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I need to calculate the output of the sequence generated by this shift register but I cannot find anywhere how to do it. Everywhere the results are just given but there is no explanation how to do compute them. I know the the sequence will repeat every 2^3-1=7 times. Could anyone explain me how to do it. Thank you.

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The answer seems to be simple:

First, we XOR the two least significant bits (LSB) to get the most significant bit. The two LSB are obtained by right-shifting. So, calculating in this way, we get the sequence 110, 111, 011, 001, 100, 010, 101.

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    $\begingroup$ The answer 110, 111, 011, 001, 100, 010, 101 is the sequence of successive states or contents of the LFSR. The sequence of outputs is 1011100... and repeats periodically with period 7. $\endgroup$ Commented Aug 26, 2012 at 23:21
  • $\begingroup$ Yes, I agree with you, assuming that the outgoing bit from the shift register is taken as the output. $\endgroup$
    – Arani
    Commented Aug 27, 2012 at 6:56

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