbf{R}}\right)}^{\mathbf{2}}}{{\left({\mathbf{\epsilon }}_{\mathbf{0}\mathbf{I}}\right)}^{\mathbf{2}}} \end{gathered}$$ …. (1)

Consider the formula for the transmission medium permittivity as

${\overline{\epsilon }}_{0T}=\frac{2}{1+\beta }{\overline{\epsilon }}_{oI}$

Determine the reflection and transmission coefficients as

Write the expression for the permittivity of the reflected region as

$$\begin{gathered} {\overline{\epsilon }}_{0R}={\left(\frac{1-\beta }{1+\beta }\right)}^2{\overline{\epsilon }}_{oI}\left\{\beta =\frac{{\mu }_1{V}_1}{{\mu }_2{V}_2}\right\} \\ \frac{{\overline{\epsilon }}_{0R}}{{\overline{\epsilon }}_{0R}}={\left(\frac{1-\beta }{1+\beta }\right)}^2 \end{gathered}$$

From above and from the equation (1) write the expression for the reflection coefficient as

$R={\left(\frac{1-\beta }{1+\beta }\right)}^2$$R={\left(\frac{1-\beta }{1+\beta }\right)}^2$

Write the expression for the transmission coefficients as

$T=\frac{{\overline{\epsilon }}_2{v}_2}{{\overline{\epsilon }}_1{v}_1}{\left(\frac{{\overline{\epsilon }}_{0R}}{{\overline{\epsilon }}_{0R}}\right)}^2$

Simplify as:

$$\begin{gathered} \frac{{\overline{\epsilon }}_2{v}_2}{{\overline{\epsilon }}_1{v}_1}=\frac{{\mu }_1}{{\mu }_2}\frac{{\overline{\epsilon }}_2{\mu }_2}{{\overline{\epsilon }}_1{\mu }_1}\frac{v_2}{v_1} \\ \frac{{\overline{\epsilon }}_2{v}_2}{{\overline{\epsilon }}_1{v}_1}=\frac{{\mu }_1}{{\mu }_2}{\left(\frac{v_1}{v_2}\right)}^2\frac{v_2}{v_1} \\ \frac{{\overline{\epsilon }}_2{v}_2}{{\overline{\epsilon }}_1{v}_1}=\frac{{\mu }_1}{{\mu }_2}\frac{v_2}{v_1} \\ \beta =\frac{{\mu }_1}{{\mu }_2}\frac{v_2}{v_1} \end{gathered}$$

Rewrite the expression for the transmission coefficient as

$T=\beta \frac{{\epsilon }_{or}^2}{{\epsilon }_I^2}$

 Step 3: Determine the proof for R +T= 1

Rewrite the expression for the transmission line coefficient as

$T=\beta {\left(\frac{2}{1+\beta }\right)}^2$

Add the expression for the reflection and transmission coefficient as

$$\begin{gathered} R+T={\left(\frac{1-\beta }{1+\beta }\right)}^2+\beta {\left(\frac{2}{1+\beta }\right)}^2 \\ =\frac{1}{{\left(1+\beta \right)}^2}\left[4\beta +{\left(1-\beta \right)}^2\right] \\ =\frac{1}{{\left(1+\beta \right)}^2}\left[4\beta +\left(1+{\beta }^2-2\beta \right)\right] \\ =1 \end{gathered}$$

Therefore, the expression for the reflection coefficient is $R={\left(\frac{1-\beta }{1+\beta }\right)}^2$and transmission coefficient is$T=\beta \frac{{\epsilon }_{or}^2}{{\epsilon }_I^2}$ . The proof for is shown R + T = 1.