CS485 Sylabus

### The 4B/5B trnasmission code

• 4B/5B code: fixing the synchronization in NRZ problem cheaply...

• The 4B/5B code:

• 4B/5B code = a very clever way to provide synchronization opportunities to the receiver by:

 Encoding 4 bits using 5 bits The encoding of 5 bits makes sure that there is a transition within the 5 bits

• Operation using 4B/5B code:

 The data is first transformed using the 4B/5B encoding scheme The encoded result is transmitted using the NRZ code as its basic transmission scheme

Note:

 When the receiver received the transmitted data, it must use the reverse mapping of the 4B/5B encoding scheme to obtain the transmitted data.

• 4B/5B mapping scheme:

 ``` Original 4 bits Transformed 4 bits ================== ======================= 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 ```

• Example using 4B5B:

• Notice that:

• After transforming the original sequence of bits, the result sequence will have:

• at most 3 consecutive 0 bits

One way to transmit 3 consecutive 0 bits is: 0010 (which is transformed to 10100) followed by 0100 (which is transformed to 01010)

 ``` Original 4 bits Transformed 4 bits ================== ======================= 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 ```

• at most 8 consecutive 1 bits

This will happen when you transmit: 0111 (which is transformed to 01111) followed by 0000 (which is transformed to 11110)

 ``` Original 4 bits Transformed 4 bits ================== ======================= 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 ```