problem 5

problem 5 - ECE 101 - Linear Systems, Winter 2009 Problem...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 5

problem 5 - ECE 101 - Linear Systems, Winter 2009 Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online