The state of a Fibonacci LFSR of degree $n$ is the first $n$ bits output by the Galois LFSR with the the same polynomial. The two constructions give the same output, but starting from a different state. That drawing uses the Fibonacci construction of a LFSR. The schematic in the question (where the first + in reading order is a XOR gate with output on the left, and any other + is just a wire) at first does not seem to be matching what's above in this answer, which applies to the so-called Galois construction of a LFSR. $$s_j \equiv c_1s_$ state in time $\mathcal O(\log i)$ rather than $\mathcal O(i)$, and linking the period of the generator to properties of $P(x)$. I know you can also get this output from the equation: (here the initial state is as in the diagram, 0,1,1,0). Then calculate it iteratively like this: 0, 1, 1, 0, 0. Obviously I can create the LFSR diagram based on the function, since this function describes where the "taps" are (see. How can I use this function f to get the output of the LFSR, given some initial state? Say you have a characteristic polynomial of an LFSR:
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |