Consider all possible 3bit datawords totally 8 Recall a pari

Consider all possible 3-bit datawords (totally 8). Recall a parity bit is a bit added to a string of binary code that indicates whether the number of 1-bits in the string is even or odd. You can use either even parity bit or odd parity bit.

a. What is the minimum Hamming distance for the code set you obtained?

b. How many bit errors can this codeset detect?

c. What is the number of legal codewords?

d. What is the number of illegal codewords?

Solution

Considering the even parity. 0-even

There will be even number of 1\'s in every code word.

Parity A0 A1 A2 are given respectively below.

0 0 0 0

1 0 0 1

1 0 1 0

0 0 1 1

1 1 0 0

0 1 0 1

0 1 1 0

1 1 1 1

We have a set of 8 mentioned above.

[a]

Hamming Distance between two binary numbers is defined as the number of binary digits differ in those 2.

Minimum hamming distance for above set is 2. Let us see the details now.

If we take 0000 and 1111, the binary digits differ are 4.

So, lets try with other 0000 and 1001, here the binary digits differ are 2.

Since we have a parity bit also, if one bit differs in last three digits, parity differs. This way there is alteast hamming distance of 2 in the above set.

[b]

1 bit error and 3 bit errors

Change in 1 bit - Can identify

Let us take 0000. If it got an error and changed to 0001, then it is clear that it is an error since the parity bit should be 1 for this.

Change it 2 bits - Can\'t identify

if 0000 gets changed to 0011 or 1001 or 0101 or 1100 we can\'t say that it is an invalid code.

Change in 3 bits - Can identify

Change in 3 bits effect parity definitely. For example 0000 got changed to 1011. Clearly parity should be 0 for it.

Change ini 4 bits- Can\'t identify

If 0000 is there and it got changed to 1111, it is a valid code word. Like this for all.

[3]

There are 8 legal codewords

Those which are mentioned above are legal. All codewords which have even number of 1\'s

[4]

Illegal codewords are 8

We know that 2⁴=16 for 4 binary digits. So, there is possibility of 16 codewords.

Illegal codewords = Possible codewords - legal codewords

= 16 - 8 = 8

All codewords which have odd number of 1\'s.