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Simplification of Boolean Functions
Another method of simplification of Boolean function is Karnaugh – Map (K-Map). This map is a diagram made of squares, each square represent one minterms, and there are several types of K- |Map depending on the number of variables in Boolean function.
1-Two – variable K-Map
2 – Three – variable K – Map
3 – Four – variable K-Map
X Y X Y
X Y X Y
M0 M
M2 M
X Y Z X Y Z X Y Z X Y Z
X Y Z XY Z X Y Z X Y Z
M0 M1 M3 M
M4 M5 M7 M
X
Y Z 0 0 01 11 10 0 8
Y
X
Y 0 1
X
X
Y Z 0 0 01 11 10 0
8
XY
ZW 0 0 01 11 10 00 08 88 80
XY
ZW 0 0 01 11 10
00 08 88 80
3 – Five and Six variables K-Map
AB
CDE
ABC
DEF
= A B C + A B C + A B C + A B C + A B C
F = C + A B
4 - F ( X,Y,Z ) = ∑ ( 0, 2, 4 , 5 , 6 )
F ( X,Y,Z ) = Z + X Y
5 – F ( X,Y,Z,W ) = ∑ ( 0, 1, 2, 4, 5, 6, 8 , 9 , 12, 13, 14 )
F ( X,Y,Z,W ) = Z + X W + Y W
X
YZ 0 0 01 11 10 0 8
XY
ZW 0 0 01 11 10
00 08 88 80
A
BC 0 0 01 11 10
0 8
6 - F = A B C + B C D + A B C D + A B C
F ( A,B,C,D ) = B D + B C + A C D
7 – F(A,B,C,D.E) = ∑ (0,2,4,6,9,11,13,15,,17,21,25,27,29,31)
F( A,B,C,D) = A B E + B E + A D E
H.W
Simplify the following functions in sum of product using K-map
1- F = X Y + X Y W + W ( X Y + X Y )
2 – F = A B D + A C D + A B + A C D + A B D
3 – F ( A, B , C, D ) = Π ( 2, 3, 6, 7, 8, 9, 10 , 11, 12, 13, 14 )
AB
CD 0 0 01 11 10 00 08 88 80
AB
CDE
2 – Product of Sums In this case the missing terms is represented by 0 in K-map and simplified to obtained F (complement of the function).
F = A B + C D + B D
And the basic function F = ( A + B ) ( C + D ) ( B + D )
Ex Simplify the function F in 1 – Sum of Products 2 – Product of Sums
X Y Z F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0
AB
CD 0 0 01 11 10
00 08 88 80
Note If the function in Product of Sums form then the complement of the function must take first and then the 0 is represented in k-map.
Ex: ( A + B + C ) ( B + D ) The function in Product of Sum form, therefore the complement is take first F = A B C + B D Then these minterms will be assign in the map by 0 because the function is complement.
Ex : Obtained the simplified expression in Product of Sums
F = ( A + B + D ) ( A + D ) ( A + B + D ) ( A + B + C + D )
Sol F = A B D + A D + A B D + A B C D
F = A B + B D + B C D
F = ( A + B ) ( B + D ) ( B + C + D )
AB
CD 0 0 01 11 10
00 08 88 80
Ex simplify the Boolean function F in 1 – Sum of Products 2 – Product of Sums F ( X,Y,Z,W) = Σ (1, 3, 7, 11, 15 ) d ( X,Y,Z,W ) = Σ ( 0, 2 , 5 )
Sol 1- Sum of Products 2 – Product of Sums
F ( X,Y,Z,W) = Z W + X Y F ( X,Y,Z,W) = W + X Z
F ( X,Y,Z,W) = W ( X + Z )
Ex Simplify the Boolean function F in 1 – Sum of Products 2 Product of Sums using don’t care condition F = A C E + A C D E + A C D E D = D E + A D E + A D E
Sol F = A C E .1 + A C D E + A C D E = A C D E + A C D E + A C D E + A C D E
D = D E ( A + A ) + A D E ( C + C ) + A D E ( C + C )
= A D E ( C + C ) + A D E ( C + C ) + A C D E + A C D E + A C D E
= A C D E + A C D E + A C D E + A C D E + A C D E + A C D E +A C D E
X 8 8 X
X 8
X X
0 X 0
AB
CD 0 0 01 11 10
00 08 88 80
AB
CD 0 0 01 11 10
00 08 88 80
1 – Sum of Products 2 – Product of Sums
S.O.P P.O.S.
F ( A,C,D,E ) = A C + C E + A C D F ( A,C,D,E ) = A C + C D + A C D
F ( A,C,D,E ) = (A + C ) (C + D ) ( A + C + D )
Ex Simplify the Boolean function F in Sum of Products using don’t care condition F = B C D + B C D + A B C D D = B C D + A B C D
X 8 X
8 X X
X 88 X
X^ X
0 X X
X 0 X
X X
X^00 X
AC
DE 0 0 01 11 10
00 08 88 80
AC
DE 0 0 01 11 10
00 08 88 80
The Block Diagram Logic Circuit
2 – Full Adder A full adder is a combinational circuit that forms the arithmetic sum of three inputs bits. It consists of three inputs and two outputs. Two of the inputs variables, X and Y, represent the two bits to be added, the third input Z , represent the carry from the previous step. The two output S (for sum) and C ( for carry ).
Input Output X Y Z C S 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 Block Diagram 1 1 1 1 1
Truth Table To find the logic equations K- map is used
S = X Y Z + X Y Z + X Y Z + X Y Z = = Z ( X Y + X Y ) + Z ( X Y + X Y ) = = Z ( X. Y ) + Z ( X Y ) = = Z ( X Y ) + Z ( X Y ) = X Y Z
X H. A.
Y
S C
S C
X Y F. A.
Z
X Y
S
C
X
YZ 0 0 01 11 10 0 1
C = X Y + X Z + Y Z
The logic curcuit
The Subtractors
1 – Half Subtractor A half subtractor is combinational circuits that subtract two bits and produce their differences. To perform (X-Y ) the truth table is: Input output X Y B D 0 0 0 0 0 1 1 1 D = difference , B = Borrow 1 0 0 1 1 1 0 0
Truth Table
X
YZ 0 0 01 11 10 0 8
S
C
X Y Z
D = X Y Z + X Y Z + X Y Z + X Y Z
= Z ( X Y + X Y ) + Z ( X Y + X Y )
= Z ( X Y ) + Z ( X Y )
= Z ( X Y ) + Z ( X Y )
= X Y Z
The logic curuit
X
YZ 0 0 01 11 10 0 8
S
C
X Y Z
Code Conversion
To convert from binary code to another code, a combinational circuit performs this transformation by means of logic gates. Ex Design a combinational circuit that convert a BCD code to Excess-3 code.
Sol The truth table consists of 4 inputs and 4 outputs
Input Output A B C D X Y Z W 0 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1 0 0 1 0 1 1 1 1 0 1 0 X = A + B D + B C 1 0 0 0 1 0 1 1 = A + B ( D + C ) 1 0 0 1 1 1 0 0 1 0 1 0 x x x x 1 0 1 1 x x x x 1 1 0 0 x x x x 1 1 0 1 x x x x 1 1 1 0 x x x x 1 1 1 1 x x x x
Y = B C D + B D + B + C
= B C D + B ( D + C )
X X
8 1 X
X X X X
X X
AB
CD 0 0 01 11 10 00 08 88 80
AB
CD 0 0 01 11 10 00 08 88 80
Ex A combinational circuit has four inputs and one output, the output equal 1 when: 1 – all the inputs are equal to 1 or 2 – non of the inputs are equal to 1 or 3 – an odd number of inputs are equal to 1. Design the logic circuit.
Sol
Input Output X Y Z W F 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 F = Y Z W + X Y W + X Y Z + X Z W 1 0 0 1 0 1 0 1 0 0 + Y Z W + X Y W + X Y Z + Y Z W 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1
XY
ZW 0 0 01 11 10 00 08 88 80
Ex Design a combinational circuit that inputs is three – bit numbers and the output is equal to the squared of the input numbers in binary?
Sol Input Output X Y Z F 5 F 4 F 3 F 2 F 1 F 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 0 1 0 1 1 0 0 1 F 0 = Z 1 1 0 1 0 0 1 0 0 1 1 1 1 1 0 0 0 1
F 3 = X Y Z + X Y Z
F 5 = X Y
X
YZ 0 0 01 11 10 0
X^8
YZ 0 0 01 11 10 0 8
X
YZ 0 0 01 11 10 0 8 X
YZ 0 0 01 11 10 0 8
F 4 = X Y + X Z
F 2 = Y Z
X
YZ 0 0 01 11 10 0 8
