Exerise Genertor polynomils of onvolutionl ode, given in binry form, re g, g j g. ) Sketh the enoding iruit. b) Sketh the stte digrm. ) Find the trnsfer funtion T. d) Wht is the minimum free distne of the ode? Solution ) Enoding iruit: b) Stte digrm: / / / input output / / / / /
) odified stte digrm inditing the weights of trnsitions (node split into two): b e d d) Stte equtions: X X X b X X X d b d X X X X b d e X. From the seond eqution we get X d X b, whih n be inserted into the third eqution: X X b X b X X X X X X X X 5 6 4 6 4 X X X X X X 5 X X. Solving for X nd inserting into the fourth eqution results in T 7 X e X 7 7 7 7 8 9...... From T we see tht d free 7. inimum free distne n lso be seen diretly from the modified stte digrm of prt. The minimumweight route e is b e. The weight of this route is d 7. free
Exerise Clulte union bound for error probbility for Convolutionl Code generted by the polynomils g j g. Solution Union bound for the error probbility is evluted s j, d di, j From the leture mteril the ode spetrum for this ode is " # For this ode the mount of bit error is d 4. We evlute the union bound s
BER - -4 simulted Union bound Exerise '() *,-. -6-8 4 6 8 SNR / '() kll m.5 Q Pi j Q mk kll m.5d ij Q mk m desribes the probbility of seleting ertin pth from the stte in the trellis nd mking m symbol errors beuse of seleting this pth. The mount of possible symbols is. In Viterbi equlizer from eh stte there re possible outgoing brnhes. By seleting one brnh we n do errors. If the pth is l trellis setions long we ould do errors in eh setion. Totl probbility of error is multiplition of probbilities for eh trnsition long the pth. kll mk m. (We multiply only l-l trnsitions sine lst L trnsitions orresponds to orret symbols otherwise the erroneous pth would not merge with orret pth). For our BPSK trnsmission = (two possible symbols) error m is. m
In one errorneous pth we do in verge j, d informtion bit errors. As for the se of Convolutionl Code for lulting the verge bit error we hve to verge over bit errors over ll the possible error pths. For simplifying our lultions we selet k to be.. For given hnnel there is no error pth with length.. There is one pth with length nd with distne Erx d, T E rx..7..7 T.9 Erx 4 T In this pth we do one error, / Where, the term /.9 normlizes the totl hnnel power to.. There is one pth with length nd with distne Erx d, T Erx..7 T.9....7 Erx. T in this pth we do two errors, / Totl error probbility is bounded s
BER - - - -4-5 -6 simulted Union bound Exerise 4 The rte / onvolutionl ode hs genertors g. g Represent this ode s: ) Shift register. b) Stte digrm. ) Tree. d) Trellis. e) trix. -7 4 6 8 SNR Solution 4 b) ) / / / / / / / / / / / / / / / /
) d) e) G Exerise 5 Represent the ode in the previous exerise s reursive systemti onvolutionl ode. Clulte the output (oded) sequene for both: systemti nd orresponding nonsystemti enoder. The informtion sequene t the u. input is
Solution 5 ) reursive systemti onvolutionl ode u z u u z u z... g z z z g z z z z 4 z z sz u u z uz uz u4z z z z z z sz z z z z 4 5 4 5 z z z z z z z z 4 5 4 5 z z z z z z z z z z s s s " b) Systemti onvolutionl ode s u z g z s u z g z ivide both sequenes with g z nd you get new sequene where one ontins the systemti bit nd the other is generted by reursive digitl filter. s u z s u z g z g z Exerise 6 The enoder strts from zero stte, ollets three bits nd dds to them prity bit lulted bsed on the previous stte nd the bit vlues. The informtion bits nd prity bits re trnsmitted. In next intervl the proess is repeted but the initil stte is equl to prity bit vlue in the next intervl. rw the trellis for given ode. (This type of ode is lled zigzg ode).
Solution 7 / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / The stte digrm / / / / Trellis of the ode / / / / Exerise 8 ' # )4 5 # 5 6 5 # 6 7 8
9 / / / b / d / / / 6576 ' # # # ) # ) # * 6'7( ( / reeived sequene # : 4 ( - - - - - - - :6' #,"; ;##"";" ""#;";# ; ; "-
" " ' " " " " ' -.47.6 -.96.54.76 Exerise 9.47 -.6 4. -4..96.84 -.84.6 -.6 -.,9.6..9 (Wiker, problem.9) Assume tht the enoder from Problem. (see the book) is used over symmetri memoryless hnnel. The bit metris re s follows. y r y 6 y 6 Find the mximum-likelihood ode word orresponding to the following reeived sequene. r,,,,,,,.. reeived sequene... -. -.8.. -.. -.8 -.8 -,6, -,4 One stge of the trellis nd the brnh outputs re shown below. The deoding proedure is illustrted in the following figure (hek it). shed lines nd solid lines indite terminting brnhes nd surviving brnhes, respetively. The survivor pth is indited with bold solid line. In se of ties, upper trnsition ws hosen. # ' # # H A? A E L A @ I A G K A? A @ A? @ A @ I A G K A? A
Exerise Consider the enoder on Figure below. esribe the deoders by the stte digrm. Assume tht the I is blok interlever. erive the output sequene k of the enoder if the input dt sequene is dk. Give lso the input sequene to the seond enoder. (The dt sequene fter interlever.) esribe wht kind of impt the two lst bits of the input sequene hve on the termintion of the trellis of eh enoder. input I output The bit sequene t the interlever output is d d k i, k The enoder stte digrm is Solution / / / / / / The enoder outputs re x x k k xk Two lst bits set the enoder to the stte. For this exmple this stte is rehed by both of the onstituent enoders. / /