solns7_600

# solns7_600 - ECE-600 Homework#7 Introduction to Digital...

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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 ] . To verify this in Matlab, we use a randomly generated signal X : clear; N = 8; X = randn(1,8); x = ifft(X,N); Y = fliplr(X); y = ifft(Y,N); n = [0:N-1]; minusnmodN = mod(-n,N); max(abs( y - x(1+minusnmodN).*exp(-j*2*pi/N*n) ))

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