5.06 Chapter 5 Master PYQ Solutions & Synthesis Puzzles

Dedicated problem-solving supplement for Chapter 5 (Sequential Logic & Flip-Flops). Contains step-by-step worked derivations for advanced 3-variable state equations, minterm state synthesis, and state diagram conversion puzzles asked in past university exams.

Related Concepts: 5.04 Synchronous Sequential Circuit Analysis & Synthesis | 5.01 Sequential Logic Fundamentals, Latches & Flip-Flops (SR & JK) | 5.05 State Reduction Algorithms & Sequence Analysis

1. 3-Variable JK State Equation Synthesis [PYQ: 2024, 2025]

PYQ Master Problem (2024, 2025): 3-Flip-Flop JK Circuit Synthesis

Question: Design a sequential circuit described by the following state equations using JK flip-flops [12 Marks]:

A(t+1) = x A B + y A’ C + x y

B(t+1) = x A C + y B’ C’

C(t+1) = x’ B + y A’ B’

1.1 Derivation Methodology

For any JK flip-flop, the characteristic equation is: Q(t+1) = J Q’ + K’ Q

To find J and K for a given target expression Q(t+1), expand the target expression into two terms: one multiplied by Q’ (which gives J) and one multiplied by Q (which gives K’).

1. Derive Inputs JA and KA for Flip-Flop A:

Target: A(t+1) = x A B + y A’ C + x y = A’ (y C + x y) + A (x B + x y)

• 1. Coefficient of A’ is JA: JA = y C + x y = y (C + x)
  2. Coefficient of A is KA’: KA’ = x B + x y = x (B + y) ⇒ KA = (x (B + y))’ = x’ + B’ y’

2. Derive Inputs JB and KB for Flip-Flop B:

Target: B(t+1) = x A C + y B’ C’ = B’ (y C’) + B (x A C)

• 1. Coefficient of B’ is JB: JB = y C’
  2. Coefficient of B is KB’: KB’ = x A C ⇒ KB = (x A C)’ = x’ + A’ + C’

3. Derive Inputs JC and KC for Flip-Flop C:

Target: C(t+1) = x’ B + y A’ B’ = C’ (x’ B + y A’ B’) + C (0)

• 1. Coefficient of C’ is JC: JC = x’ B + y A’ B’
  2. Coefficient of C is KC’: KC’ = 0 ⇒ KC = 1

2. Minterm State Equation Synthesis [PYQ: 2021, 2023]

PYQ Master Problem (2021, 2023): Minterm State Equations

Question: Design a sequential circuit whose state equations are given below [12 Marks]:

A1(t+1) = Sigma m(4, 6)

A2(t+1) = Sigma m(1, 2, 5, 6)

y(A1, A2, x) = Sigma m(3, 7)

Where inputs are A1, A2, x (3-variable domain: A1 is MSB, x is LSB).

2.1 State Minimization via K-Maps

1. For A1(t+1) = Sigma m(4, 6): Minterms m4 (100) and m6 (110): A1(t+1) = A1 . x’
2. For A2(t+1) = Sigma m(1, 2, 5, 6): Minterms m1 (001), m2 (010), m5 (101), m6 (110): Grouping m1, m5 gives A2’ x. Grouping m2, m6 gives A2 x’. A2(t+1) = A2’ x + A2 x’ = A2 XOR x
3. For Output y(A1, A2, x) = Sigma m(3, 7): Minterms m3 (011) and m7 (111): y = A2 . x

2.2 D Flip-Flop Drive Expressions

Since Q(t+1) = D:

D1 = A1 . x’

D2 = A2 XOR x

y = A2 . x