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Some Training

Some Training. State Assignment of synchronous FSM based on partitions. To encode machine M we need 3 two-block partitions such that:. Example 1. We generate closed partitions. G,H. C,D. G,E. A,C. A,B. F,H. E,F. B,D. Example continued . Gra ph of successor pairs :.

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Some Training

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  1. Some Training State Assignment of synchronous FSM based on partitions

  2. To encode machine M we need 3 two-block partitions such that: Example 1 We generate closed partitions

  3. G,H C,D G,E A,C A,B F,H E,F B,D Example continued. Graph of successor pairs:

  4. Example 1 continued A,D D,H + =2 B,F

  5. Example 1 continued Unfortunately: Thus we need one more partition :

  6. Example 1 continued t Encoding wrt1 2

  7. Example 1 continued With such encoding two excitation functions Q1’i Q2’ of this machine will depend on one internal variable, and the third one Q3’ (in the worst case) on three variables, i.e.: Q1’ = f(x,Q1) Q2’ = f(x,Q2) Q3’ = f(x,Q1,Q2,Q3) If you do not trust, please encode, calculate cost functionsQ1’, Q2’, Q3’ and check.

  8. If I am wrong I will buy beer For everybody

  9. Comment Every other encoding will lead to more complex excitation functions. In particular for binary encoding: Q1’ = f(x,Q1) Q2’ = f(x,Q1,Q2,Q3) Q3’ = f(x,Q1,Q2,Q3)

  10. Example 2 Partition compatible with inputs: Iis closed

  11. Example 2 continued Q1’ = f(Q1,Q2) wrtO Encoding wrtI Q2’ = f(Q1,Q2) Q3’ = ??? y = f(x,Q3)

  12. E counter clock Q Intuitive Encoding Example of a counter with input Enable

  13. Modulo 8 counter Transition table encoded with binary natural code Transition Table

  14. Encoded table

  15. Excitation function tablefor D flip-flops D2 = D1 = D0 =

  16. Encoded excitation tablefor T flip-flops T2 = T1 = T0 =

  17. Enable Q Q Q T T T Clock Q Q Q = T E 0 = T EQ 1 0 = = T EQ Q T Q 2 0 1 1 1 Diagram of the counter Discuss this design, how flip flops are selected, how to generalize to any number of bits

  18. Example 3 Encoding wrt1  Is not sufficient for encoding

  19. And what with remaining? We do not have to calculate excitation functions to know that the first one, D1will… Example 3 continued Let us then use closed partition: Q1’ = D1 = f(x,Q1) Unfortunately only one variable we were able to encode according to the closed partition, therefore: Q2’ = D2 = f(x,Q1,Q2,Q3) Q3’ = D3 = f(x,Q1,Q2,Q3)

  20. Example 3… May be there are more closed partitions: We will show that besides 1 is 2 Encoding wrt1 2 This is unique encoding

  21. Example 3 cont With this encoding the first excitation function Q1’of this machine will depend on one internal variable, and the second and third together (Q2’, Q3’) on two internal variables, like this: Q1’ = f(x,Q1) Q2’ = f(x,Q2,Q3) Q3’ = f(x,Q2,Q3) If you do not trust, encode and calculate excitation functions Q1’, Q2’, Q3’ and check. Do this!

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