Chapter 11, Problem 11.11P

A uniform plane wave at frequency f= 100 MHz propagates in a material having conductivity c = 3.0 S/m and dielectric constant đ�œ–_r' = 8.00. The wave carries electric field amplitude E₀ = 100 V/m. (a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric or a good conductor, (b) Calculate a, đ�›½, and $\eta$. (c) Write the electric field in pharos form, assuming x polarization and forward z travel. (d) Write the magnetic field in pharos form, (e) Write the time-average Pointing vector. S. (f) Find the 6-dB material thickness at which the wave power drops to 25% of its value on entry.

To determine

(a)

The loss tangent and whether the material is good dielectric or good conductor.

Answer to Problem 11.11P

The value of loss tangent is $67.408$ and the material is a good conductor.

Explanation of Solution

Calculation:

The line loss tangent is given by

   $\tan \theta =\frac{\sigma }{\omega \epsilon }$ ..... (1)

Here,

   $\sigma$ is the conductivity of the material.

   $\epsilon$ is the permeability of the material.

   $\omega$ is the radial frequency.

The permeability $\epsilon$ is given by,

   $\epsilon ={\epsilon }_0{\epsilon }_r$ ..... (2)

Here,

   ${\epsilon }_0$ is the absolute permittivity with standard value of $8.854\times {10}^{-12}\text{F}/\text{m}$.

   ${\epsilon }_r$ is the relative permeability of the medium.

Substitute equation (2) in equation (1).

   $\tan \theta =\frac{\sigma }{\omega \epsilon }$

   $\tan \theta =\frac{\sigma }{\omega {\epsilon }_0{\epsilon }_r}$ ..... (3)

The conversion from $\text{MHz}$ to $\text{Hz}$ is given by

   $1\text{MHz}={10}^6\text{Hz}$

The conversion of $100\text{MHz}$ is given in $\text{Hz}$ by

   $\begin{array}{c}100\text{MHz}=100\times {10}^6\text{Hz}\\ {\text{=10}}^8\text{Hz}\end{array}$

Hence, the frequency is,

   $f={10}^8\text{Hz}$

Radial frequency $\omega$ is given by

   $\omega =2\pi \times f$

Here,

   $f$ is the frequency.

Substitute ${10}^8$ for $f$ in the above equation.

   $\omega =2\pi \times {10}^8\text{rad}/\text{s}$

Substitute $2\pi \times {10}^8$ for $\omega$ , $3$ for $\sigma$ , $8.85\times {10}^{-12}$ for ${\epsilon }_0$ and $8$ for ${\epsilon }_r$ in equation (3).

   $\begin{array}{c}\tan \theta =\frac{3}{2\pi \times {10}^8\times 8.854\times {10}^{-12}\times 8}\\ =\frac{3}{141.664\pi \times {10}^{-4}}\\ =67.408\end{array}$

Since, the value is greater than 10; the material is a good conductor.

Conclusion:

Therefore, the value of loss tangent is $67.408$ .and the material is a good conductor.

To determine

(b)

The attenuation coefficient $\alpha$ , phase constant $\beta$ and wave impedance $\eta$.

Answer to Problem 11.11P

The attenuation coefficient $\alpha$ is $34.41\text{Np}/\text{m}$ , phase constant $\beta$ is $34.41\text{rad}/\text{m}$ and the wave impedance $\eta$ is $16.225\angle 45°\Omega$.

Explanation of Solution

Calculation:

The value of $\alpha$ is given by

   $\alpha =\sqrt{\frac{\omega \mu \sigma }{2}}$ ..... (4)

Here,

   $\mu$ is the permeability of the medium.

The value of $\mu$ is given by

   $\mu ={\mu }_o{\mu }_r$

Here,

   ${\mu }_0$ is the permeability of free space which has the standard value $\left(1.257\times {10}^{-6}\text{H}/\text{m}\right)$.

   ${\mu }_r$ is the relative permittivity of the medium.

Substitute $2\pi \times {10}^8$ for $\omega$ , $3$ for $\sigma$ , $1.257\times {10}^{-6}$ for ${\mu }_0$ and $1$ for ${\mu }_r$ in equation (4).

   $\begin{array}{c}\alpha =\sqrt{\frac{\omega {\mu }_0{\mu }_r\sigma }{2}}\\ =\sqrt{\frac{2\pi \times {10}^8\times 1\times 1.257\times {10}^{-6}\times 3}{2}}\\ =\sqrt{3.771\pi \times {10}^2}\\ =34.42\text{Np/m}\end{array}$

For a good conductor, the value of $\alpha$ and $\beta$ is same.

Hence, the phase constant $\beta$ is given by

   $\beta =34.41\text{rad}/\text{m}$

The value of wave impedance is given by

   $\eta =\sqrt{\frac{\omega {\mu }_0{\mu }_r}{\sigma }}$

Substitute $2\pi \times {10}^8$ for $\omega$ , $3$ for $\sigma$ , $1.257\times {10}^{-6}$ for ${\mu }_0$ and $1$ for ${\mu }_r$ in the above equation.

   $\begin{array}{c}\eta =\sqrt{\frac{\omega {\mu }_0{\mu }_r}{\sigma }}\\ =\sqrt{\frac{2\pi \times {10}^8\times 1\times 1.257\times {10}^{-6}}{3}}\\ =\sqrt{\frac{7.897\times {10}^2}{3}}\\ =16.225\Omega \end{array}$

For a good conductor, the value of impedance angle is $45°$.

The wave impedance $\eta$ is written as $16.225\angle 45°\Omega$.

Conclusion:

Thus, attenuation coefficient $\alpha$ is $34.42\text{Np/m}$ , phase constant $\beta$ is $\beta =34.42\text{rad}/\text{m}$ and wave impedance $\eta$ is $16.225\angle 45°\Omega$.

To determine

(c)

The electric field in phasor form.

Answer to Problem 11.11P

The phasor expression of electric field is $E_i={\widehat{a}}_x{E}_0{e}^{-34.31z}{e}^{j34.41z}\text{V}/\text{m}$.

Explanation of Solution

Calculation:

The electric field in phasor form is given by

   $E_i={\widehat{a}}_x{E}_0{e}^{-\alpha z}{e}^{j\beta z}$

Here,

   $E_0$ is the electric field amplitude.

   ${\widehat{a}}_x$ is unit vector in direction of positive x axis.

Substitute $100$ for $E_0$ and $34.41$ for $\alpha$ and $34.41$ for $\beta$ in the above equation.

   $E_i={\widehat{a}}_x{E}_0{e}^{-34.31z}{e}^{j34.41z}\text{V}/\text{m}$

Conclusion:

The phasor expression of electric field is $E_i={\widehat{a}}_x{E}_0{e}^{-34.31z}{e}^{j34.41z}\text{V}/\text{m}$.

To determine

(d)

The magnetic field strength in phasor form.

Answer to Problem 11.11P

The magnetic field strength in phasor form is $H_i={\widehat{a}}_{y}6.16e^{-34.41z}{e}^{-\left(j34.41z+45°\right)}\text{A}/\text{m}$.

Explanation of Solution

Calculation:

The magnetic field strength in phasor form is given by,

   $H_i={\widehat{a}}_y\frac{E_0}{\eta }{e}^{-\alpha z}{e}^{-j\beta z}{e}^{-j{\theta }_n}$

   ${\widehat{a}}_y$ is unit vector in direction of positive y direction.

Substitute $E_0$ for $100$ , $\alpha$ and $\beta$ for $34.41$ , $\eta$ for $16.22$ and ${\theta }_n$ for $45$ in the above equation.

   $\begin{array}{c}{H}_i={\widehat{a}}_y\frac{100}{16.22}{e}^{-34.41z}{e}^{-j34.41z}{e}^{-j45°}\\ H_i={\widehat{a}}_{y}6.16e^{-34.41z}{e}^{-\left(j34.41z+45°\right)}\end{array}$

Conclusion:

Thus, the magnetic field strength in phasor form is $H_i={\widehat{a}}_{y}6.16e^{-34.41z}{e}^{-\left(j34.41z+45°\right)}\text{A}/\text{m}$.

To determine

(e)

The time averaging Poynting vector.

Answer to Problem 11.11P

The time averaging Poynting vector is $S={\widehat{a}}_{z}217.97e^{-68.82z}\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

The time averaging Poynting vector is given by

   $S={\widehat{a}}_z\frac{E_0^2}{2\eta }{e}^{-2\alpha z}\cos {\theta }_n$

Here,

   ${\widehat{a}}_z$ is a unit vector in direction of positive z axis.

Substitute $E_0$ for $100$ , $\alpha$ for $34.41$ , $\eta$ for $16.22$ and ${\theta }_n$ for $45°$ in the above equation.

   $\begin{array}{c}S={\widehat{a}}_z\frac{{\left(100\right)}^2}{2\times 16.22}{e}^{-2\times 34.41z}\cos 45°\\ ={\widehat{a}}_{z}308.26\times e^{-68.82z}\times 0.707\\ S={\widehat{a}}_{z}217.97e^{-68.82z}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

The time averaging Poynting vector is $S={\widehat{a}}_{z}217.97e^{-68.82z}\text{W}/{\text{m}}^{\text{2}}$.

To determine

(f)

The6 dB material thickness at which the wave power drops to 25 % of its value on entry.

Answer to Problem 11.11P

The $6$ dB material thickness is $8.68\times {10}^{-3}\text{m}$.

Explanation of Solution

Calculation:

The power loss in dB can be given by

   $P\left(\text{dB}\right)=10\log \left(\frac{I_0}{I}\right)$

Here,

   $I$ is the intensity of the wave at the required point.

   $I_0$ is the initial intensity.

Substitute $6$ as $P\left(\text{dB}\right)$ in the above equation.

   $\begin{array}{l}6=10\log \left(\frac{I_0}{I}\right)\\ e^{0.6}=\frac{I_0}{I}\\ \frac{I_0}{I}=1.82\\ I=0.55I_0\end{array}$

The intensity of the wave is represented by the Poynting vector. In this case, the Poynting vector changes in the direction of wave propagation, with the factor of $e^{-68.82z}$ . The other factors do not change in z direction.

Taking intensity in the term of Poynting vector,

   $\begin{array}{l}217.97e^{-68.82z_1}=0.55\times 217.97e^{-68.82z_0}\\ e^{-68.82z}=0.55\times e^{-68.82z_o}\\ e^{-68.82\left(z-z_0\right)}=0.55\\ z-z_0=8.68\times {10}^{-3}\text{m}\end{array}$

Conclusion:

Thus, the 6-dB material thickness is $8.68\times {10}^{-3}\text{m}$.