Chapter 13, Problem 29P

(a)

To determine

Find the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)=10\sin \theta {\sin }^2\varphi$.

Answer to Problem 29P

The directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $2.546\sin \theta {\sin }^2\varphi$ and 2.546 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 $U_{\max }=10$.

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 $10\sin \theta {\sin }^2\varphi$ for $U\left(\theta ,\varphi \right)$ in equation (2).

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }\left(10\sin \theta {\sin }^2\varphi \right)\sin \theta d\theta d\varphi }}\\ =\frac{5}{2\pi }{\displaystyle {\int }_0^{2\pi }{\sin }^2\varphi d\varphi {\displaystyle {\int }_0^{\pi }{\sin }^2\theta d\theta }}\\ =\frac{5}{2\pi }{\left(\frac{\varphi -\frac{\sin 2\varphi }{2}}{2}\right)}_0^{2\pi }{\left(\frac{\theta -\frac{\sin 2\theta }{2}}{2}\right)}_0^{\pi }\\ =\frac{5}{8\pi }\left(2\pi -\frac{\sin 4\pi }{2}-0+\frac{\sin 2\left(0\right)}{2}\right)\left(\pi -\frac{\sin 2\pi }{2}-0+\frac{\sin 2\left(0\right)}{2}\right)\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=\frac{5}{8\pi }\left(2\pi \right)\left(\pi \right)\\ =\frac{5\pi }{4}\end{array}$

Substitute $\frac{5\pi }{4}$ for $U_{\text{ave}}$ and 10 for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{10}{\left(\frac{5\pi }{4}\right)}\\ =2.546\end{array}$

Write the general expression to calculate the directive gain $\left(G_d\right)$.

$G_d=\frac{U\left(\theta ,\varphi \right)}{U_{\text{ave}}}$        (3)

Here,

$U\left(\theta ,\varphi \right)$ is the radiation intensity.

Substitute $\frac{5\pi }{4}$ for $U_{\text{ave}}$ and $10\sin \theta {\sin }^2\varphi$ for $U\left(\theta ,\varphi \right)$ in equation (3) to find the directive gain.

$\begin{array}{c}{G}_d=\frac{10\sin \theta {\sin }^2\varphi }{\left(\frac{5\pi }{4}\right)}\\ =2.546\sin \theta {\sin }^2\varphi \end{array}$

Conclusion:

Thus, the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $2.546\sin \theta {\sin }^2\varphi$ and 2.546 respectively.

(b)

To determine

Find the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)=2{\sin }^2\theta {\sin }^3\varphi$.

Answer to Problem 29P

The directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $2.25\pi {\sin }^2\theta {\sin }^3\varphi$ and 2.25 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 $2{\sin }^2\theta {\sin }^3\varphi$ for $U\left(\theta ,\varphi \right)$ in equation (4).

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

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{2\pi }\left(-\frac{1}{3}+1-\frac{1}{3}+1\right)\left(-\frac{1}{3}+1-\frac{1}{3}+1\right)\\ =\frac{1}{2\pi }{\left(2-\frac{2}{3}\right)}^2\\ =\frac{1}{2\pi }{\left(\frac{4}{3}\right)}^2\\ =\frac{16}{18\pi }\end{array}$

Substitute $\frac{16}{18\pi }$ for $U_{\text{ave}}$ and $2$ for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{2}{\left(\frac{16}{18\pi }\right)}\\ =7.069\end{array}$

Substitute $\frac{16}{18\pi }$ for $U_{\text{ave}}$ and $2{\sin }^2\theta {\sin }^3\varphi$ for $U\left(\theta ,\varphi \right)$ in equation (3) to find the directive gain.

$\begin{array}{c}{G}_d=\frac{2{\sin }^2\theta {\sin }^3\varphi }{\left(\frac{16}{18\pi }\right)}\\ =2.25\pi {\sin }^2\theta {\sin }^3\varphi \end{array}$

Conclusion:

Thus, the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $2.25\pi {\sin }^2\theta {\sin }^3\varphi$ and $7.069$ respectively.

(c)

To determine

Find the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna for the given radiation intensity $U\left(\theta ,\varphi \right)=5\left(1+{\sin }^2\theta {\sin }^2\varphi \right)$.

Answer to Problem 29P

The directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $0.75\left(1+{\sin }^2\theta {\sin }^2\varphi \right)$ and $0.75$ respectively.

Explanation of Solution

Calculation:

From the given normalized radiation intensity, the maximum radiation intensity $U_{\max }=5$.

Write the general expression to calculate the average radiation intensity.

$U_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }U\left(\theta ,\varphi \right)}}\sin \theta d\theta d\varphi$        (5)

Substitute $5\left(1+{\sin }^2\theta {\sin }^2\varphi \right)$ for $U\left(\theta ,\varphi \right)$ in equation (5).

$\begin{array}{c}{U}_{\text{ave}}=\frac{1}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }\left[5\left(1+{\sin }^2\theta {\sin }^2\varphi \right)\right]}}\sin \theta d\theta d\varphi \\ =\frac{5}{4\pi }{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }\left(\sin \theta +{\sin }^3\theta {\sin }^2\varphi \right)}}d\theta d\varphi \\ =\frac{5}{4\pi }\left[{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }\sin \theta d\theta d\varphi }}+{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }{\sin }^3\theta {\sin }^2\varphi }}d\theta d\varphi \right]\\ =\frac{5}{4\pi }\left[{\displaystyle {\int }_0^{2\pi }d\varphi {\displaystyle {\int }_0^{\pi }\sin \theta d\theta }}+{\displaystyle {\int }_0^{2\pi }{\sin }^2\varphi d\varphi {\displaystyle {\int }_0^{\pi }{\sin }^3\theta }}d\theta \right]\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=\frac{5}{4\pi }\left[{\left(\varphi \right)}_0^{2\pi }{\left(-\cos \theta \right)}_0^{\pi }+{\left(\frac{\varphi -\frac{\sin 2\varphi }{2}}{2}\right)}_0^{2\pi }{\left(\frac{{\cos }^3\theta }{3}-\cos \theta \right)}_0^{\pi }\right]\\ =\frac{5}{4\pi }\left[2\pi \left(-\cos \pi +\cos 0\right)+\frac{1}{2}\left(2\pi -\frac{\sin 4\pi }{2}\right)\left(\frac{{\cos }^3\pi }{3}-\cos \pi -\frac{{\cos }^{3}0}{3}+\cos 0\right)\right]\\ =\frac{5}{4\pi }\left[4\pi +\pi \left(-\frac{1}{3}+1-\frac{1}{3}+1\right)\right]\\ =\frac{5}{4\pi }\left[4\pi +\frac{4\pi }{3}\right]\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{U}_{\text{ave}}=5\left(\frac{4}{3}\right)\\ =\frac{20}{3}\end{array}$

Substitute $\frac{20}{3}$ for $U_{\text{ave}}$ and 5 for $U_{\max }$ in equation (1) to find the directivity D.

$\begin{array}{c}D=\frac{5}{\left(\frac{20}{3}\right)}\\ =0.75\end{array}$

Substitute $\frac{20}{3}$ for $U_{\text{ave}}$ and $5\left(1+{\sin }^2\theta {\sin }^2\varphi \right)$ for $U\left(\theta ,\varphi \right)$ in equation (3) to find the directive gain.

$\begin{array}{c}{G}_d=\frac{5\left(1+{\sin }^2\theta {\sin }^2\varphi \right)}{\left(\frac{20}{3}\right)}\\ =0.75\left(1+{\sin }^2\theta {\sin }^2\varphi \right)\end{array}$

Conclusion:

Thus, the directive gain $\left(G_d\right)$ and directivity $\left(D\right)$ of an antenna are $0.75\left(1+{\sin }^2\theta {\sin }^2\varphi \right)$ and $0.75$ respectively.