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Unformatted text preview: ECEN 455: Assignment 7 Problems: 1. (CSE: 7.31) A Hadamard matrix is defined as a matrix whose elements are 1 and its row vectors are pairwise orthogonal. In the case where n is a power of 2, and n n Hadamard matrix is constructed by means of the recursion H 2 = bracketleftbigg 1 1 1 1 bracketrightbigg H 2 n = bracketleftbigg H n H n H n H n bracketrightbigg . Let c i denote the i th row of an n n Hadamard matrix as defined above. Show that the wave forms constructed as s i ( t ) = n summationdisplay k =1 c ik p ( t kT c ) , i = 1 , 2 ,...,n are orthogonal, where p ( t ) is an arbitrary pulse confined to the time interval t T c . The inner product of s i ( t ) and s j ( t ) is integraldisplay s i ( t ) s j ( t ) dt = integraldisplay n summationdisplay k =1 c ik p ( t kT c ) n summationdisplay l =1 c jl p ( t lT c ) dt = n summationdisplay k =1 n summationdisplay l =1 c ik c jl integraldisplay p ( t kT c ) p ( t lT c ) dt = n summationdisplay k =1 n summationdisplay l =1 c ik c jl E p kl = E p n summationdisplay k =1 c ik c jk The quantity n k =1 c ik c jk is the inner product of the row vectors c i and c j . Since the rows of the matrix H n are orthogonal by construction, we obtain integraldisplay s i ( t ) s j ( t ) dt = E p n summationdisplay k =1 c 2 ik ij = n E p ij . Thus, the waveforms s i ( t ) and s j ( t ) are orthogonal. Show that the matched filters (or crosscorrelators) for the n waveforms { s i ( t ) } can be realized by a single filter (or correlator) matched to the pulse p ( t ) followed by a set of n discretetime crosscorrelators using the code words { c i } . Consider first the signal s i ( t ) = n summationdisplay k =1 c ik p ( t kT c ) . The signal s i ( t ) has duration T = nT c and its matched filter is h ( t ) = s i ( T t ) = s i ( nT c t ) = n summationdisplay k =1 c ik p ( nT c kT c t ) = n summationdisplay l =1 c i ( n l +1) p (( l 1) T c t ) = n summationdisplay l =1 c i ( n l +1) p ( t ( l 1) T c ) 1 that is, a sequence of impulses starting at t = 0 and weighted by the mirror image sequence of { c il } . Since, s i ( t ) = n summationdisplay k =1 c ik p ( t kT c ) = p ( t ) n summationdisplay k =1 c ik ( t kT c ) the Fourier transform of the signal s i ( t ) is S i ( f ) = P ( f ) n summationdisplay k =1 c ik e j 2 fkT c and therefore, the Fourier transform of the signal matched to s i ( t ) is H ( f ) = S i ( f ) e j 2 fT = S i ( f ) e j 2 fnT c = P ( f ) n summationdisplay k =1 c ik e j 2 fkT c e j 2 fnT c = P ( f ) n summationdisplay l =1 c i ( n l +1) e j 2 f ( l 1) T c = P ( f ) F [ g ( t )] ....
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 Spring '08
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 Recursion

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