Chapter 2: Boolean Algebra and Logic Gates

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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

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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 FA
  BC with the following gate implementation can be
  shown by this truth table

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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
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Basic identities of B.A. can be proven by truth
table
demorgan
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 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)

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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

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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 xyz xyz xy
• x(yz yz) xy
• x (z(yy)) xy
• x(z.1) xy
• xz 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 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 )

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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)

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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 001011110111 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 ?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)

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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 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
  

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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

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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 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)

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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) 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)

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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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