L13 Kvs Digital Circuits 1 Full

8/12/2019 L13 Kvs Digital Circuits 1 Full

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:

r. . . r vas avaDept. of Electrical Engineering

IIT Kanpur

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Numbers

A ositional notation is commonl used to ex ress numbers

5 4 3 2 1 0( )a a a a a a a r a r a r a r a r a r = + + + + +

The decimal system has a base of 10 and uses symbols

, , , , , , , , ,

3 2 1 0=

10123.24 1 10 2 10 3 10 2 10 4 10

= + + + +

2


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Numbers

An octal number system has a base 8 and uses symbols

(0,1,2,3,4,5,6,7)

3 2 1 08(2007) 2 8 0 8 0 8 7 8= + + +

What decimal number does it represent?

1 0

8(2007) 2 512 0 64 0 8 7 8 1033= + + + =

3

810

=


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Number Symbol

0 0Numbers

exa ec ma sys em as a ase o2 2

3 3

3 2 1 0

16(2 9) 2 16 16 16 9 16BC B C= + + + 4 45 5

7 7

8 8

How do we convert it into decimal number?

1

9 9

10 A

16

09 16 11209+ =

12 C

13 D

14 E

15 F 4

16 10


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A Binary system has a base 2 and uses only two symbols 0, 1

to represent all the numbers 0

3 2 1 0

2(1101) 1 2 1 2 0 2 1 2= + + + 21 2

22 4

Which decimal number does thiscorrespond to ?

23

824 16

1 0

2(1101) 1 8 1 4 0 2 1 2 13= + + + = 26 6427 128 .

23-

2-428 256

29 512

22 21 20 2-1 - 210 1024(K)220 1048576(M)

13. 5625

2-1 2-2 2-3 2-4 2-5 2-6

0.5 0.25 0.125 0.0625 0.03125 0.015625 5


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Developing Fluency with Binary Numbers

1 1 0 0 1 = ? 25

= + + =

0.101 = ? 0.5+0.125=0.625

11.001 = ? 3+0.125=3.125

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Converting decimal to binary number

10 1 0(45) .......n nb b b=

1 11 1 045 2 2 ....... 2

n nn nb b b b

= + + +

Divide both sides by 2

1 1 022.5 2 2 ....... 2 0.5

2

n n

n nb b b b

= = + + +

1 2 01 1 022 0.5 2 2 ....... 2 0.5

n n

n nb b b b

+ = + + +

0 1b = 7


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1 2 0

1 1 022 0.5 2 2 ....... 2 0.5n n

n nb b b b

+ = + + + 0 1b =

1 2 1 0

1 2 122 2 2 ....... 2 2n n

n nb b b b

= + + +


2 3 0

1 2 111 2 2 ....... 2 0.5

2

n n

n nb b b b

= = + + +

1 0b =

2 3 1 0

1 3 211 2 2 ...... 2 2n n

n nb b b b

= + + +

3 4 0

1 3 25.5 2 2 ...... 2 0.5n n

n nb b b b

= + + + 2 1b =

3 4 1 0

1 4 35 2 2 ...... 2 2n n

n nb b b b

= + + + 8


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3 4 1 0

1 4 35 2 2 ...... 2 2n n

n nb b b b

= + + +

4 5 0

1 4 32.5 2 2 ...... 2 0.5n n

n nb b b b

= + + + 3 1b =

4 5 1 01 5 42 2 2 ...... 2 2

n nn n

b b b b

= + + +

5 6 0n n = 0b =n n

=

10 5 4 3 2 1 0

9


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e o o success ve v s on y

22 1

11 0

2 1

45 = 1 0 1 1 0 1

1 0

0 1 1044 = 1 0 1 1 0 0


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Convert (153)10 to octal number system

10 1 0 8.......n n

1 1153 8 8 ....... 8n n= + +n n


1 2 0 01 119.125 8 8 ....... 8

8 8

n n

n nb b b

= = + +

0 0.1258

=

0153

19 1

remainder

2 3 153 = (231)8

11


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. 10

(0.35) 0. .......b b b b

=

1 2

1 20.35 0 2 2 ....... 2 n

nb b b

= + + +

How do we find the b-1 b-2 coefficients?Multiply both sides by 2

1 2. ....... n

= + + 1 =

1 2 1

2 30.7 2 2 ....... 2

n

nb b b

+

= + +12


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1 2 1

2 30.7 2 2 ....... 2 n

nb b b

+

= + +

Multiply both sides by 2

2 3. ....... n

= + + 2 =

o e a + + +

1 2 2n +

3 4. ....... n =

1 33 40.8 2 ....... 2

nnb b b

+

= + 3 0b =

13


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0 . 125

x2

.

.

5x2

0 .

1. 0. = . 2

0.8125 = ?0 .

1 . 625

8125x2

25x2

1 .x2

.

1. 0x2

. . 2

14


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Binary numbers decimal 2bit 3bit 4bit 5bit0 00 000 0000 00000

Least si nificant bit or LSB

Most significant bit or MSB2 10 010 0010 00010

3 11 011 0011 00011

4 100 0100 00100

5 101 0101 00101

7 111 0111 00111

8 1000 01000

Binary digit = bit 9 1001 01001

10 1010 01010

12 1100 01100

13 1101 01101

N-bit binary number

14 1110 01110

15 1111 01111

from 0 to 2N -115


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Converting Binary to Hex and Hex to Binary Number Symbol

0(0000) 0

= 1(0001) 12(0010) 2

ex

7 6 5 4 3 2 1

7 6 5 4 3 2 1 02 2 2 2 2 2 2b b b b b b b b+ + + + + +

4(0100) 4

5(0101) 5

11 016h h= +

6(0110) 6

7(0111) 7

7 6 5 4 3 2 1 0 1 02 2 2 2 2 2 2 16h h+ + + + + + = +

9(1001) 9

10(1010) A

h1h0

10110011 1011 0011 3B= =11(1011) B

12(1100) C

ex

(110011) (11)(0011) (33)b Hex= =

14(1110) E

15(1111) F( ) (1110)(1100) (11101100)Hex bEC = = 16


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Binary Addition/Subtraction

0

0

0

1

1

0

1

11

1

0 1 1 1 11 0

1 1 0

+ 1 1 1 0

1 0 1 1 1 1 0 1 1

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Complement of a number

9s com lementDecimal system:

10s complement

9s complement of n-digit number x is 10n -1 -x

10s complement of n-digit number x is 10n -x

9s complement of 85 ? 10 1 85 99 85 14 =

s comp emen o = =

' '

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Complement of a binary number

Binary system:

2s com lement

1s complement of n-bit number x is 2n -1 -x

2s complement of n-bit number x is 2n -x

1s complement of 1011 ? =

1s complement is simply obtained by flipping a bit

1's com lement of 1001101 = ?

(changing 1 to 0 and 0 to 1)

0110010 19


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2's complement of 1010 = 1's complement of 1010+1

0101 1 0110= + =

' =

Leave all least significant 0s as they are, leave first 1 unchanged

001110

and then flip all subsequent bits

1011 0101

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Advantages of using 2s complement

Adderx1

x2

S

CY

=10 2

10 2

' =

1010

If Carry is 1; then number you get is

+1010

( ) 1020100 4

21

10100 Answer is +4


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Advantages of using 2s complement

Adder1

x2 CY

( )10 210 10106 10 = ? ( )10 26 01102's comp ement o 1010=0110

If Carry is 0; then number you get is negative

+0110Take the 2s complement of number

'01100 0100 4

22

Answer is -4


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Example

0100Y = S if Sign = 0

Y = 2's Complement of S if Sign = 1

SYx1=1010

Adder

2's Complement Sign

=

CYx2=0110

= 1 for negative numbers6

1010

1 0

1 0 1 0+ 1 0 1 0

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Example

1100 0100Y = S if Sign = 0

Y = 2's Complement of S if Sign = 1

SYx1=0110

Adder

2's Complement Sign

=

CYx2=1010

= 1 for negative numbers10

0110

0 1

0 1 1 0+ 0 1 1 0

It makes sense to use adder as a subtractor as well provided additional circuit required

for carrying out 2s complement is simple

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Representing positive and negative binary numbers

One extra bit is required to carry sign information. Sign bit = 0 represents positive

=

decimal Signed

Magnitude

0 0000

decimal Signed1s

complement

1 1110

decimal Signed2s

complement

1 1111

1 0001

2 0010

3 0011

2 1101

3 1100

4 1011

2 1110

3 1101

4 1100

4 0100

5 0101

6 0110

5 1010

6 1001

7 1000

5 1011

6 1010

7 1001

7 0111

0 1000

1 1001

3 1011

4 1100

6 1110

7 1111 25


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Example

+ 0 0 1 0+ 5

+ 2

+ 5- 2

+ 1 1 1 0

0 1 1 1+7 +3

1 0 1 1 1 0 1 1- 5+ 2- 3

+

1 1 0 1

- 5- 2- 7

+

1 0 0 1

s comp emen s = s comp emen s =

26


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Example

2s com lement is 11111001:

27

00000111= 7


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Boolean Algebra

Al ebra on Binar numbers

A variable x can take two values {0,1} 0

False

No

Low voltage

Basic operations:

1Yes

Hi h volta e

1 2AND: y = x . x

y is 1 if and only if both x1 and x2 are 1, otherwise zero

x x

0 0 00 1 0Truth Table

1 0 0

1 1 1 28


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Basic operations:

OR: = x + xy is 1 if either x1 or x2 is 1. y= 0 if and only if both variables are

zero

x1 x2 y

0 1 1

1 0 11 1 1

NOT: y = xyx

1 0 29


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Boolean AlgebraBasic Postulates

.a: x = x

P2.a: x + y = y + x

. : x . = x

P2.b: x . y = y . x

Identity element

Commutative

P3.a: x.(y+z) = x.y+x.z

P4.a: x + x = 1

P3.b: x+y.z = (x+y).(x+z)

P4.b: x . x = 0 Complement

Distributive

Basic TheoremsT1.a: x + x = x

T2.a: x + 1 = 1T1.b: x . x = x

T2.b: x . 0 = 0

T3.a: ( x ) = x

T4.a: x + (y+z) = (x+y)+zT4.b: x . (y.z) = (x.y).z

T5.a: (x+y) = x . y (DeMorgan's theorem)

T6.a: x+ x.y = x

. : x.y = x y e orga

T6.b: x.( x+y) = x 30

P1 + 0 P1 b 1


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Proving TheoremsP1.a: x + 0 = x

P2.a: x + y = y + x

P1.b: x . 1 = x

P2.b: x . y = y . x

.a: x. y z = x.y x.z

P4.a: x + x = 1

P3. : x+y.z = (x+y .(x+z

P4.b: x . x = 0

Prove T1.a: x + x = x Prove T1.b: x . x = x

x + x = x+x . .

= x+x . x+x P4.a

. = . .

= x.x + x.x P4.b

= x + x.x (P3.b) = x . (x+x ) (P3.a

= x + 0 (P4.b)

= x P1.a

= x . 1 (P4.a)

= x P1.b

31

P1 a: + 0 P1 b 1


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Proving Theorems P1.a: x + 0 = x

P2.a: x + y = y + x

P1.b: x . 1 = x

P2.b: x . y = y . x

.a: x. y z = x.y x.z

P4.a: x + x = 1

P3. : x+y.z = (x+y .(x+z

P4.b: x . x = 0

Prove : x + 1 = 1+ = + +

x + x .y = x= x . 1 + x.

x + x .y = x+y

= (x+x)+ x= x. (1+ y)

= x . 1

= x x . x y

= 1. (x+ y)

= x + x

= 1

=x= x y

DeMorgans Theorem1 2 3 1 2 3(x x x ....) x . x .x .+ + + =

1 2 3 1 2 3(x . x . x .....) ( x + x x +.....)= + 32

Si lif i ti f B l i


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Simplification of Boolean expressions

x x x .... x . x .x .+ + + =

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

1 2 2 3(x .x x .x ) ?+ =1 2 2 3

(x . x ) . (x . x )=

1 2 2 3

(x +x ) . (x + x )=

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

1 2 2 2 1 3 2 3x . x x .x +x . x x . x= + +

1 2 1 3 2 3x . x x . x x . x= + +

33

F ti f B l i bl


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Function of Boolean variables

x1

x2

1 2

0 0 0

1 0 0

1 1 0

y = 1 when x1 is 0 and x2 is 1

1 2y = .x x

Boolean expression34

Obtaining Boolean expressions from truth Table


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x1 x2 y

0 0 1 1 2y = .x xx1 x2 y

0 0 01 2y = .x x

1 0 01 1 0

1 0 11 1 0

x1 x2 y

0 0 1

1 2.x x

=0 1 01 0 01 1 1 1 2.x x

. .

35



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x1 x2 y

0 0 01 2 1 2y = . .x x x x+

1 0 11 1 0

Instead of writing expressions as sum of terms that make y equal to 1, we

can also write expressions using terms that make y equal to 0

x1 x2 y 1 2 1 2 1 2y = . . .x x x x x x+ +

0 0 10 1 11 0 11 1 0 1 2y =x x+

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x x x x

0 0 10 1 0 y =x x+

0 0 00 1 1

1 2y =x x+

1 0 1

1 1 1

1 0 1

1 1 1

1 2x x+

x1 x2 y

0 0 0 1 2 1 2y = (x ).( )x x x+ +1 0 11 1 0

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= . . . . . . . .x x x x x x x x x x x x+ + +


yx1 x2 x3

0 0 0 00 0 1 1

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

Sum of Products (SOP) form

1 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+ + + + + + + +

Product of Sum (POS) form 38

Implementing Boolean expressions


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Implementing Boolean expressions

1 2AND: y = x . x

x1

x2yAND

Why call it a gate?

x1AND y = 0

0

x1AND

1

y = x1

a e s c ose a e s open

1 2: y = x + x 1x2yOR

NOT: y = x yx39


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x1x1x2

1 2 . x21 2

x1

xyNAND

NOR: y = x + x

x1

x2OR

1 2

x1+x2

x

x2y

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= =x1 x2 y

1 2 1 2 1 2 . . 0 0 00 1 11 0 1

1 1 0

y is 1 if only one variable is 1 and the other is zero

x2

y

x1

xyXOR

x1

x1x2x2

41

Gates with more than 2 inputs


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Gates with more than 2 inputs

x

1 2 3: y = x . x . ...x yANDx2x3

x1

1 2 3: y = x + x ....x+ + x2x3

y

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3XOR: y =x x x = x . . . . . . .x x x x x x x x x x x + + +

y = 1 only if odd number of inputs is 1

42

Implementing Boolean expressions using gates


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p g p g g

x

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

=

y

C

. . .

xx

x y zy

z

xy

z

x

zS

y

yC

yz

x

x y z

z

x

y

z

x y z43

Representation of Boolean Expressions


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Representation of Boolean Expressions

x y m n erm

0 0 x . y m00 0 0

x y 1

.

1 0 x . y m21 1 x . y m31 0 11 1 0

1f = . .x y x y+ 1 1 2f = m m+ f = 1, 2

2 , , 2 = . . .x y x y x y+ +

a function44

Three variable functions


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Three variable functions

0 0 0 x . y . z m0 . .

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

m

m2m3

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

1 1 0 x . y . z

m4m5

m6 1 1 1 x . y . z

m7

=2

, ,

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

45

Product of Sum Terms Representation


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p

x y f1 x y Max term

0 1 11 0 1 0 1 x + y M11 0 x + y M2

=1

301 =

= 301 ,

46


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y zx Max. terms

M0M1

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

M2

M3

x + y + z

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

M5

M6

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

1 , , =

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

47

Simplification


48/50

p

yx1 x2 x3

0 0 0 0=

, , ,

0 1 0 00 1 1 1

x1

x2

1 0 0 01 0 1 1

1 1 0 0

3

x2

x1

1 1 1 1

y

x2

x3

x1

3

x1

x2

x3

Simplification of Boolean expression yields : y = x3 !! which does

not require any gates at all ! 48

Goal of Simplification


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= 1 2 3 1 2 3 1 2 3 1 2 3. . . . . . . .x1

x2

x3

x2

x1

y

x2

x3

x1

x3

x1

x2

x3

Goal of simplification is to reduce the complexity of gate circuit.

s requ res a we m n m ze e num er o ga es.

49

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


50/50

1 2 3 1 2 3 1 2 3 1 2 3yx1

x2

x3

x1 1 3 1 2 3 1 2 3y = . . . . .x x x x x x x x + +

y

x2

x1

x2

x3

x1

x3

x1

x2

x3y

x2

x3

x1

x3x1

x2

x3

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

- - -

gates as compared to five 3-input gates used in circuit-1

50

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