画出状态图

初态 次态
0 0 0 1 1 1
1 1 1 1 1 0
1 1 0 1 0 1
1 0 1 0 1 1
0 1 1 0 1 0
0 1 0 0 0 1
0 0 1 0 0 0

绘制次态卡诺图

00 01 11 10
0 111 000 010 001
1 xxx 011 110 101
Q2 00 01 11 10
0 1 0 0 0
1 x 0 1 1
Q1 00 01 11 10
0 1 0 1 0
1 x 1 1 0
Q0 00 01 11 10
0 1 0 0 1
1 x 1 0 1

列状态方程

令Q2=A,Q1=B，Q0=CQ_2 = A , Q_1 = B ，Q_0 = C

$Q_2^{n+1} = AB + \bar B\bar C $

Q1n+1=AC+BC+BˉCˉQ_1^{n+1} = AC +BC +\bar B\bar C

Q0n+1=C+ABˉQ_0^{n+1} = C + A \bar B

根据驱动方程 $ Q^{n+1} = J \bar Q^n + \bar K Q^n$

Q2n+1=(B+BˉCˉ)A+BˉCˉAˉQ_2^{n+1} = (B + \bar B \bar C)A + \bar B \bar C \bar A

Q2n+1=BˉC‾A+BˉCˉAˉQ_2^{n+1} = \overline {\bar B C} A + \bar B \bar C \bar A

即J2=BˉCˉ,K2=BˉCJ_2 = \bar B \bar C , K_2 = \bar B C

Q1n+1=(AC+C)B+(AC+C)BˉQ_1^{n+1} = (AC+C) B + (AC+C) \bar B

Q1n+1=CB+CBˉQ_1^{n+1} = CB + C \bar B

即J1=C,K2=CˉJ_1 = C,K_2 = \bar C

Q0n+1=ABˉC+(AB+1)CˉQ_0^{n+1} = A \bar B C + (AB + 1) \bar C

$Q_0^{n+1} = A \bar B C + \bar C $

即J0=1，K0=ABˉJ_0= 1，K_0= A \bar B

根据三组J与K的值，接入三个JK触发器，得到：

原理图