hw10_soln

# hw10_soln - 2 8.5 Next State Present State A 000 B 001 C...

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2) 8.5 Next State Present State w=0 w=1 Output z A 000 A 000 B 001 0 B 001 E 100 C 010 0 C 010 D 011 C 010 0 D 011 A 000 F 101 1 E 100 A 000 F 101 0 F 101 E 100 C 010 1 Next state equations: W Y Y W Y Y W Y Y Y D 0 2 0 1 0 1 2 2 + + = 0 1 0 1 1 Y Y W Y Y D + = W Y Y W Y Y W Y Y D 0 1 0 1 0 1 0 + + = Output: W Y Y Y z 2 0 1 + =

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3) 8.6 Next State Output W=0 W=1 Present State J1K1 J0K0 J1K1 J0K0 w=0 w=1 A 00 A 00 0d 0d B 01 0d 1d 0 0 B 01 D 11 1d d0 C 10 1d d1 0 0 C 10 D 11 d0 1d C 10 d0 0d 1 0 D 11 A 00 d1 d1 B 01 d1 d0 0 1 Equations using D flip flops 1 2 1 2 1 2 2 Y Y Y Y Y Y D = + = 0 1 2 1 2 1 2 1 2 1 2 1 Y Y Y W Y Y W Y Y W Y Y W Y Y D = + + + = [using xnor]
Output: ) 1 ( 2 1 2 1 2 W Y Y W Y Y W Y Y Z = + = [using xnor] Equations using JK flip flops W Y W Y W Y K W Y J Y K Y J = + = = = = 2 2 2 1 2 1 1 2 1 2

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3)
4) 8.7 In order to make the next state transition on a simple 1 bit value, we set 2 1 w w k = . When w1=w2: k=0 This solution uses D flip flops.

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5) 8.11 Next State Output Present w=0 w=1 w=0 w=1 A B E 0 0 B C G 0 0 C D H 0 0 D A A 1 0 E G F 0 0 F H D 0 0 G H H 0 0 H A A 0 0 8.12
8.16 Note: We are not using hot-encoding in this problem.
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hw10_soln - 2 8.5 Next State Present State A 000 B 001 C...

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