what is meaning definition of positive and negative levels Positive Level Logic System and Negative Level Logic System ?
Boolean Algebra
In boolean algebra, the binary digits are utilized to represent the two levels that occurs within digital logic circuits. A binary 1 represent a high level and a binary 0 will represent a low level in boolean equations and in terms of voltages there are two types of logic systems
(a) Positive level logic system
(b) Negative level logic system
(a) Positive Level Logic System
Out of the given two voltage levels, the more positive value is assumed as logic 1 and the more negative level as logic 0.
For example Logic 0 Logic 1
0 V 5 V
– 2 V + 3 V
– 7 V – 2 V
+ 2 V + 7 V
Negative Level Logic System
Out of the given two voltage levels, the more negative value is assumed as logic 1 and the more positive level as 0.
For Example
Logic 1 Logic 0
0 V 5 V
– 2 V + 3 V
– 7 V – 2 V
+ 2 V + 7 V
Boolean Addition
the basic rules for boolean addition are as follows
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1
Boolean addition is same as the OR operation. it differs from binary addition.
Boolean Multiplication
Multiplication in Boolean algebra follows the same basic rules govering binary multiplication.
0.0 = 0
0.1 = 0
1.0 = 0
1.1 = 1
Boolean multiplication is same as the AND operation.
Logic Expressions
Not Operation
The operation of an inverter (NOT circuit ) can be represented with symbol as follows
This expression states that the output is the complement of the input
A (input)
A (input)
AND Operation
The operation of two input AND gate can be expressed in equation form as follows
OR Operation
the operation of a two input OR gate can be expressed in equation form as follows
Rules and Laws of Boolean Algebra
Three of the basic laws of the boolean algebra are same as in ordinary the commutative laws, the associative laws and the distributive law.
Commutative Laws
The commutative law for addition of two variable
A + B = B + A
For multiplication
AB = BA
Associative laws
For addition A + (B + C) = (A + B) + C
For multiplication A (BC) = (AB) C
Distributive law the distributive law is written for three variables as follows
A (B + C) = AB + AC
Rules for Boolean Algebra
Several basic rules that are useful in manipulating and simplifying boolean algebra Expressions-
1. A + 0 = A 2. A + 1 = 1
3. A .0 = 0 4. A . 1 = A
5. A + A = A 6. A + A = 1
7. A . A = A 8. A . A = 0
9. A = A 10. A + AB = A
11. A + AB = A + B 12. (A + B) (A + C) = A + AB
De Morgan’s Theorems
De morgan’s theorems that are important part of algebra are stated in equation form as:
1. AB = A + B
2. A + B = A . B
Example 8. Apply de morgan’s therorems ot the following expression :
X = A + B + C + DE
Sol. X = A + B + C + DE
= A . B . C + DE [ A + B = A . B]
= A . B . C + (D + E) [ AB = A + B]
= A . B . C + (D + E) [D = D]
= A . B. C + (D + E)
Duality Theorem
this theorem says that starting with a boolean relation, we can derive another boolean relation called its dual.
The steps used are –
1. Changing each OR sign to an AND sign.
2. Changing each AND sign to an OR sign.
3. Complementing all 0’s and 1 ‘s.
example 9. Find the dual of expression A (B + C) + D.
Sol. Given expression is
A . (B + C) + D
OR AND AND
By replacing AND – OR
and OR – AND, we get
= [ A + B.C] D
Boolean Expressions for Gate Networks
the form of a given boolean expression indicates the type of gate network, it describes.
There are certain forms of boolean expressions that are more commonly used than others: the most important of these are the sum of products and product of sums form.
Sum of Products Form
A sum of products expression is two or more AND functions ORed together.
For example AB + BC
ABC + DEF
ABC + DEF + FGH + AFG
An important characteristics of the sum of products form is that the corresponding implementaion is always a two level gate network.
example 10. Implement the expression ABC + DE + FGHI with logic gates.
Product of Sums Form
The product of sums form can be thought of as the dual of the sum of products. it is the AND of two or more OR functions.
e.g., (A + B) (B + C)
(A + B + C) (E + F + G)
(A + B + C) (E + F) (A + E)
A POS expression can also contain a single variable term such as A (B + C + D) (E + F + G).
Example 11. Construct the following function with logic gates:
(A + B) (C + D + E ) (D + E + F + G)
Minterm and Maxterm
If each term in the sum of products form contains all the variables (literals) then the expression is known as standard sum of products form or canonical sum of products form and each individual term in standard sum of products form is called as minterm.
For example Y (A, B, C) = AB + ABC + B
To get the standard SOP form, multiply the first term by (C + C) and the third term (A + A) (C + C).
Y (A, B, C) = AB (C + C) + ABC + (A + A) B (C + C)
= ABC + ABC + ABC + ABC + ABC + ABC + ABC
= ABC + ABC + ABC + ABC + ABC + ABC
The total minterms are
ABC (111) = M7 ABC (101) = M5 ABC (100) = M4
ABC (110) = M6 ABC (001) = M1 ABC (000) = M0
It can be written as a sum of minterms as follows :
= (A, B, C) = M (0, 4, 5, 6, 7)
If each term of product of sums contains all the variables, then the expression is known as standard product of sums form or canonical product of sums form and each individual term in the standard product of sums form is called as maxterm.
For example Y (A, B, C) = (A + B) (A + B + C)
The first term of above equation has two variables, variable C is missing and hence, to get the standard POS, the term, (C) in the first term must be ORed.
Y (A, B, C) = (A + B + CC) . (A + B + C)
= (A + B+ C) ( A + B + C) (A + B + C)
The total maxterms are three, which are
(A + B + C) (010) = M2
(A + B + C) (011) = M3
(A + B + C) (001) = M1
It can be written as a product of maxterm:
Y (A, B, C) = M (1, 2, 3)
Example 12. Obtain SOP and POS expressions for the truth table of table given below.
Sol. From the given table, chosing minterms corresponding to Y = 1 in the truth table, we can write
Y = ABC + ABC + ABC + ABC
In terms of maxterms which correspond to Y = 0, we can write
Y = (A + B+ C) (A + B + C) (A + B + C) ( A + B + C)
It is observed that Y + Y = 1
So, while minterms which correspond to 1 outputs constitute SOP, the complementary minterms which corresponds to 0 outputs.
Their SOP would then constiute Y .
Y = ABC + ABC + ABC + ABC
Similarly for maxterms the output corresponding to outputs would constitute Y.
Y = (A + B + C) ( A + B + C) (A + B + C) (A + B + C)
As per SOP expression Y of the truth table the combinational circuit is drawn ahead.