7.4ClockedSynchronous State-Machine Design3.State AssignmentThe number of flip-flop needed to code sstates is the smallest integer greater than orequal to log2s.If there are some unused states, the simplestassignment s coded states to 2n possible states isto use the first s binary integers in binarycounting order. But it isn't always the best.How do we choose the best state assignmentfor a given problem? (See P570 ~571)RtumBack
How do we choose the best state assignment for a given problem? (See P570 ~571) Return Back Next 7.4 Clocked Synchronous State-Machine Design 3. State Assignment The number of flip-flop needed to code s states is the smallest integer greater than or equal to log2s. If there are some unused states, the simplest assignment s coded states to 2n possible states is to use the first s binary integers in binary counting order. But it isn’t always the best
7.4 Clocked Synchronous State-Machine DesignThere are 5 states for our problem. The numberof flip-flop is 3 (2<log25<3)We ch00se So= QiQ2Q3= 000, S1= QiQ2Q3=100, S2= QiQ2Q3= 101, S3= Q1Q2Q3= 110, S4=QiQ2Q3= 111.4.Transition/outputtableAB0100111 0zQItlQR+Qn+IQn+Q2+'Qn+Qn+IQ,tlQn+!Qn+lQn+'Qn+1Q1Q2Q3000100110111000001101101111100001011101111101001111110111110010011111111111
Return Back Next 7.4 Clocked Synchronous State-Machine Design We choose S0= Q1Q2Q3= 000, S1= Q1Q2Q3= 100, S2= Q1Q2Q3= 101, S3= Q1Q2Q3= 110, S4= Q1Q2Q3= 111. There are 5 states for our problem. The number of flip-flop is 3 (2<log25<3). 4. Transition/output table 0 1 Q1Q2Q3 AB 0 0 1 1 1 0 Z Q Q Q n 1 2 n 1 2 n 1 1 + + + 0 0 0 1 0 0 1 0 1 1 1 0 1 1 1 Q Q Q n 1 2 n 1 2 n 1 1 + + + Q Q Q n 1 2 n 1 2 n 1 1 + + + Q Q Q n 1 2 n 1 2 n 1 1 + + + 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1