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**Unformatted text preview: **ECE-600 Introduction to Digital Signal Processing Autumn 2011 Homework #3 Oct. 14, 2011 HOMEWORK SOLUTIONS #3 1. (a) When f0=1 , the effective discrete frequency = 2 Tf = 0 . 2 rad/sample, which means that the phasor progresses counterclockwise by . 2 2 = 0 . 1 revolutions per sample. (b) When f0=11 , the effective discrete frequency = 2 Tf = 2 . 2 rad/sample, which means that the phasor progresses counterclockwise by 2 . 2 2 = 1 . 1 revolutions per sample. How- ever, 1 . 1 revolutions per sample looks the same as 0 . 1 revolutions per sample, because it is impossible to see that the phasor made a complete revolution between samples. (c) When f0=-1 , the effective discrete frequency = 2 Tf =- . 2 rad/sample, which means that the phasor progresses clockwise by . 2 2 = 0 . 1 revolutions per sample. (d) When f0=9 , the effective discrete frequency = 2 Tf = 1 . 8 rad/sample, which means that the phasor progresses counterclockwise by 1 . 8 2 = 0 . 9 revolutions per sample. However, this looks the same as the previous case, because 0 . 9 revolutions counterclockwise is equivalent to 0 .....

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