Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
185 views
in General by (115k points)
closed by

The radiation pattern of an antenna in spherical co-ordinates is given by

\(F\left( \theta \right) = {\cos ^2}\theta ,0 \le \theta \le \;\pi \;/2\)

The directivity of the antenna is 


1. 10 dB
2. 12.6 dB
3. 11.5 dB
4. 18 dB

1 Answer

0 votes
by (152k points)
selected by
 
Best answer
Correct Answer - Option 1 : 10 dB

Radiation pattern \(F\left( \theta \right) = {\cos ^2}\theta ,0 \le \theta \le \;\pi \;/2\)

U(θ) = F2(θ)

Directivity of antenna is given as

\(D = \frac{{4\pi {U_{max}}}}{{{P_{rad}}}}\)

\({P_{rad}} = \mathop \smallint \limits_{\phi = 0}^{2\pi } \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {sin\theta } \right)d\theta .d\phi = 2\pi \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {\sin {\rm{\theta).(d\theta }}}\right)\)

\(= 2\pi \mathop \smallint \limits_{ - 1}^0 {t^4}dt = 2\pi \left. {\frac{{{t^5}}}{5}} \right|_{ - 1}^0 = \frac{{2\pi }}{5}\)

\({U_{max}} = {\left. {{{\cos }^4}\theta } \right|_{max}} = 1\)

⇒ Directivity \(D = \frac{{4\pi \left( 1 \right)}}{{\left( {\frac{{2\pi }}{5}} \right)}} = 10 \Rightarrow 10\log 10 = 10\;dB\)

 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...