- 2079 Bhadra · 10 marks
Sequence detector 101 using T flip-flops
Short answer
A Mealy machine with states A(00), B(01), C(10) detects overlapping '101'. T flip-flop excitation gives T1 = Q0 X' + Q1, T0 = Q0 X' + Q0' X, output Z = Q1 X.
A synchronous machine has 1-bit input 'X'. The output 'Y' goes high when input contains the message '101'. Draw the state diagram, derive the transition table (state table), excitation table and design circuit. Use only T flip-flops.
Output Y of the question is written Z (Y = Z).
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 101 arrives. Flip-flops: T.
States (what has been received so far):
- A: reset / no useful bits
- B: "1" received
- C: "10" received
State diagram (Mealy)
Arc label = input/output
A --1/0--> B --0/0--> C --1/1--> B
A --0/0--> A (self loop)
B --1/0--> B (self loop)
C --0/0--> A
State table
| Present state | Next, X=0 | Next, X=1 | Z (X=0) | Z (X=1) |
|---|---|---|---|---|
| A | A | B | 0 | 0 |
| B | C | B | 0 | 0 |
| C | A | B | 0 | 1 |
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 B), so it is self-starting.
Excitation table of T flip-flop
| Q | Q+ | T |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Transition and excitation table
| Q1 Q0 | X | Q1+ Q0+ | T1 | T0 | Z |
|---|---|---|---|---|---|
| 0 0 | 0 | 0 0 | 0 | 0 | 0 |
| 0 0 | 1 | 0 1 | 0 | 1 | 0 |
| 0 1 | 0 | 1 0 | 1 | 1 | 0 |
| 0 1 | 1 | 0 1 | 0 | 0 | 0 |
| 1 0 | 0 | 0 0 | 1 | 0 | 0 |
| 1 0 | 1 | 0 1 | 1 | 1 | 1 |
K-maps and simplified equations
T1 X=0 X=1
Q1Q0=00 0 0
Q1Q0=01 1 0
Q1Q0=11 x x
Q1Q0=10 1 1
T0 X=0 X=1
Q1Q0=00 0 1
Q1Q0=01 1 0
Q1Q0=11 x x
Q1Q0=10 0 1
Z X=0 X=1
Q1Q0=00 0 0
Q1Q0=01 0 0
Q1Q0=11 x x
Q1Q0=10 0 1
Logic diagram
+-------------+ +----------+
X ------>| next-state |---->| FF1, FF0 |--+--> Q1, Q0
+----->| logic | | (CLK) | |
| +-------------+ +----------+ |
+-------------------<--------------------+
| +-------------+
+----->| output |----> Z
X ------>| logic |
+-------------+
Gates needed (two T flip-flops with a common clock; complements are taken from the outputs or NOT gates):
- : a 2-input AND for ; direct line , combined in a 2-input OR gate.
- : a 2-input AND for ; a 2-input AND for , combined in a 2-input OR gate.
- : a 2-input AND for .
Check: input X = 0101011 gives Z = 0001010; Z = 1 at bits 4 and 6 ("10101" contains two overlapping 101s) and 0 at bit 7.
✓ Answer checked by Claude Opus 5.5 (high effort)
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