Machine from 3-state diagram using SR

The state diagram is a Mealy diagram with 3 states. To avoid confusion with the state names X, Y, Z, the input is called W and the output F. From the diagram, F = 1 only when the machine is in Z and W = 1, i.e. the machine detects three or more consecutive 1s. Flip-flops: S-R.

State diagram (given)

Arc label = input/output

 X --1/0--> Y --1/0--> Z --1/1--> Z

 X --0/0--> X  (self loop)
 Y --0/0--> X
 Z --0/0--> X

State table

State assignment

X = 00, Y = 01, Z = 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 (W=0 goes to X, W=1 goes to Z), so it is self-starting.

Excitation table of S-R flip-flop

Transition and excitation table

K-maps and simplified equations

S1          W=0  W=1
Q1Q0=00       0    0
Q1Q0=01       0    1
Q1Q0=11       x    x
Q1Q0=10       0    x

R1          W=0  W=1
Q1Q0=00       x    x
Q1Q0=01       x    0
Q1Q0=11       x    x
Q1Q0=10       1    0

S0          W=0  W=1
Q1Q0=00       0    1
Q1Q0=01       0    0
Q1Q0=11       x    x
Q1Q0=10       0    0

R0          W=0  W=1
Q1Q0=00       x    0
Q1Q0=01       1    1
Q1Q0=11       x    x
Q1Q0=10       x    x

F           W=0  W=1
Q1Q0=00       0    0
Q1Q0=01       0    0
Q1Q0=11       x    x
Q1Q0=10       0    1

$$\begin{aligned} S_1 &= Q_0 W \\ R_1 &= W' \\ S_0 &= Q_1' Q_0' W \\ R_0 &= Q_0 \\ F &= Q_1 W \end{aligned}$$

Logic diagram

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

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

• $S_1$: a 2-input AND for $Q_0 W$.
• $R_1$: direct line $W'$.
• $S_0$: a 3-input AND for $Q_1' Q_0' W$.
• $R_0$: direct line $Q_0$.
• $F$: a 2-input AND for $Q_1 W$.

Check: input W = 0111101 gives F = 0001100 — F = 1 at the 3rd and 4th consecutive 1, as the diagram requires (Z → Z on 1/1).