Microsoft Word - Digital Logic Design v_4_6a

Step 2) Determine the number of flip-flop based on the number of states
(#states = 3) ≤ 2 (#flip-flop = 2) assuming full encoding.

Step 3) Assign Unique code to each state
a: 00, b:01; C:11

Step 4) Write the excitation-input equations
The JK flip-flop excitation equation is JK  Y+ = J.Y’ + K’.Y
You may derive the general excitation equation from the characteristic table for the JK
flip-flop to obtain the excitation table for the JK flip-flop, as shown below:

Write the PS/NS table for JK, flip-flops (this intermediate step is helpful)

Draw the Composite K-map for each of the desired outputs Y1+ ,Y2 +, Z:

00

01

11

10

Y1Y2 J1 K1 J2 K2 Z

0 1 1

1 0 1

0 1 0

- - -

0 1

0 1

1 0

- -

J 1 = Y1’.Y2

K 1 = Y1 + Y2’

J 2 = Y1’

K 2 = Y1

Z = Y1’

Note: “-“ means don’t care

Y 1 Y 2 Y 1 + Y 2 + J 1 K 1 J 2 K 2 Z

0 0 0 1 0 1 1 0 0

0 1 1 1 1 0 1 0 1

1 1 0 0 0 1 0 1 1

1 0 - - - - - - -

Unused State

J K Y Y +^

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 0

Characteristic

table Output Excitation

table

Y Y+ J K^

0 0 0 -

0 1 1 -

1 0 - 1

1 1 - 0

Input Excitation

table

Note: “-“ = don’t care

J = Y +^

K = Y +’

Input-Excitation Eq.

J K Y +^

0 0 Y

0 1 0

1 0 1

1 1 Y’