Chapter 8, Problem 8.1P

(a)

To determine

The time harmonic electric field of an antenna.

Answer to Problem 8.1P

The time harmonic electric field of an antenna is $\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }$ .

Explanation of Solution

Given:

The time harmonic magnetic field $H_{\text{s}}$ is $\frac{-I_{\text{s}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}$ and the phasor driving current $I_{\text{s}}$ is $I_o{e}^{j\text{α}}$ .

Concept used:

The expression for time harmonic electric field can be written as,

  $E_{\text{s}}=-{\text{η}}_o{a}_r\times H_{\text{s}}$ ..... (1)

Calculation:

Substitute $\frac{-I_{\text{s}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}$ for $H_{\text{s}}$ in equation (1).

  $\begin{array}{c}{E}_{\text{s}}=-{\text{η}}_o{a}_r\times \left(\frac{-I_{\text{s}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\right)\\ =\frac{{\text{η}}_o{I}_{\text{s}}{\cos }^2\text{θ}}{r}\left(a_r\times a_{\text{θ}}\right)\\ =\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }\end{array}$

Therefore, the electric field is $\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }$ .

Conclusion:

Thus, the time harmonic electric field of an antenna is $\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }$ .

(b)

To determine

The time averaged power density of a wave.

Answer to Problem 8.1P

The time averaged power density of a wave is $\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }{a}_r$ .

Explanation of Solution

Concept used:

The expression for time averaged power density can be written as,

  $P\left(r,\text{θ},\varphi \right)=\frac{1}{2}\mathrm{Re}\left[E_{\text{s}}\times H_{\text{s}}^{\ast }\right]$ ..... (2)

Calculation:

From part (a), the time harmonic electric field is $\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }$ .

The conjugate of time harmonic magnetic field is calculated as,

  $\begin{array}{c}{H}_{\text{s}}^{\ast }={\left[\frac{-I_{\text{s}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\right]}^{\ast }\\ ={\left[-\frac{I_o{e}^{j\text{α}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\right]}^{\ast }\\ =-\frac{I_o{e}^{-j\text{α}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\end{array}$

Substitute $\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }$ for $E_{\text{s}}$ and $-\frac{I_o{e}^{-j\text{α}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}$ for $H_{\text{s}}^{\ast }$ in equation (2).

  $\begin{array}{c}P\left(r,\text{θ},\varphi \right)=\frac{1}{2}\mathrm{Re}\left[\frac{{\text{η}}_o{I}_{\text{s}}}{r}{\cos }^2\text{θ}{a}_{\varphi }\times \left(-\frac{I_o{e}^{-j\text{α}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\right)\right]\\ =\frac{1}{2}\mathrm{Re}\left[\frac{{\text{η}}_o{I}_o{e}^{j\text{α}}}{r}{\cos }^2\text{θ}{a}_{\varphi }\times \left(-\frac{I_o{e}^{-j\text{α}}}{r}{\cos }^2\text{θ }{a}_{\text{θ}}\right)\right]\\ =\frac{1}{2}\mathrm{Re}\left[\frac{{\text{η}}_o{I}_o^2}{r^2}{\cos }^4\text{θ}\left(a_{\varphi }\times a_{\text{θ}}\right)\right]\\ =\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }{a}_r\end{array}$

Therefore, the vector for time averaged power density is $\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }{a}_r$ .

Conclusion:

Thus, the time averaged power density of a wave is $\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }{a}_r$ .

(c)

To determine

The radiation resistance of the antenna.

Answer to Problem 8.1P

The radiation resistance of the antenna is $950\Omega$ .

Explanation of Solution

Concept used:

The total power radiated by antenna is,

  $P_{\text{rad}}={\displaystyle \iint P\left(r,\text{θ},\varphi \right)r^2\sin \text{θ}d\text{θ}d}\varphi$ ..... (3)

Calculation:

From part (a), the time averaged power density is $\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }{a}_r$ .

Substitute $\frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ}$ for $P\left(r,\text{θ},\varphi \right)$ in equation (3).

  $\begin{array}{c}{P}_{\text{rad}}={\displaystyle \iint \frac{1}{2}{\text{η}}_o\frac{I_o^2}{r^2}{\cos }^4\text{θ }\cdot r^2\sin \text{θ}d\text{θ}d\varphi }\\ =\frac{1}{2}{\text{η}}_o{I}_o^2{\displaystyle \underset{0}{\overset{\text{π}}{\int }}{\cos }^4\text{θ}\sin \text{θ}d\text{θ}}{\displaystyle \underset{0}{\overset{2\text{π}}{\int }}d\varphi }\\ =\frac{1}{2}{\text{η}}_o{I}_o^2{\displaystyle \underset{0}{\overset{\text{π}}{\int }}{\cos }^4\text{θ}\sin \text{θ}d\text{θ}}\left[2\text{π}\right]\\ =\frac{2}{5}{\text{η}}_o\text{π}{I}_o^2\end{array}$

The relation between power radiation and radiation resistance is shown below.

  $P_{\text{rad}}=\frac{1}{2}{I}_o^2{R}_{\text{rad}}$

Substitute $\frac{2}{5}{\text{η}}_o\text{π}{I}_o^2$ for $P_{\text{rad}}$ in above equation.

  $\begin{array}{c}\frac{2}{5}{\text{η}}_o\text{π}{I}_o^2=\frac{1}{2}{I}_o^2{R}_{\text{rad}}\\ R_{\text{rad}}=\frac{4}{5}{\text{πη}}_o\end{array}$

The value of intrinsic impedance is $120\text{π}$ .

So, the radiation resistance can be written as,

  $\begin{array}{c}{R}_{\text{rad}}=\frac{4}{5}\text{π}\left(120\text{π}\right)\\ =96{\text{π}}^2\\ \approx 950\Omega \end{array}$

Therefore, the value of $R_{\text{rad}}$ is $950\Omega$ .

Conclusion:

Thus, the radiation resistance of the antenna is $950\Omega$ .