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Suppose if we have a counter that counts only Even numbers and the unused states(odd numbers) treated as don't-care condition.

Current state           Next State
A  B  C  D              A  B  C  D
0  0  0  0              0  0  1  0
0  0  1  0              0  1  0  0
0  1  0  0              1  0  0  0
1  0  0  0              1  0  1  0
1  0  1  0              1  1  1  0
1  1  1  0              0  0  0  0

0  0  0  1              x  x  x  x
0  0  1  1              x  x  x  x
0  1  0  1              x  x  x  x
0  1  1  1              x  x  x  x
1  0  0  1              x  x  x  x
1  0  1  1              x  x  x  x
1  1  0  1              x  x  x  x
1  1  1  1              x  x  x  x

Above; I put the all unused states at the end.

Does this a right way? OR I must put them as normally as it's, like :

   Current state            Next State
    A  B  C  D              A  B  C  D
    0  0  0  0              0  0  1  0
    0  0  0  1              x  x  x  x
    0  0  1  0              0  1  0  0
    ........
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  • \$\begingroup\$ Since it's the values which encode the operations, not the order on paper it wouldn't typically matter. However, if you are doing a paper exercise in grouping like terms (such as in a Karnough map) you may for purposes of that need to list them in order; but a grey code order rather than a binary value one. Of course if you have a class assignment or something else with style guidelines, that could impose requirements as well. \$\endgroup\$
    – Chris Stratton
    Sep 10, 2013 at 21:55
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    \$\begingroup\$ Isn't it obvious that the "D" state variable is completely superfluous and should simply be eliminated? \$\endgroup\$
    – Dave Tweed
    Sep 10, 2013 at 22:06
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    \$\begingroup\$ @Azad - the answer is that it doesn't matter, unless there's a requirement you haven't told us about which would make it matter - I tried to explain a few possibilities along those lines. \$\endgroup\$
    – Chris Stratton
    Sep 10, 2013 at 22:09
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    \$\begingroup\$ What's 2^4??? You only have 8 valid states, which is only 2^3. The "D" state variable is always 0, so you might as well eliminate it and tie it to ground instead. \$\endgroup\$
    – Dave Tweed
    Sep 10, 2013 at 22:16
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    \$\begingroup\$ The output for D needs to be zero, but it has no relevance as a state variable. \$\endgroup\$
    – Dave Tweed
    Sep 11, 2013 at 1:58

1 Answer 1

Reset to default
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If you perform a logic minimization analysis of the four state variables A, B, C, and D and take don't cares into consideration in the Karnough map groupings you will find that the D term minimizes away as the input to any of the state transfer equations. Also you will find that the logic equation for the drive of the Q(D) flop flop becomes:

Q(D) := 0

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