Question

A- A communication channel transmits binary digits, i. e. O's and l's. The probability of error in transmission of either digit is 0.2. To reduce the chance of error, we transmit n bits for each digit as follows: a super-symbol 00....0 is transmitted instead of 0 or super symbol 1111....1 is transmitted instead of 1. a) What is the probability of error using the super-symbol strategy with 5 assuming that we receive the correct result if 3 or more of the bits are error free (this is majority decoding). b) What is the probability of error using the super-symbol strategy with n=6 assuming that we receive the correct result if 4 or more of the bits are error free (this is NOT majority decoding).

Answer 1

the probability im trmsmissem a digit I O.2 : The probaility ers or free tramsmissem f a Ligit ib o.8 the sweer-Symbol strategy Wita me5 assuming USimg that we rece ive tee Grect seswlt if 3 r more ffee biks ame errorfsee. he oecure if gorinere syamtee Them am bits. Phere probability error im bits Probability t trer im 4 bits Pro babitity ib eror im 3 or more (0.215 0.0o032 =6.81 x 6.2 = 0.00128 im 3 bils 0.00512-

Probnotit r im smissom 0.00672 USimg the smfer-symbal t ntegy idh me6 asuing that we ecine e ceve de Corre ct esult 4 ar mre the bits e0rfree me err W ocere there i errer im 3oT ,e Am bits more 6 bits0.2) probability rror im Prbalaiity of Probalaility f Paobability f 0.000064 RYnror im bits erTOr im 4 lois rOr im 3 bits l0.8x (020.004096 6.8) xt02)0. o00256 0.8)x0.10.00l 024 ererr im trombmnissom 0. 00544 7he probaleiliky f