Waleed arshad

WALEED ARSHAD

m0 m1

m3m2

(a)

y

x’y’ x’y

xyxy’1

0

yx 0 1

x

(b)

(b) x+y

1

0

yx 0 1

x

y

(a) xy

1

0

yx 0 1

x

y

1

1

11

TWO VARIABLE K-MAP

m0

m4

m1 m3 m2

m5 m6m7

y

x

x’y’z’

xy’z’

x’y’z x’yz x’yz’

xy’z xyz’xyz

yz

x

0

1

00 01 11 10

z

THREE VARIABLE MAP

y

z

yz

1

0

1 1

1 1

1 1

1 1

00 01 11 10 MAP FOR

F=x’yz+x’yz’+xy’z’+xy’z

F=x’y+xy’

Reflected CodeBinary Code Reflected Code

0000 0000

0001 0001

0010 0011

0011 0010

0100 0110

0101 0111

0110 0101

0111 0100

1000 1100

1001 1101

1010 1111

1011 1110

1100 1010

1101 1011

1110 1001

1111 1000

MAP FO R : x’yz+xy’z’+xyz+xyz’=yz+xz’

yz

x00 01 11 10

0

1 11

1

1

z

y

A

BC

A00 01 11 10

0

1

C

BC B

1

1 1

11

1

MAP FOR : A’C+A’B+AB’C+BC=C+A’B

yz

x00 01 11 10

0

1

y

x

F(x,y,z)=Σ(0,2,4,5,6)=z’+xy’

1

1

1 1

1

m0 m1 m3 m2

m6

m14

m10m11m9m8m8

m12 m13 m15

m7m5m4

(a)y

z

w

00

01

11

10

00 01 11 10wxyz

x

w’x’y’z’

w’xy’z’

wxy’z’

wx’y’z’

w’x’y’z w’x’yzw’x’yz w’x’yz’

w’xyz’

wxyz’

wx’yz’wx’yzwx’y’z

wxy’z wxyz

w’xyzw’xy’z

FOUR VARIBALE MAP

1 square represents a term of 4 literals.2 adjacent squares represent a term of 3 literals.4 adjacent squares represent a term of 2 literals.8 adjacent squares represent a term of 1 literal.16 adjacent squares represent the function equal to 1

MAP FOR : F(w,x,y,z)=Σ(0,1,2,4,5,6,8,9,12,13,14)=y’+w’z’+xz’

yyz

z

00

01

11

10

10110100wx

x

w

1

1

1

1 1

1

1

1 1

1

1

CCD

D

00

01

11

10

10110100AB

B

A

1 1 1

1

1

MAP FOR : A’B’C’+B’CD’+A’BCD’+AB’C’=B’D’+B’C’+A’CD’

1 1

A

C

AB

00

01

11

10

000 001 011 010 110 111 101 100

CDE

B

DE

E

FIVE -VARIABLE MAP

0 1 3 2 6 77 5 4

8 9 11 10 15 13 12

282931==

30

14

26272524

16 17 19 18 22 23 21 20

B

ABC

DEF

000 001 011 010 110 111 101 100

000

001

011

010

0 1 3 2 6 77 5 4

8 9 11 10 15 13 12

282930 3126272524

16 17 19 18 22 23 21 20

110

111

111

101

100

C

C

D

A

E FF

48

56

49 51 50 54 55 53 52

57 59 58 62 63 61 60

40 41 43 42 46 47 45 44

32 33 35 34 38 39 37 36

SIX VARIABLE MAP

14

No of adjacent squares & literals in a term 2k adjacent squares=n-k literalsn-variable mapk 2k n=2 n=3 n=4 n=5 n=6 n=7 0 1 2 3 4 5 6 71 2 1 2 3 4 5 60 4 0 1 2 3 4 53 8 0 1 2 3 44 16 0 1 2 35 32 0 1 26 64 0 17 128 0

11

1111

1111

111 1

CDE

000 001 011 010 110 111 101 100

C

B

AB

00

01

11

10

E D E

A

F=BE+AD’E+A’B’E’

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

ExampleF(A , B, C, D) =∑(0,1,2,5,8,9,10)

(i) Simplify it in:

(ii) Sum of products

(iii) Product of sums

AB

(i) Sum of products

F=B’D’+B’C’+A’C’D

F’ =AB+CD+BD’

(ii) F=(A’+B’)(C’+D’)(B’+D)

00

01

11

10A

11CD00 01 10

1011

0000

0010

1011

C

D

B

B’

D’

C’

A’

D

F

A’

B’

D’C’

D

F

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

GATE IMPLEMENTATION

X Y ZF

0 0 00

0 0 11

0 1 00

0 1 11

1 0 01

1 0 10

1 1 01

1 1 10

TRUTH TABLE OF

FUNCTION F

F(x,y,z)= (1,3,4,6)

F(x,y,z)= (1,3,4,6)

F(x,y,z) = (0,2,5,7)

1001

01100

1

00 01 11 10

z

yzx

x

y

Sum of products

F=x’z+xz’)

Product of sums

F’=xz+x’z’

F=(x’+z’)(x+z)

NAND and NOR implementation

xyz

F=(xyz)’

AND invert

xyz

F=(x’+y’+z’)=(xyz)’

Invert-OR

F=(x+y+z)’

OR-invert

(a) TWO GRAPHIC SYMBOLS FOR NAND GATE

xyz F=x’y’z’=(x+y+z)’

Invert-AND

(b) TWO GRAPHIC SYMBOLS FOR NOR GATE

x’x x x’

Buffer-invert AND-invert

x x’

OR-invert

THREE GRAPHIC SYMBOLS FOR INVERTOR

A

B

CD

E

AB

C

DE’

F

F

(a)

(b)

NAND Implementation

Simplified Boolean function

F=AB+CD+E

Convert AND gates with at least two literals to NAND gates at first level.

THREE WAYS TO IMPLEMENT

F=AB+CD+E

Draw a single NAND gate at level-2

&

NAND gate for a term with single

literal

(c)

A

B

CD

E

F

F={(AB)’.(CD)’.E’}

F=AB+CD+D

x’

y’

z’

x

F

Two-level implementation

1000

0001

z

00 01 11 10

yyz

0

1

x F(x,y,z)= (0,6)

F=x’y’z’+xyz’

F’=x’y+xy’+z

(a) MAP SIMPLIFICATION IN SUM OF PRODUCTS

y

z’

(b) F=x’y’z’+xyz’

X’

y

xY’

z

F’

Three level implementation

F

(c) F’=x’y+xy’+z

IMPLEMENTATION OF THE FUNCTION WITH NAND GATES

F(x,y,z) = (0,6)

NOR Implementation

F

(C+D)

A

B

C

D

E

(A+B)Simplified Boolean function

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

F

A

B

C

D

E’

Convert OR gates with at least two inputs to NOR gates at first level

(a)

(b)

THREE WAYS TO IMPLEMENT

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

A

B

C

D

E

(C)

Draw a single NOR gate at level 2 & a NOR gate for a term with single literal.

Example

F(n, y, z) = {(0, 6)F = x’y’z’ + xyz’

F’ = x’y + xy’ + zx

y’

x’

y

z’

F

(a) F = (x + y’) (x’ + y)z’

x

y

z

x’

y’

z

F

F’

(b) F’ = (x + y + z) (x’ + y’ + z)

IMPLEMENTATION WITH NOR GATES

y

0100

0100

01X0

X11X

01

11

10

YZ00 01 11 10

w

x

z

wx

00

F(w,x,y,z)=

(1,3,7,11,15)

D(w,n,y,z)=(0,2,5)

Combing 1’s and X’s

F=w’z+yz

or

F=w’x’+yz

YZy

0100

0100

01X0

X11X

01

11

10

00 01 11 10

w

x

z

wx

00

Combing 0’s and X’s

F’=z’+wy’

F=z(w’+y)

EXAMPLES WITH DON’T CARE CONDITION

DETERMINE THE PRIME-IMPLICANTS OF THE FUNCTION

F(w,x,y,z)=∑(1,4,6,7,8,9,10,11,15)

(a) (b) (c)

0001

1

0100

4

1000

8

01106

10019

101010

01117

10 1111

111115

1, 9(8)*

4, 6(2)*

8, 9(1)

8, 10(2)

6, 7(1)*

9, 11 (2)

10,11 (1)7, 15 (8)*

11,15 (4)*

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

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

PRIME-IMPLICANTSDecimal Binary

w x y zTerm

1,9 (8)

4,6 (2)

6,7 (1)

7,15 (8)

11,15 (4)

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

- 0 0 1

0 1 - 0

0 1 1 -

- 1 1 1

1 - 1 1

1 0 - -

X’ y’ z

W’ x z’

w’ x y

X y z

w y z

W x’

F=x’y’z+w’xz’+w’xy+xyz+wyz+wx’

(ii) Selection of

PRIME-IMPLICANTS

1 4 6 7 8 9 10 11 15

x’y’z 1,9 X X

w’xz’ 4,6 X X

w’xy 6,7 X X

xyz 7,15 X X

wyz 11,15 X X

wx’ 8,9,10,11 X X X X

F=x’y’z+w’xz’+wx’+xyz

Essential Prime-implicants

1111

1

111

1

01

11

10

YZ00 01 11 10

w

x

z

wx

00

F(w,x,y,z)=∑(1,4,6,7,8,9,10,11,15)

y

MAP FOR THE FUNCTION OF

F = x’y’z+w’xz’+xyz+wx’

THE TABULATION METHOD (Q-M Mehthod)Example:

F(w,x,y,z) = ∑(0,1,2,8,10, 11,14,15)

(i) DETERMINATION OF PRIME-IMPLICANTS(a) (b) (c)

wxyz wxyz wxyz

0 0000 0,1 000-*

0,2 00-00,2,8,10 -0-0*

0,8,2,10 -0-0*

1 0001 0,8 -0002 0010

10,11,14,15 1-1-*

10,11,14,15 1-1-*

8 1000 2,10 -010 8,10 10-010 1010

10,11 101-11 1011 10,14 1-1014 1110

11,15 1-1115 11111 14,15 111-

Prime-implicants

F=w’x’y’+x’z’+wy

(ii) Selection of

PRIME-IMPLICANTS

0 1 2 8 10 11 14 15

w’x’y’ 0,1 X X

x;z’ 0,2,8,10 X X X X

wy 10,11,14,15 X X X X

Essential Prime-implicants : F=w’x’y’+x’z’+wy

1 11

1 1

1 1 1

00 01 11 10

Y

10

11

01

00

wx

yz

W

X

F= w’x’y’ + x’z’+wy

MAP FOR THE FUNCTION

RULES FOR NAND and NOR IMPLEMENTATION

Case

Function to

Simplify

Standrad From to

Use

How to drive

Implement with

Number of Levels

to F

(a)

(b)

(c)

(d)

F

F

F

F

Sum of Products

Sum of Products

Product

of Sums Product

of Sums

Combine 1,s in map

Combine 0, s

In map

Complement F in (b)

Complement F in (a)

NAND

NAND

NOR

NOR

2

3

2

3

AB

CD

F

F = (AB + CD)’

F = (A B)’ . (C D)’

(a) Wired – AND in open Collector TTL NAND gates

C

AB

D

F

F = [(A + B ) ( C + D)]’

F = (A + B)’ + (C + D)’

(b) Wired – OR in ECL gates

AB

CD

E

AB

CD

E

F F

(AND – OR - INVERT) OR – AND - INVERT

WIRED LOGIC

AB

CD

E

F

(a) AND - NOR (b) AND - NOR

(c) NAND - AND

AND – OR – INVERT CIRCUITS F = (AB + CD + E)’(Non degenerate form)

AND - - NOR & NAND – AND are equivalent forms

NONDEGENERATE FORMSCOMMON GATE : AND, OR, NAND, NORIF AT LEVEL 1 : ONE TYPE OF GATE AT LEVEL 2 : ONE TYPE OF GATE

* POSSIBLE COMBINATIONS = 16

8 OF THESE COMBINATIONS ARE SAID TO BE DEGENERATE FORMS BECAUSE THEY DEGENERATE TO A SINGLE OPERATION.

FOR EXAMPLE : NAND – OR AND – AND OPERATION NAND – NOR OR – OR OPERATION NOR – AND AND – NAND OPERATION NOR – NAND OR – NOR OPERATION8 OF THESE COMBINATION S ARE SAID TO BE NON – DEGENERATE

FORMS BECAUSE THEY PRODUCE AN IMPLEMENTATION IN SUM OF PRODUCTS OR PRODUCT OF SUMS.

FOR EXAMPLE :AND – OR OR – ANDNAND – NAND NOR – NORNOR – OR NAND – ANDOR – NAND AND - NOR

AB

C

D

E

F

(a) OR - NAND

AB

C

D

E

F

(b) OR - NAND

CONTED

AB

C

D

E

F

(C) NOR - OR

OR – AND – INVERT F = [(A + B ) (C + D ) E]’(Non – degenerate form)

OR – NAND & NOR – OR ARE EQUIVALENT FORMS

X’

y

X

y’

z

F

X’y

Xy’

z

F

Example

AND - NOR NAND - AND

F = x’ y + x y’ + z

(a) F = (x’ y + x y’ + z)’

1 0 0 0

0 0 0 1

0

1

00 01 11 10

F’ = x’ y + x y’ + z

F = x’ y z’ + x y z’

xyz

z

x’y’

F

xyz

z

x’y’

F

OR - NAND NAND - OR

F = x’ y’ z’ + x y z’

F = (x + y + z) (x’ + y’ +z)

(b) F = [(x + y + z) (x’ + y’ + z)]’

TWO – LEVEL IMPLEMENTATION