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Minimum-Cost Boolean Expressions and Their Corresponding Circuits, Assignments of Electrical and Electronics Engineering

The University of Tennessee (UT)Electrical and Electronics Engineering

The minimum-cost sop and pos expressions for various boolean functions, along with their corresponding circuits implemented using nand and nor gates. It also includes examples where multiple minimum-cost pos forms exist.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

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2.40. The minimum-cost SOP expression for the function f(x1, x2, x3) = !m(3,4,6,7) is
f=x1x3+x2x3
The corresponding circuit implemented using NAND gates is
f
x1
x2
x3
2.41. A minimum-cost SOP expression for the function f(x1, x2, x3) = !m(1,3,4,6,7) is
f=x1x2+x1x3+x1x3
The corresponding circuit implemented using NAND gates is
f
x1
x2
x3
2.42. The minimum-cost POS expression for the function f(x1, x2, x3) = !m(3,4,6,7) is
f= (x1+x3)(x2+x3)
The corresponding circuit implemented using NOR gates is
f
x1
x2
x3
2-13
2.43. The minimum-cost POS expression for the function f(x1, x2, x3) = !m(1,3,4,6,7) is
f= (x1+x3)(x1+x2+x3)
The corresponding circuit implemented using NOR gates is
f
x1
x2
x3
2.44. The simplest SOP expression is derived as
f=x1x3+x1x2+x1x2+x2x3
=x1x3+x2x3+x1x2+x1x2+x1x2
=x1x3+x2x3+x1(x2+x2) + x2(x1+x1)
=x1x3+x2x3+x1+x2
=x1+x2
2.45. The simplest SOP expression is derived as
f=x1x2x3+x1x3+x2x3+x1x2x3
= (x1+x2)x3+ (x1+x2)x3+x1x2x3
=x3+x1x2x3
=x3+x1x2
2.46. The simplest POS expression is derived as
f=x2+x1x3+x1x3
=x2(x1+x3)(x1+x3)
=x2(x1x3+x3x1)
=x2x1x3+x2x3x1
T hen
f=x2x1x3+x2x3x1
=x2x1x3·x2x3x1
= (x2+x1+x3)(x2+x3+x1)
2-14
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2.40. The minimum-cost SOP expression for the function f (x 1 , x 2 , x 3 ) =

m(3, 4 , 6 , 7) is f = x 1 x 3 + x 2 x 3 The corresponding circuit implemented using NAND gates is f x 1 x 2 x 3 2.41. A minimum-cost SOP expression for the function f (x 1 , x 2 , x 3 ) =

m(1, 3 , 4 , 6 , 7) is f = x 1 x 2 + x 1 x 3 + x 1 x 3 The corresponding circuit implemented using NAND gates is f x 1 x 2 x 3 2.42. The minimum-cost POS expression for the function f (x 1 , x 2 , x 3 ) =

m(3, 4 , 6 , 7) is f = (x 1 + x 3 )(x 2 + x 3 ) The corresponding circuit implemented using NOR gates is f x 1 x 2 x 3 2- 2.43. The minimum-cost POS expression for the function f (x 1 , x 2 , x 3 ) =

m(1, 3 , 4 , 6 , 7) is f = (x 1 + x 3 )(x 1 + x 2 + x 3 ) The corresponding circuit implemented using NOR gates is f x 1 x 2 x 3 2.44. The simplest SOP expression is derived as f = x 1 x 3 + x 1 x 2 + x 1 x 2 + x 2 x 3 = x 1 x 3 + x 2 x 3 + x 1 x 2 + x 1 x 2 + x 1 x 2 = x 1 x 3 + x 2 x 3 + x 1 (x 2 + x 2 ) + x 2 (x 1 + x 1 ) = x 1 x 3 + x 2 x 3 + x 1 + x 2 = x 1 + x 2

Chapter 4

4.1. SOP form: f = x 1 x 2 + x 2 x 3 POS form: f = (x 1 + x 2 )(x 2 + x 3 ) 4.2. SOP form: f = x 1 x 2 + x 1 x 3 + x 2 x 3 POS form: f = (x 1 + x 3 )(x 1 + x 2 )(x 2 + x 3 ) 4.3. SOP form: f = x 1 x 2 x 3 x 4 + x 1 x 2 x 3 x 4 + x 2 x 3 x 4 POS form: f = (x 1 + x 4 )(x 2 + x 3 )(x 2 + x 3 + x 4 )(x 2 + x 4 )(x 1 + x 3 ) 4.4. SOP form: f = x 2 x 3 + x 2 x 4 + x 2 x 3 x 4 POS form: f = (x 2 + x 3 )(x 2 + x 3 + x 4 )(x 2 + x 4 ) 4.5. SOP form: f = x 3 x 5 + x 3 x 4 + x 2 x 4 x 5 + x 1 x 3 x 4 x 5 + x 1 x 2 x 4 x 5 POS form: f = (x 3 + x 4 + x 5 )(x 3 + x 4 + x 5 )(x 2 + x 3 + x 4 )(x 1 + x 3 + x 4 + x 5 )(x 1 + x 2 + x 4 + x 5 ) 4.6. SOP form: f = x 2 x 3 + x 1 x 5 + x 1 x 3 + x 3 x 4 + x 2 x 5 POS form: f = (x 1 + x 2 + x 3 )(x 1 + x 2 + x 4 )(x 3 + x 4 + x 5 ) 4.7. SOP form: f = x 3 x 4 x 5 + x 3 x 4 x 5 + x 1 x 4 x 5 + x 1 x 2 x 4 + x 3 x 4 x 5 + x 2 x 3 x 4 + x 2 x 3 x 4 x 5 POS form: f = (x 3 + x 4 + x 5 )(x 3 + x 4 + x 5 )(x 1 + x 2 + x 3 + x 4 + x 5 ) 4.8. f =

m(0, 7) f =

m(1, 6) f =

m(2, 5) f =

m(0, 1 , 6) f =

m(0, 2 , 5) etc. 4.9. f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 4.10. SOP form: f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 + x 1 x 3 x 4 + x 2 x 3 x 4 POS form: f = (x 1 + x 2 + x 3 )(x 1 + x 2 + x 4 )(x 1 + x 3 + x 4 )(x 2 + x 3 + x 4 )(x 1 + x 2 + x 3 + x 4 ) The POS form has lower cost. 4.11. The statement is false. As a counter example consider f (x 1 , x 2 , x 3 ) =

m(0, 5 , 7). Then, the minimum-cost SOP form f = x 1 x 3 + x 1 x 2 x 3 is unique. But, there are two minimum-cost POS forms: f = (x 1 + x 3 )(x 1 + x 3 )(x 1 + x 2 ) and f = (x 1 + x 3 )(x 1 + x 3 )(x 2 + x 3 ) 4- 4.5. SOP form: f = x 3 x 5 + x 3 x 4 + x 2 x 4 x 5 + x 1 x 3 x 4 x 5 + x 1 x 2 x 4 x 5 POS form: f = (x 3 + x 4 + x 5 )(x 3 + x 4 + x 5 )(x 2 + x 3 + x 4 )(x 1 + x 3 + x 4 + x 5 )(x 1 + x 2 + x 4 + x 5 ) 4.6. SOP form: f = x 2 x 3 + x 1 x 5 + x 1 x 3 + x 3 x 4 + x 2 x 5 POS form: f = (x 1 + x 2 + x 3 )(x 1 + x 2 + x 4 )(x 3 + x 4 + x 5 ) 4.7. SOP form: f = x 3 x 4 x 5 + x 3 x 4 x 5 + x 1 x 4 x 5 + x 1 x 2 x 4 + x 3 x 4 x 5 + x 2 x 3 x 4 + x 2 x 3 x 4 x 5 POS form: f = (x 3 + x 4 + x 5 )(x 3 + x 4 + x 5 )(x 1 + x 2 + x 3 + x 4 + x 5 ) 4.8. f = ∑ m(0, 7) f = ∑ m(1, 6) f = ∑ m(2, 5) f = ∑ m(0, 1 , 6) f = ∑ m(0, 2 , 5) etc. 4.9. f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 4.10. SOP form: f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 + x 1 x 3 x 4 + x 2 x 3 x 4 POS form: f = (x 1 + x 2 + x 3 )(x 1 + x 2 + x 4 )(x 1 + x 3 + x 4 )(x 2 + x 3 + x 4 )(x 1 + x 2 + x 3 + x 4 ) The POS form has lower cost. 4.11. The statement is false. As a counter example consider f (x 1 , x 2 , x 3 ) = ∑ m(0, 5 , 7). Then, the minimum-cost SOP form f = x 1 x 3 + x 1 x 2 x 3 is unique. But, there are two minimum-cost POS forms: f = (x 1 + x 3 )(x 1 + x 3 )(x 1 + x 2 ) and f = (x 1 + x 3 )(x 1 + x 3 )(x 2 + x 3 ) 4-