Hamming code and correct the seven bits

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julian403

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I have to do a correction circuit bit7, using hamming. I did the next:



As you can see, there is 4 bits, so there must to be 3 redundant bits. I did the 3 C and the parity bits are b1 , b2 and b4.

But, how the receiver circuit can correct the bit 7?, after the transmittance in the receiver

I think b7 is:

b7= b4 x b5x b6x or b7= b2xb3xb6 or b7=b1xb3xb5

where x is xor gates.

or maybe I must use the Cs bits to correct the 7 bit ?
 
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So, one times I generate in receptor the 3 bits (which if the 3 bits are 0 there is not error) to correct the seven bits I must put a and gates with that 3 bits (because 1-1-1 are seven) and the output of the and gates make the parity (with xor gates) whit the seven bits. right?
 

The receiver generates a syndrome word, from which by using a bitwise XOR operation, the error can be corrected. You must consider the fact that only correctable errors will generate a correct output word at receiver side.
 

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