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Digital Circuits and Logic Design



                   Notes         The Figure 10.19 shows two possible assignments. Both assign S0 to 00 and place S3’ and S4’
                                 adjacent to each other.
                                               Let us consider more complicated case.
                                              Figure 10.20: The State Diagram of the 4-bit String Recognizer

                                                                                 Reset
                                                                        S0
                                                                  0/0         1/0


                                                              S1                  S2
                                                                     1/0   1/0
                                                            0/0                    0/0



                                                              S3’                 S4’


                                                           0, 1/0      0/0         1/0

                                                              S7’                S10’

                                                          0, 1/0            0/1       1/0



                                 Applying the guidelines yields the following set of assignment constraints:
                                   Highest priority  :  (S3’, S4’), (S7’, S10’);

                                 Medium priority  :  (S1, S2), (S3’, S4’), (S7’, S10’);
                                   Lowest priority  :  0/0: (S0, S1, S2, S3’, S4’, S7’);

                                             1/0  :  (S0, S1, S3’, S4’, S7’, S10’);
                                 The Figure 10.20 shows two alternative assignments that meet most of these constraints:
                                                      Figure 10.21: Two Alternative Assignments

                                          Q1Q0                              Q1Q0
                                               00    01     11   10       Q2     00     01    11    10
                                       Q2
                                           0   S0                            0   S0
                                           1                                 1

                                          Q1Q0                               Q1Q0
                                       Q2      00    01     11   10        Q2     00    01    11    10
                                           0   S0          S3′                0   S0

                                           1               S4′                1   S7′              S10′

                                         Q1Q0                               Q1Q0
                                               00    01     11   10               00    01    11    10
                                       Q2                                  Q2
                                           0   S0          S3′   S7′          0   S0          S3′
                                           1               S4′   S10′         1   S7′         S4′   S10′
                                                                                                    Contd...

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