Karnaugh -Map , reduction of Boolean Expressions
- 1. Minimization of SOP and POS Expressions using Karnaugh map Prepared by- Dr. Sutapa Mukherjee
- 2. Different variable K Map 0 1 0 A B A B 1 A B A B A B 00 01 11 10 0 A B C A B C A B C A B C 1 A B C A B C A B C A B C A BC Two Variable Map Three Variable Map
- 3. Four variable K Map 00 01 11 10 00 A B C D A B C D A B C D A B C D 01 A B C D A B C D A B C D A B C D 11 A B C D A B C D A B C D A B C D 10 A B C D A B C D A B C D A B C D AB CD 00 01 11 10 00 A+B+C+D A+B+C+D A+B+C+D A+B+C+D 01 A+B+C+D A+B+C+D A+B+C+D A+B+C+D 11 A+B+C+D A+B+C+D A+B+C+D A+B+C+D 10 A+B+C+D A+B+C+D A+B+C+D A+B+C+D CD AB POS Form SOP Form
- 4. Min term distributions in K- map CD 00 01 11 10 00 m0 m1 m3 m2 01 m4 m5 m7 m6 11 m12 m13 m15 m14 10 m8 m9 m11 m10 AB
- 5. Reduce the Expression f=∑m(1,5,6,12,13,14)+d(2,4) CD 00 01 11 10 00 m0 1 m1 m3 X m2 01 X m4 1 m5 m7 1 m6 11 1 m12 1 m13 m15 1 m14 10 m8 m9 m11 m10 AB 2 3 1 fmin=BC+A CD+BD
- 6. Problem • Reduce the expression f=∑m(1,5,6,12,13,14)+d(2,4) • The given expression in the POS form is f=∏M(0,3,7,8,9,10,11,15)∏d(2,4) 00 01 11 10 00 0 M0 M1 0 M3 X M2 01 X M4 M5 0 M7 M6 11 M12 M13 0 M15 M14 10 0 M8 0 M9 0 M11 0 M10 AB CD F=(A+B)(C+D)(B+D) 2 3 1