Discussion
Calculate the minimum sampling rate to avoid aliasing when a continuous-time signal is given by \(x(t) = 20\cos 500\pi t\).
Answer: C
Write the signal as \(x(t) = 20\cos(2\pi f t)\) with \(2\pi f = 500\pi\), so the frequency is
\[ f = 250\ \text{Hz} \]
To avoid aliasing the sampling rate must be at least twice the highest frequency (the Nyquist rate).
\[ f_s = 2f = 500\ \text{Hz} \]
The value 250 Hz is the signal frequency itself, which is only the Nyquist frequency, and 125 Hz and 375 Hz are not related to it. The answer is 500 Hz, option C.
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