3. 3 3
3 3 3
3 3
5 5
5 5 5
5 5
T S D T D T add D T D T both side
T D T D T S D T D T D T D T
T D T D T R because R T D T D T
Proof
4.
5. If we D15 get then we stop as we
have 15 bit data
6. T = 101110001101100
D3T= 000101110001100
D5= 000001011100011
R= 101010100000111=S
Descrambler
Putting values of F, F2 and F3
S = 101010100000111
D3S = 000101010100000111
D5S = 00000101010100000111
D6S = 000000101010100000111
D9S = 000000000101010100000111
D10S =0000000000101010100000111
D11S =00000000000101010100000111
D13S =0000000000000101010100000111
D15S =000000000000000101010100000111
T = 101110001101100
T = 101110001101100(Transmitted)
Data sent = Data Received
*Note that the input sequence contains the periodic sequence 10101010 ··· , as well as a long string of 0s. The scrambler
output effectively removes the periodic component, as well as the long string of 0s. The input sequence has 15 digits.
The scrambler output up to the 15th digit only is shown, because all the output digits beyond 15 depend on input digits
beyond 15, which are not given.
