Discussion
An AM envelope varies between \( 15 \ \mathrm{V} \) and \( 5 \ \mathrm{V} \). The modulation index is
Answer: D
From the envelope extremes:
\[ \begin{aligned} \mu &= \frac{V_{max}-V_{min}}{V_{max}+V_{min}} \\ &= \frac{10}{20} = 0.5 \end{aligned} \]
so 0.5, or fifty per cent modulation. 0.33 is the ratio of the minimum to the maximum. 3.0 is the ratio of maximum to minimum, which is greater than one and so cannot be an index for an undistorted signal. 0.67 comes from dividing the difference by the maximum alone. Reading the index off an oscilloscope trace this way is the standard bench measurement.
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