1. 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

2. 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

3. 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

4. Universal NAND Truth Table (T.T)  A B Y 1 Math expression Y= A+B1
Math expression Y= A+B
Universal NAND
Universal NOR

5. 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.

6. 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

7. 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

8. 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=

9. 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.

10. Logic Function Representation2.
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

11. 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

12. 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

17. 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:

18. 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?

19. Q14)Applying the distributive law to the expression2 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.