Question

if the present state (abc) is 110, and the input x = 0, what will be the next state if the flip flops input functions are ja = b x, ka = 1, jb = a + c x, kb = x c + c x, jc = a x + a b x, kc = x. 111 001 010 011

Explanation:

Step1: Calculate JA

Given present state ABC = 110 and x = 0. First, find $B'$. Since B = 1, $B'=0$. Then $JA = B'x=0\times0 = 0$. And $KA = 1$. For a JK - flip - flop, when $JA = 0$ and $KA = 1$, the next state of A, $A_{next}=0$.

Step2: Calculate JB

$JB=A + C'x$. Here, A = 1, C = 0 so $C' = 1$, and x = 0. Then $C'x=1\times0 = 0$. So $JB=1+0 = 1$. $KB=xC'+Cx=0\times1 + 0\times0=0$. For a JK - flip - flop, when $JB = 1$ and $KB = 0$, the next state of B, $B_{next}=1$.

Step3: Calculate JC

$JC=Ax+A'B'x$. A = 1, x = 0, $A' = 0$, $B' = 0$. So $Ax=1\times0 = 0$, $A'B'x=0\times0\times0 = 0$. Then $JC=0+0 = 0$. $KC=x = 0$. For a JK - flip - flop, when $JC = 0$ and $KC = 0$, the next state of C, $C_{next}=0$.

Answer:

010