Discussion
If a wave of frequency 100 MHz propagates in the free space, then value of propagation constant will be :
Answer: D
The propagation constant is \(j\,2.09\ \text{m}^{-1}\).
In lossless free space the attenuation constant is zero, so the propagation constant \(\gamma = \alpha + j\beta\) is purely imaginary, with \(\beta = \omega/c\).
\[ \begin{aligned} \beta &= \frac{2\pi f}{c} = \frac{2\pi \times 10^{8}}{3 \times 10^{8}} \\ &\approx 2.09\ \text{rad/m} \end{aligned} \]
So \(\gamma = j\,2.09\ \text{m}^{-1}\), matching a wavelength of 3 m. \(j\,1.05\) leaves out the factor 2 in \(2\pi\), and \(j\,0.03\ \text{nm}^{-1}\) has the wrong unit. The constant is finite, so infinity is wrong.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development