Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 30)
30.
The analog signal given below is sampled at 1200 samples per second. m(t) = 2 sin 120 pt + 3 sin 240 pt + 4 sin 360 pt + 5 sin 480 pt + 6 sin 600 pt + 7 sin 720 pt. The folding (maximum) frequency is __________ .
Answer: Option
Explanation:
Folding frequency (max.) =
= 600 Hz.
Discussion:
2 comments Page 1 of 1.
Jude said:
8 years ago
So the folding frequency minimum would be?
Ankit said:
9 years ago
Please explain it.
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