Draw transition/excitatiob table for state machine

[leonne]

Homework Statement

draw transition/excitatiob table for state machine with jk flip flop. using the tables given

Homework Equations

excitation table for jk ff
q qn j l
000x
011x
10x1
11x0

x is don't care

table
ps ns ns
x=0 x=1
00 10 10
01 10 11

The Attempt at a Solution

Im just not sure how to do this. I know how to do it for a d ff and when they give a circuit but no idea about this

the solution is
x=0 x=1
q1q0 Q1Q0 J1K1 J0K0 Q1Q0 J1K1 J0K0
00 10 1x 0x 10 1x 0x
01 10 1x x1 11 1x x0

I know how they get the q1q0 and Q1Q0 but no idea how they get the jk it said to use excitation table for jk but don't see how that has anything to do with this.

 

[KataKoniK]

Hello,

The excitation table for a JK flip flop is used to determine the inputs (J and K) for the next state based on the current state and the desired next state. In this case, the tables given provide the current state (q1q0) and the desired next state (Q1Q0) for both possible inputs (x=0 and x=1).

To fill in the rest of the table, you need to use the excitation table for a JK flip flop that is given in the homework statement. This table provides the values for J and K based on the current state (q) and the next state (qn).

For example, for the first row of the table (x=0, q1q0=00), the next state (Q1Q0) is 10. To determine the inputs (J1K1 and J0K0) for the next state, we look at the excitation table for a JK flip flop. We see that when q=0 and qn=1, the inputs should be J=1 and K=x. Similarly, when q=0 and qn=0, the inputs should be J=x and K=0. Therefore, for the first row, the inputs for the next state of Q1Q0=10 would be J1=1, K1=x, J0=x, and K0=0.

I hope this helps to clarify how to fill in the remaining values in the table. Let me know if you have any further questions. Good luck!