Solutions Communications technology II WS 2009/2010 - UniversitÃ¤t ...

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“Communications technology II” – Solutions WS 2009/2010 82 ViterbiSolution to exercise 6 (vit01):(a), (b) S0 = {−1 − j}, S1 = {−1 + j}, S2 = {+1 − j}, S3 = {+1 + j}S0S1S3S4d(1)=1+jd(2)=1-jd(3)=-1-jd(4)=-1+jd(5)=-1-jz 0,0 = −2 − 2jz 0,1 = −2z 0,2 = −2jz 0,3 = 0z 1,0 = −2z 1,1 = −2 + 2jz 1,2 = 0z 1,3 = 2jz 2,0 = −2jz 2,1 = 0z 2,2 = 2 − 2jz 2,3 = 2z 3,0 = 0z 3,1 = 2jz 3,2 = 2z 3,3 = 2 + 2jc)λ 1 = |˜z 0,3 − (1.5 + j)| 2 = | + 1 + j − (1.5 + j)| 2 = 0.25λ 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 = 2.5i=1
9 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 7 (vit08):a) Trellis element for 2nd-order channel:Trellis diagram for a complete sequence:EinschwingphaseAusschwingphaseS {-1,-1}01620214292828 182721S {-1,1}14171217281718S {1,-1}24162181613S {1,1} 382524924s(i)1.5 -0.5 -1.5 2.5 -2.5 0.5b) Table with the (not standadized) partly 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
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