Logic Gates By Dr.Mazin Alzewary 2017-2018 Al-Mamoon University College Computer Science Department Second Year Architecture and Logic Gates Logic Gates By Dr.Mazin Alzewary 2017-2018

Basic Logic Gates AND Gate Symbol Truth Table (T.T)  the table describe relation between inputs and outputs (Gate function) A B Y 1 Math Expression Y= A.B OR Gate Symbol Truth Table (T.T)  A B Y 1 Math Expression Y= A+B NOT Gate Symbol Truth Table (T.T) A Y 1 Math Expression Y= A Buffer Gate Symbol Truth Table (T.T)  A Y 1

Math expression Y= A NAND Gate Symbol Truth Table (T.T)  A B Y 1 Math Expression Y= A.B NOR Gate Symbol Truth Table (T.T)  A B Y 1 Math Expression Y= A+B XOR Gate Symbol Truth Table (T.T)  A B Y 1 Math Expression Y= A+B = A.B + A.B XNOR Gate Symbol

Universal NAND Truth Table (T.T)  A B Y 1 Math expression Y= A+B 1 Math expression Y= A+B Universal NAND Universal NOR

Boolean Expression Consist of logical (0 or 1) variable A, B, C, etc and logical operations (through gates) such as * (AND), + (OR), complement(NOT) Minterms are AND terms with every variable present in either true or complemented form such as A.B, B.C.D Maxterms are OR terms with every variable in true or complemented form such as (A+B), (B+C) There are two type of logical expression Sum of Product (SOP) Y= A.B Z= A.C + D X= ABC + E Product of Sum (POS) M=(A+B)(B+C) Q 1 Convert the logical expression to logical gates or to digital circuit 1. Y=A.B +C 2. Y= A.B’ + B.C Q2) Convert the logical gates to Boolean expression 1. 2.

Simplification Using Boolean Algebra (Rules) Logical Algebra (Rules) Q1 Prove that A + AB = A A+AB= A(1+B) =A.1 =A Q2 Prove that A + AB= A + B =(A+AB)+AB =(AA+AB)+AB =AA+AB+AA+AB =(A+A)(A+B) =A+B Q3 Prove that (A + B) (A + C)=A+AB =AA+AC+AB+BC =A+AC+AB+BC =A(1+C)+AB+BC =A*1+AB+BC =A(1+B)+BC =A*1+BC =A+BC Boolean Expression Simplification Simplification Using Boolean Algebra (Rules) Order of execution is Parentheses NOT

AND OR Q1) Simplify the expression using logical rules Y=ABC Q2) Simplify the expression using logical rules Y=AB(C+D) Q3) Simplify the expression using logical rules Y=(A+B+C)D Q4) Simplify the expression using logical rules Y=[AB(C+BD)+AB]C=(ABC+ABBD+AB)C =(ABC+0+AB)C =(ABC+AB)C =ABCC+ABC =ABC+ABC =BC(A+A) =BC.1 =BC Q5) Simplify the expression using logical rules and draw circuit AB+AC+ABC

2. Simplification Using Karnaugh Map (K-MAP) Karnaugh Map Take each minterm from expression and find the suitable cell in the MAP (AB, AB, etc for two variable ) and write 1 for that location Write 0 to the rest location Build block for the ones such that Contiguous cells The block is 2,4,8,etc location Choose bigger block first The cell (1) could be contain in more than one block Build Boolean term for each block The simplified expression is the sum of all sub expression Q1) Simplify using karnaugh map 1. Y= 2. Y=

Q2) Reduce the expression using K-MAP X=ABC+ABC+ABC+ABC+ABC Using SOP POS a. b. Q3) Reduce the expression using K-MAP Y=ABCD+ABCD+ ABCD+ABCD+ ABCD+ABCD+ABCD using 1. SOP 2 POS 1.

Logic Function Representation 2. Logic Function Representation 1. Logic Function as Boolean Expression Form Y=AB + AB + AB Y= ABC + ABC + ABC + ABC + ABC And these could be solved using K-Map But if we have Y=AB+ABC+ABC The term AB not include all three variable and how could we insert it in the map If we multiply the AB by (C+C) to get AB=AB*1 = AB*(C+C) = ABC + ABC Then the new expression could be Y=ABC + ABC + ABC + ABD And could be solved using K-Map 9

2. Logic Function as Truth Table Form X(A,B,C) = (1,4,5,6,7) A Input B C Output Y 1 Y= ABC + ABC +ABC + ABC + ABC The standard map for any function could be mansion bellow

Q) Simplify the following using K-Map X(A,B,V,D)= (1,4,8,9,11,12,13,15) Don’t Care There is situation that the value of cell not affect the expression( the same if it is 0 or 1) this situation is called don’t care (X) and could be used as 0 or 1 in building the block (the function do not care for output 0 or 1) Q) Simplify the Boolean function given in the truth table A Input B C Output Y 1 X

Questions Q1) Simplify the following Boolean expression using Karnaugh map Y=AB+AB+AB Y= ABC + ABC + ABC + ABC + ABC + ABC Y= ABC + ABC + ABC + ABC + ABC + ABC Y= ABCD + ABCD + ABCD + ABCD + ABCD + ABCD Y= AB + ABC + ABC + ABC + ABC + ABC Y = A+ ABC + ABC + ABC + ABC + ABC + ABC Y= ABC + BC + C + A + ABC + ABC Q2) Convert the following SOP expression to an equivalent POS expression. Q3) Determine the values of A, B, C, and D that make the sum term equal to zero. Q4) Which of the following expressions is in the sum-of-products (SOP) form? (A + B)(C + D) (A)B(CD) AB(CD) AB + CD Q5) Derive the Boolean expression for the logic circuit shown below:

Q6) From the truth table below, determine the standard SOP expression. One of De Morgan's theorems states that difference between: . Simply stated, this means that logically there is no a NOR and an AND gate with inverted inputs a NAND and an OR gate with inverted inputs an AND and a NOR gate with inverted inputs a NOR and a NAND gate with inverted inputs Q8) Applying DeMorgan's theorem to the expression A. , we get _. B. C. D. Q9) Which output expression might indicate a product-of-sums circuit construction? A. Q10) For the SOP expression A. 1 , how many 1s are in the truth table's output column?

Q14)Applying the distributive law to the expression 2 C. 3 D. 5 Q11) How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z) A. 1 B. 2 C. 4 D. 5 Q12) Determine the values of A, B, C, and D that make the product term A. A = 0, B = 1, C = 0, D = 1 B. A = 0, B = 0, C = 0, D = 1 C. A = 1, B = 1, C = 1, D = 1 D. A = 0, B = 0, C = 1, D = 0 Q13) AC + ABC = AC A. True equal to 1. B. False Q14)Applying the distributive law to the expression A. , we get . B. C. D.