Digital Circuits (2)

8/22/2019 Digital Circuits (2)

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Representation of Boolean Expressions

x y min term

0 0 x . y m00 1 x . y m11 0 x . y m21 1 x . y m3

0 0 00 1 11 0 11 1 0

x y f1

1f = . .x y x y 1 1 2f = m m 1f = (1,2)

2f = (0,2,3) ? 2f = . . .x y x y x y

A minterm is a product that contains all the variables used in a function


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Three variable functions

y z min terms

0 0 0 x . y . z0 0 1 x . y . z0 1 0 x . y . z0 1 1 x . y . z1 0 0 x . y . z

1 0 1 x . y . z1 1 0 x . y . z1 1 1 x . y . z

x

m0m1m2m3m4m5m6m7

2f = (1,4,7) ?

2f = . . . . . .x y z x y z x y z


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Product of Sum Terms Representation

0 0 00 1 11 0 11 1 0

x y f1

1f =(x ).( )y x y

x y

0 0 x + y M00 1 x + y M11 0 x + y M21 1 x + y M3

Max term

1 1 2f = .M M 1f = (1,2)

2f = (0,3) ?

2f =(x ).( )y x y


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y zx Max. terms

M0M1M2M3M4M5M6M7

0 0 0 x + y + z0 0 1 x + y + z0 1 0 x + y + z0 1 1 x + y + z1 0 0 x + y + z

1 0 1 x + y + z1 1 0 x + y + z1 1 1 x + y + z

1f = (1,5,7) ?

2f =(x ).( ).( )y z x y z x y z


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Simplification

yx1 x2 x3

0 0 0 00 0 1 10 1 0 00 1 1 1

1 0 0 01 0 1 11 1 0 01 1 1 1

1 2 3 1 2 3 1 2 3 1 2 3y= . . . . . . . .x x x x x x x x x x x x

y= (1,3,5,7)

y

x1

x2

x3

x1

x2

x3

x2

x3

x1

x2

x3

x1

Simplification of Boolean expression yields : y = x3 !! which does not require any

gates at all !


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Goal of Simplification

1 2 3 1 2 3 1 2 3 1 2 3y= . . . . . . . .x x x x x x x x x x x x

y

x1

x2

x3

x1

x2

x3

x2

x3

x1

x2

x3

x1

Goal of simplification is to reduce the complexity of gate circuit. This requires that we

minimize the number of gates. Since number of gates depends on number of minterms,

one of the goals of simplification is to minimize the number of minterms in SOP

expression


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1 2 3 1 2 3 1 2 3 1 2 3y= . . . . . . . .x x x x x x x x x x x x

y

x1

x2

x3

x1

x2

x3

x2

x3

x1

x2

x3

x1

1 3 1 2 3 1 2 3y= . . . . .x x x x x x x x

y

x1

x2

x3

x2

x3

x1

x1

x3

This circuit is simpler not just because it uses 4 gates instead of 5 but also because

circuit-2 uses one 2-input and three 3-input gates as compared to five 3-input gates

used in circuit-1

SN

SPx1

SN

x1

x2

SPx2

VDD = 5V

SN

SPx1

SN

x1

SP

VDD = 5V

x2

SN

SPx2

x3

x3

yy

2-input NAND3-input NAND


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Goal of Simplification

1. Minimize number of product terms

2. Minimize number of literals in each term

In the SOP expression:

Simplification Minimization


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Minimization

1 2 3 1 2 3 1 2 3 1 2 3y= . . . . . . . .x x x x x x x x x x x x

1 3 2 2 1 3 2 2y= . .( ) . .( )x x x x x x x x

1 3 1 3y= . .x x x x

1 1 3y=( ).x x x

3y= x

Principle used: x + x = 1


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f = . . .x y x y x y

Apply the Principle: x + x = 1 to simplify

f = .( ) .x y y x y

f = .x x y

How do we simplify further?

f = . . . . . . .x y x y x y x y x y x y x y

Principle used : x + x = x

f = . . . .

. ( ) ( ).

x y x y x y x y

x y y x x y x y


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Simplify1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

1 2 3 4 1 2 3 4

. . . . . . . . . . . .

. . . . . .

f x x x x x x x x x x x x x x x x

x x x x x x x x

Principle: x + x = 1 and x + x = x

Need a systematic and simpler method for applying these two principles

Karnaugh Map (K map) is a popular technique for carrying out simplification

It represents the information in problem in such a way that the twoprinciples become easy to apply


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K-map representation of truth table

x y min term

0 0 x . y m00 1 x . y m11 0 x . y m21 1 x . y m3

0

1

1

0

m0

m3

xy

m2

m1

0 0 0

0 1 11 0 11 1 0

x y f1

0

1

1

0xy

0 1

1 0


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2f = (0,2,3) 01

1

0xy

0 1

1 1

0

1

1

0xy

1

1 0

0

f = . .x y x y


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3-variable K-map representation

y z min terms

0 0 0 x . y . z

0 0 1 x . y . z0 1 0 x . y . z0 1 1 x . y . z1 0 0 x . y . z1 0 1 x . y . z1 1 0 x . y . z

1 1 1 x . y . z

x

m0

m1m2m3m4m5m6

m7

yz

x 00 01 11

0

1

10m0 m1 m3 m2

m4 m5 m6m7

0 0 0 00 0 1 10 1 0 00 1 1 11 0 0 01 0 1 11 1 0 0

1 1 1 1

x y z f yz

x 00 01 11

0

1

10

0 1 1 0

0 1 1 0


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yz

x 00 01 110

1

101 0

0 1 1 0

01

f = . . . . . . . .x y z x y z x y z x y z

4 i bl K t ti


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w x y z

0 0 00 0 0 1

0

min terms

m0m1

0 0 1 0 m2

0 0 11 m3

1 1 1 0 m14

1 1 1 1 m15

4-variable K-map representation

yzwx 01 11

11

01

00

00 10

10

0 1 3 2

4 5 7 6

12 13 15 14

8 9 1011

yz

1 01wx 01 11

11

01

00 0

00 10

0 1 1 0

1 0 10

1 0 0010

f =w. . . w. . . . . . . . .

. . . . . . . . .

x y z x y z w x y z w x y z

w x y z w x y z w x y z

Minimization using Kmap


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Minimization using Kmap

2f = (2,3)0

1

1

0xy

0

1 1

0

f = . .x y x y

f = .( )x y y

f = x

Combine terms which differ in only one bit position. As a result, whatever is

common remains.


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0

1

1

0xy

10

0 1

f = . .x y x y

f = ( . ).x x y f = y

0

1

1

0xy

1 0

1 0

f = y

0

1

1

0xy

1

00

1

f =x


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2f = (0,2,3) 01

1

0xy

0 1

1 1

f = . . .x y x y x y

f = .( ) .

.

x y y x y

x x y

f = . .

( ).

x x y x y

x x x y

x y

The idea is to cover all the 1s with as few and as simple terms as possible

3 i bl i i i ti


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yz

x 00 01 11

0

1

101 0

0 1 1 0

01

3-variable minimization

f = . . . . . . . .x y z x y z x y z x y z

.x z

.y z

f = . . . .x y z y z x z

3 i bl i i i ti


21/49

yz

x 00 01 11

0

1

10

0 1 1 0

1 00 1


f = . . . . . . . .x y z x y z x y z x y z

.x z

.x z

f = . .x z x z

3 i bl i i i ti


22/49

yz

x 00 01 11

0

1

10

1 1

00

1 1

0 0


f = . . . . . . . .x y z x y z x y z x y z

.x y.x y

f = . .x y x y

f = .( )x y y x

yzx 00 01 11

0

1

10

1 1

00

1 1

0 0

x


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yzx 00 01 11

0

1

10

1 1 1 1

0 0 0 0 x

yzx 00 01 11

0

1

10

1 1

0 01 1

0 0

z

yzx 00 01 11

0

1

10

1

1

0 0

0 0

1

1

z

yzx 00 01 11

0

1

10

1

1

0 01

1 1 1z

x

f =x z


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yz

x 00 01 11

0

1

100 0

1 1 1

0

0

0

Can we do this ?

Note that each encirclement should represent a single product term. In this case it

does not.

f = . . . . . .

. .

x y z x y z x y z

x y x z

We do not get a single product term. In general we cannot make groups of 3 terms.


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yz

x 00 01

0

1

0 0

1 1

0

0

010 11

0

Can we use kmap with the following ordering of variables?

Can we combine these two terms into a single term ?

f = . . . .

.( . . )

x y z x y z

x y z y z

Note that no simplification is possible. Kmap requires information to be represented


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yz

x 00 01

0

1

00

0

10 11

0

1 1

0 0

These two terms can be combined into a single term but it is not easy to show

that on the diagram.

f = . . . .

.( ). .

x y z x y z

x y y z x z

Kmap requires information to be represented in such a way that it is easy to apply

the principle 1x x

variable minimization


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yz

1 01wx 01 11

11

01

00 000 10

0 1 1 0

1 0 10

1 0 0010. .w x z

-variable minimization

. .w y z

. .w y z

f = . . . . . . . . . . . .w y z w x z w y z w x y z w x y z

But is this the simplest expression ?


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yz

1 01wx 01 11

11

01

00 0

00 10

0 1 1 0

1 0 10

1 0 0010

. . . . . . . .w x y z w x y z w x z

yz

1 01wx 01 11

11

01

00 0

00 10

0 1 1 0

1 0 10

1 0 0010

. . . . . . . .w x y z w x y z x y z



29/49

yz

1 01wx 01 11

11

01

00 000 10

0 1 1 0

1 0 10

1 0 0010. .w x z


. .w y z

. .w y z

f = . . . . . . . . . .w y z w x z w y z w x z x y z

. .w x z

. .x y z

Is this the best that we can do ?

Cover the 1s with minimum number of terms


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yz

1 01wx

01 11

11

01

00 0

00 10

0 1 1 0

1 0 10

1 0 0010

f = . . . .

. . . . . .

w y z w x z

w y z w x z x y z

yz

1 01wx

01 11

11

01

00 0

00 10

0 1 1 0

1 0 10

1 0 0010

Cover the 1 s with minimum number of terms

f = . . . .

. . . .

w y z w x z

w x z x y z



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yz

01wx 01 11

11

01

00 0

00 10

1 0

0 0

1 0 010

1

0

0

0 0

1


yz

01wx 01 11

11

01

00 0

00 10

1 0

0 0

1 0 010

1

0

0

0 0

1

f = . . . . . .w x y w x z w y z

f = . . . . . .w x y w x z x y z

Groups of 4


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Groups of 4yz

0wx 01 11

11

01

00

00 10

1

0

010

1

0

0 0

0

1 1

0 0

1

1

1

.w x

.y z

yz

0wx 01 11

11

01

00

00 10

1

10

0

0 0

0

1

0 0

1

1

0 0

1

1

0

.x z

.w z


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yz

0wx 01 11

11

01

00

00 10

10

00

0 0

0

1 0 0 1

1 0 0 1

0 0

yzwx 01 11

11

01

00

00 10

10

0

0 0

0 0

01 1

1 1

0 0

0 0

0 0

yz

0wx 01 11

11

01

00

00 10

10

0 0

0

0 01

0 0

1

1 1

0 0

0 0

yz

0wx 01 11

11

01

00

00 10

10

0 0

0

0 0

1

01 1

1

0

0

0

00

.x z .x z

.x z??

Groups of 8


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Groups of 8

yz

0wx 01 11

11

01

00

00 10

1

10

0

0

1

0 0

1

1

1

1

1

1

0

00 z

yz

0wx 01 11

11

01

00

00 10

1

10

0

1

0 0

1 1

1 1

1 1

0 0

0 0

x

yzwx 01 11

11

01

00

00 10

10

1 1

1 1

0 0

0 0

1

1

1

1

0 0

0 0

z

yzwx 01 11

11

01

00

00 10

10

1

1

1

1

0 0

0 0

1 1

1 1

0 0

0 0

x

Examples


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Examples

yz

wx01 11

11

01

00

00 10

1

10

0

1

0

1 1

1 1

1 1

0

0 0

1 1

1

yz

wx01 11

11

01

00

00 10

1

10

0

0

1 1

1 1

1 1

0

0 0

1 1

1

0

Dont care termsb d


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Don t care terms

0 0 0

0 0 0 1

0

0 0 1 0

0 0 11

a b c d

1 1

0

0 1

0 1

1 0 0

1 0

1 0

10

1 0 0 0

0 0 1

0 1 0

0 11

1 1

1

1

1 0 01 0

1 0

1

1

1

1

11

1

1

0 1 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0

y0y1y2y3y4y5y6y7y8y9

0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 1

1 0 0 0 0 0 0 0 0 0

x x x x x x x x x x

x x x x x x x x x x

x x x x x x x x x x

x x x x x x x x x x

x x x x x x x x x x

x x x x x x x x x x

a

b

c

d

y0y1y2y3

y9

Decimal

Decoder

01 11

11

01

0000 10

10 0

ab

cd

0 0

0 0 0 0

x x x x

x x

10

0

Y3

3 . . .y a b c d


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01 11

11

01

00

00 10

10 0

abcd

0 0

0 0 0 0

x x x x

x x

10

0

Y3

Dont care terms can be chosen as 0 or 1. Depending on the problem, we can

choose the dont care term as 1 and use it to obtain a simpler Boolean expression

3 . .y b c d

Dont care terms should only be included in encirclements if it helps in obtaining a

larger grouping or smaller number of groups.

Minimization of Product of Sum Terms using Kmap


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Minimization of Product of Sum Terms using Kmap

0

1

1

0xy

0 1

1 1

f = . .

( ).

x x y x y

x x x y

x y

01

1

0x

y

0 1

1 1

f = x y

0

1

1

0xy

10

0 1

f = y


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0

1

1

0xy

1 0

1 0

f = y

0

1

1

0xy

1

00

1

f =x

yzx 00 01 11

0

1

10

0 1 1 0

1 00 1

.x zx z

f =( . ).( )x z x z f = . .x z x z


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yzwx 01 11

11

01

00

00 10

1

10

0

1

0

1

1 1

1 1

0 0

1 1

0

0

0

x y z x y z

w y z

w x

f =( ).( ).( ).( )x y z x y z w y z w x

Example


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yzwx 01 11

11

01

00

00 10

10

1

1

1 1

1

10

0 x

x

0

x

x1

1

1

p

Obtain the minimized PoS by suitably using dont care terms

f =( ).( ).( )x w z x w y y z

Design Flow x


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Design Flow

System Description

Truth Table

Boolean Expression

MinimizedBoolean Expression

Gate Netlist

y

zfsystem

0 0 0 00 0 1 10 1 0 00 1 1 11 0 0 01 0 1 11 1 0 0

1 1 1 1

x y z f

f = . . . . . . . .x y z x y z x y z x y z

f = . .x z x z

x

y

y

z

z

x

C

Mapping of Boolean expression to a Network of gates available in the


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Library of available Gates Cost

Inverter 1

Two input NAND 2

Three input NAND 3

AND-OR-Invert 3CABY

y

x1

x2

x3

x1

x2

x3

x2

x3

x1

x2

x3

x1

library

1 2 3 1 2 3 1 2 3 1 2 3y= . . . . . . . .x x x x x x x x x x x x

Implementation using only NAND gates


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.x x x .x y.x y

x

yf = .x y x y

A SoP expression is easily implemented with NAND gates. f = . . .a b c d g h

a

b

c

d

g

h

f

a

b

c

d

g

h

f


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a

b

c

d

g

h

f

a

b

c

d

g

h

f

a

b

c

d

g

h

f

There is a one-to-one mapping between AND-OR network and NAND network

Often there is lot of further optimization that can be done


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Consider implementation of XOR gate . .f A B A B

. . . .

( ) ( )

f A B B B A B A A

B A B A A B

Implementation using only NOR gates


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x x x x y x y

x

yf = .x y x y

To implement using NOR gates, it is easiest to start with minimized Boolean expression

in POS form

f =( ).( ).( )a b c d g h

f =( ).( ).( )a b c d g h


48/49

a

b

cd

g

h

f

a

b

c

d

g

h

f

a

b

c

d

g

h

f

There is a one-to-one mapping between OR-AND network and NOR network

To implement SoP expression using NOR gates, determine first the corresponding PoS


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expression and then follow the procedure outlined earlier

Implement f(x,y,z)= . . using NOR gatesx z x z

yzx 00 01 11

0

1

10

0 1 1 0

1 00 1 f =( . ).( )x z x z

x

x

z

zf

x

zf

z

x

Documents

Digital Circuits (2)