Chapter 13, Problem 28P

(a)

To determine

Calculate $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)={\sin }^{2}2\theta$.

Answer to Problem 28P

The $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.5333, 1, and 1.875 respectively.

Explanation of Solution

Calculation:

Write the general expression to calculate the directivity (D).

$D=\frac{U_{\max }}{U_{\text{ave}}}$        (1)

Here,

$U_{\max }$ is the maximum radiation intensity, and

$U_{\text{ave}}$ is the average radiation intensity.

From the given normalized radiation intensity, the maximum radiation intensity $U_{\max }=1$.

Write the general expression to calculate the average radiation intensity.

$U_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }U\left(\theta ,\varphi \right)}}\sin \theta d\theta d\varphi$        (2)

Substitute ${\sin }^{2}2\theta$ for $U\left(\theta ,\varphi \right)$ in equation (2).

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }{\sin }^{2}2\theta \sin \theta d\theta d\varphi }}\\ =\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }d\varphi {\displaystyle {\int }_0^{\pi }{\left(2\sin \theta \cos \theta \right)}^2\sin \theta d\theta }}\left\{\because \sin 2\theta =2\sin \theta \cos \theta \right\}\\ =\frac{1}{4\pi }{\left(\varphi \right)}_0^{2\pi }{\displaystyle {\int }_0^{\pi }{\left(2\sin \theta \cos \theta \right)}^2\sin \theta d\theta }\\ =\frac{4}{4\pi }\left(2\pi \right){\displaystyle {\int }_0^{\pi }{\left(\sin \theta \cos \theta \right)}^2\sin \theta d\theta }\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=\frac{4}{4\pi }\left(2\pi \right){\displaystyle {\int }_0^{\pi }{\left(\sin \theta \cos \theta \right)}^2\sin \theta d\theta }\\ =2{\displaystyle {\int }_0^{\pi }{\left(\sin \theta \cos \theta \right)}^{2}d\left(-\cos \theta \right)}\\ =2{\displaystyle {\int }_0^{\pi }\left({\cos }^4\theta -{\cos }^2\theta \right)d\left(\cos \theta \right)}\\ =2{\left(\frac{{\cos }^5\theta }{5}-\frac{{\cos }^3\theta }{3}\right)}_0^{\pi }\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=2{\left(\frac{{\cos }^5\theta }{5}-\frac{{\cos }^3\theta }{3}\right)}_0^{\pi }\\ =2\left(-\frac{1}{5}+\frac{1}{3}-\frac{1}{5}+\frac{1}{3}\right)\\ =0.5333\end{array}$

Substitute $0.5333$ for $U_{\text{ave}}$ and $1$ for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{1}{0.5333}\\ =1.875\end{array}$

Conclusion:

Thus, the $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.5333, 1, and 1.875 respectively.

(b)

To determine

Calculate $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)=4{\csc }^2\theta$.

Answer to Problem 28P

The $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.5493, 4, and 7.2819 respectively.

Explanation of Solution

Calculation:

From the given normalized radiation intensity, the maximum radiation intensity $U_{\max }=4$.

Write the general expression to calculate the average radiation intensity.

$U_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{\pi }{\displaystyle {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}U\left(\theta ,\varphi \right)}}\sin \theta d\theta d\varphi$        (3)

Substitute $4{\csc }^2\theta$ for $U\left(\theta ,\varphi \right)$ in equation (3).

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{\pi }{\displaystyle {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}4{\csc }^2\theta \sin \theta d\theta d\varphi }}\\ =\frac{1}{\pi }{\left(\varphi \right)}_0^{\pi }{\displaystyle {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{{\sin }^2\theta }\sin \theta d\theta }\left\{\because \csc \theta =\frac{1}{\sin \theta }\right\}\\ =\frac{1}{\pi }\left(\pi \right){\displaystyle {\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{\sin \theta }d\theta }\\ ={\left(\frac{\ln \left(1-\cos \theta \right)-\ln \left(\cos \theta +1\right)}{2}\right)}_{\frac{\pi }{3}}^{\frac{\pi }{2}}\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}={\left(\frac{\ln \left(1-\cos \theta \right)}{2}-\frac{\ln \left(\cos \theta +1\right)}{2}\right)}_{\frac{\pi }{3}}^{\frac{\pi }{2}}\\ =\left(\frac{\ln \left(1-\cos \frac{\pi }{2}\right)}{2}-\frac{\ln \left(\cos \frac{\pi }{2}+1\right)}{2}-\frac{\ln \left(1-\cos \frac{\pi }{3}\right)}{2}+\frac{\ln \left(\cos \frac{\pi }{3}+1\right)}{2}\right)\\ =-\frac{\ln \left(1-\frac{1}{2}\right)}{2}+\frac{\ln \left(\frac{1}{2}+1\right)}{2}\\ =-\frac{\ln \left(\frac{1}{2}\right)}{2}+\frac{\ln \left(\frac{3}{2}\right)}{2}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=-\frac{\ln \left(\frac{1}{2}\right)}{2}+\frac{\ln \left(\frac{3}{2}\right)}{2}\\ =0.5493\end{array}$

Substitute $0.5493$ for $U_{\text{ave}}$ and 4 for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{4}{0.5493}\\ =7.2819\end{array}$

Conclusion:

Thus, the $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.5493, 4, and 7.2819 respectively.

(c)

To determine

Calculate $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)=2{\sin }^2\theta {\sin }^2\varphi$.

Answer to Problem 28P

The $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.3333, 2, and 6 respectively.

Explanation of Solution

Calculation:

From the given normalized radiation intensity, the maximum radiation intensity $U_{\max }=2$.

Write the general expression to calculate the average radiation intensity.

$U_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{\pi }{\displaystyle {\int }_0^{\pi }U\left(\theta ,\varphi \right)}}\sin \theta d\theta d\varphi$        (4)

Substitute $2{\sin }^2\theta {\sin }^2\varphi$ for $U\left(\theta ,\varphi \right)$ in equation (4).

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{\pi }{\displaystyle {\int }_0^{\pi }2{\sin }^2\theta {\sin }^2\varphi \sin \theta d\theta d\varphi }}\\ =\frac{1}{2\pi }{\displaystyle {\int }_0^{\pi }{\sin }^2\varphi d\varphi }{\displaystyle {\int }_0^{\pi }{\sin }^3\theta d\theta }\\ =\frac{1}{2\pi }{\left(\frac{\varphi -\frac{\sin 2\varphi }{2}}{2}\right)}_0^{\pi }{\left(\frac{{\cos }^3\theta }{3}-\cos \theta \right)}_0^{\pi }\\ =\frac{1}{2\pi }\left(\frac{\pi -\frac{\sin 2\pi }{2}}{2}-\frac{0-\frac{\sin 2\left(0\right)}{2}}{2}\right)\left(\frac{{\cos }^3\pi }{3}-\cos \pi -\frac{{\cos }^{3}0}{3}+\cos 0\right)\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{2\pi }\left(\frac{\pi -0}{2}-0\right)\left(-\frac{1}{3}+1-\frac{1}{3}+1\right)\\ =\frac{1}{2\pi }\left(\frac{\pi }{2}\right)\left(-\frac{2}{3}+2\right)\\ =\frac{1}{2}\left(\frac{1}{2}\right)\left(\frac{4}{3}\right)\\ =0.3333\end{array}$

Substitute $0.3333$ for $U_{\text{ave}}$ and 2 for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{2}{0.3333}\\ =6\end{array}$

Conclusion:

Thus, the $U_{\text{ave}}$, $U_{\text{max}}$, directivity (D) of an antenna are 0.3333, 2, and 6 respectively.