1 gate-level minimization although truth tables representation of a function is unique, it can be...

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Gate-level Minimization

Although truth tables representation of a function is unique, it can be expressed algebraically in different formsThe procedure of simplifying Boolean expressions (in 2-4) isdifficult 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 FunctionFact: Any function can be expressed as sum of minterms K-map method can be seen as a pictorial form of the truth table.

m0 m1

m2 m3

xy

'' yx yx'

'xy xy

0 1

1

0

y

x

Two-variable map

2

xy

'' yx yx'

'xy xy

0 1

1

0

y

x

Two-variable K-MAP

xy

xyF 1

0 1

1

0

y

x

xy 0 1

1

0

y

x1 1 1

1

xyxyyx

mmmF

'' 3212

3

xy 0 1

1

0

y

x 1 1

1

yxF 2

The three squares can be determined from the intersectionof variable x in the second row and variable y in the second column.

xy

'' yx yx'

'xy xy

0 1

1

0

y

x

Two-variable K-MAP

4

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

Three-Variable K-Map

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2 3 4 5( , , , ) ' ' ' ' ' '

' ( ') '( ') ' '

F m m m m x yz x yz xy z xy z

x y z z xy z z x y xy

Example 1


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Example 2


)7,6,4,3(),,( zyxFSimplify:

m0 m1 m3 m2

m4 m5 m7 m6

'xz 'xzyz

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)6,5,4,2,0(),,( zyxFExample 3

Simplify:

m0 m1 m3 m2

m4 m5 m7 m6

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)6,4,2,0(),,( zyxFExample 3

Simplify:

m0 m1 m3 m2

m4 m5 m7 m6

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Example 4


Given: BCCABBACACBAF '''),,(

(a) Express F in sum of minterms. (b) Find the minimal sum of products using K-Map

BCAABCAABC

CAB

BCABCACCBA

CBABCABBCA

')'(

'

''')'('

''')'('

)7,5,3,2,1(

''''''),,(

ABCCABBCABCACBACBAF(a)

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Example 4 (continued) )7,5,3,2,1(),,( CBAF

m0 m1 m3 m2

m4 m5 m7 m6

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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

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Four-Variable K-Map

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Four-Variable K-Map

Example 5

'''' xzzwyF

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

1

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Four-Variable K-Map

Example 6

'''''' CDACBDBF

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

'''

'''''''''

CBA

CABBCDACDBCBA Represented by 0001 or 0000

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• 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

Prime Implicants

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

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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

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'AB CD BD

' 'F AB CD BD

Simplify the following Boolean function in:(a) sum of products (b) product of sums

( , , , ) (0,1,2,5,8,9,10)F A B C D Combining the one’s:

Combining the zero’s:

' ' ' ' ' 'F B D B C A C D

Taking the the complement:

( ') '

( ' ')( ' ')( ' )

F F

A B C D B D

SUM of PRODUCT and PRODUCT OF SUM

(a)

(b)

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SUM OF PRODUCT (SOP) PRODUCT OF SUM (POS)

SOP and POS gate implementation

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

ab

c

F

Implementation of Boolean Functions

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• 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

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NAND IMPLEMENTATION

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TWO LEVEL

IMPLEMENT-ATION

F=AB+CDF=(A’B’)’+(C’D’)’

F=[(AB)’.(CD)’]’=AB+CD

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F(X,Y,Z)=(1,2,3,4,5,7) SUM OF PRODUCT

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COVERT AND TO NAND WITH AND INVER.

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

CHAPTER 4

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• SIMPLIFICATION WITH TABULATION METHOD DO IT ON BOARD

1 gate-level minimization although truth tables representation of a function is unique, it can be...

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