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Mealy detector for 110 using JK

Short answer

Use three states A(00), B(01), C(10) for a Mealy machine detecting overlapping 110. With JK flip-flops: J1 = Q0 X, K1 = X', J0 = Q1' X, K0 = 1, output Z = Q1 X'.

Using Mealy circuit with JK flip-flops design a synchronous sequence detector that produces output Z = 1 when it detects the serial input X = 110.

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 110 arrives. Flip-flops: J-K.

States (what has been received so far):

  • A: reset / last bit 0
  • B: "1" received
  • C: "11" received (stays in C on more 1s)

State diagram (Mealy)

Arc label = input/output

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

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

State table

Present stateNext, X=0Next, X=1Z (X=0)Z (X=1)
AAB00
BAC00
CAC10

State assignment

A = 00, B = 01, C = 10 (flip-flop outputs Q1Q0Q_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 A, X=1 goes to C), so it is self-starting.

Excitation table of J-K flip-flop

QQ+JK
000x
011x
10x1
11x0

Transition and excitation table

Q1 Q0XQ1+ Q0+J1K1J0K0Z
0 000 00x0x0
0 010 10x1x0
0 100 00xx10
0 111 01xx10
1 000 0x10x1
1 011 0x00x0

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    0

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

K0          X=0  X=1
Q1Q0=00       x    x
Q1Q0=01       1    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
J1=Q0XK1=X′J0=Q1′XK0=1Z=Q1X′\begin{aligned} J_1 &= Q_0 X \\ K_1 &= X' \\ J_0 &= Q_1' X \\ K_0 &= 1 \\ 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′Q' outputs or NOT gates):

  • J1J_1: a 2-input AND for Q0XQ_0 X.
  • K1K_1: direct line X′X'.
  • J0J_0: a 2-input AND for Q1′XQ_1' X.
  • K0K_0: tied to logic 1.
  • ZZ: a 2-input AND for Q1X′Q_1 X'.

Check: input X = 1110110 gives Z = 0001001; Z = 1 at bits 4 and 7.

✓ Answer checked by Claude Opus 5.5 (high effort)

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