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Gate-level Minimization. Although truth tables representation of a function is unique, it can be expressed algebraically in different forms The procedure of simplifying Boolean expressions (in 2-4) is

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Gate-level Minimization

  • Although truth tables representation of a function is unique, it can be expressed algebraically in different forms

  • The procedure of simplifying Boolean expressions (in 2-4) is

    difficult since it lacks specific rules to predict the successive steps in the simplification process.

  • Alternative: Karnaugh Map (K-map) Method.

    • Straight forward procedure for minimizing Boolean Function

    • Fact: Any function can be expressed as sum of minterms

    • K-map method can be seen as a pictorial form of the truth table.

y

y

x

m0

m1

m2

m3

x

Two-variable map


Two-variable K-MAP

y

y

x

x

y

y

y

y

x

x

x

x


Two-variable K-MAP

y

y

y

y

x

x

x

x

The three squares can be determined from the intersection

of variable x in the second row and variable y in the second

column.


Three-Variable K-Map

  • Any two adjacent squares differ by only one variable.

  • M5 is row 1 column 01.  101= xy’z=m5

  • Since adjacent squares differ by one variable (1 primed, 1 unprimed)

    • From the postulates of Boolean algebra, the sum of two minterms in adjacent squares can be simplified to a simple AND

    • For example m5+m7=xy’z+xyz=xz(y’+y)=xz



Three-Variable K-Map

Example 2

Simplify:


Three-Variable K-Map

Example 3

Simplify:


Three-Variable K-Map

Example 3

Simplify:


Three-Variable K-Map

Example 4

Given:

(a) Express F in sum of minterms.

(b) Find the minimal sum of products using K-Map

(a)


Three-Variable K-Map

Example 4 (continued)


Three-variable K-Map: Observations

  • One square represents one minterm  a term of 3 literals

  • Two adjacent squares  a term of 2 literals

  • Four adjacent squares  a term of 1 literal

  • Eight adjacent squares  the function equals to 1



Four-Variable K-Map

Simplify F(w,x,y,z) = S(0,1,2,4,5,6,8,9,12,13,14)

Example 5

1


Four-Variable K-Map

Example 6

Simplify F(A,B,C,D) =

Represented by 0001 or 0000


Prime Implicants

  • Need to ensure that all Minterms of function are covered

  • But avoid any redundant terms whose minterms are already covered

  • Prime Implicant is product Term obtained by combining maximum possible number of adjacent squares

  • If a minterm in a square is covered by only prime implicant then ESSENTIAL PRIME IMPLICANT

Non Essential prime implicant CD, B’C, AD and AB’

Essential prime implicant BD and B’D’


Four-variable K-Map: Observations

  • One square represents one minterm  a term of 4 literals

  • Two adjacent squares  a term of 3 literals

  • Four adjacent squares  a term of 2 literal

  • Eight adjacent squares  a term of 1 literal

  • sixteen adjacent squares  the function equals to 1


SUM of PRODUCT and PRODUCT OF SUM

Simplify the following Boolean function in:

(a) sum of products (b) product of sums

Combining the one’s:

(a)

Combining the zero’s:

Taking the the complement:

(b)


SOP and POS gate implementation

PRODUCT OF SUM (POS)

SUM OF PRODUCT (SOP)


Implementation of Boolean Functions

  • Draw the logic diagram for the following function: F = (a.b)+(b.c)

a

b

F

c


  • Implement a circuit

    • 2 Level

    • More than two level

    • SOP

    • POS

  • Implement a circuit using OR and Inverter Gates only

  • Implement a circuit using AND and Inverter Gates only

  • Implement a circuit using NAND Gates only

  • Implement a circuit using NOR Gates only



F=AB+CD

F=(A’B’)’+(C’D’)’

F=[(AB)’.(CD)’]’=AB+CD

TWO LEVEL

IMPLEMENT-ATION



CHAPTER 4

COVERT AND TO NAND WITH AND INVER.

CONVERT OR TO NAND WITH INVERT OR. SINGLE BUBBLE WITH INVERTER



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