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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
5
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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
7
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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
12
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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.
13
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Example 4
Construct a Truth Table for the logical functions
at points C, D and Q in the following circuit .
14
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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
19
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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
21
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/