Discussion
Page Shift Keying (PSK) Method is used to modulate digital signal at 9600 bps using 16 level. Find the line signals and speed (i.e. modulation rate).
Answer: A
The modulation rate is 2400 bauds.
With 16 levels each signal element carries \(\log_2 16 = 4\) bits, so the baud rate is the bit rate divided by 4:
\[ \text{baud rate} = \frac{9600}{\log_2 16} \]
\[ = \frac{9600}{4} = 2400 \]
1200 bauds (B) would need 8 bits per signal element, which means 256 levels. 4800 bauds (C) comes from taking 2 bits per element, which is 4 levels, and 9600 bauds (D) would be right only if each element carried one bit.
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