I"m trying to create a circuit chart that synchronizes to the Mealy chart I produced for the following problem, however I"m not sure how numerous flip-flops I should use.

You are watching: How many flip flops per state

Problem:

A data present receives serial data the 1 bit, synchronised through a clock pulse. Produce a Mealy State diagram that:

Starts from an initial state (IS).Outputs xy = 01 when it recognises the bit-sequence 1001 and returns to IS.Outputs xy = 10 when it recognises the bit-sequence 011 and returns to IS.In any kind of other case, the mechanism should return xy = 00.

Mealy diagram (Corrected):(This is the chart I came up with)

Based top top the Mealy diagram above, I would certainly say 3 flip-flops room needed, because 100 demands 3 bits to be represented. Is there, however, a means to implement it through less? If so, why?

digital-logic flipflop chart state-machines
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edited jan 17 "18 in ~ 15:47
asked january 17 "18 in ~ 10:50
user174667user174667
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\$n \$ flip-flops can represent \$2^n \$ states. The variety of bits required to represent all the claims will be the variety of flip-flops needed to implement that state machine. So for this state diagram, 4 flip-flops are needed due to the fact that one that the says is 1001, which demands 4 bits. However through state reduction techniques like row corresponding method, succeeding partitioning method, implicitly chart, and also state assignment/encoding techniques, no. Of claims (and no.of bits/state) come implement a state maker can it is in reduced. In turn no. Of flip-flops essential are reduced.

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edited january 17 "18 in ~ 12:47
answered january 17 "18 in ~ 12:29

Mitu RajMitu Raj
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The names of the states do not recognize the number of FFs required. The number counts on how plenty of states over there are and what sort of encoding you pick to use (e.g., binary vs. One-hot). Binary encoding needs \$\lceillog_2N\rceil\$ FFs, if one-hot needs \$N\$ FFs.

A Mealy device also needs a FF for each output.

In any kind of case, your diagram is incorrect. It just recognizes the sequences 10011 and 0111. It likewise fails to uncover either the the order if the initial bit is not correct — for example, the succession 11011 has 011, but your an equipment won"t uncover it.

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edited jan 17 "18 at 12:46
answered january 17 "18 at 12:30

Dave TweedDave Tweed
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