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1. Consider a sampled signal y(t) = 5 x 10-6 x(t) where x(t) = 10 cos (8p x 103)t and t s = 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
(A): 10 cos(8p x 103)t
(B): 5 x 10-1 cos(8p x 103)t
(C): 5 x 10-5 cos (8p x 103)t
(D): 5 x 10-6 cos (8p x 103)t
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