Chapter 13, Problem 18P

(a)

To determine

Find the corresponding $H_s$ field for the given electric field $E_s=\frac{\cos 2\theta }{r}{e}^{-j\beta r}{a}_{\theta }\text{V}/\text{m}$.

Answer to Problem 18P

The corresponding $H_s$ field is $\frac{\cos 2\theta }{120\pi r}{e}^{-j\beta r}{a}_{\varphi }\text{A}/\text{m}$.

Explanation of Solution

Calculation:

Write the general relationship between the electric and magnetic field intensity.

$E_s=\eta H_s$        (1)

Rearrange the equation (1) as follows,

$H_s=\frac{E_s}{\eta }$

$H_s=\frac{E_s}{120\pi }\left\{\because \eta =120\pi \right\}$        (2)

Since, the cross product of the unit vectors is,

$a_E\times a_H=a_k\to a_{\theta }\times a_H=a_r$

From the above expression,

$a_H=a_{\varphi }$        (3)

Using equation (3), the equation (2) becomes,

$\begin{array}{c}{H}_s=\frac{\frac{\cos 2\theta }{r}{e}^{-j\beta r}}{120\pi }{a}_{\varphi }\text{A}/\text{m}\\ =\frac{\cos 2\theta }{120\pi r}{e}^{-j\beta r}{a}_{\varphi }\text{A}/\text{m}\end{array}$

Conclusion:

Thus, the corresponding $H_s$ field is $\frac{\cos 2\theta }{120\pi r}{e}^{-j\beta r}{a}_{\varphi }\text{A}/\text{m}$.

(b)

To determine

Calculate the radiated power.

Answer to Problem 18P

The radiated power is $7.778\text{mW}$.

Explanation of Solution

Calculation:

Write the general expression to calculate the average power.

$P_{\text{avg}}=\frac{{\left|E_s\right|}^2}{2\eta }$        (4)

Here,

$E_s$ is the electric field intensity, and

$\eta$ is the intrinsic impedance.

Substitute $\frac{\cos 2\theta }{r}{e}^{-j\beta r}{a}_{\theta }$ for $E_s$ in equation (4).

$\begin{array}{c}{P}_{\text{avg}}=\frac{{\left|\frac{\cos 2\theta }{r}{e}^{-j\beta r}{a}_{\theta }\right|}^2}{2\eta }\\ =\frac{{\left|\frac{\cos 2\theta }{r}\left(\cos \beta r-j\sin \beta r\right)a_{\theta }\right|}^2}{2\eta }\\ =\frac{{\cos }^{2}2\theta }{2\eta r^2}\left\{\because {\sin }^2\theta +{\cos }^2\theta =1\right\}\end{array}$

Write the general expression to calculate the time-average radiated power.

$P_{\text{rad}}={\displaystyle {\int }_S{P}_{\text{avg}}dS}$        (5)

Where,

$dS$ is the differential surface which is equal to $r^2\sin \theta d\theta d\varphi$.

Substitute $\frac{{\cos }^{2}2\theta }{2\eta r^2}$ for $P_{\text{avg}}$ and $r^2\sin \theta d\theta d\varphi$ for $dS$ in equation (5) with integration limits.

$\begin{array}{c}{P}_{\text{rad}}={\displaystyle {\int }_S\left(\frac{{\cos }^{2}2\theta }{2\eta r^2}\right)r^2\sin \theta d\theta d\varphi }\\ =\frac{1}{2\eta }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }{\cos }^{2}2\theta \sin \theta d\theta d\varphi }}\\ =\frac{1}{2\left(120\pi \right)}{\left(\varphi \right)}_0^{2\pi }{\displaystyle {\int }_0^{\pi }{\cos }^{2}2\theta \sin \theta d\theta }\left\{\because \eta =120\pi \right\}\\ =\frac{1}{2\left(120\pi \right)}\left(2\pi \right){\displaystyle {\int }_0^{\pi }{\cos }^{2}2\theta \sin \theta d\theta }\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{P}_{\text{rad}}=\frac{1}{120}{\displaystyle {\int }_0^{\pi }{\cos }^{2}2\theta \sin \theta d\theta }\\ =\frac{1}{120}{\displaystyle {\int }_0^{\pi }{\left({\cos }^2\theta -{\sin }^2\theta \right)}^2\sin \theta d\theta }\left\{\because {\cos }^{2}2\theta ={\cos }^2\theta -{\sin }^2\theta \right\}\\ =\frac{1}{120}{\displaystyle {\int }_0^{\pi }{\left({\cos }^2\theta -\left(1-{\cos }^2\theta \right)\right)}^2\sin \theta d\theta }\left\{\because {\sin }^2\theta =1-{\cos }^2\theta \right\}\\ =-\frac{1}{120}{\displaystyle {\int }_0^{\pi }{\left(2{\cos }^2\theta -1\right)}^{2}d\left(\cos \theta \right)}\end{array}$

Simplify the equation as follows,

$P_{\text{rad}}=-\frac{1}{120}{\displaystyle {\int }_0^{\pi }{\left(2{\cos }^2\theta -1\right)}^{2}d\left(\cos \theta \right)}$

$P_{\text{rad}}=-\frac{1}{120}{\displaystyle {\int }_0^{\pi }\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)d\left(\cos \theta \right)}$        (6)

Reduce the equation as follows,

$\begin{array}{c}{P}_{\text{rad}}=-\frac{1}{120}{\left(\frac{4{\cos }^5\theta }{5}-\frac{4{\cos }^3\theta }{3}+\cos \theta \right)}_0^{\pi }\\ =-\frac{1}{120}\left(\frac{4{\cos }^5\pi }{5}-\frac{4{\cos }^3\pi }{3}+\cos \pi -\frac{4{\cos }^{5}0}{5}+\frac{4{\cos }^{3}0}{3}-\cos 0\right)\\ =-\frac{1}{120}\left(-\frac{4}{5}+\frac{4}{3}-1-\frac{4}{5}+\frac{4}{3}-1\right)\\ =7.778\times {10}^{-3}\text{W}\end{array}$

$P_{\text{rad}}=7.778\text{mW}\left\{\because 1\text{m}={10}^{-3}\right\}$

Conclusion:

Thus, the radiated power is $7.778\text{mW}$.

(c)

To determine

Find the fraction of the total power radiated in the belt.

Answer to Problem 18P

The fraction of the total power radiated in the belt is $76.78\%$.

Explanation of Solution

Calculation:

Refer to Part (b),

$P_{\text{rad}}=-\frac{1}{120}{\displaystyle {\int }_0^{\pi }\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)d\left(\cos \theta \right)}$

The range of $\theta$ must vary from 60  to 120 . Therefore, the above expression can be rewritten as follows,

$\begin{array}{c}{P}_{\text{rad}}=-\frac{1}{120}{\displaystyle {\int }_{60°}^{120°}\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)d\left(\cos \theta \right)}\\ =-\frac{1}{120}{\left(\frac{4{\cos }^5\theta }{5}-\frac{4{\cos }^3\theta }{3}+\cos \theta \right)}_{60°}^{120°}\\ =-\frac{1}{120}\left(\frac{4{\cos }^{5}120°}{5}-\frac{4{\cos }^{3}120°}{3}+\cos 120°-\frac{4{\cos }^{5}60°}{5}+\frac{4{\cos }^{3}60°}{3}-\cos 60°\right)\\ =-\frac{1}{120}\left(-0.025+0.16667-0.5-0.025+0.16667-0.5\right)\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{P}_{\text{rad}}=5.972\times {10}^{-3}\text{W}\\ =5.972\text{mW}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Therefore, the fraction of the total power radiated in the belt is,

$\left(\begin{array}{l}\text{Fraction of the total power}\\ \text{ radiated in the belt}\end{array}\right)=\frac{\text{Power}\text{radiated}\text{for}60°\text{to}120°}{\text{Power}\text{radiated}\text{for}0\text{to}\pi }\times 100$        (7)

Refer to Part (b),

Substitute $7.778\text{mW}$ for Power radiated for 0 to $\pi$ and $5.972\text{mW}$ for Power radiated for 60  to 120  in equation (7).

$\begin{array}{c}\text{Fraction of the total power radiated in the belt}=\frac{5.972\text{mW}}{7.778\text{mW}}\times 100\\ =0.7678\times 100\\ =76.78\%\end{array}$

Conclusion:

Thus, the fraction of the total power radiated in the belt is $76.78\%$.