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I have a state transtion table that looks like this:

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And the binary equation obtained from this is:

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I don't have any example in my text book that solves this table with Karnaugh map. The text book just states that it can be done by inspection and I am confused about the process.

Can someone please help me covert this to Karnaugh map and solve it?

Peter Cordes
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goxarad784
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    Make two 4 input maps from the first 4 columns, one for S'1 and one for S'0. The "can be solved by inspection" means that since both next state variables only have a single `1` you can immediately read off the result without a map. – Jester Mar 22 '21 at 00:47

1 Answers1

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There two tricks, basically-- 1: Convert the given "don't care" Xs to 1s and 0s (see "[1]" below for first given "don't care" X) 2: Note that the outputs are never both 1 (see resultant "don't care" Xs below) Truth Table, Completed:

S1 S0 A B  S'1 S'0
0  0  0 0  0   0 [1]
0  0  0 1  0   0 [1]
0  0  1 0  0   1
0  0  1 1  0   1
0  1  0 0  0   0
0  1  0 1  1   0
0  1  1 0  0   0
0  1  1 1  1   0
1  0  0 0  0   0
1  0  0 1  0   0
1  0  1 0  0   0
1  0  1 1  0   0
1  1  0 0  X   X
1  1  0 1  X   X
1  1  1 0  X   X
1  1  1 1  X   X

This results in the following S'1 Karnaugh Map:

S1S0\AB 00  01  11  10

00      0   0   0   0

10      0   0   0   0
           _______
11      X | X   X | X 
          |       |    
01      0 | 1   1 | 0
          ---------

This results in a minimized Sum Of Products of:

S'1 = SoB

S'0 is determined similarly.

Andrew
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