Chapter 10, Problem 61P

(a)

To determine

Find the wave frequency.

Answer to Problem 61P

The wave frequency is $\underset{\_}{9\times {10}^8\text{rad}/\text{s}}$.

Explanation of Solution

Calculation:

Write the expression for given incident electric field in medium-1 (air).

$E_i=10\sin \left(\omega t+3z\right)a_x\text{V}/\text{m}$        (1)

From the expression, the values of magnitude of incident electric field at $z=0$ and phase constant of medium-1 are $10\text{V}/\text{m}$ and $3\text{rad}/\text{m}$, respectively.

$\begin{array}{l}{E}_{io}=10\text{V}/\text{m}\\ {\beta }_1=3\text{rad}/\text{m}\end{array}$

Consider the expression for phase constant of medium-1.

${\beta }_1=\omega \sqrt{\mu \epsilon }$

Rewrite the expression for the given data.

${\beta }_1=\omega \sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon ={\epsilon }_{\text{o}}\end{array}\right\}$

Rearrange the expression for wave frequency.

$\omega =\frac{{\beta }_1}{\sqrt{{\mu }_{\text{o}}{\epsilon }_{\text{o}}}}$

Substitute $3\text{rad}/\text{m}$ for ${\beta }_1$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}\omega =\frac{3\text{rad}/\text{m}}{\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}\\ =9\times {10}^8\text{rad}/\text{s}\end{array}$

Conclusion:

Thus, the wave frequency is $\underset{\_}{9\times {10}^8\text{rad}/\text{s}}$.

(b)

To determine

Find the wavelength of the signal in the air.

Answer to Problem 61P

The wavelength of the signal in the air is $\underset{\_}{2.094\text{m}}$.

Explanation of Solution

Calculation:

Consider the expression to find the wavelength of the signal in air (medium-1).

${\lambda }_1=\frac{2\pi }{{\beta }_1}$

Substitute $3\text{rad}/\text{m}$ for ${\beta }_1$.

$\begin{array}{c}{\lambda }_1=\frac{2\pi }{3\text{rad}/\text{m}}\\ \cong 2.094\text{m}\end{array}$

Conclusion:

Thus, the wavelength of the signal in the air is $\underset{\_}{2.094\text{m}}$.

(c)

To determine

Find the loss tangent and intrinsic impedance of the ocean.

Answer to Problem 61P

The loss tangent and intrinsic impedance of the ocean are $\underset{\_}{6.288}$ and $\underset{\_}{16.71\angle 40.48°\Omega }$, respectively.

Explanation of Solution

Calculation:

Consider the expression for loss tangent for nonmagnetic media.

$\tan \left(2{\theta }_{{\eta }_2}\right)=\frac{\sigma }{\omega \epsilon }$        (2)

Here,

$\sigma$ is the conductivity and

$\epsilon$ is the permittivity.

Rewrite the Equation (2) for ocean surface (medium-2).

$\tan \left(2{\theta }_{{\eta }_2}\right)=\frac{\sigma }{\omega \left(80{\epsilon }_{\text{o}}\right)}\left\{\because \epsilon =80{\epsilon }_{\text{o}}\right\}$

Substitute $4\text{S}/\text{m}$ for $\sigma$, $9\times {10}^8\text{rad}/\text{s}$ for $\omega$, and $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}\tan \left(2{\theta }_{{\eta }_2}\right)=\frac{4\text{S}/\text{m}}{\left(9\times {10}^8\text{rad}/\text{s}\right)\left(8\text{0}\right)\left(\frac{{10}^{-9}}{36\pi }\right)}\\ \cong 6.288\end{array}$

Rearrange the expression for phase angle of the intrinsic impedance of the ocean surface.

$\begin{array}{c}{\theta }_{{\eta }_2}=0.5{\tan }^{-1}\left(6.288\right)\\ \cong 40.48°\end{array}$

Consider the expression to find the magnitude of the intrinsic impedance of the ocean surface.

$\left|{\eta }_2\right|=\frac{\sqrt{\frac{\mu }{\epsilon }}}{{\left[1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2\right]}^{\frac{1}{4}}}$

Rewrite the expression for the given data.

$\left|{\eta }_2\right|=\frac{\sqrt{\frac{{\mu }_{\text{o}}}{80{\epsilon }_{\text{o}}}}}{{\left[1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2\right]}^{\frac{1}{4}}}$

Substitute 6.288 for $\left(\frac{\sigma }{\omega \epsilon }\right)$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}\left|{\eta }_2\right|=\frac{\sqrt{\frac{4\pi \times {10}^{-7}\text{H}/\text{m}}{\left(80\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}}}{{\left[1+{\left(6.288\right)}^2\right]}^{\frac{1}{4}}}\\ \cong 16.71\Omega \end{array}$

Conclusion:

Thus, the loss tangent and intrinsic impedance of the ocean are $\underset{\_}{6.288}$ and $\underset{\_}{16.71\angle 40.48°\Omega }$, respectively.

(d)

To determine

Find the reflected and transmitted electric fields.

Answer to Problem 61P

The reflected and transmitted electric fields are $\underset{\_}{9.35\sin \left(\omega t-3z+179.7°\right)a_x\text{V}/\text{m}}$ and $\underset{\_}{\underset{\_}{0.857e^{43.94z}\sin \left(\omega t+51.48z+38.89°\right)a_x\text{V}/\text{m}}}$, respectively.

Explanation of Solution

Calculation:

Consider the expression for reflected electric field.

$E_r=E_{ro}\sin \left(\omega t-3z+\angle \Gamma \right)a_x$        (3)

Here,

$E_{ro}$ is the magnitude of reflected electric field at $z=0$.

Consider the expression for reflected electric field at $z=0$.

$E_{ro}=\Gamma E_{io}$        (4)

Here,

$E_{io}$ is the magnitude of incident electric field at $z=0$, which is $10\text{V}/\text{m}$, and

$\Gamma$ is the reflection coefficient.

Write the expression to find the reflection coefficient.

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$        (5)

Here,

${\eta }_1$ is the intrinsic impedance for medium-1 and

${\eta }_2$ is the intrinsic impedance for medium-2.

As the medium-1 is air (free space), the value of ${\eta }_1$ is $120\pi \Omega$.

${\eta }_1=120\pi \Omega$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $16.71\angle 40.48°\Omega$ for ${\eta }_2$ in Equation (5).

$\begin{array}{c}\Gamma =\frac{16.71\angle 40.48°\Omega -120\pi \Omega }{16.71\angle 40.48°\Omega +120\pi \Omega }\\ \cong 0.935\angle 179.7°\end{array}$

Substitute $0.935\angle 179.7°$ for $\Gamma$ and $10\text{V}/\text{m}$ for $E_{io}$ in Equation (4).

$\begin{array}{c}{E}_{ro}=\left(0.935\angle 179.7°\right)\left(10\text{V}/\text{m}\right)\\ =9.35\angle 179.7°\text{V}/\text{m}\end{array}$

Substitute $9.35\angle 179.7°\text{V}/\text{m}$ for $E_{ro}$ in Equation (3).

$E_r=9.35\sin \left(\omega t-3z+179.7°\right)a_x\text{V}/\text{m}$

Consider the expression for transmitted electric field.

$E_t=E_{to}{e}^{\alpha z}\sin \left(\omega t+{\beta }_{2}z+\angle \tau \right)a_x$        (6)

Here,

$E_{to}$ is the magnitude of transmitted electric field at $z=0$ and

${\beta }_2$ is the phase constant in medium-2.

Consider the expression for attenuation constant for medium-2.

${\alpha }_2=\omega \sqrt{\frac{\mu \epsilon }{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2}-1\right]}$

Rewrite the expression for the given data.

${\alpha }_2=\omega \sqrt{\frac{{\mu }_{\text{o}}\left(80{\epsilon }_{\text{o}}\right)}{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2}-1\right]}$

Substitute $9\times {10}^8\text{rad}/\text{s}$ for $\omega$, 6.288 for $\left(\frac{\sigma }{\omega \epsilon }\right)$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\alpha }_2=\left(9\times {10}^8\text{rad}/\text{s}\right)\sqrt{\frac{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(80\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}{2}\left[\sqrt{1+{\left(6.288\right)}^2}-1\right]}\\ \cong 43.94\text{Np}/\text{m}\end{array}$

Consider the expression for phase constant for medium-2.

${\beta }_2=\omega \sqrt{\frac{\mu \epsilon }{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2}+1\right]}$

Rewrite the expression for the given data.

${\beta }_2=\omega \sqrt{\frac{{\mu }_{\text{o}}\left(80{\epsilon }_{\text{o}}\right)}{2}\left[\sqrt{1+{\left(\frac{\sigma }{\omega \epsilon }\right)}^2}+1\right]}$

Substitute $9\times {10}^8\text{rad}/\text{s}$ for $\omega$, 6.288 for $\left(\frac{\sigma }{\omega \epsilon }\right)$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}{\beta }_2=\left(9\times {10}^8\text{rad}/\text{s}\right)\sqrt{\frac{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(80\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}{2}\left[\sqrt{1+{\left(6.288\right)}^2}+1\right]}\\ \cong 51.48\text{rad}/\text{m}\end{array}$

Consider the expression to find the magnitude of transmitted electric field at $z=0$.

$E_{to}=\tau E_{io}$        (7)

Here,

$\tau$ is the transmission coefficient.

Consider for the expression for transmission coefficient.

$\tau =\frac{2{\eta }_2}{{\eta }_2+{\eta }_1}$

Substitute $120\pi \Omega$ for ${\eta }_1$ and $16.71\angle 40.48°\Omega$ for ${\eta }_2$.

$\begin{array}{c}\tau =\frac{\left(2\right)\left(16.71\angle 40.48°\Omega \right)}{16.71\angle 40.48°\Omega +60\pi \Omega }\\ \cong 0.0857\angle 38.89°\end{array}$

Substitute $0.0857\angle 38.89°$ for $\tau$ and $10\text{V}/\text{m}$ for $E_{io}$ in Equation (7).

$\begin{array}{c}{E}_{to}=\left(0.0857\angle 38.89°\right)\left(10\text{V}/\text{m}\right)\\ \cong 0.857\angle 38.89°\text{V}/\text{m}\end{array}$

Substitute $0.857\angle 38.89°\text{V}/\text{m}$ for $E_{to}$, $43.94\text{Np}/\text{m}$ for ${\alpha }_2$, $51.48\text{rad}/\text{m}$ for ${\beta }_2$, and $9\times {10}^8\text{rad}/\text{s}$ for $\omega$ in Equation (6).

$E_t=0.857e^{43.94z}\sin \left(\omega t+51.48z+38.89°\right)a_x\text{V}/\text{m}$

Conclusion:

Thus, the reflected and transmitted electric fields are $\underset{\_}{9.35\sin \left(\omega t-3z+179.7°\right)a_x\text{V}/\text{m}}$ and $\underset{\_}{\underset{\_}{0.857e^{43.94z}\sin \left(\omega t+51.48z+38.89°\right)a_x\text{V}/\text{m}}}$, respectively.