--- title: "Machine from 4-state diagram using SR" subject: "Digital Logic" chapter: 9 key: "semester_3/digital_logic:state-diagram-four-state-sr" source: https://www.bctengineer.com/q/semester_3/digital_logic/state-diagram-four-state-sr site: BCT Engineering --- # Machine from 4-state diagram using SR Digital Logic · Chapter 9: Sequential Machines - Asked in: 2081 Baisakh, 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-four-state-sr - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-9 ## Question Design a synchronous sequential machine from the given state diagram. Use SR flip-flops for realization. [Figure: state diagram with four states 00, 01, 10, 11, edges labelled input/output: 00→00 on 1/1; 00→01 on 0/0; 01→01 on 0/1; 01→10 on 1/0; 10→10 on 1/0; 10→11 on 0/0; 11→11 on 0/1; 11→00 on 1/0] ## Short answer States are directly coded 00, 01, 10, 11. SR flip-flop equations: S1 = Q1' Q0 X, R1 = Q1 Q0 X, S0 = X', R0 = X; output Z = Q0 X' + Q1' Q0' X. ## Answer The diagram is a Mealy machine with 4 states already coded as $Q_1 Q_0$ = 00, 01, 10, 11, so no extra state assignment is needed. Input X, output Z, two **S-R** flip-flops ($Q_1$ = MSB). Each state stays in itself on one input value and moves to the next state on the other. #### State diagram (given) ```text Arc label = input/output 00 --0/0--> 01 --1/0--> 10 --0/0--> 11 --1/0--> 00 00 --1/1--> 00 (self loop) 01 --0/1--> 01 (self loop) 10 --1/0--> 10 (self loop) 11 --0/1--> 11 (self loop) ``` #### State table | Present state | Next, X=0 | Next, X=1 | Z (X=0) | Z (X=1) | |---|---|---|---|---| | 00 | 01 | 00 | 0 | 1 | | 01 | 01 | 10 | 1 | 0 | | 10 | 11 | 10 | 0 | 0 | | 11 | 11 | 00 | 1 | 0 | #### 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 | 1 | | 0 1 | 0 | 0 1 | 0 | x | x | 0 | 1 | | 0 1 | 1 | 1 0 | 1 | 0 | 0 | 1 | 0 | | 1 0 | 0 | 1 1 | x | 0 | 1 | 0 | 0 | | 1 0 | 1 | 1 0 | x | 0 | 0 | x | 0 | | 1 1 | 0 | 1 1 | x | 0 | x | 0 | 1 | | 1 1 | 1 | 0 0 | 0 | 1 | 0 | 1 | 0 | #### K-maps and simplified equations ```text S1 X=0 X=1 Q1Q0=00 0 0 Q1Q0=01 0 1 Q1Q0=11 x 0 Q1Q0=10 x x R1 X=0 X=1 Q1Q0=00 x x Q1Q0=01 x 0 Q1Q0=11 0 1 Q1Q0=10 0 0 S0 X=0 X=1 Q1Q0=00 1 0 Q1Q0=01 x 0 Q1Q0=11 x 0 Q1Q0=10 1 0 R0 X=0 X=1 Q1Q0=00 0 x Q1Q0=01 0 1 Q1Q0=11 0 1 Q1Q0=10 0 x Z X=0 X=1 Q1Q0=00 0 1 Q1Q0=01 1 0 Q1Q0=11 1 0 Q1Q0=10 0 0 ``` $$ \begin{aligned} S_1 &= Q_1' Q_0 X \\ R_1 &= Q_1 Q_0 X \\ S_0 &= X' \\ R_0 &= X \\ Z &= Q_0 X' + Q_1' Q_0' X \end{aligned} $$ #### 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 3-input AND for $Q_1' Q_0 X$. - $R_1$: a 3-input AND for $Q_1 Q_0 X$. - $S_0$: direct line $X'$. - $R_0$: direct line $X$. - $Z$: a 2-input AND for $Q_0 X'$; a 3-input AND for $Q_1' Q_0' X$, combined in a 2-input OR gate. **Check:** from 00, X = 10101 gives states 00 → 00 → 01 → 10 → 11 → 00 and Z = 10000, matching the diagram. $S_1 R_1$ and $S_0 R_0$ are never both 1. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is no separate state assignment step needed? A. The diagram's states are already coded 00, 01, 10, 11 B. S-R flip-flops need none C. Only one state is used D. The output is the state Answer: A. The given diagram labels the states with their codes Q1Q0. (Key point: State assignment) 2. In state 11 with X=1 (next state 00), what are S1, R1? A. S1 = 1, R1 = 0 B. S1 = 0, R1 = 1 C. S1 = x, R1 = 0 D. S1 = 0, R1 = 0 Answer: B. Both flip-flops go from 1 to 0, so S = 0 and R = 1. (Key point: Flip-flop going 1 to 0) 3. Which output equation was found? A. Z = Q1 X B. Z = Q0 X + Q1 X' C. Z = Q0 X' + Q1' Q0' X D. Z = Q1' Q0 X Answer: C. Z = 1 for states 01 and 11 on X=0 and for 00 on X=1. (Key point: Output) 4. Which S-R input combination must never occur? A. S = 0, R = 0 B. S = 1, R = 0 C. S = x, R = 0 D. S = 1, R = 1 Answer: D. S = R = 1 is invalid for an S-R flip-flop; the answer checks it never happens. 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