Question

A message is sent over a noisy channel. The message is a sequence 1,2,... .*n Of n bits (ri E 0, 1)). Since the channel is noisy, there is a chance that any bit might be corrupted, resulting in an error (a 0 becomes a 1 or vice versa). Assume that the error events are independent. Let p be the probability that an individual bit has an error (0 < p < 1/2). Let yì,U2, . . . ,Un be thereceived message (so ui = Zi İf there is no error in that bit, but-: 1-Zi İf there is an error there). To help detect errors, the nth bit is reserved for a parity check: an is defined to be 0 if zi +T2 + +Zn-i İs even, and l if zit z2 + . . . + zn-1 įs odd, when the message is received, the recipient checks whether yn has the same parity as y +y.t the parity is wrong, the recipient knows that at least one error occurred; otherwise, the recipient assumes that there were no errors. (a) For n5,p 0.1, what is the probability that the received message has errors which go undetected? (b) For general n and p, write down an expression (as a sum) for the probability that the received message has errors which go undetected

Answer 1

Solution :-

(Note: Assuming parity bit (x_n) will not be in error at receiver side.)

Parity bit is defined on whether x₁+x₂+x₃+ ....x_n-1 is odd or even. So if at receiver side while parity checking if any even no. of bits (let's say 2,4,6,..) are in error then also parity check will not be able to detect and these bit streams' errors remain undetected.

a) For n=5 and p=0.1

So total 4 information bits along with last parity bit were transmitted to receiver. Now while error detection if any 2 out of 4 bits or all 4 bits are in error then this error will go undetected. Let's calculate exact probability...

Prob of undetected errors in msg nsg - (4(0.1) (0.9))*(0.1)*- 6 0.12 0.92 +0.140.04860.0001 0.0.0487

b) For generalised n and p

{Now if the assumption is not true that parity bit may be in error then

need to be added in the above equation. }

2j-1 n-1-2j j=1