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ece290HWK #9 Solution (Count)

# ece290HWK #9 Solution (Count) - HWK#9 Solution Problem 9.1...

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Unformatted text preview: HWK #9 Solution Problem 9.1 111 -> 011 -> 001 -> 000 -> 100 -> 110 -> 111 -> ... a. 010 and 101 are illegal states. If the counter starts in one of these states, it will be stuck in the loop 010 -> 101 -> 010 -> 101 -> ... and will never reach the normal counting sequence. b. For all states except 100 and 101, this modi¡cation clearly does not affect DC (the added term is 0 for all other states). For 100, DC is already 1, so there is no change to any state in our normal counting sequence. For 101, however, the next state is changed from 010 to 110. This is in the normal counting sequence, and so any starting state will lead to the sequence. (010 requires two clock cycles, as it still passes through 101 before reaching 110). c. Problem 9.2 State diagram: a. Next-state/excitation table: The state table below shows the current state CBA, the next state C+B+A+, and the ¢ip-¢op excitation inputs T C T B T A . C B A C+ B+ A+ T C T B T A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x x x x x x 1 1 1 1 1 1 1 1 1 1 x x x x x x 1 1 1 1 1 1 b. Boolean expressions: We want minimal SOP expressions for T C , T B , and T A as de¡ned in the above table. Each T is a function of C, B, and A. By putting the truth table information into K-maps, we easily obtain the expressions below. BA 00 01 11 10 C 1 1 x 1 1 1 1 x 1 T C = 1 BA 00 01 11 10 C 1 x 1 x 1 T B = A'C' BA 00 01 11 10 C 1 x 1 1 1 x 1 T A = A'B' + C Circuit: c. Augmented state diagram We know the state progression when the counter starts from any of states 111, 010, 100, 001, 101, 000. We want to examine how the counter we designed behaves if it gets into any of the "don't care" states 011, 110. If the counter is in state CBA = 011, then T C T B T A = 100 and so the next state is 111. If the counter is in state CBA = 110, then T C T B T A = 101 and so the next state is 011....
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ece290HWK #9 Solution (Count) - HWK#9 Solution Problem 9.1...

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