sol.HW6 - 1 EE 7615 1 Solutions to Problem Set 6 11,22,10 a...

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1 EE 7615 Solutions to Problem Set # 6 11,22,10 1) a) K = 3 , L = 4 . The code rate is R c = K/L = 3 / 4 . b) Binary source Channel encoder Binary ASK mod. + Z Z Z 1 2 3 s s s s 1 2 3 4 s(t) N (t) W r(t) f( T -t) L t=T L r r r r 1 2 3 4 Vector Decision rule s s s s 1 2 3 4 T L t f( T -t) L Z Z Z 1 2 3 Inverse of the encoder The binary ASK modulator multiplies each s i by a delayed version of the a unit energy pulse such as φ ( t ) plotted above. T L = T/L = KT b /L = . 75 T b = . 75 R b . The vector decision rule is ˆ s = s i if || r - s i || ≤ || r - s i || for all j 6 = i The encoding rule is given by Z 1 , Z 2 , Z 3 s 000 s 1 001 s 2 010 s 3 011 s 4 100 s 5 101 s 6 110 s 7 111 s 8 The decoding rule is the inverse of the table above. c) P ( E ) 1 2 K 8 X i =1 X j 6 = i Q ˆ d ij 2 N 0 ! where d ij = || s i - s j || . Now for this signal set for each i there is one j such that d ij = 4 E c and for all other signals d ik = 2 2 E c . Thus P ( E ) 1 2 K 2 K " Q ˆ 4 E c 2 N 0 ! + (2 K - 2) Q ˆ 2 2 E c 2 N 0 !# = Q ˆ 2 2 E c N 0 ! +6 Q ˆ 2 s E c N 0 ! d) The signal set is biorthogonal with energy 4 E c . The exact formula for P ( E ) is given in the text.
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