boolean algebra part 2 april 73

7/31/2019 Boolean Algebra Part 2 April 73

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Boolean Algebra

Lectures 13

1


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Carry out simple operations with Boolean

Algebra,

Construct truth tables,

Convert switch circuit /

logic gates

intoBoolean Addition and Multiplication.

Recognise

properties of Boolean Algebra

2


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A

system

may

have

more

than

two

inputsand

the

Boolean

expression

for

a

three

-

input

OR

-

function

having

elements

A

,

B

andC

is

A

+

B

+

C

.

Similarly,

a

three

-

input

AND

-

function

iswritten

as

A

B

C

.

The

equivalent

electrical

circuits

and

truth

tables

for

three

-

input

OR

and

AND

-

functionsare

shown

in

Figs

.

1

(a)

and

(b)

respectively(next

slide)

.

3


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Input Output

Fig 1(a) The OR

-

function electrical circuit and truth table

Input

A B C

Output

Z = A + B + C0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1 4


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Output Input

Input

A B C

Output

Z = A l B l C0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 01 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

Fig 1(b) The AND

-

function electrical circuit and truth

table

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Example

1

Derive

the

Boolean

expression

and

construct

a

truth

table

for

the

switching

circuit

shown

in

Figure

2

.

1 2

3 4

65 7 8

6


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The switches between 1 and 2 in Figure 2 arein series and have a Boolean expression of B .A. The parallel circuit 1 to 2 and 3 to 4 have aBoolean expression of (B . A + ). The

parallel circuit can be treated as a singleswitching unit, giving the equivalent ofswitches 5 to 6, 6 to 7 and 7 to 8 in series .Thus the output is given by :

B B A

Solution :

B

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1 2 3 4 5 6 7

A B B . A 0 0 0 1 1 1 1

0 1 0 0 0 1 0

1 0 0 1 1 0 0

1 1 1 0 1 0 0

. + = ( + )

The truth table is as shown above . Columns 1 and 2give all the possible combinations of switches A and

B. Column 3 is the and -function applied to columns 1and 2, giving B.A. Column 4 is , that is, the oppositeof column 2. Column 5 is the or -function applied tocolumns 3 and 4. Column 6 is , that is the oppositeto column 1. The output is column 7 and is obtainedby applying the and -function to columns 4, 5 and 6.

Solution :

8


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l l l l

l l l l

l l

A

Input OutputB l l l l

B l l

l l

9

Example 2

Derive the Boolean expression .


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A B C C A C A A B A B C Z = A C + A B C

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

CBABACAZ

0

0

0

0

1

0

1

0

1

1

1

1

0

0

0

0

0

0

1

1

0

0

0

0

0

0

1

0

0

0

0

0

0

0

1

1

1

0

1

0

CBABACAZ

10


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Example 3

Derive the Boolean expression and constructa truth table for the switching circuit shown inFigure 3.

l l

1 2

5 6

89

7

3 4

11


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l

The parallel circuit 1 to 2 and 3 to 4 gives (A +) and this is equivalent to a single switching

unit between 7 and 2. The parallel circuit 5 to6 and 7 to 2 gives C + (A + ) and this isequivalent to a single switching unit between 8and 2. The series circuit 9 to 8 and 8 to 2gives the output

Solution :

B

B

B

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1 2 3 4 5 6 7

A B C +

0 0 0 1 1 1 0

0 0 1 1 1 1 00 1 0 0 0 0 0

0 1 1 0 0 1 1

1 0 0 1 1 1 0

1 0 1 1 1 1 0

1 1 0 0 1 1 1

1 1 1 0 1 1 1

= [ + + ] + +

Solution :

The truth table is as shown above. Columns 1, 2 and 3 give all thepossible combinations of A, B and C. Column 4 is and is theopposite to column 2. Column 5 is the or -function applied to columns1 and 4, giving (A + ). Column 6 is the or -function applied to column3 and 5 giving C + (A + ). The output is given in column 7 and isobtained by applying the and- function to columns 2 and 6.

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Example 4

Construct a Truth Table for the logical functions

at points C, D and Q in the following circuit .

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From the truth table above, column C represents the output functionfrom the NAND gate and column D represents the output functionfrom the Ex -OR gate . Both of these two output expressions thenbecome the input condition for the Ex -NOR gate at the output . It canbe seen from the truth table that an output at Q is present when any ofthe two inputs A or B are at logic 1. The only truth table that satisfiesthis condition is that of an OR Gate . Therefore, the whole of the abovecircuit can be replaced by just one single 2-input OR Gate .

Solution :

15


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Example 5

Find the Boolean algebra expression for the following

system.

The system consists of an AND Gate, a NOR Gate and finallyan OR Gate. The expression for the AND gate is A . B, andthe expression for the NOR gate is + . Both theseexpressions are also separate inputs to the OR gate which isdefined as A+B. Thus the final output expression is given as

on the next slide) : 16


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The output of the system is given asQ = (A . B ) + ( + )

Solution :

17


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Associative

(A . B) . C = A . (B. C)

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

Distributive

A . (B + C) = A . B + A . C

A + (B . C) = (A + B) . (A +C)

Commutative

A . B = B . A

A + B = B + A

Laws of Boolean Algebra

18


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Rules of Boolean Algebra

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Prepared by : Department of General Studies

(Mathematics)Brunei Polytechnic 2012

Lecturers :Abdul Nafri H. Hussin

Chan Koo Kiat Hamizah Md. Taha Kam Boon Long,

Lim Li Yan,Lim Ting Hing,

Mes Lina H. Husin.20


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Resources

Bird, J. (2010). Engineering Mathematics (6th Edition),Elsevier Ltd, Oxford, UK.www.kerryr.net

www.eskimo.comwww.livinginternet.comwww.opamp_electronics.comwww.asic-world.com/digitalwww.electroics-tutorials.ws http://www.kpsec.freeuk.com/gates.htm#nor http://www.electronics-tutorials.ws/boolean/bool_6.html

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http://www.electroics-tutorials.ws/http://www.kpsec.freeuk.com/gates.htmhttp://www.kpsec.freeuk.com/gates.htmhttp://www.kpsec.freeuk.com/gates.htmhttp://www.electroics-tutorials.ws/http://www.electroics-tutorials.ws/http://www.electroics-tutorials.ws/

boolean algebra part 2 april 73

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