Mealy detector for 010 using JK

A Mealy machine is used (output depends on present state and input), so fewer states are needed. Overlapping sequences are allowed: after a detection the machine keeps the useful tail of the input. One input X and one output Z; Z = 1 in the clock period in which the last bit of 010 arrives. Flip-flops: J-K.

States (what has been received so far):

• A: reset / no useful bits
• B: "0" received
• C: "01" received

State diagram (Mealy)

Arc label = input/output

 A --0/0--> B --1/0--> C --0/1--> B

 A --1/0--> A  (self loop)
 B --0/0--> B  (self loop)
 C --1/0--> A

State table

State assignment

A = 00, B = 01, C = 10 (flip-flop outputs $Q_1 Q_0$). The unused code 11 is taken as don't care (x) in all K-maps. The flip-flops are cleared to 00 at power-on (start state). Checking the final equations, from 11 the machine goes to a valid state on the next clock (X=0 goes to B, X=1 goes to A), so it is self-starting.

Excitation table of J-K flip-flop

Transition and excitation table

K-maps and simplified equations

J1          X=0  X=1
Q1Q0=00       0    0
Q1Q0=01       0    1
Q1Q0=11       x    x
Q1Q0=10       x    x

K1          X=0  X=1
Q1Q0=00       x    x
Q1Q0=01       x    x
Q1Q0=11       x    x
Q1Q0=10       1    1

J0          X=0  X=1
Q1Q0=00       1    0
Q1Q0=01       x    x
Q1Q0=11       x    x
Q1Q0=10       1    0

K0          X=0  X=1
Q1Q0=00       x    x
Q1Q0=01       0    1
Q1Q0=11       x    x
Q1Q0=10       x    x

Z           X=0  X=1
Q1Q0=00       0    0
Q1Q0=01       0    0
Q1Q0=11       x    x
Q1Q0=10       1    0

$$\begin{aligned} J_1 &= Q_0 X \\ K_1 &= 1 \\ J_0 &= X' \\ K_0 &= X \\ Z &= Q_1 X' \end{aligned}$$

Logic diagram

           +-------------+     +----------+
 X  ------>| next-state  |---->| FF1, FF0 |--+--> Q1, Q0
    +----->| logic       |     |  (CLK)   |  |
    |      +-------------+     +----------+  |
    +-------------------<--------------------+
    |      +-------------+
    +----->| output      |----> Z
 X  ------>| logic       |
           +-------------+

Gates needed (two J-K flip-flops with a common clock; complements are taken from the $Q'$ outputs or NOT gates):

• $J_1$: a 2-input AND for $Q_0 X$.
• $K_1$: tied to logic 1.
• $J_0$: direct line $X'$.
• $K_0$: direct line $X$.
• $Z$: a 2-input AND for $Q_1 X'$.

Check: input X = 1010101 gives Z = 0001010; Z = 1 at bits 4 and 6 (overlapping 01010). After 010 the tail "0" is kept, so C goes to B on 0.