--- title: "Sequence detector 011 using RS flip-flops" subject: "Digital Logic" chapter: 9 key: "semester_3/digital_logic:detector-011-rs" source: https://www.bctengineer.com/q/semester_3/digital_logic/detector-011-rs site: BCT Engineering --- # Sequence detector 011 using RS flip-flops Digital Logic · Chapter 9: Sequential Machines - Asked in: 2075 Chaitra, 11 marks - Video: [Sequence Detector (Example)](https://www.youtube.com/watch?v=HXG_YPVNIsM) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/detector-011-rs - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-9 ## Question Design a synchronous sequential machine such that it gives output Z=1 if input contains the sequence of message 011 and it retains in its own state in other condition giving output zero. Use RS-Flip-Flop. ## Short answer Use a Mealy machine with states A(00), B(01), C(10). Z=1 only when 011 completed (C with input 1). S1=Q0 X, R1=Q0', S0=X', R0=X, Z=Q1 X. Two S-R flip-flops. ## Answer The machine stays in its own state when the input does not advance the sequence (A on 1, B on 0), and gives Z = 0 except when 011 is completed. 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: **S-R**. States (what has been received so far): - **A**: reset, or last bit 1 not part of a sequence (stays in A on 1) - **B**: "0" received (stays in B on further 0s) - **C**: "01" received #### State diagram (Mealy) ```text 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 $Q_1 Q_0$). 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). #### 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 1 | 0 | x | 1 | 0 | 0 | | 0 0 | 1 | 0 0 | 0 | x | 0 | x | 0 | | 0 1 | 0 | 0 1 | 0 | x | x | 0 | 0 | | 0 1 | 1 | 1 0 | 1 | 0 | 0 | 1 | 0 | | 1 0 | 0 | 0 1 | 0 | 1 | 1 | 0 | 0 | | 1 0 | 1 | 0 0 | 0 | 1 | 0 | x | 1 | #### K-maps and simplified equations ```text S1 X=0 X=1 Q1Q0=00 0 0 Q1Q0=01 0 1 Q1Q0=11 x x Q1Q0=10 0 0 R1 X=0 X=1 Q1Q0=00 x x Q1Q0=01 x 0 Q1Q0=11 x x Q1Q0=10 1 1 S0 X=0 X=1 Q1Q0=00 1 0 Q1Q0=01 x 0 Q1Q0=11 x x Q1Q0=10 1 0 R0 X=0 X=1 Q1Q0=00 0 x Q1Q0=01 0 1 Q1Q0=11 x x Q1Q0=10 0 x Z X=0 X=1 Q1Q0=00 0 0 Q1Q0=01 0 0 Q1Q0=11 x x Q1Q0=10 0 1 ``` $$ \begin{aligned} S_1 &= Q_0 X \\ R_1 &= Q_0' \\ S_0 &= X' \\ R_0 &= X \\ Z &= Q_1 X \end{aligned} $$ #### Logic diagram ```text +-------------+ +----------+ 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 $Q'$ outputs or NOT gates): - $S_1$: a 2-input AND for $Q_0 X$. - $R_1$: direct line $Q_0'$. - $S_0$: direct line $X'$. - $R_0$: direct line $X$. - $Z$: a 2-input AND for $Q_1 X$. **Check:** input X = 1011011 gives Z = 0001001; Z = 1 at bits 4 and 7. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which transitions keep the machine in its own state without output? A. A on 0 and B on 1 B. A on 1 and B on 0 C. C on 1 and A on 0 D. All states on 1 Answer: B. Extra 1s while no 0 has been seen keep A, and extra 0s keep B (the last 0 may start 011). (Key point: Staying in own state) 2. In state C ("01") with X = 1, what happens? A. Z = 1 and next state A B. Z = 0 and next state A C. Z = 1 and next state B D. Z = 1 and next state C Answer: A. 01 plus 1 completes 011, so Z = 1. The last 1 gives no useful tail, so the machine returns to A. (Key point: Detection) 3. How is the unused code 11 treated? A. As state C B. As don't care (x) in the K-maps, with flip-flops cleared to 00 at start C. As a second reset state with Z = 1 D. It is removed by using three flip-flops Answer: B. Three states use codes 00, 01, 10. The code 11 never occurs, so its K-map entries are don't cares. (Key point: Unused code) 4. For a 1 to 0 transition, which S and R are used? A. S = 1, R = 0 B. S = x, R = 0 C. S = 0, R = 1 D. S = 0, R = x Answer: C. To reset an S-R flip-flop S = 0 and R = 1. 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