Transcript Slide 1

Chapter 2: Boolean Algebra and
Logic Gates
Topics in this Chapter:
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Boolean Algebra
Boolean Functions
Boolean Function Simplification
Canonical and Standard Forms
Minterms and Maxterms
Building Boolean Function from the Truth Table
Conversion between Canonical Forms
Digital Logic Gates
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Boolean Algebra
•Purpose of BA is to facilitates design and analysis of digital circuits.
•For example the value of boolean function F=A + BC’ with the
following gate implementation can be shown by this truth table:
A B C BC A+BC
A
B
C
A+BC
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Boolean Algebra
• Basic identities of boolean algebra
1. X + 0 = X
2. X1 = X
identity
3. X + 1 = 1
4. X0 = 0 base
5. X + X = X
6. XX = X idempotence
7. X + X ’ = 1excluded middle 8. XX’ = 0 non contradiction
9. (X ’)’ = X involution
10. X + Y = Y + X
11. XY = YX commutative
12. X+(Y+Z ) = (X+Y )+Z 13. X(YZ ) = (XY )Z associative
14. X(Y+Z ) = XY + XZ 15. X+(YZ ) = (X+Y )(X+Z ) distributive
16. (X + Y ) = X Y 
17. (XY)’ = X+Y  demorgan
18. X + XY = X
19. X.(X+Y) = x
absorbtion
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Basic identities of B.A. can be proven by truth table:
(X + Y ) = X Y 
X+(YZ ) = (X+Y )(X+Z )
XY (X + Y ) X Y  XYZ YZ X+(YZ )X+Y X+Z (X+Y )(X+Z )
00
1
1
000 0
0
0
0
0
01
0
0
001 0
0
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1
0
10
0
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010 0
0
1
0
0
11
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011 1
1
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100 0
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101 0
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110 0
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demorgan
111 1
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distributive
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Boolean Algebra
• For each algebraic expression the dual of of
algebraic expression achieved by interchanging
AND and OR operators and replacing 0’s and 1’s.
• Parallel columns illustrate duality principle. The
duality principle states that if E1 and E2 are
Boolean expressions then
E1= E2  dual (E1)=dual (E2)
where dual(E) is the dual of E
• Note: 15-17 have no counterpart in ordinary
algebra.
• Other handy identity.
X+X’Y=X+Y (15, 7 and 2)
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Boolean Algebra
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By using boolean algebra rules, a simpler expression
may be obtained
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Operator Precedence: when evaluating boolean
expression , order of precedence is:
1- Parentheses
2-NOT
3-AND
4-OR
For example :Look at DeMorgan truth table first (X+Y) is
computed then complement of (X+Y).
But for X’.Y’ first the complement of X and complement of
Y is computed and then the result is ANDed
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Boolean Function Simplification
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Boolean function simplification
• It means by manipulation of B.A. reducing the
number of terms and literals in the function. For
example:
f= x’y’z + x’yz + xy’
= x’(y’z + yz) + xy’
= x’ (z(y’+y)) + xy’
= x’(z.1) +xy’
= x’z + xy’
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The Consensus Theorem
Theorem. XY +YZ + X Z = XY + X Z
Proof. XY +YZ + X Z
= XY + (X + X )YZ + X Z
2,7
= XY + XYZ + X YZ + X Z
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= XY(1 + Z ) + X Z(Y + 1) 2,11,14
= XY + X Z
3,2
Dual. (X + Y )(Y + Z )(X  + Z ) = (X + Y )(X  + Z )
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Complement of a Function
There are two ways for doing that:
1) Using DeMorgan’s theorem
2) Taking the dual of the function and
complement each literal
For example complements of
x’yz’ + x’y’z = (x+y’+ z)(x+ y+z’)
x(y’z’ + yz) = x’ + (y+z)(y’ + z’)
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Canonical and standard Forms
• The sum of products is one of two standard forms for Boolean
expressions.
sum-of-products-expression = p-term + p-term ... + p-term
p-term = literal 
literal    literal
– example. X Y Z + X Z + XY + XYZ
• A minterm is a product term that contains every variable, in either
complemented or un-complemented form.
– example. in expression above, X Y Z is minterm, but X Z is not
• A sum of minterms expression is a sum of products expression in
which every term is a minterm.
– example: X Y Z + X YZ + XYZ  + XYZ is sum of minterms
expression that is equivalent to expression above.
– shorthand : list minterms numerically, so X Y Z + X YZ +
XYZ  + XYZ becomes 001+011+110+111 or Sm (1,3,6,7)
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Canonical and standard Forms
• The product of sums is the second standard form for Boolean
expressions.
product-of-sums-expression = s-term  s-term ...  s-term
s-term = literal + literal + 
+ literal
– example. (X +Y +Z )(X +Z )(X +Y )(X +Y +Z )
• A maxterm is a sum term that contains every variable, in
complemented or uncomplemented form.
– example. in exp. above, X +Y +Z is a maxterm, but X +Z is not
• A product of maxterms expression is a product of sums expression in
which every term is a maxterm.
– example. (X +Y +Z )(X +Y+Z )(X+Y+Z )(X+Y+Z ) is product of
maxterms expression that is equivalent to expression above.
– shorthand : list maxterms numerically: so, (X +Y +Z )(X +Y+Z)
(X+Y+Z )(X+Y+Z ) becomes 110+100+001+000 or
– P M(6,4,1,0)
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How to build the boolean function
from truth table
• One way is to find a minterms or standard products by
ANDing the terms of the n variable, each being primed if
it is 0 and unprimed if it is 1. A boolean function can be
formed by forming a minterm for each combination of
variables that produce 1 in the function and then taking
OR of all those forms.
• Another way is by finding maxterms or standard sums
by OR term of the n variables, with each variable being
unprimed if corresponding bit is 0 and primed if it is 1. A
boolean function can be formed as a product of
maxterms for each combination of variables that produce
0 in the function and then form And of all those forms
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For example:
x y
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
z
0
1
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0
1
function f1 function f2
0
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1
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minterms maxterms
m0
M0
m1
M1
m2
M2
m3
M3
m4
M4
m5
M5
m6
M6
m7
M7
Sum of minterms
f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7
f2= x’yz + xy’z + xyz’ + xyx = m3 + m5 + m6 + m7
Product of maxterms
f1= (x+y+ z)(x+y’+z)(x+y’+z’)(x’+ y + z’)(x’ + y + z) = M0M2M3M5M6
f2= (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z) = M0M1M2M4
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Conversion between Canonical Forms
• To convert it to the product of maxterms
F= xy + x’z = (xy + x’)(xy +z) =
(x’ + x) (x’+ y)(x+z)(y+z) = (x’+y)(x+z)(y+z)
it is in the form of products of sums (P.O.S)
but we want the product of maxterms. So
= (x’+y+(z.z’)) (x + z + (y.y’))(y +z + (x.x’))
= (x’ + y + z)(x + z + y)(x+ z + y’)(x’ + y +z’)
= M0M2M4M5=∏(0,2,4,5)
• Easier than this is by using the truth table
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Conversion between Canonical Forms
• By reading from a truth table the two canonical
forms ( sum of minterms and product of
maxterms) can be easily obtained.
• A boolean function can be converted to the
canonical form. For example:
F= xy + x’z (is in form of the sum of the products
S.O.P) by doing
= (z+ z’)xy + x’z(y+y) = xyz + xyz’ + x’zy + x’zy’
it can be converted to sum of minterms. Using the
truth table the sum of minterms can be shown by
m1+m3+m6+m7 or ∑(1,3,6,7)
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Conversion between Canonical Forms
• In general to convert from one canonical form to another,
interchange the symbol ∑ and ∏ and list those numbers
missing from the total number of minterms or maxterms
which is 2n,where n is number of variables.
• To prove the correctness of the above conversion method
lets consider the following example:
F(x,y,z) = m1+m3+m6+m7= ∑(1,3,6,7)
We know the complement of F (presented in the form of
sum of the minterms) is the minterms that makes F to be
zero, Thus
F’(x,y,z) = (∑(1,3,6,7))’= (m0 + m2 + m4 + m5)
F = (F’(x,y,z))’ = (m0 + m2 + m4 + m5)’
Using Demorgan’s =m0’m2’m4’m5’=
since each m’j = Mj thus M0M2M4M5 = ∏(0,2,4,5)
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Standard forms
•
Sometimes boolean functions are shown as standard
forms. For example:
F1 =y’ + xy + x’y’z’ ( sum of products)
F2 = x(y’ + z) (x’ + y + z’) (product of sums)
the product and sum can be used to make the gate
structure consist of AND and OR gates
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Sometimes boolean function can be shown in non
standard forms:
F3= AB + C(D + E) can be changed to AB + CD + CE
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Different forms results different level of implementation
of logical gates (see the next slide)
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