solns7_600

solns7_600 - ECE-600 Introduction to Digital Signal...

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Unformatted text preview: ECE-600 Introduction to Digital Signal Processing Autumn 2010 Homework #7 Nov. 19, 2010 HOMEWORK SOLUTIONS #7 1. For this problem, we can use the IDFT definition plus a variable change ( l = N − 1 − k ): y [ n ] = 1 N N − 1 summationdisplay k =0 Y [ k ] e j 2 π N nk = 1 N N − 1 summationdisplay k =0 X [ N − 1 − k ] e j 2 π N nk = 1 N N − 1 summationdisplay l =0 X [ l ] e j 2 π N n ( N − 1 − l ) = e j 2 π N n ( N − 1) 1 N N − 1 summationdisplay l =0 X [ l ] e j 2 π N l ( − n ) . We can’t claim that the previous expression equals e − j 2 π N n ( N − 1) x [ − n ] because this would involve illegal time indices − n = 0 , − 1 , . . . , − N + 1. But, since we can write − n = (− n ) N + qN for some integer q , we have y [ n ] = e j 2 π N n ( N − 1) 1 N N − 1 summationdisplay l =0 X [ l ] e j 2 π N l ( qN + (− n ) N ) = e − j 2 π N n 1 N N − 1 summationdisplay l =0 X [ l ] e j 2 π N l (− n ) N = e − j 2 π N n x [ (− n ) N ] .....
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This note was uploaded on 03/05/2011 for the course ECE 600 taught by Professor Clymer,b during the Fall '08 term at Ohio State.

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solns7_600 - ECE-600 Introduction to Digital Signal...

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