Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 25 (Q.No. 16)
16.
Consider a sampled signal y(t) = 5 x 10-6 x(t)
where x(t) = 10 cos (8p x 103)t and ts = 100 μ sec. when y(t) is passed through an ideal low-pass filter with a cut off frequency of 5 kHz, the output of the filter is
where x(t) = 10 cos (8p x 103)t and ts = 100 μ sec. when y(t) is passed through an ideal low-pass filter with a cut off frequency of 5 kHz, the output of the filter isDiscussion:
2 comments Page 1 of 1.
Ram charan said:
9 years ago
Y(t) = 5 * 10^-6x(t),
Here x(t) = 10cos(8p*10^3)t,
= 5 * 10^-6 * 10cos(8p * 10^3)t,
= 5*10^-5 cos(8p*10^3)t.
Here x(t) = 10cos(8p*10^3)t,
= 5 * 10^-6 * 10cos(8p * 10^3)t,
= 5*10^-5 cos(8p*10^3)t.
(2)
Vini said:
1 decade ago
Can be check easily by amplitude.
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