Question

The far-zone power density radiated by a helical antenna is approximated as:

\({{\rm{\vec W}}_{{\rm{rad}}}} = {{\rm{\vec W}}_{{\rm{average}}}} \approx \widehat {{{\rm{a}}_{\rm{r}}}}{{\rm{C}}_0}\frac{1}{{{{\rm{r}}^2}}}{\cos ^4}{\rm{\theta }}\)

The radiated power density is symmetrical with respect to φ and exists only in the upper

hemisphere: 0 ≤ θ ≤ π/2 ; 0 ≤ ϕ ≤ 2π; C₀ is a constant. The power radiated by the antenna (in watts) and the maximum directivity of the antenna, respectively, are

1. 1.5C₀,10dB
2. 1.256C₀,10dB
3. 1.256C₀,12dB
4. 1.5C₀,12dB

Answer 1

Correct Answer - Option 2 : 1.256C₀,10dB

Concept: 

Power Radiated can be calculated from average power density by integrating it over the given surface.

Application:

Given that \({{\rm{W}}_{{\rm{rad}}}} = {{\rm{W}}_{{\rm{avg}}}} = \frac{1}{{{{\rm{r}}^2}}}{\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}.\overrightarrow {{{\rm{a}}_{\rm{r}}}}\)

Power radiated \(= \smallint {{\rm{W}}_{{\rm{rad}}}}.{\rm{d\vec S}}\)

But \({\rm{d\vec S}} = {{\rm{r}}^2}{\rm{sin\;\theta \;d\theta d}}\phi .\overrightarrow {{{\rm{a}}_{\rm{r}}}}\)

Power radiated = \(\smallint \smallint \frac{1}{{{{\rm{r}}^2}}}{\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}.{{\rm{r}}^2}{\rm{sin\;\theta \;d\theta d}}\phi\)

Where \(0 \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}{\rm{\;and\;}}0 \le \phi \le 2{\rm{\pi }}\)

\(\begin{array}{l} \Rightarrow {\rm{Power\;Radiated}},{{\rm{P}}_{{\rm{rad}}}} = {\rm{\;}}\smallint \smallint {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}.{\rm{sin\;\theta \;d\theta d}}\phi \\ = {{\rm{C}}_0}\left( {2{\rm{\pi }}} \right)\smallint {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{\rm{sin\theta }}.{\rm{d\theta }}:{\rm{where\;}}0 \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}\\ = {{\rm{C}}_0}\left( {2{\rm{\pi }}} \right) \times \frac{{3 \times 1 \times 1}}{{5 \times 3 \times 1}} = 1.256{{\rm{C}}_0}. \end{array}\)

\({\rm{Radiation\;intensity}} = {\rm{U}} = {{\rm{r}}^2}.{{\rm{W}}_{{\rm{rad}}}} = {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}\)

Directivity \({{\rm{D}}_0} = 4{\rm{\pi }} \times \frac{{{{\rm{U}}_{{\rm{max}}}}}}{{{{\rm{P}}_{{\rm{rad}}}}}} = 4{\rm{\pi }} \times \frac{{{{\rm{C}}_0}}}{{1.256{{\rm{C}}_0}}} = 10.005\)

Directivity in \(\left( {{\rm{dB}}} \right) = {\rm{\;}}10{\rm{\;lo}}{{\rm{g}}_{10}}10.005 = 10\)