- 2068 Baisakh · 12 marks
Sequence detector 110 using SR flip-flops
Short answer
Mealy machine, input MSB first: states A (reset), B (1 received), C (11 received). SR flip-flops, assignment A=00, B=01, C=10. Equations: S1 = Q0 X, R1 = X', S0 = Q1' Q0' X, R0 = Q0. Output Z = Q1 X'.
Design a synchronous state machine with the following specification: a) No. of input: 1 b) No. of output: 1 c) The output of the machine is to be set high when the data in the input is 110 in sequence, starting from the MSB (Use SR flip-flop).
"Starting from the MSB" means the bits arrive serially in the order 1, 1, 0 (MSB first).
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: S-R.
States (what has been received so far):
- A: reset / last bit 0
- B: "1" (first bit, MSB) received
- C: "11" received
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 state | Next, X=0 | Next, X=1 | Z (X=0) | Z (X=1) |
|---|---|---|---|---|
| A | A | B | 0 | 0 |
| B | A | C | 0 | 0 |
| C | A | C | 1 | 0 |
State assignment
A = 00, B = 01, C = 10 (flip-flop outputs ). 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 S-R flip-flop
| Q | Q+ | S | R |
|---|---|---|---|
| 0 | 0 | 0 | x |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | x | 0 |
Transition and excitation table
| Q1 Q0 | X | Q1+ Q0+ | S1 | R1 | S0 | R0 | Z |
|---|---|---|---|---|---|---|---|
| 0 0 | 0 | 0 0 | 0 | x | 0 | x | 0 |
| 0 0 | 1 | 0 1 | 0 | x | 1 | 0 | 0 |
| 0 1 | 0 | 0 0 | 0 | x | 0 | 1 | 0 |
| 0 1 | 1 | 1 0 | 1 | 0 | 0 | 1 | 0 |
| 1 0 | 0 | 0 0 | 0 | 1 | 0 | x | 1 |
| 1 0 | 1 | 1 0 | x | 0 | 0 | x | 0 |
K-maps and simplified equations
S1 X=0 X=1
Q1Q0=00 0 0
Q1Q0=01 0 1
Q1Q0=11 x x
Q1Q0=10 0 x
R1 X=0 X=1
Q1Q0=00 x x
Q1Q0=01 x 0
Q1Q0=11 x x
Q1Q0=10 1 0
S0 X=0 X=1
Q1Q0=00 0 1
Q1Q0=01 0 0
Q1Q0=11 x x
Q1Q0=10 0 0
R0 X=0 X=1
Q1Q0=00 x 0
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
Logic diagram
+-------------+ +----------+
X ------>| next-state |---->| FF1, FF0 |--+--> Q1, Q0
+----->| logic | | (CLK) | |
| +-------------+ +----------+ |
+-------------------<--------------------+
| +-------------+
+----->| output |----> Z
X ------>| logic |
+-------------+
Gates needed (two S-R flip-flops with a common clock; complements are taken from the outputs or NOT gates):
- : a 2-input AND for .
- : direct line .
- : a 3-input AND for .
- : direct line .
- : a 2-input AND for .
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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