AProblem 3: 15 Points (10, 5) 16-bit messages are transmitted using a Hamming co
ID: 3700694 • Letter: A
Question
AProblem 3: 15 Points (10, 5) 16-bit messages are transmitted using a Hamming code with 5 check bits to ensure that the receiver can detect and correct single-bit errors. Show the bit pattern transmitted for this message 1011,1101,1111, 0011 using odd parity. Now change the bit in the transmitted message counting from the left end that corresponds to the last digit of your jag number plus 3 (if the digit is zero, then it becomes three, or if the digit is nine, it becomes twelve) and see whether the change is detected.Explanation / Answer
message to be transmitted = 1011110111110011
The parity bits/ check bits will be present in the places of powers of 2 i.e
P0 P1 1 P2 0 1 1 P3 1 1 0 1 1 1 1 P4 1 0 0 1 1
P0 will look after the bit numbers (1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21) i.e P0 1 0 1 1 0 1 1 1 0 1
As it is odd parity number of 1,s should be odd so, P0 will be 0.
P1 will look after the bit numbers (2 , 3 , 6 , 7 , 10 , 11 , 14 , 15 , 18 , 19) i.e P1 1 1 1 1 0 1 1 0 0
As it is odd parity number of 1,s should be odd so, P1 will be 1.
P2 will look after the bit numbers (4 , 5 , 6 , 7 , 12 , 13 , 14 , 15 , 20 , 21) i.e P2 0 1 1 1 1 1 1 1 1
As it is odd parity number of 1,s should be odd so, P2 will be 1.
P3 will look after the bit numbers (8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 ) i.e P3 1 1 0 1 1 1 1
As it is odd parity number of 1,s should be odd so, P3 will be 1.
P4 will look after the bit numbers (16 , 17 , 18 , 19 , 20 , 21) i.e P4 1 0 0 1 1
As it is odd parity number of 1,s should be odd so, P4 will be 0.
So, the transmitted message will be ;
0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 0 0 1 1
Lets say we change the bit number 4 from 1 to 0.Now the received message will be ;
0 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 1 0 0 1 1
Now in the receiver side check bit P0 will look after the bits (1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21) so, the total number of 1's should be odd.
If we see the received message the total number of 1's in the above bits is 7 so there is no error here.
Now in the receiver side check bit P1 will look after the bits(2 , 3 , 6 , 7 , 10 , 11 , 14 , 15 , 18 , 19) so, the total number of 1's should be odd.
If we see the received message the total number of 1's in the above bits is 7 so there is no error here.
Now in the receiver side check bit P2 will look after the bits (4 , 5 , 6 , 7 , 12 , 13 , 14 , 15 , 20 , 21) so, the total number of 1's should be odd.
If we see the received message the total number of 1's in the above bits is 8 so there is error in any one of these bits.
Now in the receiver side check bit P3 will look after the bits (8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 ) so, the total number of 1's should be odd.
If we see the received message the total number of 1's in the above bits is 7 so there is no error.
Now in the receiver side check bit P4 will look after the bits (16 , 17 , 18 , 19 , 20 , 21) so, the total number of 1's should be odd.
If we see the received message the total number of 1's in the above bits is 3 so there is no error.
Now in the bits of P2 where there is error we have the bits (4 , 5 , 6 , 7 , 12 , 13 , 14 , 15 , 20 , 21)
From P1 bit number 6 , 7 , 14 , 15 dosen't have error so, we are left with bits (4 , 5 , 12 , 13 , 20 , 21)
From P0 bit number 5 , 13 , 21 dosen't have error so, we are now left with bits (4 , 12 , 20)
From P3 bit number 12 will not have error so, we are left with bits (4 , 20)
Finally from P4 bit number 20 will not have error so, finally we are left with bit 4 which has an error.
So, the receiver was able to detect the one bit error in bit number 4.