--- title: "Machine from 3-state diagram using SR" subject: "Digital Logic" chapter: 9 key: "semester_3/digital_logic:state-diagram-xyz-sr" source: https://www.bctengineer.com/q/semester_3/digital_logic/state-diagram-xyz-sr site: BCT Engineering --- # Machine from 3-state diagram using SR Digital Logic · Chapter 9: Sequential Machines - Asked in: 2075 Asoj, 10 marks - Video: [Design of Clocked Sequential Circuit | for given State Diagram | using JK Flip flops](https://www.youtube.com/watch?v=zxThLeS2sk8) — Learn with Dr. Shobha Nikam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/state-diagram-xyz-sr - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-9 ## Question Design a synchronous sequential machine from the state diagram given below. Use S-R Flip-Flop. [Figure: state diagram with three states X, Y, Z, edges labelled input/output: X→X on 0/0; X→Y on 1/0; Y→X on 0/0; Y→Z on 1/0; Z→Z on 1/1; Z→X on 0/0] ## Short answer Assign states X=00, Y=01, Z=10. Output F=1 only in Z with W=1 (three consecutive 1s). S1=Q0 W, R1=W', S0=Q1' Q0' W, R0=Q0, F=Q1 W. Two S-R flip-flops. ## Answer 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) ```text 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 | Present state | Next, W=0 | Next, W=1 | F (W=0) | F (W=1) | |---|---|---|---|---| | X | X | Y | 0 | 0 | | Y | X | Z | 0 | 0 | | Z | X | Z | 0 | 1 | #### 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 | 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 | W | Q1+ Q0+ | S1 | R1 | S0 | R0 | F | |---|---|---|---|---|---|---|---| | 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 | 0 | | 1 0 | 1 | 1 0 | x | 0 | 0 | x | 1 | #### K-maps and simplified equations ```text 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 ```text +-------------+ +----------+ 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does this state diagram detect? A. The sequence 011 only B. Three or more consecutive 1s (output 1 on each 1 after the second) C. Exactly two 1s D. The sequence 101 Answer: B. Z loops on itself with W = 1 and gives F = 1, so every 1 after two consecutive 1s outputs 1. (Key point: What the machine does) 2. Why is F = Q1 W? A. F is 1 in state Y (01) with W = 0 B. F is 1 in state X with W = 1 C. F is 1 only in state Z (10) with W = 1, and 11 is a don't care D. F is 1 for every W = 1 Answer: C. Only the arc Z to Z on 1/1 has an output of 1, and Q1 = 1 only in state Z. (Key point: Output equation) 3. For a 1 to 1 transition in an S-R flip-flop, what are S and R? A. S = 0, R = 1 B. S = 1, R = 0 C. S = 1, R = 1 D. S = x, R = 0 Answer: D. To hold a 1, R must be 0 and S may be anything (0 or 1), so S = x. (Key point: S-R excitation) 4. What happens from the unused code 11? A. The machine locks up B. It goes to a valid state on the next clock (X on W=0, Z on W=1), so it is self-starting C. It goes to state Y D. F stays 1 forever Answer: B. Checking the final equations shows 11 moves to X when W = 0 and to Z when W = 1. 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