Simplify the following Boolean functions to a minimum number of literals'

1

• Example
• Simplify the following Boolean functions to a
  minimum number of literals.
• x(xy)
• xx xy By Distributive law
• 0 xy By 2b
• xy By 1a
• 2. xxy
• (xx)(xy) By Distributive law
• 1(xy) By 2a
• xy By 1b

2

• 3. (xy)(xy)
• x yy By Distributive law
• x
• 4. xyxz yz
• xy xzyz(xx) By 2a
• xyxz xyz xyz By distributive
  law
• xy(1z) xz(1y) By associative
  law
• xy xz

3

• Example
• Find the complement of the functions F1 xyz
  xyz and
• F2 x(yzyz) by applying DeMorgans theorem
  as many times as necessarily
• F1 (xyz xyz )
• (xyz)(xyz)
• (xyz)(xyz)
• F2 x(yzyz)
• x ( yz yz)
• x (yz). (yz)
• x ( y z)(yz)

4

• Synthesis Using AND, OR and NOT
• As was previously stated, synthesis is the design
  of a network that implements a desired functional
  behavior.
• Design Example
• Suppose that we want to synthesize a circuit that
  has two switches x and y. The required functional
  behavior of the circuit is that the output must
  be equal to 0 if the switch x is closed (x0 )
  and y is open (y1) otherwise the output must be
  1.
• Answer
• We can show the function by summing all product
• terms that correspond to those rows that value
  of
• function is 1
• f x y x y xy

5

• Continue on example
• We can try to simplify the above expression by
  manipulating it using the various theorems and
  properties.
• According to the theorem xxx we can replicate
  any term in a logical sum expression.
• Replicating the xyterm yields
• f x y x y x y xy
• Using the associative property we can group
  elements to obtain
• f ( x y x y ) ( x y xy)
• Using the distributive property
• f y( x x ) x( y y )
• Applying the theorem xx1 we get
• f y . 1 x . 1
• Finally, using the theorem x . 1x leads to
• f y x
• Which can again be implemented using our standard
  gates. Obviously, the cost of implementation of
  simplified function is much less than the first
  one.

6

• We can conclude that
• Any function can be implemented by using a
  product (AND) term for each row of the truth
  table for which the function is equal to 1. The
  sum (OR) of these terms realizes the function.
• The above example illustrated that different
  algebraic expressions can yield the same output.
  However, there are two standard ways of
  expressing a given function the standard form
  and the canonical form. (which will be discuss on
  next slide)

7

• Conical and Standard form of function
• The standard form of an expression is either a
  sum-of-products
• (SoP) or a product-of-sums (PoS).
• SoP - A Boolean expression containing ORing the
  product terms (prouduct terms are variables that
  are ANDed together)
• f xy x z (SoP) standard form
• PoS - A Boolean expression containing ANDing the
  sum terms (sum terms are variables that are ORed
  together).
• f ( x y )( x z ) (PoS) standard form
• A function that is represented as a mix of PoS
  and SoP is nonstandard form of function.
• f y ( x z ) xy nonstandard form

8

• If the terms in a standard form reference all
  of the variables, then the expression is in
  canonical form.
• When an expression is in a canonical SoP form
  the terms in the expression are referred to as
  minterms.
• A minterm is product term in which all the
  variable
• appear exactly once, either in Complemented or
• uncomplemented form.
• There are 2n distinct minterms for n variables.
  For example if there are two variables, x and y,
  there would be four minterms
• x'y', xy, x'y, and xy.
• For a given row of the truth table, the minterm
  is formed by including x if x1 and by including
  x if x0.

9

• mj represents a minterm where the j denotes the
  decimal equivalent of the binary combination.

10

• Likewise, when an expression is in a canonical
  PoS form the terms in the expression are referred
  to as maxterms.
• A maxterm is sum term in which all the variables
  appear exactly once, either in complemented or
  uncomplemented form.
• Mj represents a maxterm,
• where j denotes the decimal
• equivalent of the binary
• combination

11

• Each maxterm is obtained by ORing (sum) each
  variable.
• Note Each maxterm is the complement of its
  corresponding minterm, and vice versa.
• Any Boolean function can be expressed
  algebraically from a given truth table by forming
  a logical sum of all the minterms that produce a
  1 in the function (f) column. Such an expression
  is called a sum of minterms.
• Similarly, any Boolean function can be expressed
  algebraically from a given truth table by forming
  the logical product( ANDing) of all the maxterms
  that produce a 0 in the function (f column).
  Such an expression is called a product of
  maxterms.

12

• Example
• Determine the function which corresponds to the
  following truth table
• f1 has a value of 1, for minterms m2 , m4, m5 ,
  and m7.
• f1 xy z x y z x y z xyz
• m2 m4 m5 m7
• This can be further abbreviated as
• f 1 (x,y,z) S (2,4,5,7)
• Similarly,
• f2 x y z x y z x y z xyz
• m0 m1 m4 m7
• This can be further abbreviated as
• f2 (x,y,z) S (0,1,4 ,7)

13

• Now, lets consider the complement of the
  functions. By forming a minterm for each
  combination that produces a 0 and then ORing
  those terms, we can obtain the complement of f1.
• (f 1) x y z x y z xyz xyz m 0
  m 1 m 3 m 6
• If we take the complement of f 1 , we obtain
  the original function f1 however in PoS form.
• ((f 1) ) (x y z x y z xyz
  xyz)
• f 1 ( x y z ) ( x y z) (
  x y z ) ( x y z )
• M0. M1. M3. M6
• f1 ( x, y, z) P (0, 1 , 3 , 6 )
• Similarly, the function f2 can be read from the
  table as
• f 2 (x,y,z) ( x y z ) ( x y z ) (
  x y z ) ( x y z )
• M 2. M 3. M 5. M 6
• f2 (x , y, z) P ( 2 , 3 ,5 , 6 )

14

• Note
• A function that includes all the 2n minterms is
  equal to logic 1.
• From the design example (page 4 ), it can be
  seen that there is a direct relationship between
  a function expressed in canonical form and a
  truth table. As a result it is useful to have
  functions expressed in canonical form.

15

• Procedures to represent a function in (SoP) and
  (PoS)
• To find the sum of product of a given function
  from truth table
• Make the truth table for the function
• Look at those rows that function is 1
• Write down the corresponding product terms and
  sum them together to find sum of products (SoP)
• To find the product of sums
• From truth table find out complement of function
  ,f, (Find the complement of function by
  obtaining SoP of the terms that ).
• find out (f) which will result in f but with
  product of sum (PoS) form

16

• Example
• Given the function f AB AC , find its
  representation in both sum of minterm and product
  of maxterm.
• Make the truth table for function by putting the
  value of function to 1 for those terms that AB1
  or AC 1
• Finding the SoP form of function
• f ABCABCABCABC
• Find the complement of function
• by obtaining SoP of the terms that
• are 0 on the f column
• fABCABCABCABC
• To get the function back with PoS
• representation get (f) the result
• would be
• (f) f (ABC)(ABC)(ABC)(ABC)