Question

The E-field pattern of an antenna, independent of $\varphi$ varies as follows
$E=1,0\le \theta \le {45}^{\prime }$
$=0,45<\theta \le {90}^{\prime }$
$=\frac{1}{2},{90}^{\prime }<\theta \le {180}^{\prime },$

Then the directivity of the antenna is

• A. 7.4
• B. 1.85
• C. 3.7
• D. 37

Answer 1

The correct option is C 3.7

$U(\theta ,\varphi )=W_{nd}{r}^2$

$U(\theta ,\varphi )=\frac{E^2}{2\eta }{r}^2$

$P_{\text{rad}}={\oint }_{s}U(\theta ,\varphi )\sin \theta d\theta \varphi$

$={\int }_{\theta =0}^{2x}{\int }_{6=0}^{\frac{x}{4}}\frac{1}{2\eta }{r}^2\sin \theta d\theta d\varphi +{\int }_{\theta =0}^{2\pi }{\int }_{\partial =\frac{\pi }{2}}^{\pi }\frac{r^2}{8\eta }\sin \theta d\theta d\varphi$

$=\frac{r^2}{2\eta }{[-\cos \theta ]}_0^{\frac{\pi }{4}}{\left[\varphi \right]}_0^{2\pi }+\frac{r^2}{8\eta }{[-\cos \theta ]}_{{\frac{\pi }{2}}^{\pi }}[\varphi {]}_0^{2\pi }$

$\frac{r^2\pi }{\eta }\times 0.29+\frac{r^2\pi }{\eta }\times 0.25=0.54\left(\frac{\pi r_2}{\eta }\right)$

$p_{\text{rad}}=0.54\left(\frac{\pi r^2}{\eta }\right)$

$U_{\text{max}}=\frac{r^2}{2\eta }$

$D=4\pi \frac{U_{\text{max}}}{P_{\text{rad}}}=4\pi \frac{\frac{r^2}{2\eta }}{0.54\pi \frac{r^2}{\eta }}=\frac{2}{0.54}=3.7$