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Contents:

Why simplification?

The Map Method

Two, Three, Four and Five variable Maps.

Simplification of two, three, four and five variable Boolean function by Map method.

Product of sums and Sum of products simplification.

NAND and NOR implementation.

Simplification of Boolean Functions

Course Instructor

Mohammed Abdul kader

Assistant Professor, EEE, IIUC

The complexity of the digital logic gates that implement a Boolean function is directly

related to the complexity of the algebraic expression from which the function is

implemented.

Although the truth table representation of a function is unique but the algebraic

expression appears in many different form.

Before implementation of logic circuit, the choice of simplest boolean expression from

different representation results the minimum number of gates and less complexity in

the digital circuit.

There are different ways of simplification of Boolean function. In this section we will

discuss the “Map Method” of simplifying Boolean function.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 2

Simplification of Boolean Functions 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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The map method provides a simple straight forward procedure for minimizing Boolean

functions.

The map method, first proposed by Veitch and slightly modified by Karnaugh, is also

known as the Veitch diagram or the Karnaugh map.

Karnaugh maps provide an alternative way of simplifying logic circuits.

Instead of using Boolean algebra simplification techniques, you can transfer logic

values froma Boolean statement or a truth table into a Karnaugh map.

The arrangement of 0's and 1's with in the map helps you to visualize the logic

relationships between the variables and leads directly to a simplified Boolean

statement.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 3

The Map Method 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 4

Two Variable Map

There are four minterms for two variables; hence the map consists of fours quares, one of

each minterm.

F=xy F=x+y = xy+xy+xy = m1+ m2 + m3

Representation of

functions in two

variable map.

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 5

Three Variable Map

There are eight minterms for three variables, i.e. the map consist of eight squares.

Minterms are not arranged in binary sequence but in a sequence similar to gray

code/reflected code.

There are four squares where each variable is equal to 1 and four where each is equal

to 0.

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 6

Three Variable Map

F = m5 + m7 = xyz + xyz = xz (y+y) = xz

Understanding the usefulness of map for simplification of Boolean function

F = m1 + m7 = xyz + xyz

Isolated ‘1’ in map

Adjacent Pair/ 1’s in two adjacent square

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 7

Three Variable Map

F = 1

Understanding the usefulness of map for simplification of Boolean function

F = m1 + m3 + m5 + m7

= xyz + xyz+ xyz + xyz

= xz (y+y) +xz (y+y) = xz +xz = z (x+x) = z

Adjacent Quad/ 1’s in four adjacent square

Adjacent Octet/ 1’s in eight adjacent square

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 8

Three Variable Map: Examples

Example 3-2: Simplify the Boolean function

F= xyz + xyz + xyz + xyz Solution :

yz

xz

So, after simplification, F = yz+ xz

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 9

Three Variable Map: Examples

Example 3-3: Simplify the Boolean function

F= AC+ AB+ ABC+BC Solution :

F = AC+ AB+ ABC+BC =ABC+ ABC +ABC+ ABC+ ABC+ABC+ABC

C

AB

So, after simplification, F = C+ AB

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 10

Three Variable Map: Examples

Example 3-4: Simplify the Boolean function

F (x,y,z)= ∑ (0, 2, 4, 5, 6)

Solution :

So, after simplification, F = z+ xy

xy

z

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 11

Four Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 12

Four Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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The map for Boolean function of four binary

variables has 16 minterms and the squares assigned to

each.

The rows and columns are numbered in a reflected

code sequence, with only one digit changing value

between two adjacent rows and columns.

The minterm corresponding to each square can be

obtained from the concatenation of the row number

with the column number.

The combination of adjacent squares that is useful

during the simplification process is easily determined

from inspection of the four variable map-

wxyz: 1010 (10 in binary), so

this square represent m10

One square represents one minterm, giving a term of four literals.

Two adjacent squares represent a term of three literal.

Four adjacent squares represent a term of two literals.

Eight adjacent squares represent a term of one literals.

Sixteen adjacent squares represent the function equal to 1.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 13

Four Variable Map: Examples

Example 3-5: Simplify the Boolean function

F (x,y,z)= ∑ (0, 1,2, 4, 5, 6, 8, 9, 12, 13, 14)

Solution :

So, after simplification, F = y+ wz+x z

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 14

Four Variable Map: Examples

Example 3-6: Simplify the Boolean function

F = AB C +B CD + A BCD +AB C Solution :

So, after simplification, F = BD + BC+A C D

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 15

Five Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 16

Five Variable Map: Examples 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Example 3-7: Simplify the Boolean function

F (A, B, C, D, E) = ∑ (0, 2, 4, 6, 9, 11, 13, 15, 17, 21, 25, 27, 29, 31)

Solution :

ADE

BE

A BE

So, after simplification, F = BE+AD E + ABE

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 17

Product of Sums Simplification 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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All previous examples are in sum-of-products form [F = BE+AD E + ABE ]

How to obtain the product-of-sum form

* Simplify F in the form of sum of products. [If we mark the empty

squares by 0’s and combine them into valid adjacent squares, we obtained a

simplified expression of the complement of the function, i.e. of F]

* Apply DeMorgan's theorem F = (F ')

* F': sum of products => F : product of sums

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 18

Example 3-6: Simplify the Boolean function in (a) sum of products and (b) product of

sums

F (A, B, C, D, E) = ∑ (0, 1, 2, 5, 8, 9, 10)

Solution :

F = AB + CD + BD

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So, F = (A +B ) (C +D ) (B + D ) by DeMorgan theorem

(b) product of sums simplification

(a) Sum of products simplification

F = B D + B C + A C D

Product of Sums Simplification (Cont.)

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 19

Product of Sums Simplification (Cont.)

Gate implementation of the function Example 3-8

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(b) F = (A +B ) (C +D ) (B + D ) (a) F = B D + B C + A C D

Sum of products product of sums

Simplify the Boolean function F(A,B,C,D)=Π(0,1,2,3,4,10,11)

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 20

Product of Sums Simplification (Cont.)

Example: Simplify the Boolean function F(A,B,C,D)=Π (0,1,2,3,4,10,11)

101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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AB

CD

From Map we get,

F = A B + B C + A C D

Using DeMorgan theorem-

F = (A B + B C + A C D )

= (A + B ) ( B +C ) ( A + C + D )

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 21

NAND and NOR Implementation

Why NAND and NOR implementation?

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Digital circuits are frequently constructed with NAND/NOR rather than with AND/OR gates.

NAND and NOR gates are easier to fabricate with electronic components than AND/OR.

Cheaper(lower cost) and faster(less delay).

Any Boolean function can be constructed using only NAND or only NOR gates. That’s why NAND

and NOR are known as universal gates.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 22

NAND and NOR Implementation

Implementation of basic gates by NAND gate

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x x

x y

(xy) ((xy)) = xy

x

y

x

y

(xy) = x+y

NOT gate by NAND

gate

AND gate by NAND

gate

OR gate by NAND

gate

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 23

NAND and NOR Implementation

Implementation of basic gates by NOR gate

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x x

x y

(x+y) ((x+y)) = x+y

x

y

x

y

(x+y) = xy

NOT gate by NOR gate

OR gate by NOR gate

AND gate by NOR

gate

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 24

NAND and NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Two graphic symbols for NAND gate

Two graphic symbols for NOR gate

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 25

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

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Implementation F=AB+CD+E by NAND gate only.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 26

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

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Rules for obtaining the NAND logic diagram from a Boolean function

1. Simplify the function and express it in sum of products.

2. Draw a NAND gate for each product term of the function that has at least two literals. The inputs

to each NAND gate are the literals of the term. This constitutes a group of first-level gates.

3. Draw a single NAND gate (using the AND-invert or invert-OR graphic symbol) in the second

level, with inputs coming from outputs of the 1st level.

4. A term with a single literal requires an inverter in the first level or may be complemented and

applied as an input to the second-level NAND gate.

Note: If we simplify the function combining 0’s in a map, we obtain the simplified expression of the

complement of the function in sum of product. The complement of the function can then be

implemented with two levels of NAND gates using the rules stated above. If the normal output is

desired, it would be necessary to insert a one-input NAND gate.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 27

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

Example 3-9: Implement the following function with NAND gates

F(x,y,z) = ∑ (0, 6)

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 28

NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

Implement the function F= (A+B) (C+D) E with NOR gates

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 29

NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

Example 3-9: Implement the following function with NAND gates

F(x,y,z) = ∑ (0, 6)

Sum of Product

Product of sums

Sum of Product

Product of sums

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 30

Don’t-Care Conditions 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

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You don’t always need all 2𝑛 input combinations in an n-variable function.

–If you can guarantee that certain input combinations never occur.

–If some outputs aren’t used in the rest of the circuit.

A four bit decimal code, for example, has six combinations which are not used. Any

digital circuit using this code operates under the assumption that these unused

combinations will never occur as long as the system is working properly.

The unused combinations is known as don’t care conditions and can be used on the map

to provide further simplification of the boolean expression.

It should be realized that a don’t care minterm is a combination of variables whose

logical value is not specified. It cannot be marked with a 1 or, 0 in the map as it is not

specified as 0 or 1. To distinguish the don’t care condition from 1’s and 0’s, an X is used.

Thus, an X inside a square in the map indicates that we don’t care whether the value of 0

or 1 is assigned to F for the particular minterms.

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 31

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

Example 3-12: Simplify F(w,x,y,z) = Σ(1,3,7,11,15) which has the don’t care conditions

d(w,x,y,z) = Σ (0,2,5)

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 32

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

Exercise. 3-18 : Implement the following function with either NAND or NOR gates. Use

only four gates.

F = w’xz+w’yz+x’yz’+wxy’z and d= wyz

NAND Implementation

Sum of products:

Combine 1’s and some of X’s

F= x y+xz

wx

yz

x

y

z

F

x (x y)

(x z)

Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 33

NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010

NOR Implementation

Products of sums:

Combine 0’s and some of X’s

F = x y +xz

So, F= (x+y) (x +z)

wx

yz

x

z

y

x (x +z)

(x +y )

F