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Sequence detector for 110

Short answer

Mealy machine with states A (reset), B (1 received), C (11 received). D flip-flops, assignment A=00, B=01, C=10. Equations: D1 = Q0 X + Q1 X, D0 = Q1' Q0' X. Output Z = Q1 X'.

Design a sequential machine that has one serial input and one output z. The machine is required to give an output z = 1 when the input X contains the message 110.

No flip-flop type is specified, so D flip-flops are used (simplest: D = next state).

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: D.

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 D flip-flop

QQ+D
000
011
100
111

Transition and excitation table

Q1 Q0XQ1+ Q0+D1D0Z
0 000 0000
0 010 1010
0 100 0000
0 111 0100
1 000 0001
1 011 0100

K-maps and simplified equations

D1          X=0  X=1
Q1Q0=00       0    0
Q1Q0=01       0    1
Q1Q0=11       x    x
Q1Q0=10       0    1

D0          X=0  X=1
Q1Q0=00       0    1
Q1Q0=01       0    0
Q1Q0=11       x    x
Q1Q0=10       0    0

Z           X=0  X=1
Q1Q0=00       0    0
Q1Q0=01       0    0
Q1Q0=11       x    x
Q1Q0=10       1    0
D1=Q0X+Q1XD0=Q1′Q0′XZ=Q1X′\begin{aligned} D_1 &= Q_0 X + Q_1 X \\ D_0 &= Q_1' Q_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 D flip-flops with a common clock; complements are taken from the Q′Q' outputs or NOT gates):

  • D1D_1: a 2-input AND for Q0XQ_0 X; a 2-input AND for Q1XQ_1 X, combined in a 2-input OR gate.
  • D0D_0: a 3-input AND for Q1′Q0′XQ_1' Q_0' X.
  • ZZ: a 2-input AND for Q1X′Q_1 X'.

Check: input X = 0110110 gives Z = 0001001; Z = 1 at bits 4 and 7.

✓ Answer checked by Claude Opus 5.5 (high effort)

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