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Unformatted text preview: ECE 101  Linear Systems, Winter 2009 Problem Set # 5 Solutions (send comments/questions to [email protected]) Problem 1 (7.21 (c, d, and f) OW2) 7.21 (c) Here Im { X ( jω ) } 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 ( jω ) = X * ( jω ). Thus, X ( jω ) = 0 for  ω  > 5000. The Nyquist rate for the given signal is 2 × 5000 π = 10000 π . Therefore, in order to be able to recover x ( t ) from x p ( t ), the sampling period must at most be T max = 2 π/ 10000 π = 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 ( jω ) = 0 for  ω  > ω 1 , then X ( jω ) * X ( jω ) = 0 for  ω  > 2 ω 1 . Therefore, in this part, X ( jω ) = 0 for  ω  > 7500 π . The Nyquist rate for this signal is 2 × 7500 π = 15000 π . Therefore, in order to recover x ( t ) from x p ( t ), the sampling period must at most be...
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This note was uploaded on 02/24/2010 for the course ECE ECE25 taught by Professor Bill lin during the Spring '09 term at UCSD.
 Spring '09
 bill lin

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