1. Karnaugh Map (K-Map) By Dr. M. Khamis Mrs. Dua’a Al Sinari

2. C ONTINUE LOGIC GATES Logic gate: NAND (the complement of the product) NOR (the complement of the sum) XOR (Exclusive OR) F=(xy)’, or F=(x  y)’ F=(x + y)’ F=xy’+ x’y Operation: Expression: June 10th, 2008 1 Number Systems and Boolean Algebra Truth Table:

3. G ATE -L EVEL M INIMIZATION The complexity of the digital logic gates is related to the complexity of the algebraic expression. The map method provides a simple, straightforward procedure for minimizing Boolean functions. The map method is also known as Karnaugh map or k-map. Two-variable Map: June 10th, 2008 2 Number Systems and Boolean Algebra m 1 X’ y m 0 X’ y’ m 3 X y m 2 X y’ m1m1 m0m0 m3m3 m2m2 y X 0 1 01 y X m 1 +m 2 +m 3 =x’y + xy’ +xy = x+y

4. TWO-VARIABLE MAP: m11m11 m0m0 m31m31 m21m21 y X 0 1 01 y X m1m1 m0m0 m31m31 m2m2 y X 0 1 01 y X (a) xy (b) x+y Three-VARIABLE MAP: m 2 x’ yz’ m 3 x’ yz m 1 x’ y’z m 0 x’ y’z’ m 6 xyz’ m 7 x yz m 5 x y’z m 4 x y’z’ yz X 0 1 0001 y X 11 10 z m 5 + m 7 = xy’z+ xyz = xz(y’ + y) =xz

5. EXAMPLE(1): Simplify the Boolean function F(x,y,z) = ∑ (2,3,4,5) m21m21 m31m31 m1m1 m0m0 m6m6 m7m7 m51m51 m41m41 yz X 1 0001 y X 11 10 z 0 xy’ x’y m 3 + m 2 = x’yz + x’yz’=x’y(z+z’)=x’y m 4 + m 5 =.................??...............=xy’ F(x,y,z)=∑(2,3,4,5)= x’y + xy’

6. TRAINING Try this one: (simplify the Boolean function) F(x,y,z) = ∑ (3,4,6,7) Solution: after representing k-map the simplification will be as shown F= yz + xz’ F(x,y,z) = ∑ (0,2,4,5,6) Solution: after representing k-map the simplification will be as shown F= z’ + xy’

7. EXAMPLE(2): Let the Boolean function F= A’C + A’B + AB’C + BC A) Express this function as a sum of minterms. B) Find the minimal sum-of-products expression. Solution: A) F(A,B,C)= ∑(1,2,3,5,7) B) F= C + A’B m21m21 m31m31 m11m11 m0m0 m6m6 m71m71 m51m51 m4m4 BC A 1 0001 B A 11 10 C 0

8. F OUR - VARIABLE MAP : m 2 W’X’YZ’ m 3 W’X’YZ m 1 W’X’Y’Z m 0 W’X’Y’Z’ m 6 W’XYZ’ m 7 W’XYZ m 5 W’XY’Z m 4 W’XY’Z’ m 14 WXYZ’ m 15 WXYZ m 13 WXY’Z m 12 WXY’Z’ m 10 WX’YZ’ m 11 WX’YZ m 9 WX’Y’Z m 8 WX’Y’Z’ YZ WX 01 0001 Y W 11 10 Z 00 11 10 X

9. The combination of adjacent squares that is useful during the simplification process is easily determined from inspection of the four-variable map: One square represents a term with four literals. Two adjacent squares represent a term with three literals. Four adjacent squares represent a term with two literals. Eight adjacent squares represent a term with one literal. Sixteen adjacent squares produce a function that is always equal to 1. F ACTS FOR SIMPLIFICATION :

10. E XAMPLE (3): Simplify the Boolean function F(w,x,y,z)=∑(0,1,2,4,5,6,8,9,12,13,14) w’y’z’ + w’yz’ =w’z’ xy’z’ + xyz’ =xz’ F(w,x,y,z)= y’+ w’z’+ xz’ m21m21 m3m3 m11m11 m01m01 m61m61 m7m7 m51m51 m41m41 m 14 1 m 15 m 13 1 m 12 1 m 10 m 11 m91m91 m81m81 YZ WX 01 0001 Y W 11 10 Z 00 11 10 X w’y’z’ xy’z’ Y’ xyz’ w’yz’

11. E XAMPLE (4): Simplify the Boolean function: F=A’B’C’ +B’CD’ + A’BCD’ + AB’C’ TRY TO REPRESENT IT YOURSELF ON K-MAP Solution: F= B’D’ + B’C’ + A’CD’