Discussion
A pulse train has pulses \( 1 \ \mathrm{\mu s} \) wide. The bandwidth needed is roughly
Answer: B
The usual rule takes the bandwidth as the reciprocal of twice the pulse width:
\[ \begin{aligned} B \approx \frac{1}{2\tau} &= \frac{1}{2 \times 10^{-6}} \\ &= 500 \ \mathrm{kHz} \end{aligned} \]
so about 500 kHz. One megahertz is the reciprocal of the width alone, a more conservative figure some texts prefer. The other values follow from a factor of four either way. Narrow pulses demand wide channels, which is the resource every pulse scheme spends and the reason bandwidth rather than power usually limits them.
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