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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
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