Solutions Communications Technology II WS 2011/2012

More documents

Recommendations

Info

“Communications Technology II” – Solutions WS 2011/2012 82 ViterbiSolution to exercise 6 (vit01):a) S0 = {−1−j},S1 = {−1+j},S2 = {+1−j},S3 = {+1+j}S0z 0,0 = −2−2j z 2,0 = −2jz 0,1 = −2 z 2,1 = 0z 0,2 = −2j z 2,2 = 2−2jz 0,3 = 0 z 2,3 = 2z 1,0 = −2 z 3,0 = 0z 1,1 = −2+2j z 3,1 = 2jz 1,2 = 0 z 3,2 = 2z 1,3 = 2j z 3,3 = 2+2jS1S3S4d(1)=1+jd(2)=1-jd(3)=-1-jd(4)=-1+jd(5)=-1-jb)c)λ 1 = |z 0,3 −(0.5+j)| 2 = |0−(0.5+j)| 2 = 1.25λ 2 = |z 3,2 −(2+0.5j)| 2 = |2−(2+0.5j)| 2 = 0.25λ 3 = |z 2,0 −(−3j)| 2 = |−2j −(−3j)| 2 = 1λ 4 = |z 0,1 −(−2)| 2 = |−2−(−2)| 2 = 0λ 5 = |z 1,0 −(−2+j)| 2 = |−2−(−2+j)| 2 = 15∑λ i = 3.5i=1
9 WS 2011/2012 “Communications Technology II” – SolutionsSolution to exercise 7 (vit08):a) Trellis element for 2 nd order channel:Trellis diagram for a complete sequence:ramp-up phaseS0{-1,-1}16 202142928slow down phase28 182721S {-1,1}14171217281718S {1,-1}24162181316S {1,1}3s(i)2598 24 241.5 -0.5 -1.5 2.5 -2.5 0.5b) Table with the (not standardized) partial path costs:P 1.5 -0.5 -1,5 2.5 -2.5 0.5z 0,−1 = -2.5 16 4 1 25 0 9z 0,1 = -0.5 4 0 1 9z 1,−1 = -1.5 0 16 1 4z 1,1 = 0.5 4 4z 2,−1 = -0.5 0 1 9 4z 2,1 = 1.5 4 9 1z 3,−1 = 0.5 4 4 9z 3,1 = 2.5 16 0
Page 1: SolutionsCommunicationsTechnologyII
Page 4 and 5: “Communications Technology II”
Page 14: “Communications Technology II”
Page 17 and 18: 13 WS 2011/2012 “Communications T
Page 24: “Communications Technology II”

Solutions Communications Technology II WS 2011/2012

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?