00:01
Okay, so from the given diagram, we know ja equals a plus b x bar and ka equals x bar, and j .b equals i0s bar plus i1s, a bar plus b bar x, and kb equals x.
00:26
So for any jk flip lock, q plus equals jq, bar plus k bar q and for flip -top a we have q a bar plus equals j a q a bar plus k a bar q a bar q a plus q b x bar q a bar q a bar plus x double bar q a x bar q b x bar plus q a x q a x q a x q a x q a x plus q b and for flip flop b you have q b j b b b b b plus k b bar plus k b bar q b b b b b b b b b b b b b b b x q b bar qb which then simplifies to qb x plus q a bar qb bar and now we'll do a state table so we have the present state qa qb and then input input x and then next state q a plus and qb 0 -0 -0 -0 -1 -11 -1 -0 -0 -1 -0 -1 -0 -1 -1 -1 -1 -0 -1 -1 -1 -1 -0 -1 -1 -1 okay so we have 0 -0 -0 -1 -0 -1 and 11 -1 -0 -1 -0 and 1 -1 -0 -0 -1 -0 and 1 -0 -0 and the 4 -c the state diagram 0 -0 -1 -11 -0 so you have this direction x equals 1 this direction x equals 0 and x equals 1 and this direction x equals 0 and x equals 0 and x equals 0 0 x equal 0 x equals 0 x equals 1 and then the same thing x equals 1 and uh for 2 so we have 2 sequence are required 1 1 -1 -10 0 0 -0 -1 this one and so this is when x equals 0 and 0 and 0 0 and 1 1 1 x equals 1.
04:16
Or combined, you get 0 ,0 ,0, 1, 1, and 1 0, x equals 0, x equals 1, x equals 1, x equals 0, x equals 1, x equals 0, and 1, and 1, and then x equals 0.
04:46
And stable would be you have the input x and the present state qa qb and then next state qa plus qb plus and then flip -lop input j -a k -a j -b kb so you have 0 -0 -0 -1 -1 -1 -1 -0 011, 011, 011, 0101, and then 0, excuse me, 10101, 01011, and 10101, and 1010111, and 110010111, and then 100101, and then 1201, 0, 0 ,0, 0 110, x, x, x, 1 ,0 ,000.
05:57
Xx10 and then 1x1 1x 1 x 1 x 1 x 0 x x 0 x 1...