Chapter 13, Problem 5P

(a)

To determine

Prove that the radiation resistance is $R_{\text{rad}}=20{\pi }^2{\left[\frac{\mathcal{l}}{\lambda }\right]}^2$.

Explanation of Solution

Calculation:

Using the mentioned equation given in the textbook,

$\begin{array}{c}{A}_{zs}=\frac{e^{-j\beta r}}{4\pi r}{\displaystyle {\int }_{-\frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}{I}_{\text{o}}\left(1-\frac{2\left|z\right|}{\mathcal{l}}\right)e^{j\beta z\cos \theta }dz}\\ =\frac{e^{-j\beta r}}{4\pi r}{I}_{\text{o}}\left[{\displaystyle {\int }_{-\frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}\left(1-\frac{2\left|z\right|}{\mathcal{l}}\right)\cos \left(\beta z\cos \theta \right)dz+j{\displaystyle {\int }_{-\frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}\left(1-\frac{2\left|z\right|}{\mathcal{l}}\right)\sin \left(\beta z\cos \theta \right)dz}}\right]\end{array}$

Consider the real part of $A_{zs}$. Therefore,

$\begin{array}{c}{A}_{zs}=\frac{e^{-j\beta r}}{4\pi r}{I}_{\text{o}}\left[{\displaystyle {\int }_{-\frac{\mathcal{l}}{2}}^{\frac{\mathcal{l}}{2}}\left(1-\frac{2\left|z\right|}{\mathcal{l}}\right)\cos \left(\beta z\cos \theta \right)dz}\right]\\ =\frac{e^{-j\beta r}}{4\pi r}2I_{\text{o}}\left[{\displaystyle {\int }_0^{\frac{\mathcal{l}}{2}}\left(1-\frac{2z}{\mathcal{l}}\right)\cos \left(\beta z\cos \theta \right)dz}\right]\\ =\frac{I_{\text{o}}{e}^{-j\beta r}}{2\pi r}\left(\frac{2}{\mathcal{l}}\right)\left\{\frac{1-\cos \left[\beta \left(\frac{\mathcal{l}}{2}\right)\cos \theta \right]}{{\beta }^2{\cos }^2\theta }\right\}\end{array}$

$A_{zs}=\frac{I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}{\beta }^2{\cos }^2\theta }\left[1-\cos \left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)\right]$        (1)

The general expression to calculate the electric field intensity in relationship with the magnetic vector potential.

$E_s=-j\omega \mu A_s$

From the above expression,

$E_{\theta s}=j\omega \mu \sin \theta A_{zs}=j\beta \eta \sin \theta A_{zs}$        (2)

Substitute equation (1) in equation (2)

$E_{\theta s}=j\beta \eta \sin \theta \left\{\frac{I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}{\beta }^2{\cos }^2\theta }\left[1-\cos \left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)\right]\right\}$

$E_{\theta s}=\left(\frac{j\eta I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}}\right)\left\{\frac{\sin \theta \left[1-\cos \left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)\right]}{\beta {\cos }^2\theta }\right\}$        (3)

In equation (3), if $\frac{\beta \mathcal{l}}{2}\ll 1$ then $\cos \left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)=1-\frac{{\left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)}^2}{2!}$. Therefore, equation (3) becomes,

$\begin{array}{c}{E}_{\theta s}=\left(\frac{j\eta I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}}\right)\left\{\frac{\sin \theta \left[1-\left(1-\frac{{\left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)}^2}{2!}\right)\right]}{\beta {\cos }^2\theta }\right\}\\ =\left(\frac{j\eta I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}}\right)\left\{\frac{\sin \theta \left[\frac{{\left(\frac{\beta \mathcal{l}}{2}\cos \theta \right)}^2}{2}\right]}{\beta {\cos }^2\theta }\right\}\\ =\left(\frac{j\eta I_{\text{o}}{e}^{-j\beta r}}{\pi r\mathcal{l}}\right)\left(\frac{\beta {\mathcal{l}}^2}{8}\sin \theta \right)\end{array}$

$E_{\theta s}=\frac{j\eta I_{\text{o}}}{8\pi r}\beta \mathcal{l}{e}^{-j\beta r}\sin \theta$        (4)

Write the general expression for the average radiated power.

$P_{\text{ave}}=\frac{{\left|E_{\theta s}\right|}^2}{2\eta }$        (5)

Here,

$\eta$ is the intrinsic impedance, and

$E_{\theta s}$ is the electric field intensity.

Substitute equation (4) in (5).

$\begin{array}{c}{P}_{\text{ave}}=\frac{{\left|\frac{j\eta I_{\text{o}}}{8\pi r}\beta \mathcal{l}{e}^{-j\beta r}\sin \theta \right|}^2}{2\eta }\\ =\frac{\frac{{\eta }^2{I}_{\text{o}}^2}{64{\pi }^2{r}^2}{\beta }^2{\mathcal{l}}^2{\sin }^2\theta }{2\eta }\\ =\frac{\eta I_{\text{o}}^2}{128{\pi }^2{r}^2}{\beta }^2{\mathcal{l}}^2{\sin }^2\theta \end{array}$

Write the general expression to calculate the radiated power.

$P_{\text{rad}}={\displaystyle {\int }_S{P}_{\text{ave}}dS}$        (6)

Here,

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

Substitute $\frac{\eta I_{\text{o}}^2}{128{\pi }^2{r}^2}{\beta }^2{\mathcal{l}}^2{\sin }^2\theta$ for $P_{\text{ave}}$ and $r^2\sin \theta d\theta d\varphi$ for $dS$ in equation (6) with integration limits.

$\begin{array}{c}{P}_{\text{rad}}={\displaystyle {\int }_S\frac{\eta I_{\text{o}}^2}{128{\pi }^2{r}^2}{\beta }^2{\mathcal{l}}^2{\sin }^2\theta r^2\sin \theta d\theta d\varphi }\\ ={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^{\pi }\frac{\eta I_{\text{o}}^2}{128{\pi }^2{r}^2}{\beta }^2{\mathcal{l}}^2{\sin }^2\theta r^2\sin \theta d\theta d\varphi }}\\ ={\left(\varphi \right)}_0^{2\pi }{\displaystyle {\int }_0^{\pi }\frac{\eta I_{\text{o}}^2}{128{\pi }^2}{\beta }^2{\mathcal{l}}^2{\sin }^3\theta d\theta }\\ =2\pi {\displaystyle {\int }_0^{\pi }\frac{\eta I_{\text{o}}^2}{128{\pi }^2}{\beta }^2{\mathcal{l}}^2{\sin }^3\theta d\theta }\end{array}$

Simplify the equation as follows,

$\begin{array}{c}{P}_{\text{rad}}=2\pi {\displaystyle {\int }_0^{\pi }\frac{\left(120\pi \right)I_{\text{o}}^2}{128{\pi }^2}{\left(\frac{2\pi }{\lambda }\right)}^2{\mathcal{l}}^2{\sin }^3\theta d\theta }\left\{\because \beta =\frac{2\pi }{\lambda },\eta =120\pi \right\}\\ =\frac{120I_{\text{o}}^2}{64}{\left(\frac{2\pi }{\lambda }\right)}^2{\mathcal{l}}^2{\displaystyle {\int }_0^{\pi }{\sin }^3\theta d\theta }\end{array}$

$P_{\text{rad}}=\frac{480{\pi }^2{I}_{\text{o}}^2}{64}{\left(\frac{\mathcal{l}}{\lambda }\right)}^2{\displaystyle {\int }_0^{\pi }{\sin }^3\theta d\theta }$        (7)

On solving the integration, ${\displaystyle {\int }_0^{\pi }{\sin }^3\theta d\theta }=\frac{4}{3}$. Therefore, equation (7) becomes,

$P_{\text{rad}}=\frac{480{\pi }^2{I}_{\text{o}}^2}{64}{\left(\frac{\mathcal{l}}{\lambda }\right)}^2\left(\frac{4}{3}\right)$

$P_{\text{rad}}=10{\pi }^2{I}_{\text{o}}^2{\left(\frac{\mathcal{l}}{\lambda }\right)}^2$        (8)

Consider the expression to calculate the radiated power.

$P_{\text{rad}}=\frac{1}{2}{I}_{\text{o}}^2{R}_{\text{rad}}$        (9)

Here,

$I_{\text{o}}$ is the peak value of input current, and

$R_{\text{rad}}$ is the radiation resistance.

Equating equation (8) and (9).

$\begin{array}{l}\frac{1}{2}{I}_{\text{o}}^2{R}_{\text{rad}}=10{\pi }^2{I}_{\text{o}}^2{\left(\frac{\mathcal{l}}{\lambda }\right)}^2\\ \frac{1}{2}{R}_{\text{rad}}=10{\pi }^2{\left(\frac{\mathcal{l}}{\lambda }\right)}^2\\ R_{\text{rad}}=20{\pi }^2{\left(\frac{\mathcal{l}}{\lambda }\right)}^2\end{array}$

Conclusion:

Thus, the radiation resistance $R_{\text{rad}}=20{\pi }^2{\left[\frac{\mathcal{l}}{\lambda }\right]}^2$ is proved.

(b)

To determine

Find the length $\left(\mathcal{l}\right)$ of the dipole.

Answer to Problem 5P

The length $\left(\mathcal{l}\right)$ of the dipole is $\mathcal{l}=0.05\lambda$.

Explanation of Solution

Calculation:

Refer to part (a),

The radiation resistance for a short dipole antenna is,

$R_{\text{rad}}=20{\pi }^2{\left[\frac{\mathcal{l}}{\lambda }\right]}^2$

Substitute 0.5 for $R_{\text{rad}}$ in above equation to find the length of the dipole $\mathcal{l}$.

$\begin{array}{l}0.5=20{\pi }^2{\left[\frac{\mathcal{l}}{\lambda }\right]}^2\\ {\left[\frac{\mathcal{l}}{\lambda }\right]}^2=\frac{0.5}{20{\pi }^2}\\ \frac{\mathcal{l}}{\lambda }=\sqrt{\frac{0.5}{20{\pi }^2}}\\ \mathcal{l}=0.05\lambda \end{array}$

Conclusion:

Thus, the length $\left(\mathcal{l}\right)$ of the dipole is $\mathcal{l}=0.05\lambda$.