Logic Gate Boolean KMap Laws and Rules of

Logic Gate Boolean K-Map

Laws and Rules of Boolean Algebra § กฎของพชคณตบล น กฎการสลบท : AA + B = B + A+B A B AB = BA A B AB B A B+A BA

กฎการจดหม : A + (B + C) = (A + B) + C A B C A+(B+C) B+C A A+B B C (A+B)+C A(BC) = (AB)C A B C A(BC) BC A B C AB (AB)C

Applying De. Morgan’s Theorems Example Apply De. Morgan’s theorems to each of the following expressions: X Y X+Y = XY

Example Apply De. Morgan’s theorems to the expressions

Boolean Analysis of Logic Circuits § นพจนของวงจรลอจก C D B CD B+CD A(B+CD) A X=A(B+CD)

A A=1 B=1 0 0 0 0 1 1 1 1 B 0 0 0 0 1 1 1 1 C 0 0 1 1 D 0 1 0 1 X=A(B+CD) 0 0 0 1 1 1 A = 1 และ B = 1 A = 1 และ CD = 1 C=1 และ D = 1

Simplification using Boolean Algebra Example Simplify the following Boolean expression : AB+A(B+C)+B(B+C)

Example Simplify the following Boolean expression :

Converting of General Expression to SOP การแปลงผนนพจนทวไปใหอยในรป SOP Example Convert the following Boolean expressions to SOP form: (C(A+B)+D)E = (C+AB+D)E = CE+ABE+DE

Example Convert the Boolean expression into standard SOP form : X = ABE+DE = ABE(D+D)+(A+A)(B+B)DE = ABDE+(AB+AB+AB+AB)DE = ABDE+ABDE+ABDE+ABDE

Example Convert the Boolean expression into standard POS form : B(A+C)(A+B+C) X = B(A+C)(A+B+C) = (AA+B+CC)(A+BB+C)(A+B+C) = (A+B+CC)(A+B+C)(A+B+C) = (A+B+C)(A+B+C)(A+B+C)(A+B+C)

Example Develop a truth table for SOP expression AB+ABC X = AB+ABC = AB(C+C)+ABC = ABC+ABC+ABC

Example Determine the truth table for the standard POS expression เปน 0 เมอ พจนผลบวกพจนใดพจนหนงเปน 0 จะได -> A=0, B=0, C=0 X -> A=0, B=1, C=0 -> A=0, B=1, C=1 -> A=1, B=0, C=1 -> A=1, B=1, C=0

Example From the truth table, determine the standard SOP expression and the standard POS expression A 0 0 1 1 B 0 0 1 1 C 0 1 0 1 X 0 0 0 1 1

AB CD 00 01 11 10 ตาราง4 ตวแปร 00 01 11 10

§ เซลชดตดกน AB CD 00 00 01 11 10 (Cell Adjacent) 01 11 10

Example Map the following standard SOP texpression on a Karnaugh map : 0011 0100 1101 1111 1100 CD AB 00 01 11 10 00 0 1 1 0 0 0 11 1 0 10 0 1 0001 1010

Example Map the SOP expression on a Karnaugh map: X 00 X 10 XX 110 X 1010 0001 1000 1101 1001 1100 1011 00 1 1 0 0 0 0 11 1 1 0 0 10 1 1 1 0000 0001 1000 1001 1011 CD AB 00 01 11 10 1

Example Write the minimum SOP expression for the Karnaugh map CD AB 00 01 11 10 00 0 0 1 1 01 1 10 0 1 0 0 B

Example Write the minimum SOP expression for the Karnaugh map CD AB 00 01 11 10 00 0 0 1 0 01 1 0 11 0 1 10 0 1 0 0

Example Use a Karnaugh map to minimize the following SOP expression: 101 011 000 BC 00 01 11 10 0 1 1 1 0 A 1 1 1 0 0 100

Example Map the following standard POS expression on a Karnaugh map: 1100 1011 0011 1111 CD AB 00 01 11 10 00 1 1 0 0 01 1 1 11 0 1 10 1 1 0010

Example Use a Karnaugh map to minimize the following POS expression 000 001 010 BC 00 01 11 10 0 0 A 1 1 0 011 110

Example Using a Karnaugh map, convert the following POS expression into a minimum POS expression, and a minimum SOP expression CD AB 00 01 11 10 00 1 0 0 0 SOP : 01 0 1 1 1 POS : 11 0 1 10 1 1