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2.4  = 4000 rad/sec, so f = 4000/(2) = 2000/ Hz, and T = /2000 sec. Therefore, five periods cover 5/2000 = /400 sec. For this signal, the Nyquist sampling rate is 2(2000/) = 4000/ Hz, so the sampling rate in this case is fS = 4(4000/) = 16000/ Hz. Therefore, the total number of samples collected in five periods is (16000/ samples/sec )(/400 sec) = 40. 2.5  = 2500 rad/sec, so f = 2500/(2) = 1250 Hz, and T = 1/1250 sec. Therefore, five periods cover 5/1250 sec. For this signal, the Nyquist sampling rate is 2(1250) = 2500 Hz, so the sampling rate in this case is fS = 7/8(2500) = 2187.5 Hz. The total
number of samples collected is (2187.5 samples/sec)(5/1250 sec) = 8.75. This means only eight samples are collected while five periods of the analog signal elapse. In fact, for this signal, an integer number of samples can never be collected in an integer number of analog periods.
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发表于 2016-10-27 19:52:10
发表于 2019-11-21 22:20:29
