hwsolB25

# hwsolB25 - S3:s3 = P^4*s<> S3 = F^-4*S v4 = 1-1 1-1 1-1...

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Sheet1 Page 1 HW 25 Solution _______________ s = [ a b c d e f g h ].' S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].' S1: R*s <--> R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].' s1 = P*R*s <--> F^(-2)*R*S = S1 The entries of R*S are multiplied by those of the k = 8-1 = 6th sinusoid, given by v6 = [ 1 -j -1 j 1 -j -1 j ].' (r = sqrt(2)). Therefore S1 = [ 4 (-2+j) 5j 1+3*j -6 1-3*j -5*j -2-j ].' (The last three entrees can be obtained by conjugate symmetry, since s1 is real also.) S2: s2 = R^2*P*s <--> S2 = R^2*F^(-1)*S = F^(-1)*S S is multiplied entry by entry by the k = 7rd sinusoid, given by v7 = [ 1 (1-j)*r/2 -j (-1-j)*r/2 -1 (-1+j)*r/2 j (1+j)*r/2 ].' Therefore S2 = [ 4 (1+3*j)*r/2 5 (-1-2*j)*r 6 (-1+2*j)*r 5 (1-3*j)*r/2 ].' (Again, conjugate symmetry may be used to speed up the computation.)

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Unformatted text preview: S3:s3 = P^4*s <--> S3 = F^(-4)*S v4 = [ 1 -1 1 -1 1 -1 1 -1].' S3 = [ 4 1-2*j 5*j -3-j -6 -3+j -5*j 1+2*j].' S4: s4 = F^2*s <--> S4 = P^2*S. Therefore S4 = [ -5j -1-2j 4 -1+2j 5j 3+j -6 3-j ].' S5: s5 = (F^2 + F^(-2))*s <--> S5 = ( P^(-2)+P^2 )*S. Sheet1 Page 2 S5 = [ 5*j 3+j -6 3-j -5*j -1-2j 4 -1+2*j].' + [ -5*j -1-2j 4 -1+2*j 5*j 3+j -6 3-j ].' = [ 0 2-j -2 2+j 0 2-j -2 2+j ].' S6: s6 = s - P^4*s <--> S6 = S - F^(-4)*S Therefore S6 = S - S3 = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].' - [ 4 1-2*j 5*j -3-j -6 -3+j -5*j 1+2*j].' = [ 0 -2+4j 0 6+2*j 0 6-2*j 0 -2-4*j ].' S7: s7 = s - R*s <--> S7 = S - R*S. Therefore S7 = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].' - [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].' = [ 0 4*j 10*j 2*j 0 -2*j -10*j -4*j ].'...
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## This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB25 - S3:s3 = P^4*s<> S3 = F^-4*S v4 = 1-1 1-1 1-1...

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