Thread: Algorithmic state machine Truth Table equations?

  1. #1
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    Algorithmic state machine Truth Table equations?

    ok so im afraid o ask this hear but i have gone to every book i could find in 2 different libraries and asked everyone that i could get my hands on who knows anything about this sort of stuff and googled th question a million different ways and just cant find anything. i know this is a programming forum but as a beginner i have received so much help here i was hoping you guys could do it for me again and this related to programming kinda.

    anyways so i have this truth table for an algorithmic state machine with 2 states (present and next), 3-inputs (xyz), and 3-outputs (jkl) and the last part of all this is derive some boolean equations from this "Truth table" and im just lost can some please tell me how i can get these equations???

    Thank you soo much, going on 3 days trying to figure out this last problem just need a nudge in the right direction
    and if there is away to upload a jpg. of the truth table i can post that up to but i dont know how to do that :/ but for no i got a link if you want to take a look at.

    again thank you

    http://i1230.photobucket.com/albums/...Truthtable.jpg

  2. #2
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    I think you might be misinterpreting the chart a bit. The way I read it, you are given a set of input conditions (X, Y, Z values), a state transition chart (the previous and next state columns), and an list of output, give the current state and input. I think what you are after is basically a Mealy machine. The Symbol number, next to the A, B, C columns (or NA, NB, NC) is the state number (S0, S1, etc), corresponding to one of the circles on a Mealy diagram.

    Let's look at the first row. It describes one rule: when you are in state (A, B, C) = (0, 0, 0), and you receive input (X, Y, Z) = (0, *, *), you transition to state (NA, NB, NC) = (0, 0, 1) and the outputs are (J, K, L) = (1, 0, 1). I think the N in NA just means Next, as in on the next iteration, (A, B, C) = (0, 0, 1).

    EDIT: In a slightly more programming way, this would be expressed as "If X == 0 && A == 0 && B == 0 && C == 0 then NA = 0, NB = 0, NC = 1, J = 1, K = 0, L = 1".

    So you may want to put these values into a chart to make a Karnaugh map. You will have as "inputs" X, Y, Z, A, B and C since those six values determine the next state and output. You will want to allow for all combinations of X, Y, Z, A, B and C (2^6 = 64), even though many of them are "don't care" values, so you can properly draw your boxes on the K-map.

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