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Chapter 2: Boolean Algebra and Logic Gates

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Title: Chapter 2: Boolean Algebra and Logic Gates


1
Chapter 2 Boolean Algebra and Logic Gates
  • Topics in this Chapter
  • 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

2
Boolean Algebra
  • Purpose of BA is to facilitates design and
    analysis of digital circuits.
  • For example the value of boolean function FA
    BC with the following gate implementation can be
    shown by this truth table

3
Boolean Algebra
  • Basic identities of boolean algebra

1. X 0 X 3. X 1 1 5. X X X
7. X X 1excluded middle 9. (X ) X
involution 10. X Y Y X 12. X(YZ ) (XY
)Z 14. X(YZ ) XY XZ 16. (X Y )? X ?Y
? 18. X XY X
2. X1 X identity 4. X0 0
base 6. XX X idempotence 8. XX 0
non contradiction 11. XY YX
commutative 13. X(YZ ) (XY )Z
associative 15. X(YZ ) (XY )(XZ )
distributive 17. (XY) X?Y ? demorgan 19.
X.(XY) x absorbtion
4
Basic identities of B.A. can be proven by truth
table
demorgan
distributive
5
Boolean Algebra
  • For each algebraic expression the dual of of
    algebraic expression achieved by interchanging
    AND and OR operators and replacing 0s and 1s.
  • 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.
  • XXYXY (15, 7 and 2)

6
Boolean Algebra
  • By using boolean algebra rules, a simpler
    expression may be obtained
  • 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
    (XY) is computed then complement of (XY).
  • But for X.Y first the complement of X and
    complement of Y is computed and then the result
    is ANDed

7
Boolean Function Simplification
8
Boolean function simplification
  • It means by manipulation of B.A. reducing the
    number of terms and literals in the function. For
    example
  • f xyz xyz xy
  • x(yz yz) xy
  • x (z(yy)) xy
  • x(z.1) xy
  • xz xy

9
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 14
  • 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 )

10
Complement of a Function
  • There are two ways for doing that
  • Using DeMorgans theorem
  • Taking the dual of the function and complement
    each literal
  • For example complements of
  • xyz xyz (xy z)(x yz)
  • x(yz yz) x (yz)(y z)

11
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 001011110111 or
    Sm (1,3,6,7)

12
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 ?YZ )(XYZ ?)(XYZ )
    is product of maxterms expression that is
    equivalent to expression above.
  • shorthand list maxterms numerically so, (X ?Y
    ?Z )(X ?YZ) (XYZ ?)(XYZ ) becomes
    110100001000 or
  • P M(6,4,1,0)

13
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

14
  • For example
  • x y z function f1 function f2
    minterms maxterms
  • 0 0 0 0
    0 m0 M0
  • 0 0 1 1
    0 m1 M1
  • 0 1 0 0
    0 m2 M2
  • 0 1 1 0
    1 m3 M3
  • 1 0 0 1
    0 m4 M4
  • 1 0 1 0
    1 m5 M5
  • 1 1 0 0
    1 m6 M6
  • 1 1 1 1
    1 m7 M7
  • Sum of minterms
  • f1 xyz xyz xyz m1 m4 m7
  • f2 xyz xyz xyz xyx m3 m5 m6
    m7
  • Product of maxterms
  • f1 (xy z)(xyz)(xyz)(x y z)(x y
    z) M0M2M3M5M6
  • f2 (xyz)(xyz)(xyz)(xyz) M0M1M2M4

15
Conversion between Canonical Forms
  • To convert it to the product of maxterms
  • F xy xz (xy x)(xy z)
  • (x x) (x y)(xz)(yz) (xy)(xz)(yz)
  • it is in the form of products of sums (P.O.S)
    but we want the product of maxterms. So
  • (xy(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

16
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 xz (is in form of the sum of the
    products S.O.P) by doing
  • (z z)xy xz(yy) xyz xyz xzy
    xzy
  • it can be converted to sum of minterms. Using
    the truth table the sum of minterms can be shown
    by m1m3m6m7 or ?(1,3,6,7)

17
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) m1m3m6m7 ?(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 Demorgans m0m2m4m5
  • since each mj Mj thus M0M2M4M5 ?(0,2,4,5)

18
Standard forms
  • Sometimes boolean functions are shown as standard
    forms. For example
  • F1 y xy xyz ( 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
  • Sometimes boolean function can be shown in non
    standard forms
  • F3 AB C(D E) can be changed to AB
    CD CE
  • Different forms results different level of
    implementation of logical gates (see the next
    slide)

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