Solve the equation 2P-1 < N 2P, for N = 4. It is seen by inspection that P = 2 is the solution of this equation, as 2 < 4 4, or 21 < 4 22. We need two flip-flops. Step 3 Assign a Unique P-Bit Binary Number to Each State – As P = 2, we are to assign a unique 2-bit number to each state. The assignments specified in the problem statement are A = 00, B = 01, C = 10, and D = 11. The flip-flops will be labeled Y1 and Y0, so the state is labeled as Y1Y0. Step 4 Generate the Transistion Table with Output – PS NS / Z PS NS / Z X = 0 X = 1 X = 0 X = 1 A 00 00 / 0 01 / 0 A 00 00 / 0 01 / 0 B 01 00 / 0 10 / 0 B 01 00 / 0 10 / 0 C 10 11 / 0 10 / 0 C 10 11 / 0 10 / 0 D 11 00 / 0 00 / 1 D 11 00 / 0 01 / 1 Overlap Not Allowed With Overlap Step 4a Generate the Output Table and Output Equation – This is the same for both the overlap and non-overlap case. PS Z Y1Y0 X = 0 X = 1 The only output occurs when X = 1, Y1 = 1, and Y0 = 1. 00 0 0 01 0 0 10 0 0 The output equation is thus Z = X Y1 Y0, a product term. 11 0 1 Step 5 Separate the Transition Table into Two Tables, One for Each Flip-Flop – Step 6 Decide on the Type of Flip-Flops to Use – The problem specifies that JK be used. Q(T) Q(T + 1) J K 0 0 0 d The table at right is the excitation table for a JK 0 1 1 d flip-flop. We shall need to use the excitation table in 1 0 d 1 order to deduce the inputs to each flip-flop that will 1 1 d 0 make the circuit behave in the desired manner. Step 7 Derive an Input Table for Each Flip-Flop – Here we shall do first the no-overlap case and then the overlap case. Without Overlap
Solve the equation 2P-1 < N 2P, for N = 4. It is seen by inspection that P = 2 is the solution of this equation, as 2 < 4 4, or 21 < 4 22. We need two flip-flops. Step 3 – Assign a Unique P-Bit Binary Number to Each State As P = 2, we are to assign a unique 2-bit number to each state. The assignments specified in the problem statement are A = 00, B = 01, C = 10, and D = 11. The flip-flops will be labeled Y1 and Y0, so the state is labeled as Y1Y0. Step 4 – Generate the Transistion Table with Output PS NS / Z X = 0 X = 1 A 00 00 / 0 01 / 0 B 01 00 / 0 10 / 0 C 10 11 / 0 10 / 0 D 11 00 / 0 01 / 1 PS NS / Z X = 0 X = 1 A 00 00 / 0 01 / 0 B 01 00 / 0 10 / 0 C 10 11 / 0 10 / 0 D 11 00 / 0 00 / 1 Overlap Not Allowed With Overlap Step 4a – Generate the Output Table and Output Equation This is the same for both the overlap and non-overlap case. PS Z Y1Y0 X = 0 X = 1 00 0 0 01 0 0 10 0 0 11 0 1 The only output occurs when X = 1, Y1 = 1, and Y0 = 1. The output equation is thus Z = X Y1 Y0, a product term. Step 5 – Separate the Transition Table into Two Tables, One for Each Flip-Flop Step 6 – Decide on the Type of Flip-Flops to Use The problem specifies that JK be used. Q(T) Q(T + 1) J K 0 0 0 d 0 1 1 d 1 0 d 1 1 1 d 0 The table at right is the excitation table for a JK flip-flop. We shall need to use the excitation table in order to deduce the inputs to each flip-flop that will make the circuit behave in the desired manner. Step 7 – Derive an Input Table for Each Flip-Flop Here we shall do first the no-overlap case and then the overlap case. Without Overlap
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