Forms of Expression Representation

• Let's take a look at a simple 2 input truth table.

  a	b	a·b	a·b	a·b	ab
  0	0	1	0	0	0
  0	1	0	1	0	0
  1	0	0	0	1	0
  1	1	0	0	0	1

• Note that each of these terms produce exactaly one 1.
• Remember a + 0 = a
• So we could form a ⊕ b = a·b + a·b

  • a	b	a·b	a·b	a ⊕ b
    0	0	0	0	0
    0	1	1	0	1
    1	0	0	1	1
    1	1	0	0	0

• We call each of the terms in the first table minterms
  • A minterm is a product term where each variable in the expression appears exactaly once.
  • These correspond to the rows in a truth table.
  • The only operators are the and and not
  • Not is only applied to individual terms.
  • term a is positive if the corresponding input value is 1 and a if the input value is 0.
• For an expression with n input values, there are $2^n$ minterms.
  • Each corresponds to the binary pattern.
  • In three bits m₀ corresponds to input 000 and term a·b·c
  • In three bits m₅ corresponds to input 101 and term a·b·c

  • a	b	a⊕b	Minterm
    0	0	0
    0	1	1	ab, m1
    1	0	1	ab, m2
    1	1	0

• If F(a,b) = a⊕ b
  • The above truth table is one way to represent the function
  • The sum of products F(a,b) = ab + ab is another.
  • Sometimes you see F(a,b) = m₁ + m₂
  • Or even F(a,b) = Σm(1,2)
  • We cound notice that F(a,b) = (a+b)(a + b). This is calld the product of sums.
• The last term might require a bit more explinantion
  • The maxterms like the minterms are formed of
    • Exactaly one instance of each variable
    • only or and not
    • The not is only applied to individual terms.
    • These are rows in the table where the answer is false.
    • Designated M_i where i is formed as above.
    • For the table M₀ = a + b, M₃ = a + b

    • a	b	a⊕b	Minterm
      0	0	0	a + b, M0
      0	1	1
      1	0	1
      1	1	0	a + b, M3

    • Let's build the table for the minterms

      a	b	a+b	a + b		(a+b)( a + b)
      0	0	0	1	0
      0	1	1	1	1
      1	0	1	1	1
      1	1	1	0	0

    • We could also expand (a+b)( a + b)

       (a + b)( a + b) = a·a + a·b + b·a + b·b
                       =  1  + a·b + b·a + 1 
                       =  a·b + b·a

  • so F(a,b) = M₀·M₃
  • This can also be expressed by F(a,b) = ΠM(0,3)
  • And the general table:

  • a	b	a + b	a + b	a + b	a + b
    0	0	0	1	1	1
    0	1	1	0	1	1
    1	0	1	1	0	1
    1	1	1	1	1	0

• Note:
  • The sum of products is called Disjunctive Normal Form
  • The product of sums is called Conjunctive Normal Form
• There are two more forms that the book (Kumar) refers to as "Canonical"
  • Disjunctive Canonical Form is a sum of produts where all terms contains all variables.
  • Conjunctive Canonical Form is a product of sums where each factor contains all variables.
  • For F(a,b) = a⊕b our two expressions were in Canonical form
    • F(a,b) = a·b + b·a
    • F(a,b) + (a + b)( a + b)
  • Let's examin F(a,b,c,d) = a(c+d)+cbd
    • We can build the table if you wish,
      • How many rows? (2⁴ = 16)

      • a	b	c	d	c	c + d	a(c + d)	cbd	a(c+d)+cbd
        0	0	0	0	1	1	0	0	0
        0	0	0	1	1	1	0	0	0
        0	0	1	0	0	0	0	0	0
        0	0	1	1	0	1	0	0	0
        0	1	0	0	1	1	0	0	0
        0	1	0	1	1	1	0	1	1
        0	1	1	0	0	0	0	0	0
        0	1	1	1	0	1	0	0	0
        1	0	0	0	1	1	1	0	1
        1	0	0	1	1	1	1	0	1
        1	0	1	0	0	0	0	0	0
        1	0	1	1	0	1	1	0	1
        1	1	0	0	1	1	1	0	1
        1	1	0	1	1	1	1	1	1
        1	1	1	0	0	0	0	0	0
        1	1	1	1	0	1	1	0	1

    • a(c+d)+cbd = ac+ ad + cbd 

    • Note each term is a product (and)
    • And the final is a sum of these (or)
    • This is a sum of products.
    • F(a,b,c,d) = a(c+d)+cbd
    • The SOP is F(a,b,c,d) = ac+ ad + cbd
    • We could convert this to DCF

    • ac+ ad + cbd 
         = ac(b + b) + ad + cbd 
         = acb + ac·b + ad + cbd 
         = abc + ab·c + ad + cbd 
         = abc(d + d) + ab·c + ad + cbd 
         = abcd + abc·d + ab·c + ad + cbd 
         = m13 + m12 + ab·c + ad + cbd 
         = m13 + m12 + ab·c(d + d) + ad + cbd 
         = m12 + m13  + ab·cd + ab·c·d + ad + cbd 
         = m12 + m13 + m9 + m8 + ad + cbd 
         = m8 + m9 + m12 + m13 + ad(c + c) + cbd(a + a) 
         = m8 + m9 + m12 + m13 + adc + adc + cbda + cbda
         = m8 + m9 + m12 + m13 + adc + adc + abcd + abcd
         = m8 + m9 + m12 + m13 + adc + adc + m13 + m5 
         = m5 + m8 + m9 + m12 + m13 + acd + acd
         = m5 + m8 + m9 + m12 + m13 + acd(b+b) + acd(b+b)
         = m5 + m8 + m9 + m12 + m13 + acdb+acdb + acdb+ acdb
         = m5 + m8 + m9 + m12 + m13 + abcd + abcd + abcb+ ab·cd
         = m5 + m8 + m9 + m12 + m13 + m15 + m11 + m13+ m9
         = m5 + m8 + m9 + m11 + m12 + m13 + m15
         = Σm(5,8,9,11,12,13,15)
                     

  • Note from this we can easily determine that F(a,b,c,d) = ΠM(0,1,2,3,4,7,6,10,14)
    • But I have no desire to create this expression.