Discussion
A receiver's intermediate filter is \( 180 \ \mathrm{kHz} \) wide at 10.7 MHz. Its fractional bandwidth is about
Answer: D
Dividing bandwidth by centre frequency:
\[ \frac{0.18}{10.7} = 0.017 \]
so about 1.7 per cent, which a ceramic filter achieves comfortably. The same 180 kHz at 98 MHz would be under 0.2 per cent, which is why the selectivity is provided after translation rather than at the signal frequency. That comparison is the numerical form of the superheterodyne's whole justification, and it is worth working once for each service.
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