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Unformatted text preview: ECE 101  Linear Systems, Winter 2009 Problem Set # 5 Solutions (send comments/questions to kquest@ucsd.edu) Problem 1 (7.21 (c, d, and f) OW2) 7.21 (c) Here Im { X ( j ) } is not specified. Therefore, the Nyquist rate for the signal x ( t ) cannot be determined. This implies that one cannot guarantee that x ( t ) would be recoverable from x p ( t ). 7.21 (d) Since x ( t ) is real, it follows that X ( j ) = X * ( j ). Thus, X ( j ) = 0 for   > 5000. The Nyquist rate for the given signal is 2 5000 = 10000 . Therefore, in order to be able to recover x ( t ) from x p ( t ), the sampling period must at most be T max = 2 / 10000 = 2 10 4 sec. Since the sampling period used is T = 10 4 < T max , x ( t ) can be recovered from x p ( t ). 7.21 (f) If X ( j ) = 0 for   > 1 , then X ( j ) * X ( j ) = 0 for   > 2 1 . Therefore, in this part, X ( j ) = 0 for   > 7500 . The Nyquist rate for this signal is 2 7500 = 15000 . Therefore, in order to recover x ( t ) from x p ( t ), the sampling period must at most be...
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 Spring '09
 bill lin

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