Module 2 logic gates

MODULE 2 MCA-102 DIGITAL SYSTEMS & LOGIC DESIGN ADMN 2014-‘17

Dept. of Computer Science And Applications, SJCET, Palai Page 1

Logic Gates

Basic logical operators are the logic functions AND, OR and NOT.

Logic gates implement logic functions.

The three basic logical operations are:

1. AND

2. OR

3. NOT

1. AND Gates

Is denoted by a dot (·).

Fig 2.1 truth table and logical symbol for AND gate

2. OR Gates

Is denoted by a plus (+).

Fig 2.2 logical symbol and truth table for OR gate

3. NOT Gates

is denoted by an over bar ( ¯ ),

Fig 2.3 logical symbol and truth table for NOT gate



Fig 2.4 timing diagram for AND, OR, NOT gates

Problem: implement the following logic function using AND, OR, NOT gates

F=X+Y’Z

Solution:

Fig 2.5 Truth table and logic diagram for F= X+Y’Z

NAND Gates

combination of AND and NOT gate

Fig 2.6 logical symbol and truth table for NAND gate



NOR Gate

combination of OR and NOT gate

Fig 2.7 logical symbol and truth table for NOR gate

XOR (Exclusive-OR) Gate

xy

x Å yx y + x y

Fig 2.8 logic symbol and truth table for XOR gate

XNOR (Exclusive-NOR Gate)

xy

x Å y

x � yx y + x y

Fig 2.9 logic symbol and truth table for XNOR gate

Boolean algebra

a useful mathematical system for specifying and transforming logic functions

Boolean Functions: is an expression formed with binary variables, the 2 two binary operators OR

and AND, and unary operator NOT, parentheses, and an equal sign.

For a given value of the variables, the function can be either 0 or 1.

A Boolean function has:

i. At least one Boolean variable,

ii. At least one Boolean operator, and

iii. At least one input from the set {0, 1}.

It produces an output that is also a member of the set {0, 1}.

All Boolean functions through AND and NOT operations



Fig 2.10 Boolean functions through AND and NOT

All Boolean functions through NAND function

Fig 2.11 Boolean functions through NAND gate

All Boolean functions through NOR function

Fig 2.12 Boolean functions through NOR gate



Fig 2.13 Boolean algebra Laws

A Boolean expression can be represented in

i. Minterms

ii. Maxterms

iii. SOP (Sum Of Products)

iv. POS (Product of Sums)

i. Minterms

Are AND terms with every variable present in either true or complemented form.

Evaluates to ‘1’ for a specific combination

Fig 2.14 Minterm table



Sum of Minterms(SOM)

Fig 2.15 Example for SOM

ii. Maxterms

Are OR terms with every variable in true or complemented form.

Evaluates to ‘0’ for a specific combination

Fig 2.15 maxterm table

Product-Of-Maxterm (POM)

Fig 2.16 example for POM



iii. SOP (Sum of Products)

equations are written as an OR of AND terms

Implementation of this form is a two-level network of gates such that:

The first level consists of n-input AND gates

The second level is a single OR gate

This form often can be simplified so that the corresponding circuit is simpler.

Iv.Product of Sum (POS)

equations are written as an AND of OR terms

))()(( CBBACAF

Minterms and maxterms are related



Converting Between Standard Forms

Example



Example 1.

Karnaugh-Map

• A Karnaugh map is a graphical method used to obtain the most simplified form of an expression in

a standard form (Sum-of-Products or Product-of-Sums).

• The simplest form of an expression is the one that has the minimum number of terms with the least

number of literals (variables) in each term.

• An n-variable K-map has 2n cells with each cell corresponding to a row of an n-variable truth table.

• K-map cells are labeled with the corresponding truth-table row.

• K-map cells are arranged such that adjacent cells correspond to truth rows that differ in only one bit

position (logical adjacency).

• For the case of 2 variables, we form a map consisting of 22=4 cells as shown in Figure



Fig 2.17 2 variable K-Map

3 variables Karnaugh map


4 variables Karnaugh map


The Karnaugh map is completed by entering a '1‘(or ‘0’) in each of the appropriate cells.

Within the map, adjacent cells containing 1's (or 0’s) are grouped together in twos, fours, or eights.

C B(0,4)f BA (4,5)f B(0,1,4,5)f A(0,1,2,3)f

BC00

0

01

1

11 10A

1 0 0 0

1 0 0 0

BC00

0

01

1

11 10A

0 0 0 0

1 1 0 0

BC00

0

01

1

11 10A

1 1 1 1

0 0 0 0

BC00

0

01

1

11 10A

1 1 0 0

1 1 0 0

C A(0,4)f CA (4,6)f C A(0,2)f C(0,2,4,6)f

BC00

0

01

1

11 10A

0 1 1 0

0 0 0 0

BC00

0

01

1

11 10A

0 0 0 0

1 0 0 1

BC00

0

01

1

11 10A

1 0 0 1

1 0 0 1

BC00

0

01

1

11 10A

1 0 0 1

0 0 0 0

Fig 2.20 examples of simplication of 3 variable k-map



DCB(0,8)f DCB(5,13)f DBA(13,15)f DBA(4,6)f

CA(2,3,6,7)f DB)(4,6,12,14f CB)(2,3,10,11f DB(0,2,8,10)f

CD00

00

01

01

11

11

10

10

AB

1 0 0 0

0 0 0 0

0 0 0 0

1 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 0 0

0 1 0 0

0 1 0 0

0 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 0 0

0 0 0 0

0 1 1 0

0 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 0 0

1 0 0 1

0 0 0 0

0 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 1 1

0 0 1 1

0 0 0 0

0 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 0 0

1 0 0 1

1 0 0 1

0 0 0 0

CD00

00

01

01

11

11

10

10

AB

0 0 1 1

0 0 0 0

0 0 0 0

0 0 1 1

CD00

00

01

01

11

11

10

10

AB

1 0 0 1

0 0 0 0

0 0 0 0

1 0 0 1

Fig 2.21 examples for 4 variable K-Map

Example:

F(W,X,Y,Z)= ∏ M(0,2,3,4,5,6)

= Σm(1,7,8,9,10,11,12,13,14,15)

F(W,X,Y,Z)= W + XYZ + X’Y’Z



Don't Cares

In certain cases some of the minterms may never occur or it may not matter what happens if they

do.In such cases we fill in the Karnaugh map with and X meaning don't care. When minimizing an

X is like a "joker" X can be 0 or 1 - whatever helps best with the minimization.

Ex:

Simplifies to B if x=1

Ex: simplify the expression using K-map for the truth table given below



Ex: Find a minimal SOP expression from the following 4 variable K-Map

M(A,B,C,D) = B D + A’ B’

Problem 1: Determine the minimum-cost SOP and POS expressions for the function f(x1, x2, x3, x4) =

_∑m(4, 6, 8, 10, 11, 12, 15) + D(3, 5, 7, 9).

Fig 2.22 Karnaugh maps for problem 1



Problem2: Use Karnaugh maps to find the minimum-cost SOP and POS expressions for the function

f(x1, . . . , x4) = x1x3x4 + x3x4 + x1x2x4 + x1x2x3x4

assuming that there are also don’ t-cares defined as D = ∑(9, 12, 14).

Fig 2.23 Karnaugh maps for problem 2

K-map with five variables

A 5-variable K-Map will have 25 = 32 cells. A function F which has maximum decimal value of 31,

can be defined and simplified by a 5-variable Karnaugh Map.

Fig 2.24 5 variable k-Map



Example 1

Given function, F = Σ (1, 3, 4, 5, 11, 12, 14, 16, 20, 21, 30)

Since, the biggest number is 30, we need to have 5 variables to define this function.

Fig 2.25 5 variable k-map for example 1

Applying rules of simplifying K-Map, there is no octet. There is one quad that is obtained by

visualizing Second Square on first, there are 4 adjacent cells – 4,5,20 and 21. The octet is highlighted by a

blue connecting line. There are 5 pairs. Similar to quad, there is one pair between two squares which is

highlighted by the blue connecting line.

(4, 5, 20, 21) – B’CD’ (Since A & E are the changing variables, it is eliminated)

(12, 14) – A’BCE’ (Since D is the changing variable, it is eliminated)

(14, 30) – BCDE’ (Since A is the changing variable, it is eliminated)

(3, 11) – A’C'DE (Since B is the changing variable, it is eliminated)

(16, 20) – AB’D'E’ (Since C is the changing variable, it is eliminated)

(1, 3) – A’B'C’E (Since D is the changing variable, it is eliminated)

Thus, F = B’CD’ + A’BCE’ + BCDE’ + A’C'DE + AB’D'E’ + A’B'C’E

Example 2

Given function, F = Σ (0, 2, 3, 5, 7, 8, 11, 13, 17, 19, 23, 24, 29, 30)




Fig 2.26 solution for example 2

Applying rules of simplifying K-Map, there is no octet. First we need to look for quads within each

of the squares. There are none but there is a quad between two squares that is obtained by visualizing

Second Square on first, there are 4 adjacent cells – 3, 7, 19 and 23. This quad is highlighted by blue

connecting line. There are 6 pairs, out of which two are between two squares, highlighted by blue

connecting line.

(3, 7, 19, 23) - B’DE (Since A & C are the changing variables, they are eliminated)

(3, 11) – A’C'DE (Since B is the changing variables, it is eliminated)

(1, 2) – A’B'C’E’ (Since D is the changing variables, it is eliminated)

(5, 7) – A’B'CE (Since D is the changing variables, it is eliminated)

(17, 19) – AB’C'E (Since D is the changing variables, it is eliminated)

(13, 29) – BCD’E (Since A is the changing variables, it is eliminated)

(8, 24) – BC’D'E’ (Since A is the changing variables, it is eliminated)

There is 1 in cell 30, which cannot be looped with any adjacent cell, hence it cannot be simplified

further and left as it is.30 – ABCDE’

Thus, F = B’DE + A’C'DE + A’B'C’E’ + A’B'CE + AB’C'E + BCD’E + BC’D'E’ + ABCDE’



Example 3

Given function, F = Σ (0, 1, 2, 3, 8, 9, 16, 17, 20, 21, 24, 25, 28, 29, 30, 31)


Fig 2.27 solution for example 3

Applying rules of simplifying K-Map, there are 2 octets. First one is in square 2 circled in red.

Another octet is between 2 squares highlighted by blue connecting lines. There are 2 quads between each

of the squares.

(16, 17, 20, 21, 28, 29, 24, 25) – AD’ (Since B, C and E are changing variables, they are eliminated)

(0, 1, 8, 9, 16, 17, 24, 25) – C’D’ (Since A, B and E are changing variables, they are eliminated)

(0, 1, 2, 3) – A’B'C’ (Since D and E are changing variables, they are eliminated)

(28, 29, 30, 31) – ABC (Since D and E are changing variables, they are eliminated)

Thus, F = AD’ + C’D’ + A’B'C’ + ABC

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Module 2 logic gates