Digital logic Design for Boolen Algebra

7/23/2019 Digital logic Design for Boolen Algebra

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Boolean Algebra

Boolean Expressions and Theorems:

1. A. 0 = 0 7. A + A = 1 13. A ( B C) = (A B) C

2. A. 1 = A

3. A. A = A

8. A + A = A

9. A + AB= A

14. (A + B) +C = (A )+(B+ C)

=4. A. A = 0

5. A + 0 = A

10. A+ A B = +

11. A + B = B+ A

.

De- Morgan’s Law :

6. A + 1 = 1 12. A . B = B . A

1. A B = A + B 2. A + B = A B

EC214 Digital Logic DesignJ Ravindranadh , Associ. Prof 1



Simplify the given boolean expression

BC BC A ABC ++=

)1( ++= A BC ABC

BC ABC +=

=

BC =

Z XY Y X XYZ f ++=

Y X Z Z XY ++= )(

Y X XY +=

Y =

z y x z y x yz x z y x f +++=

z z y x y y z x +++= )()(


=




=

B AC C ABC C B A

=

++++= )()(

A B A

B B A B A

+=

++= )(

B+=

yz z x xy f ++=

x x x x

x x yz z x xy

+++=

+++= )(

z z x z xy +++= )1()1(


z x xy +=




+++=

C A B B AB B A AA ++++= ))((

C B A AC

C B A B AC

ABC C B A AAC A AB B A A A AA

+=

++=

+++++=

)1(

)(

AC

B AC

=

+= )1(

D D B A B A f ))(( +++=

)( ++++= D D A BB B A AB A A

)1(

+=

++=

+++=

D AB D B

D AB A D B

D D A D B D B A D AB


DB )1( =+= A D B




=

C B A B A ABC B A B AA AB

C A B A ABC A B A AB

++=++=

++=++=

)( )(

B B A B A ABC B A AB

+=

+=++=

)( )1(

=

).)()(( AC BC C A B A f ++=

−

C B BC AC B AC BC B +++++=+++=

C BC C AC C B A ABC AC A B A A BC BC B A B B A

C A B BC AC B A

++++++++=

++++= ))((r

C B A B B AC

C B A ABC B AC B A

+++=

+++=

)1()(


+=




BCD ABC A BC f +++=

BC C A +=

f= )( C A B A + B A+ )( C B A ++

DC A D BC BC B AF ++++++=

))((

DC B+=

z x x z x xF ++++=

= 1




Logic Gates

Logicgate

Booleanexpression

Truthtable

Logicdiagram

IEEEsymbol

Switching logic

OR Y=A B

A B Y

0 0 0

0 1 1

1 0 1 0 1

1 1 1

A B Y

AND

Y A B 0 0 0

0 1 0

1 0 0

1 1 1

AND

Y=A . B




Logic

gate

Boolean

expression Truth

Logic

diagram

IEEE

symbol

Switching logic

table

NOT

Y=A

A Y

0 1

1 0

Y=A BOR

A B Y

0 0 1

1 0

0 1 0

1 0 0

1 1 0

Y=A . BAND

A B Y

0 0 1

0 1 1

1 0 1 0 1

1 1 0




Logic Gates

Logicgate

Booleanexpression

Truthtable

Logicdiagram

IEEEsymbol

Switching logic

A B Y

0 0 0

0 1 1

Y=A B A B

EX-OR

1 0 1

1 1 0

A B Y

Y=A B A BX 0 0 1

0 1 0

1 0 0

1 1 1

Y=A B A BX-

NOR




Universal Gates

The NAND and NOR gates are universal gates.

All other gates/functions can be implemented by NOR or NAND gates. Sothey are called universal gates. ( or) A universal gate is a gate which canimplement any Boolean function without need to use any other gate type.

Advanta e of NAND and NOR ates are economical and easier to fabricate

The basic gates NAND and NOR are used in all IC digital logic families.




NAND Gate is a Universal Gate mp emen ng an nver er s ng on y a e

Implementing AND Gate Using only NAND Gates

Implementing OR Gate Using only NAND Gates




Implementing EX-OR Gate Using only NAND Gate

Implementing EX-NOR Gate Using only NAND Gate




NOR Gate is a Universal Gate mp emen ng an nver er s ng on y a e

Implementing OR Gate Using only NOR Gates

Implementing AND Gate Using only NOR Gates




Implementing EX-NOR Gate Using only NOR Gate

Implementing EX-OR Gate Using only NOR Gate




Minterms & Maxterms

Minterms Minterms is a product terms that contain all variable present in either true

or complemented form.

n

Maxterms

Maxterms is a sum terms that contain all variable present in true orcomplemented form.


or n var a es t ere n maxterms



For Minterms:

• “1” means the variable is “Not Complemented” and• “0” means the variable is “Complemented”.

For Maxterms:

• “ ” “ ”


• “1” means the variable is “Complemented”.



Canonical form Boolean functions expressed as a sum of minterms or product of maxterms

are said to be in canonical form.

Example: Express the boolean function F= A + B’ C in a sum of minterms.

unc on as ree var a e , an

First term A is missing two variable

A= A(B+B’) = AB + AB’ still missing the one variable

= ’ = ’ ’ = ’ ’ ’ ’ ’

Second term B’C missing one variable

B’C= (A +A’)B’C = A B’ C + A’ B’ C

Combining all terms F= A + B’ C

F= A B C + A B’ C’ + A B’ C + A B’ C’ + A B’ C + A’ B’ CF= A’ B’ C + A B’ C’ + A B’ C + A B C’ + A B C

= m1 + m4 + m5 + m6 + m7

or no a on: , , = , , , ,




Example: Express the Boolean function F= x y + x’ z in a product of maxtermform.

F= x y + x’ z = ( x y + x’ ) ( x y + z )

= ( x + x’ ) ( x’ + y) ( x + z ) ( y + z)

= ( x’ + y) ( x + z ) ( y + z )

The function has three variable x, y, and z. Each OR term is missing one variable

x + y = x + y +z z = x + y + z x + y + z

( x + z ) = ( x + z +y y’) = ( x + y + z ) ( x + y’ + z )

( y + z ) = ( y + z +xx’) = ( x + y + z) ( x’ + y +z)

Combining all the terms and removing those that appear more than once.

F = ( x + y + z ) ( x + y’ + z) ( x’ + y + z) ( x’ + y + z’)

= M0 M2 M4 M5

Short notation

F = Π ( 0, 2, 4, 5)




Express the function Y= A + B’ C in (a) canonical SOP (b) canonical POS

Y = Σ 1 4 5 6 7 Y = Π ( 0, 2, 3 )

Express the following functions in a sum of minterms and product of maxterms

F(A,B,C,D) = D (A’ + B) + B’ D

F=Σ (1,3,5,7,9,11,13,15) F= Π ( 0,2,4,6,8,10,12,14)

F( w, x,y,z) = y’ z + w x y’ + w x z’ + w’ x’ z

F=Σ 1 3 5 9 12 13 14 F= Π 0 2 4 6 7 8 10 11 15

F( A,B,C) = ( A’ + B ) ( B’ + C)

F= Π ( 2,4,5,6) F=Σ (0,1,3,7)

F = ( x y + z) ( y + x z )

, , , , , ,




Standard forms

Standard form configuration , the terms that form the Boolean functionmay contain one, two, or any number of literals.

There are two types of standard forms : the sum of products and product ofsums

Example : Function expressed in sum of products ( S O P)

F1 = y’ + x y + x’ y z’

Example : Function expressed in product of sum ( P O S )

F2 = x ( y’ +z) ( x +y + z’)

All canonical forms are standard form but standard forms are not




Karnaugh Maps In 1953, Maurice Karnaugh invented a graphical way of visualizing and then

simplifying Boolean expressions.

s grap ca represen a on , now nown as arnaug map, or - ap.

-values of a Boolean function

K-map can be of two forms

Sum – of-Product (SOP) Form

Product – of- Sum ( POS) Form




Rules for K-map Simplification using Sum of Products Form ( SOP)

Groupings can contain only 1s; no 0s

Groups can be formed only at right angles ; diagonal groups are not allowed

The number of 1s in a group must be a power of 2- even if it contain asingle 1

The groups must be made as large as possible

Groups can overlap and wrap around the sides of the K-map




Two variable K-ma

B 0 1

A0 m0 m1

0 A’B’ A’B

’ 1 m2 m3

Sim lif the Boolean ex ression in S O P Y A B = Σ 2 3

B

A 0 1

0

1 1 1

F = A




Simplify the Boolean expression in S O P Y(A,B)= Σ (0, 1, 3)

B

A

0 1 1 Y = A’ + B

Simplify the expression Y= Σ ( 0, 2, 3)

B

A 0 1

1 1 1 Y = A + B’




Simplify the boolean expression Y ( A,B)= Σ ( 1, 2 )

B

A 0 1

0 1

1 1 Y= A' B + A B' = (A B)

Simplify the expression Y(A,B) = Σ ( 0, 1,2, 3, )

B

A 0 1

0 1 1

1 1 1 Y = 1




Three Variable K-map

B C

A

00 01 11 10

0 m m m m

1 m4 m5 m7 m6

, , , , ,

y z 00 01 11 10

0 1 Y= y z + x z + x y




Simplify the expression Y( A,B,C)= Σ ( 2, 3, 4, 5 )

A

0 1 1

' ‘

1 1 1

Simplify the expression f( x, y, z)= Σ ( 0,1,2, 3, 6, 7 )

y z 00 01 11 10 x

0 1 1 1 1 f = x' + y




Simplify the expression Y(A,B,C)= Σ ( 1, 3, 4, 6 )

A

0 1 1

' '

1 1 1

Simplify the expression Y= Σ ( 0, 2, 4, 6, 7 )

A

0 1 1 Y= C' + A B

1 1 1 1




Simplify the expression Y= Σ ( 1, 2, 4, 7 )

A

0 1 1

' ' ' ' ' ‘1 1 1

= (A B C)

Simplify the expression (a) P O S (b) S O P f = Π ( 0, 1, 2, 3, 4, 6 )

A

0 0 0 0 0

A

0

1 0 0

f= (A) (C )

1 1 1




Simplify the expression in P O S Y= Π ( 0, 1, 4, 5, 6)

A

0 0 0

’ 1 0 0 0

Simplify the expression in P O S f = Π ( 1, 2, 3, 5, 6, 7 )

A

0 0 0 0f= (B’) (C ‘)

1 0 0 0




Simplify the expression (a) S O P (b) P O S Y= B C + A C’ +A B + A B C

A

0 1

B C A

00 01 11 10

0 0 0 0

1 1 1 1

Y = B C + A C’

1 0

= ’

Simplify the expression ( a) S O P ( b) P O S Y= A’ B + B C’ + B’ C’

A

0 1 1 1

B C A

00 01 11 10

0 0

1 1 1

= ’ '

1 0 0

’ ’ ’


=



Simplify given boolean expression Y= Π ( 0,1,2,3,6) (a) S O P (b ) P O S

A

0

’ 1 1 1 1

A

0 0 0 0 0 Y= A ( B’ + C )

1 0




Four Variable K-map

A B

00 m0 m1 m3 m2

01 m

4

m

5

m

7

m

6

11 m12 m13 m15 m14

10 m8 m9 m11 m10 C D

A B

00 01 11 10

Simplify the expression in S O P

Y= Σ ( 1, 4, 5, 6, 12, 14)

00 1

01 1 1 1

Y= B D’ + A’ C’ D11 1 1

10




Simplify the expression in S O P Y= Σ ( 1,2,3,5,7,9,10,11,13,15)

C D 00 01 11 10 A B

00 1 1 1

01 1 1

11 1 1

Y= D + B’ C

C D

A B

00 01 11 1010 1 1 1

Y= Σ ( 1, 3, 4,5, 9, 10, 11, 12, 13 ) 00 1 1

01 1 1

Y= B C’+ B’ D + A B’ C 11 1 1

10 1 1 1




Simplify the expression Y= Σ ( 0, 2, 5, 7, 8, 10, 13, 15)

C D 00 01 11 10 A B

00 1 1

01 1 1

11 1 1

Y= B D + B’ D’

C D

A B

00 01 11 1010 1 1

Y= Π ( 0,1, 4, 5, 8, 9, 10, 11, 14, 15) 00 0 0

01 0 0

Y= ( A + C ) ( A’ + C’) ( A’ + B) (or) (B + C ) 11 0 0

10 0 0 0 0




Simplify the expression Y= Π ( 0, 1, 4, 5, 6, 8, 9,12, 13,14)

C D 00 01 11 10 A B

00 0 0

01 0 0 0

11 0 0 0

Y= C ( B’ + D)

C D

A B

00 01 11 1010 0 0

Y= Π ( 1, 3, 5, 7, 8, 9, 12, 13) 00 0 0

01 0 0

Y= ( A’ + C ) ( A + D’) 11 0 0

10 0 0




Obtain (a) minimal sum of product (b) product of sum Y= Σ ( 0, 1, 2, 5, 8,9, 10)

C D 00 01 11 10 A B

00 1 1 1

01 1

11

Y= B’ D’ + A’ C’ D + B’ C’

C D

A B

00 01 11 1010 1 1 1

00 0

01 0 0 0

Y= ( A’ + B’ ) ( C’ + D’ ) ( B’ + D ) 11 0 0 0 0

10 0




Obtain (a) minimal sum of product (b) product of sum

Y= A B D + A’ C’ D’ + A’ B + A’ C D’ + A B’ D’

C D

A B

00 01 11 10

00 1 1

01 1 1 1 1

Y= B’ D’ + A’ D’ + B D +

(or) A’ B

C D

A B

00 01 11 1011 1 1

10 1 100 0 0

01

Y= ( B + D’) ( A’ + B’ + D ) 11 0 0

10 0 0




Obtain (a) minimal sum of product (b) product of sumF(A,B,C,D) = Π ( 1,3,7,9,11,15 )

C D 00 01 11 10 C D 00 01 11 10

00 1 1

00 0 0

11 1 1 1 11 0

10 1 1 10 0 0

’ ’ ’

F= D’ + B C’




Five Variable K-map

D E 00 01 11 10 D E 00 01 11 10

A=0 A=1

00 m0 m1 m3 m2

B C

00 m16 m17 m19 m18

m4 m5 m7 m6

11 m12 m13 m15 m14

01 m20 m21 m23 m22

11 m28 m29 m31 m30

10 m8 m9 m11 m10 10 m24 m25 m27 m26




Simplify the boolean fuction

F(A,B,C,D,E) = Σ ( 0,2,4,6,9,11,13,15,17,21,25,27,29,31)

D EB C

00 01 11 10 D E 00 01 11 10

A=0 A=1

00 1 1

01 1 1

00 1

01 1

11 1 1 11 1 1

= ’ ’ ’ ’





F(A,B,C,D,E) = Σ ( 2,3,10,11,12,13,16,17,18,19,20,21,26,27 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00 1 1

01

00 1 1 1 1

01 1 1

11 1 1

10 1 1

11

= ’ ’ ’ ’ ’





F(A,B,C,D,E) = Σ ( 8,9,10,11,13,15,16,18,21,24,25,26,27,30,31 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00

01

00 1 1

01 1

11 1 1

10 1 1 1 1

11 1 1

= ’ ’ ’ ’ ’ ’





F(A,B,C,D,E) = Π ( 0,1,4,8,12,13,15,16,17,23,29,31 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00 0 0

01 0

00 0 0

01 0

11 0 0 0

10 0

11 0 0

= ’ ’ ’ ’ ’ ’ ’





F(A,B,C,D,E) = Π ( 0,1,4,5,6,18,19,22,23 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00 0 0

01 0 0 0

00 0 0

01 0 0

11

10

11

= ’ ’ ’ ’ ’




DON’T- CARE CONDITIONS. on care erms are ose w c are no e compu sory erms ,w e so v ng

k'map we have to consider those don't care terms if required otherwise we don't

have to care about those term.2. Don’t care will help to minimize the boolean expressions.

’ ‘ ’

Simplify the Boolean function F( w,x,y,z) = Σ ( 1,3,7,11,15 ) + d Σ( 0,2,5 )

. on care represen

w x

00 X 1 1 X’

01 X 1

11 1

46

10 1

Simplify the expression in S O P Y= Σ ( 4 5 6 8 9) + d Σ ( 3 7 10 11 14 15 )



Simplify the expression in S O P Y= Σ ( 4,5,6,8,9) + d Σ ( 3, 7, 10,11,14,15 )

C D 00 01 11 10

A B

00 X

01 1 1 X 1

11 X X

Y= A’ B + A B’

= A B

C D

A B

00 01 11 1010 1 1 X X

Y= Σ ( 0,3,4,5,7 ) + Φ Σ ( 8,9,10,11,12,13,14,15) 00 1 1

01 1 1 1

Y= C’ D’ + C D + B C’ ( or) B D 11 X X X X

10 X X X X


Simplify the expression in P O S F= Π ( 4, 6, 8, 9,10, 12, 13,14) + d Σ ( 0,2,5 )



Simplify the expression in P O S F= Π ( 4, 6, 8, 9,10, 12, 13,14) + d Σ ( 0,2,5 )

Y Z 00 01 11 10

W X

00 X X

01 0 X 0

11 0 0 0

F= Z ( W’ + Y )

C D

A B

00 01 11 1010 0 0 0

Y= Π ( 1,2,6, ) + Φ d ( 8,9,10,11,12,13,14,15) 00 0 0

01 0

Y= ( C’ + D) ( B + C + D’ ) 11 X X X X

10 X X X X


Simplify the boolean function (a) S 0 P (b) P O S




F(A,B,C,D) = Σ ( 1,5,6,12,13,14) + d Σ (2,4)

C D

A B

00 01 11 10

01 X 1 1

F= B C’ + B D’ + A’ C’ D

C D

A B

00 01 11 10

11 1 1 1

10

00 0 0 X

01 X 0

Y= ( C’ + D’) ( A’ + B) ( B + D ) 11 0

10 0 0 0 0





F(A,B,C,D,E) = Σ ( 1, 5,9,11,13,15,25,2729,31) + d( 3, 16,17,20,21)

D EB C

00 01 11 10 D E 00 01 11 10

A=0 A=1

00 1 x

01 1

00 x x

01 x x

11 1 1 11 1 1

= ’





A=0 A=1

, , , , , , , , , , , , , , , , ,

B C

00 1 1 1

D E

B C

00 01 11 10

00 X X 1

01

11 1 1 X

01

10 1 110 1 X 1

F= C’ D’ + A’ B C E + B’ C’ E + A C’ E




A=0 A=1

B C00 0

D E

B C

00 01 11 10

00 X X 0

01 0 0 0 0

11 0 X

01 0 0 0 0

10 0 0 10 X 0

F= ( B + C’) ( D’ + E ) ( A’ + C’ ) ( C’ + D + E ) ( A + B’ + C + D’)




Simplify the minimum sum of product and minimum product of sum for given

Boolean function

F(A,B,C,D,E) = Σ ( 0,1,2,6,7,9,10,15,16,18,20,21,27,30) + Σd ( 3, 4, 11,12,19 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00 1 1 x 1

01 x 1 1

00 1 x 1

01 1 1

11 x 1

10 1 x 1

11 1

= ’ ’ ’ ’ ’ ’ ’ ’ ’ ’ ’


+ A’ D E + A’ B’ D (or) A’ B’ E’



Simplify the minimum product of sum for given

Boolean function

F(A,B,C,D,E) = Π ( 5,8,13,14,17,22,23,24,25,26,28,29,31) + Π d ( 3, 4, 11,12,19 )

A=1

D EB C

00 01 11 10 D EB C

00 01 11 10

00 x

01 x 0

00 0 x

01 0 0

11 x 0 0

10 0 x

11 0 0 0

F= ( A + C’ + D ) ( B’ + D + E ) ( A + B’ + C’ + E ) ( A’+ B’ + C’ + E’ )’ ’ ’ ’ ’ ’ ’




Quine – McCluskey Method

Simplify the following boolean function by using the tabulation method

F = Σ ( 0,1,2,8,10,11,14,15 )(b) c

w x y z

0 0 0 0 0

w x y z

0,1 0 0 0 __

0,2 0 0 __ 0

w x y z

0,2,8,10 __ 0 __ 0

1 0 0 0 1

2 0 0 1 0

8 1 0 0 0

0,8 __ 0 0 0

2,10 __ 0 1 0

8 10 1 0 0

, , , __ __

10,11,14,15 1 __ 1 __

10 1 0 1 0

11 1 0 1 1

__

10,11 1 0 1 __ 10,14 1 __ 1 0

11,15 1 __ 1 1

10,14,11,15 1 __ 1 __

’ ’ ’ ’ ’


15 1 1 1 114,15 1 1 1 __

Simplify the following Boolean function by using the tabulation method



F = Σ ( 0,1,2,8,10,11,14,15 )

(a)

w x y z

(b) (c )

0 0 0 0 0

1 0 0 0 1

0,1 (1)0,2,8,10 ( 2, 8)

0 8 2 10 2 8

0,2 (2)

2 0 0 1 0

8 1 0 0 0 10,11,14,15 (1, 4 )

0,8 (8)

2,10 (8)

10 1 0 1 0

1 ,14,11,15 1, 4

8,10 (2)

10,11 (1)11 1 0 1 1

14 1 1 1 0

F= w’ x’ y’ + x’ z’ + w y10,14 (4)11,15 (4)


Determine the prime-implicants , essential prime implicants and simplifiedexpression



expressionF( w,x,y,z) = Σ (1,4,6,7,8,9,10,11,15 )

(a)

w x y z

(b) (c )

1 0 0 0 1

4 0 1 0 0

1,9 (8)8,9,10,11 ( 1, 2 )

8 10 9 11 2 1

4,6 (2)

8 1 0 0 0

6 0 1 1 0

8,9 (1)

8,10 (2)

10,11 (1)

9 1 0 0 1

10 1 0 1 0

,

9,11 (2)

7 0 1 1 1

11 1 0 1 1

7,15 (8)

11,15 (4)


The prime implicants consist of all the unchecked terms in the table

Th i i li t ’ ’ ’ ’ ’ d ’



The prime implicants are x’ y’ z, w’ x z’, w’ x y, x y z, w y z, and w x’

prime-implicant table

1 4 6 7 8 9 10 11 15

1,9 x’ y’ z

4,6 w’ x z’

’

X X

X X

X ,

7,15 x y z

11,15 w y z

X X

X X

, ,1 ,11 w x

The rime im licants that cover minterms with a sin le cross in theircolumn are called essential prime implicants .

x’ y’ z , w’ x z’ and w x’


Simplified expression F= x’ y’ z + w’ x z’ + w x’ + x y z




pF( A,B,C,D,E) = Σ (0,1,4,5,6,7,12,18,19,22,23,28,31)

(a) A B C D E

0 0 0 0 0 0

(b) (c )

0,1 1

1 0 0 0 0 1 0,1,4,5 ( 1, 4 )

4 0 0 1 0 0 0,4,1,5 ( 4, 1 )

4 5 6 7 1 2

0,4 (4)

1,5 (4)

4 5 1

P4

6 0 0 1 1 0

12 0 1 1 0 0

5 7 2 6,7,22,23 ( 1, 16 )

4,6,5,7 ( 2, 1 )

4,6 (2)

4,12 (8) P1

P5

P6

19 1 0 0 1 1

22 1 0 1 1 0

6,7 (1)

6,22 (16)

12,28 (16) P2

7 0 0 1 1 16,22,7,23 ( 16, 1 )

18,19,22,23 ( 1, 4 )

18,22,19,23 ( 1, 4 )

P7

23 1 0 1 1 17,23 (16)

18,19 (1)

28 1 1 1 0 0

18,22 (4)

59

31 1 1 1 1 1 19,23 (4)

22,23 (1)

23,31 (8) P3


The prime implicants are A’CD’E’ BCD’E’ ACDE A’B’D’ A’B’C B’CD AB’D



The prime implicants are A’CD’E’ , BCD’E’ , ACDE , A’B’D’ , A’B’C , B’CD , AB’D


0 1 4 5 6 7 12 18 19 22 23 28 31

X X

X X

X X

4,12 A’CD’E’

12,28 BCD’E’

X X X X

X X X X

,

0,1,4,5 A’B’D’

4,5,6,7 A’B’C

X X X X

,7, , B D

18,19,22,23 AB’D

Simplified expression

Essential prime implicants A’B’D’ + AB’D + ACDE + BCD’E’


F= A’B’D’ + AB’D + ACDE + BCD’E’ + A’B’C (or) B’CD




F = Σ (1,2,12,13,15,17,18,19,20,21,23,24,25,27,29,31)

(a) A B C D E

1 0 0 0 0 1

(b)

1,17 (16) P1

(c )

12 0 1 1 0 0

17,19,21,23 ( 2, 4 )

17 1 0 0 0 1

,12,13 (1) P3

17,19 (2)17,21 (4)17 25 8

17,19,25,27 ( 2, 8 )

17 21 25 29 4 817,21,19,23 ( 4, 2)

20 1 0 1 0 0

24 1 1 0 0 0

18,19 (1) P4

20,21 (1) P5

13 15 2

13,15,29,31 ( 2, 16 ) P7

19,23,27,31 ( 4, 8 )

24,25 (1) P6

17,25,19,27 ( 8, 4 )17,25,21,29 ( 8, 4 )

19 1 0 0 1 1

21 1 0 1 0 125 1 1 0 0 1

13,29 (16)19,23 (4)19,27 (8)21,23 (2)

, , , ,25,27,29,31 ( 2, 4 )

(d)

25,29,27,31 ( 4, 2 )

15 0 1 1 1 123 1 0 1 1 1 25,29 (4)

, , , , ,

25,27,29,31 P8

17,19,25,27 ( 2, 8 ,4)

25,27 (2)21,29 (8)

61

29 1 1 1 0 1

31 1 1 1 1 1

23,31 (8)

27,31 (4)

29,31 (2)

, , ,17,21,25,29 ( 4, 8,2 )

19,23,27,31


The prime implicants are B’C’D’E B’C’DE’ A’BCD’ A’B’C’D AB’CD’ ABC’D’ BCE AE



The prime implicants are B C D E, B C DE , A BCD , A B C D, AB CD , ABC D , BCE, AE

prime-inmplicant table

1 2 12 13 15 17 18 19 20 21 23 24 25 27 29 31

X

X

1,17 B’C’D’E

2,18 B’C’DE’

’ ’

X X

X X

,

18,19 AB’C’D

20,21 AB’CD’

X X X X

X X X X X X X X

4, 5 AB D

13,15,29,31BCE

17,19,21,23

Essential prime implicants and Simplified expression

25,27,29,31 AE

EC214 Digital Logic Design

J Ravindranadh , Associ. Prof 62

F= B’C’D’E + B’C’DE’ + A’BCD’ + AB’CD’ + ABC’D’ + BCE +AE


F ( ) Σ (1 3 4 5 9 10 11) Σ ( 6 8 )



F ( w,x,y,z)= Σ (1,3,4,5,9,10,11) + Σ φ ( 6,8 )

(a)

w x y z

(b) (c )

1 0 0 0 1

4 0 1 0 0

1,3 (2)

1,3,9,11 ( 2, 8)1,9,3,11 ( 8,2)

1,5 (4) P1

P4

8 1 0 0 0

3 0 0 1 1

1,9 (8)

4,5 (1) P2

8,9,10,11 ( 1,2)

8,10,9,11 ( 2,1)P5

6 0 1 1 0

9 1 0 0 1

8,10 (2)

5 0 1 0 1

, 3

8,9 (1)

11 1 0 1 1

3,11 (8)

9,11 (2)

10 1 0 1 0



,


The prime implicants are w’ y’ z , w’ x y’, w’ x z’, x’ z, w x’



The prime implicants are w y z , w x y , w x z , x z, w x


1 3 4 5 9 10 11

X X

X X

1,5 w’y’z

4,5 w’ x y’

’ ’

X X X X

X X X

,

1,3,9,11 x’ z

8,9,10,11 w x’

The rime im licants that cover minterms with a sin le cross in theircolumn are called essential prime implicants .

x’ z , w x’

Simplified expression F = x’ z + w x’ + w’ x y’




F ( w x y z) Σ (2 3 4 7 9 11 12 13 14) + Σ φ ( 1 10 15 )



F ( w,x,y,z)= Σ (2,3,4,7,9,11,12,13,14) + Σ φ ( 1,10,15 )

(a)

w x y z

(b) (c )

1 0 0 0 1

2 0 0 1 04 0 1 0 0

1,3 (2)

1,3,9,11 ( 2, 8)1,9,3,11 ( 8,2)

1,9 (8)2,3 (1)2 10 8

3 0 0 1 1

10 1 0 1 0

9 1 0 0 1

4,12 (8)

3,7 (4)

2,3,10,11 ( 1,8)

2,10,3,11 ( 8,1)

7 0 1 1 1

,9,11 (2)

10,11 (1)

12 1 1 0 0

9,13 (4)

,7,11,15 4,

3,11,7,15 ( 4, 8)

9,11,13,15 ( 4, 2)

,

13 1 1 0 114 1 1 1 0

12,13 (1)12,14 (2)

7,15 (8)

, , , ,

10,11,14,15 ( 1, 4)10,14,11,15 ( 1, 4)12,13,14,15 ( 1, 2)

65

15 1 1 1 1 ,

13,15 (2)

14,15 (1)

12,14,13,15 ( 1, 2)


The prime implicants are B C’ D’ , B’D, B’C, CD,AD,AC,AB



p p , , , , , ,


2 3 4 7 9 11 12 13 14

X X

X X X

4,12 B C’ D’

1,3,9,11 B’D

B’C

X X X

X X X

, , , B C

3,7,11,15 CD

9,11,13,15 AD

, , , AC

12,13,14,15 X X X

Essential primeimplicants B’C , BC’D’,CD



Simplified expression F = B’C + BC’D’+CD +AB + AD


F = Σ (8 12 13 18 19 21 22 24 25 28 31) + Σ d ( 1 2 3 6 7 11 )



F = Σ (8,12,13,18,19,21,22,24,25,28,31) + Σ d ( 1,2,3,6,7,11 )

(a) A B C D E

1 0 0 0 0 1

(b)

1,3 (2) P3

(c )

3 0 0 0 1 16 0 0 1 1 0

,

2,18 (16)8,12 (4)

, , , ,8 0 1 0 0 0 2,6 (4)

2,3,18,19 ( 1, 16 ) P10

2 6 3 7 4 1

18 1 0 0 1 0

24 1 1 0 0 0

,

3,7 (4) P4

3,11 (8) P5

3 19 16

2,6,18,22 ( 4, 16 ) P11

2,18,3,19 ( 16, 1 )

11 0 1 0 1 1

13 0 1 1 0 1

6,7 (1)

6,22 (16) 8,12,24,28 ( 4, 16 ) P12

2,18,6,22 ( 16, 4 )

21 1 0 1 0 1 P1

22 1 0 1 1 012,28 (16) P7

, 6

18,19 (1)

8,24,12,28 ( 16, 4 )

67

28 1 1 1 0 0

31 1 1 1 1 1 P2

24,25 (1) P8

,,

24,28 (4)

The prime implicants are AB’CD’E, ABCDE, A’B’C’E, A’B’DE, A’C’DE,A’BCD’,BCD’E’, ABC’D’,A’B’D,B’C’D,B’DE’,BDE’

prime implicant table




8 12 13 18 19 21 22 24 25 28 31

X

31 ABCDE

1,3 A’B’C’E

X X

3,7 A’B’DE

3,11 A’C’DE

’ ’

X X

X X

,

12,28 BCD’E’

24,25 ABC’D’

X X

X X

2,3,6,7 A’B’D

2,3,18,19 B’C’D

2,6,18,22 ,B’DE’

68

X X X X

8,12,24,28 BD’E’

The prime implicants that cover minterms with a single cross in theircolumn are called essential prime implicants .



Essential prime implicants are AB’CD’E, ABCDE,A’BCD’, ABC’D’,B’C’D, BDE’

Sim lified ex ression

AB’CD’E + ABCDE + A’BCD’ + ABC’D’+ B’C’D+ BDE’ + B’DE’





N ND Implementation

A B C = A B C A + B + C =

mp emen e o ow ng unc on n wo eve =

A A

D

CF

C

F



Implement the following function in two level NAND F= x’ y + x y’ + z



x’

y

x’

y x’ y

y’

x

z’ y’

x

z y’

x F

z’

Implement the following function in two level NAND



Multi level NAND Implementation

Implement the following function in multi level NAND



=

= (AB’+A’B)(C+D’)

72

Implement the following function in multi level NAND f = a'c'd+bc'd+bcd'+acd'



= c'd(a'+b)+cd'(a+b)

level 1level 2level 3



Implement the following function in multi level NAND f = (AB+C)(D+E+FG) + H



level 1level 2level 3level 4





NOR Implementation

Two level NOR Implementation

A NOR gate equivalent logic gate

A + B + C A B C = A + B + C=

Implement the following function in two level NOR F= ( A + B ) ( C + D)



Implement the following function in multi level NOR f = a'c'd+bc'd+bcd'+acd'

= c'd(a'+b)+cd'(a+b)





Implement the following function in multi level NOR f = (AB+C)(D+E+FG) + H



level 4 level 5



Digital logic Design for Boolen Algebra

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