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Unit 10: Clocked Sequential Circuits



                    Since, these are not in the same block in P2 then at least one of the states in (CEFG) is   Notes
                    not equivalent to the others.
                     State F must be different from C, E and G because its 1- successor, D, is in a different
                    block than E, C, and G.
               •  Therefore, P = (ABD)(CEG)(F).
                          3
               •  At this point, we know that state F is unique since it is in a block by itself.
               •  P = (ABD)(CEG)(F).
                  3
               •  The process repeats yielding the following:
                    0-successors of (ABD) are (BDB).
                     A, B and D are still considered equivalent.
                    1-successors of (ABD) are (CFG), which are not in the same block,

                     B cannot be equivalent to A and D since F is in a different block than C and G.
                    The 0- and 1-successors of (CEG) are (FFF) and (ECG)
                     C, E, and G must still be considered equivalent.
               •  Thus, P  = (AD)(B)(CEG)(F).
                       4
               •  If we repeat the process to check the 0- and 1-successors of the blocks (AD) and (CEG), we
                 find that
                    P  = (AD)(B)(CEG)(F).
                      5
               •  Since P  = P  and no new blocks are generated, it follows that states in each block are
                           4
                       5
                 equivalent.
                    A and D are equivalent.
                    C, E, and G are equivalent.

               •  The state table can be rewritten, removing the rows for D, E and G and replacing all
                 occurrences of D with A and all occurrences of E or G with C.
               •  The resulting state table is as follows:
                                    Table 10.10: Resulting State Table

                                               Next state  Output
                                  Present state
                                               w = 0, w = 1  Z
                                       A        B     C      1
                                       B        A     F      1
                                       C        F     C      0
                                       F        C     A      0



                          Always clock has low rise/fall times, then both pMOS and nMOS may conduct.

            10.5 State Assignment

            A 2n-state finite state machine has n!-different possible ways to assign state variable values to
            states.  Thus, a 16-state finite state machine (which requires at least 4 state variables) has 24 (4!)
            different assignment of states to state variable values.  A 1024-state finite state machine (which
            requires at least 10 state variables) has over 3 million (10!) state assignments.  If the state is not
            densely encoded in the fewest number of bits, even more encodings are possible!


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