Counter Design 2

S a ab r a ab 00 01 11 10 00 01 11 10 x 1 x 1 1 1 1 1

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S A AB R A AB 00 01 11 10 00 01 11 10 x 0 0 1 0 × x 0 × 0 1 0 1 0 1 × 0 1 × 0 0 1 S B AB R B AB 00 01 11 10 00 01 11 10 x 0 1 0 0 1 x 0 0 1 1 0 1 1 × 0 0 1 0 0 1 × S A = A’B R A = x’ AB + x B’ S B = x ’B’ + A’B’ R B = x B + AB D A AB D B AB 00 01 11 10 00 01 11 10 x 0 0 1 0 1 x 0 1 0 0 1 1 0 1 1 0 1 1 1 0 0 D A = A’B + x B + x ’AB’ D B = x ’B’ + x A’ J A AB K A AB 00 01 11 10 00 01 11 10 x 0 0 1 × × x 0 × × 1 0 1 0 1 × × 1 × × 0 1 J B AB K B AB 00 01 11 10 00 01 11 10 x 0 1 × × 1 x 0 × 1 1 × 1 1 × × 0 1 × 0 1 × J A = B K A = x ’B + x B’ = x B J B = x ’ + A’ K B = x ’ + A T A AB T B AB 00 01 11 10 00 01 11 10 x 0 0 1 1 0 x 0 1 1 1 1 1 0 1 0 1 1 1 0 1 0 T A = A’B + x ’B + x AB’ T B = x ’ + A’B’ + AB Each of these counters has been implemented in a single LogicWorks simulation, which has been attached with this section of the notes. Try each one to verify that it works correctly. The various binary probes that have been placed on the state outputs allow you to see how the counter is progressing through the desired sequence of states. You may want to probe the flip-flop inputs to observe the way in which they ensure that the correct sequence of states occurs. END OF PROBLEM 2
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