CS 121 Digital Logic Design

CS 121Digital Logic

Design

Gate-Level Minimization

Chapter 3

Outline3.1 Introduction3.2 The Map Method3.3 Four-Variable Map3.4 Product of sums simplification3.5 Don‘t Care Conditions3.7 NAND and NOR Implementaion3.8 Other Two-Level Implementaion3.9 Exclusive-OR function

3.1 Introduction (1-1)

Gate-Level Minimization refers to the design task of finding an optimal gate-level implementation of the Boolean functions describing a digital circuit.

Notes about simplification of Boolean expression: Minimum number of terms and literals in each term

(minimum number of gates and inputs in the digram). Reduce the complexity of the digital gates. The simplest expression is not unique.

Simplification Methods: Algebraic minimization lack on specific rules. (section

2.4). Karnaugh map or K-map.

A Karnaugh map is a graphical tool for assisting in the general simplification procedure.

Combination of 2, 4, … adjacent squares

The relation is:Logic circuit ↔ Boolean function ↔ Truth table ↔ K-map ↔ conical form ↔ satndrad form.

Conical form: ( sum of minterms , product of maxterms.

Standrad form: ( simplifier : sum of product , product of sum

3.2 The Map Method (1-12)

Number of sequares (minterms) is , where n is the number of variables.

So in tow-variable map there are 4 squares(minterms).

3.2 The Map Method (2-12)Two-variable

maps:

Y’ Y

X

X’


Rules for K-map:o We can reduce functions by circling 1’s in the K-mapo Each circle represents a minterm reductiono Following circling, we can deduce minimized and-or

form. Rules to considero Every cell containing a 1 must be included at least

once.o The largest possible “power of 2 rectangle” must be

enclosed.o The 1’s must be enclosed in the smallest possible

number of rectangles.

Example 1:F(X,Y) = XY’ + XY

Two-Variable maps (cont.)


From the map, we see thatF (X,Y) = X.Note: There are implied 0s in other boxes.

This can be justified using algebraic manipulations:F(X,Y) = XY’ + XY

= X(Y’ +Y) = X.1

= X

11 X

Example 2:G(x,y) = m1 + m2 + m3



G(x,y) = m1 + m2 + m3

= X’Y + XY’ + XY

From the map, we can see that :

G = X + Y

1

11 X

Y

Example 3:F = Σ(0, 1)



Using algebraic manipulations:

F = Σ(0,1) = x’y + x’y’ = x’ (y+y’) = x’

11 X’

x y F0 0 10 1 11 0 01 1 0

3 variables 8 squares ( minterms). On a 3-variable K-Map:

◦ One square represents a minterm with three variables◦ Two adjacent squares represent a product term with two variables◦ Four “adjacent” terms represent a product term with one variables◦ Eight “adjacent” terms is the function of all ones (logic 1).

3.2 The Map Method (7-12)Three-variable

maps:

using algebraic manipulations: F = X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’ = Z’ (X’Y’ + X’Y + XY’ + XY)

= Z’ (X’ (Y’+Y) + X (Y’+Y)) = Z’ (X’+ X)

= Z’

3.2 The Map Method (8-12)Three-variable maps

(cont.):Example 1:F(X,Y) = X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’

Y Z

x 1

1

1

1

Example 2:F=AB’C’ +ABC +ABC +ABC + A’B’C + A’BC’

three-Variable maps (cont.)


From the map, we see thatF=A+BC +BC

B CA

1

1

1

11 1

00 01 11 1001

Example 4 :F (x, y, z)= Σm (2, 3, 6, 7)



using algebraic manipulations:

F(x , y, z) = x’yz + xyz + x’yz’ + xyz’

= yz (x’ + x) + yz’ (x’ + x)

= yz + yz’ = y (z + z’) = y

y zx

1 1

11

00 01 11 1001

Y

Example (3-1) , (3-2) :



Example (3-3) , (3-4) :



3.3 Four-Variables Map (1-9)

4 variables 16 squares ( minterms). On a 4-variable K-Map: Two adjacent squares represent a term of three literals. Four adjacent squares represent a term of two literals. Eight adjacent squares represent a term of one literal. Note: The larger the number of squares combined, the

smaller the number of literals in the term.


Flat Map Vs. Torus


Example 1 (3-5) :F(w,x,y,z) = ∑ ( 0,1,2,4,5,6,8,9,12,13,14)

y z

w x 1

00 01 11 1000011110

1

1 1

1 1

1 1

1

1

1Y’

W’YZ’

XYZ’

F = y‘ + w‘yz‘ + xyz‘


Example 2 (3-6) :F = A’B’C’ + B’CD’ + A’BCD’ + AB’C’

C D

A B1

00 01 11 101

F = B‘D‘ + B‘C‘ + A‘CD‘

00011110

1

1

11 1 B’D’

B’C’

A’CD’


Simplification using Prime Implicantso A Prime Implicant is a product term obtained by

combining the maximum possible number of adjacent squares in the map .

o If a minterm in a square is covered by only one prime implicant, that implicant is called Essential .

o Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map.

o Notes:• Two adjacent 1’s form prime implicant, if they are not within a

group of four adjacent squares.• Four adjacent 1’s form prime implicant, if they are not within

a group of eight adjacent squares and so on.

3.3 Four-Variables Map (6-9)Simplification using Prime ImplicantsExample 1:F(A,B,C,D) = ∑ (0,2,3,5,7,8,9,10,11,13,15)

Minterms covered by single prime implicant

DB

CB

1 1

1 1

1 1

B

D

A

1 11 1

1

ESSENTIAL Prime Implicants

C

BD

CD

BD

DB

1 1

1 1

1 1

B

C

D

A

1 1

1 1

1

AD

BA

3.3 Four-Variables Map (7-9)Simplification using Prime ImplicantsExample 1:F(A,B,C,D) = ∑ (0,2,3,5,7,8,9,10,11,13,15)

o Essential prim implicants:BD , B’D’o Prime implicant: CD , B’C, AD , AB’.o The minterms that not cover by essential implicants are: m3, m9,

m11.The simplified expression is optained from the sum of the

essential implicants and other prime implicants that may be needed to cover any remaining minterms.

o So this function can be written with these ways:• F = BD + B’D’ + CD + AD• F = BD + B’D’ + CD + AB’• F = BD + B’D’ + B’C + AD• F = BD + B’D’ + B’C + AB’

3.3 Four-Variables Map (8-9)Simplification using Prime ImplicantsExample 2:F(W,X,Y,Z) = ∑ (0,2,3,8,9,10,11,12,13,14,15)

X

Y

Z

W

1 1 1

1 1 1 1

1 1 11

W

X’Y

X’Z’ Note: that all

of these prime implicants are

essential .

3.3 Four-Variables Map (9-9)Simplification using Prime ImplicantsExample 3:F(W,X,Y,Z) = ∑ (0,2,3,4,7,12,13,14,15)

X

Y

Z

W

1 1 1

1 1

1 1 1 1

WX

W’Y’Z’

W’X’Y

W’YZ

XYZXY’Z

’

• Essential: WX• Prime: XYZ ,

XY’Z’ , W’Y’Z’, W’YZ, W’X’Y , W’X’Z’

W’X’Z’

3.5 Producut-of-Sum simplification (1-9)

1. Mark with 1’s the minterms of F.2. Mark with 0’s the minterms of F’.

3. Circle 0’s to express F’.4. Complement the result in step 3 to

obtain a simplified F in product-of-sums form.

3.5 Producut-of-Sum simplification (1-9)Example 1:Simplify :F= ∑(0,1,2,5,8,9,10) in Product-of-Sums Form

B

C

D

A

1 1 1

1

1 1 1

0

0 0 0

0 0 0 0

0

CD

ABBD’

• F’ = AB + CD + BD’• F = (F’)’ = (A’+B’) + (C’+D’) +

(B’+D)

3.5 Producut-of-Sum simplification (1-9)

Example 2:Simplify :F(x, y, z) =(0, 2, 5,7)in Product-of-Sums Form

y zx

0

00 01 11 1001 0

0 0

XZ

X’Z’

• F’ = XZ + X’Z’• F = (F’)’ = (X’+Z’) +

(X+Z)

3.6 Don't Cares Condition (1-4)

Sometimes a function table or map contains entries for which it is known:

The input values for the minterm will never occur, or The output value for the minterm is not used. Functions that have unspecified outputs for some input combinations are

called incompletely specified functions. In these cases, the output value is defined as a “don't care” ( an “x”

entry) assumed to be either 0 or 1. The choice between 0 and 1 is depending on the way the incompletely

specified function is simplied. By placing “don't cares” in the function table or map, the cost of the logic

circuit may be lowered.


Example : A logic function having the binary codes

for the BCD digits as its inputs. Only the codes for 0 through 9 are used.

The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”


Example (3.9) :F(W,X,Y,Z) = ∑ (1,3,7,11,15)d(W,X,Y,Z) = ∑ (0,2,5)

X

Y

Z

W

1 1 xx

1

0

x

1

1

0

0

0 0

0

0

0

F = YZ + W’Z

X

Y

Z

W

1 1 xx

1

0

x

1

1

0

0

0 0

0

0

0

F = YZ + W’X’


Example (3.9) :F(W,X,Y,Z) = ∑ (1,3,7,11,15)d(W,X,Y,Z) = ∑ (0,2,5)

X

Y

Z

W

1 1 xx

1

0

x

1

1

0

0

0 0

0

0

0

F’ = Z’ + WY’F = Z ( W’ + Y)

CS 121 Digital Logic Design

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