quiz1soln - t ? (Recall 1 KHz = 1000 Hz = 1000 s-1 )...

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Quiz 1, MATH 464, 6 April 2010 Your Name: _____________________________________ (1) [3 pts] A band-limited signal is sampled at its critical (Nyquist) sampling rate of 100 KHz . Thus in one-second interval 100,000 samples are collected. If only 1000 samples are used to compute the signal at some time in the middle of that one- second interval, the reconstruction error is 10 -4 . Are there enough samples to guarantee a reconstruction error 20 times smaller? Solution The reconstruction error is upper bounded by N E t e 2 | ) ( | That is, it decays like N 1 . To guarantee a decrease of the error by a factor of 20, the reconstruction formula must use 20 2 =400 times more samples. This means 400,000 samples. However we only have 100,000 samples. Hence there are NOT enough samples to GUARANTEE a reconstruction error 20 times smaller. (2) [3 pts] What is the maximum sampling period of a 50 KHz band-limited signal so that we can perfectly compute the signal at any time
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Unformatted text preview: t ? (Recall 1 KHz = 1000 Hz = 1000 s-1 ) Solution The maximum sampling period for perfect reconstruction is T 1/(2*50,000) = 10-5 s = 10 s (3) [4pts] A 10 KHz band-limited signal is sampled at Nyquist rate. The only nonzero samples are: = = = ms t ms t t x 2 . 1 2 . 1 2 ) ( Compute x(0) (that is, x(t) for t=0 ms ). Recall 1 ms = 10-3 s . Solution The sampling period is T=1/(20,000) = 0.05 ms . Note that both -1.2 ms and 0.2 ms are integer multiples of the sampling period 0.05 ms . The reconstruction formula yields: ) 4 ( sinc ) 24 ( sinc 2 05 . 2 . sinc ) 2 . ( 0.05 (-1.2)-sinc ) 2 . 1 ( ) ( = + = = + = = t x t x x Hence x(0)=0. Alternatively, note that t=0 is one of the sampling times and the problem states that the sample value is zero. Hence x(0)=0....
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This note was uploaded on 05/05/2010 for the course MATH 464 taught by Professor Staff during the Spring '08 term at Maryland.

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quiz1soln - t ? (Recall 1 KHz = 1000 Hz = 1000 s-1 )...

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