- 2080 Bhadra · 10 marks
Sequence detector 011 using T flip-flops
Short answer
Three states A(00), B(01), C(10) detect overlapping 011. T flip-flop inputs: T1 = Q1 + Q0 X, T0 = Q0 X + Q0' X', output Z = Q1 X.
A asynchronous state machine has one bit serial input. The Y of the machine is set to be high when the input contains the message 011. Draw the state diagram for this machine and design the circuit. Use T Flip-Flop.
Since flip-flops and a clock are used, the machine is designed as a clocked (synchronous) Mealy machine; the 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 011 arrives. Flip-flops: T.
States (what has been received so far):
- A: reset / no useful bits (a 1 that does not follow "0", or a completed 011)
- B: "0" received
- C: "01" received
State diagram (Mealy)
Arc label = input/output
A --0/0--> B --1/0--> C --1/1--> A
A --1/0--> A (self loop)
B --0/0--> B (self loop)
C --0/0--> B
State table
| Present state | Next, X=0 | Next, X=1 | Z (X=0) | Z (X=1) |
|---|---|---|---|---|
| A | B | A | 0 | 0 |
| B | B | C | 0 | 0 |
| C | B | A | 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 B, X=1 goes to A), 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 1 | 0 | 1 | 0 |
| 0 0 | 1 | 0 0 | 0 | 0 | 0 |
| 0 1 | 0 | 0 1 | 0 | 0 | 0 |
| 0 1 | 1 | 1 0 | 1 | 1 | 0 |
| 1 0 | 0 | 0 1 | 1 | 1 | 0 |
| 1 0 | 1 | 0 0 | 1 | 0 | 1 |
K-maps and simplified equations
T1 X=0 X=1
Q1Q0=00 0 0
Q1Q0=01 0 1
Q1Q0=11 x x
Q1Q0=10 1 1
T0 X=0 X=1
Q1Q0=00 1 0
Q1Q0=01 0 1
Q1Q0=11 x x
Q1Q0=10 1 0
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 = 0011011 gives Z = 0001001; Z = 1 at bits 4 and 7.
✓ Answer checked by Claude Opus 5.5 (high effort)
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