Electronics and Communication Engineering - Signals and Systems - Discussion
Discussion Forum : Signals and Systems - Section 1 (Q.No. 23)
23.
The analog signal m(t) is given below m(t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt, the Nyquist sampling rate will be
Answer: Option
Explanation:
m (t) = 4 cos 100 pt + 8 sin 200 pt + cos 300 pt
Nyquist sampling freq fs ≤ 2fm where fm is highest frequency component in given signal and highest fm in 3rd part
2pfmt = 300 pt
fm = 150 Hz
fs = 2 x 150 p 300 Hz
.
Discussion:
1 comments Page 1 of 1.
Pranav said:
9 years ago
fs<= 2m? Explain.
(1)
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