Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 15', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | o 0 000 00 0 011 1 111 0 101V- 01__ 10 -A 1 100 ------- 11 o 1 001 3 In the next figure we draw two frames of the trellis associated with the code. Solid lines indicate an input equal to 0 whereas dotted lines correspond to an input equal to 1. 4 The diagram used to find the transfer function is shown in the next figure. Xa DNJ Using the flow graph results we obtain the system Xc D3NJXa DNJXb Xb D2JXc D2JXd Xd DNJXc DNJXd Xa D2JXb Eliminating Xb Xc and Xd results in T D N J X Xa To find the free distance of the code we set N D7NJ 3 1 - DNJ - D3NJ2 T1 D T D N J w J 1 J 1 in the transfer function so that D7 - ---- D7 D8 D9 1 - D - D3 Hence dfree 7 5 Since there is no self loop corresponding to an input equal to 1 such that the output is the all zero sequence the code is not catastrophic. 278 Problem Using the diagram of Figure we see that there are only two ways to go from state Xa to state Xa with a total number of ones sum of the exponents of D equal to 6. The corresponding transitions are Path 1 D2 D D D2 Xa x Xc x Xd x XbDX Xa Path 2 D2 D D D2 Xa x Xc x Xb x Xc x XbX Xa These two paths correspond to the codewords C1 0 0 1 0 1 0 1 1 0 0 0 0 . C2 0 0 0 1 0 0 0 1 1 1 0 0 . Problem 1 The state transition diagram and the flow diagram used to find the transfer function for this code are depicted in the next figure. Xa Thus XC Xb Xd X DNJXa D2 NJXb DJXc D2JXd NJXC DNJXd DJXb and by eliminating Xb XC and Xd we obtain T D N J X Xa D3NJ 3 1 - DNJ - D3NJ2 To find the transfer function of the code in the form T D N we set J 1 in T D N J . Hence D3N 1 - DN - D3N 2 To find the free distance of the code we set N 1 in the transfer function T D N so that D3 T1 D T D N w 1 ----- 3 D3 D4 D5 2D6 . 1 D D Hence dfree 3 279 3 An upper bound on the bit error probability when hard decision decoding is used is given by Pb k kS N 1 D ự4p 1-p Since ửT D N _ ử D3N _ D3 ửN N 1 dN 1 - D D3 N N 1 1 - D D3 2 with k 1 p 10 6 we obtain Pb D3 1 - D D3 2 X 10-9
