hwsolB24 - S3:Same as S2 S4: s4 = F^4*s...

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Sheet1 Page 1 HW 24 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^(-1)*R*S = S1 The entries of R*S are multiplied by those of the k = 8-1 = 7th sinusoid, given by v7 = [ 1 (1-j)*r/2 -j (-1-j)*r/2 -1 (-1+j)*r/2 j (1+j)*r/2 ].' (r = sqrt(2)). Therefore S1 = [ 4 (-3-j)*r/2 -5 (-2-j)*r 6 (-2+j)*r -5 (-3+j)*r/2 ].' (The last three entrees can be obtained by conjugate symmetry, since s1 is real also.) S2: s2 = P^(-3)*s <--> S2 = F^3*S S is multiplied entry by entry by the k = 3rd sinusoid, given by v3 = [ 1 (-1+j)*r/2 -j (1+j)*r/2 -1 (1-j)*r/2 j (-1-j)*r/2 ].' Therefore S2 = [ 4 (-1-3j)*r/2 5 (1+2j)*r 6 (1-2j)*r 5 (-1+3j)*r/2 ].' (Again, conjugate symmetry may be used to speed up the computation.)
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Unformatted text preview: S3:Same as S2 S4: s4 = F^4*s &lt;--&gt; S4 = P^4*S. Therefore S4 = [ -6 3-j -5*j -1-2j 4 -1+2j 5*j 3+j ].' S5: s5 = (P + P^(-1))*s &lt;--&gt; S5 = (F^(-1)+F)*S. This is the same as multiplying the entries of S by those of T Sheet1 Page 2 2*real(v7) = [ 2 r 0 -r -2 -r 0 r ].' (same as 2*cos(pi*k/4)). Therefore S5 = [ 8 (-1+2j)*r 0 -(3+j)*r 12 -(3-j)*r 0 (-1-2j)*r ].' S6: s6 = s + F^4*R*s &lt;--&gt; S6 = S + P^4*R*S Therefore S6 = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].' S + [ -6 3+j +5*j -1+2j 4 -1-2j -5*j 3-j ].' = [ -2 2+3j 10j 2+3j -2 2-3j -10j 2-3j ].' S7: s7 = (F + F^(-1))*s &lt;--&gt; S7 = (P^(-1) + P)*S. Therefore S7 = [ -1+2j 5*j 3+j -6 3-j -5*j -1-2j 4 ].' S + [-1-2j 4 -1+2j 5*j 3+j -6 3-j -5*j ].' = [ -2 4+5*j 2+3*j -6+5*j 6 -6-5*j 2-3*j 4-5*j].'...
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hwsolB24 - S3:Same as S2 S4: s4 = F^4*s...

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