Solutions Communications technology II WS 2009/2010 - Universität ...

13 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 10 (vit15):a) (from p.556) upper path: d(i) = 0 / lower path: d(i) = 1true data sequence (bold): d(i) = 0 1 1 0 0b) estimated sequence (dashed): ˆd(i) = 1 0 1 0 00001101101010101error vector:e = [∆d(i 0 ), . . . , ∆d(i 0 + L f − l − 1] TL f = 4; l = 2; → L f−l−1 = 4 − 2 − 1 = 1An error vector of length 2 results:⇒e = [∆d(i 0 ), ∆d(i 0 + 1)] Te = [−1, 1] Tc) The structure of the AK-matrix for a channel of order 2 is given on p.577. To find thismatrix we have to determine the energy-ACF of the error vector first:r ee (0) = ∑ νr ee (1) = ∑ νe(ν)e(ν + 0) = 2e(ν)e(ν + 1) = −1r ee (λ) = ∑ ν⎡⇒ R E ee =⎢⎣e(ν)e(ν + λ) = 0 f”ur λ ≥ 22 −1 0−1 2 −10 −1 2⎤⎥⎦d) Eigenvalue-equation, cf. Eq.(14.5.48a):det(R E ee − λI) = 0Eq.(14.5.59a): (r ee (0) − λ) 2 − 2|r ee (1)| 2 = 0⇒ λ min = 2 − √ 2S/N-loss: γmin 2 = 2 − √ 2 ˆ= 2.3 dB