In: Electrical Engineering
Design a system with one input called “Up” such that when “Up” equals 1, the machine counts 2, 3, 4, 5, 6, and repeat. If the “Up” input equals 0, the machine counts down 6, 5, 4, 3, 2, and repeat.
Show the State Transition Diagram, State Transition table, K-maps, and equations assuming JK flip flops and AND/OR/NOT combinational logic. Make sure to indicate on your State Transition Diagram what happens if the system initially is in one of the unused states (0, 1, or 7).
Input |
Present State |
Next State |
JK Flip Flop Inputs |
|||||||||
Up |
Q2 |
Q1 |
Q0 |
Q2+ |
Q1+ |
Q0+ |
J2 |
K2 |
J1 |
K1 |
J0 |
K0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
X |
1 |
X |
0 |
X |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
X |
1 |
X |
X |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
X |
X |
0 |
0 |
X |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
X |
X |
0 |
X |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
X |
1 |
1 |
X |
1 |
X |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
X |
0 |
0 |
X |
X |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
X |
0 |
X |
1 |
1 |
X |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
X |
0 |
X |
0 |
X |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
X |
1 |
X |
0 |
X |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
X |
1 |
X |
X |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
X |
X |
0 |
1 |
X |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
X |
X |
1 |
X |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
X |
0 |
0 |
X |
1 |
X |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
X |
0 |
1 |
X |
X |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
X |
1 |
X |
0 |
0 |
X |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
X |
1 |
X |
0 |
X |
1 |
When Down Counting, i.e. Up = 0, then unused states 0,1,7 transits to state 6
When UP Counting, i.e. Up = 1, then unused states 0,1,7 transits to state 2
Excitation Table of JK Flip Flop
Q |
Q+ |
J |
K |
0 |
0 |
0 |
X |
0 |
1 |
1 |
X |
1 |
0 |
X |
1 |
1 |
1 |
X |
0 |
LogiSim is used to design the circuit as per K-Map deduced equations for JK Flops.