simplification of boolean functions - wordpress.com froma boolean statement or a truth table into...
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Contents:
Why simplification?
The Map Method
Two, Three, Four and Five variable Maps.
Simplification of two, three, four and five variable Boolean function by Map method.
Product of sums and Sum of products simplification.
NAND and NOR implementation.
Simplification of Boolean Functions
Course Instructor
Mohammed Abdul kader
Assistant Professor, EEE, IIUC
The complexity of the digital logic gates that implement a Boolean function is directly
related to the complexity of the algebraic expression from which the function is
implemented.
Although the truth table representation of a function is unique but the algebraic
expression appears in many different form.
Before implementation of logic circuit, the choice of simplest boolean expression from
different representation results the minimum number of gates and less complexity in
the digital circuit.
There are different ways of simplification of Boolean function. In this section we will
discuss the “Map Method” of simplifying Boolean function.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 2
Simplification of Boolean Functions 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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The map method provides a simple straight forward procedure for minimizing Boolean
functions.
The map method, first proposed by Veitch and slightly modified by Karnaugh, is also
known as the Veitch diagram or the Karnaugh map.
Karnaugh maps provide an alternative way of simplifying logic circuits.
Instead of using Boolean algebra simplification techniques, you can transfer logic
values froma Boolean statement or a truth table into a Karnaugh map.
The arrangement of 0's and 1's with in the map helps you to visualize the logic
relationships between the variables and leads directly to a simplified Boolean
statement.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 3
The Map Method 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 4
Two Variable Map
There are four minterms for two variables; hence the map consists of fours quares, one of
each minterm.
F=xy F=x+y = xy+xy+xy = m1+ m2 + m3
Representation of
functions in two
variable map.
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 5
Three Variable Map
There are eight minterms for three variables, i.e. the map consist of eight squares.
Minterms are not arranged in binary sequence but in a sequence similar to gray
code/reflected code.
There are four squares where each variable is equal to 1 and four where each is equal
to 0.
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 6
Three Variable Map
F = m5 + m7 = xyz + xyz = xz (y+y) = xz
Understanding the usefulness of map for simplification of Boolean function
F = m1 + m7 = xyz + xyz
Isolated ‘1’ in map
Adjacent Pair/ 1’s in two adjacent square
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 7
Three Variable Map
F = 1
Understanding the usefulness of map for simplification of Boolean function
F = m1 + m3 + m5 + m7
= xyz + xyz+ xyz + xyz
= xz (y+y) +xz (y+y) = xz +xz = z (x+x) = z
Adjacent Quad/ 1’s in four adjacent square
Adjacent Octet/ 1’s in eight adjacent square
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 8
Three Variable Map: Examples
Example 3-2: Simplify the Boolean function
F= xyz + xyz + xyz + xyz Solution :
yz
xz
So, after simplification, F = yz+ xz
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 9
Three Variable Map: Examples
Example 3-3: Simplify the Boolean function
F= AC+ AB+ ABC+BC Solution :
F = AC+ AB+ ABC+BC =ABC+ ABC +ABC+ ABC+ ABC+ABC+ABC
C
AB
So, after simplification, F = C+ AB
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 10
Three Variable Map: Examples
Example 3-4: Simplify the Boolean function
F (x,y,z)= ∑ (0, 2, 4, 5, 6)
Solution :
So, after simplification, F = z+ xy
xy
z
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 11
Four Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 12
Four Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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The map for Boolean function of four binary
variables has 16 minterms and the squares assigned to
each.
The rows and columns are numbered in a reflected
code sequence, with only one digit changing value
between two adjacent rows and columns.
The minterm corresponding to each square can be
obtained from the concatenation of the row number
with the column number.
The combination of adjacent squares that is useful
during the simplification process is easily determined
from inspection of the four variable map-
wxyz: 1010 (10 in binary), so
this square represent m10
One square represents one minterm, giving a term of four literals.
Two adjacent squares represent a term of three literal.
Four adjacent squares represent a term of two literals.
Eight adjacent squares represent a term of one literals.
Sixteen adjacent squares represent the function equal to 1.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 13
Four Variable Map: Examples
Example 3-5: Simplify the Boolean function
F (x,y,z)= ∑ (0, 1,2, 4, 5, 6, 8, 9, 12, 13, 14)
Solution :
So, after simplification, F = y+ wz+x z
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 14
Four Variable Map: Examples
Example 3-6: Simplify the Boolean function
F = AB C +B CD + A BCD +AB C Solution :
So, after simplification, F = BD + BC+A C D
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 15
Five Variable Map 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 16
Five Variable Map: Examples 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Example 3-7: Simplify the Boolean function
F (A, B, C, D, E) = ∑ (0, 2, 4, 6, 9, 11, 13, 15, 17, 21, 25, 27, 29, 31)
Solution :
ADE
BE
A BE
So, after simplification, F = BE+AD E + ABE
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 17
Product of Sums Simplification 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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All previous examples are in sum-of-products form [F = BE+AD E + ABE ]
How to obtain the product-of-sum form
* Simplify F in the form of sum of products. [If we mark the empty
squares by 0’s and combine them into valid adjacent squares, we obtained a
simplified expression of the complement of the function, i.e. of F]
* Apply DeMorgan's theorem F = (F ')
* F': sum of products => F : product of sums
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 18
Example 3-6: Simplify the Boolean function in (a) sum of products and (b) product of
sums
F (A, B, C, D, E) = ∑ (0, 1, 2, 5, 8, 9, 10)
Solution :
F = AB + CD + BD
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So, F = (A +B ) (C +D ) (B + D ) by DeMorgan theorem
(b) product of sums simplification
(a) Sum of products simplification
F = B D + B C + A C D
Product of Sums Simplification (Cont.)
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 19
Product of Sums Simplification (Cont.)
Gate implementation of the function Example 3-8
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(b) F = (A +B ) (C +D ) (B + D ) (a) F = B D + B C + A C D
Sum of products product of sums
Simplify the Boolean function F(A,B,C,D)=Π(0,1,2,3,4,10,11)
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 20
Product of Sums Simplification (Cont.)
Example: Simplify the Boolean function F(A,B,C,D)=Π (0,1,2,3,4,10,11)
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AB
CD
From Map we get,
F = A B + B C + A C D
Using DeMorgan theorem-
F = (A B + B C + A C D )
= (A + B ) ( B +C ) ( A + C + D )
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 21
NAND and NOR Implementation
Why NAND and NOR implementation?
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Digital circuits are frequently constructed with NAND/NOR rather than with AND/OR gates.
NAND and NOR gates are easier to fabricate with electronic components than AND/OR.
Cheaper(lower cost) and faster(less delay).
Any Boolean function can be constructed using only NAND or only NOR gates. That’s why NAND
and NOR are known as universal gates.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 22
NAND and NOR Implementation
Implementation of basic gates by NAND gate
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x x
x y
(xy) ((xy)) = xy
x
y
x
y
(xy) = x+y
NOT gate by NAND
gate
AND gate by NAND
gate
OR gate by NAND
gate
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 23
NAND and NOR Implementation
Implementation of basic gates by NOR gate
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x x
x y
(x+y) ((x+y)) = x+y
x
y
x
y
(x+y) = xy
NOT gate by NOR gate
OR gate by NOR gate
AND gate by NOR
gate
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 24
NAND and NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Two graphic symbols for NAND gate
Two graphic symbols for NOR gate
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 25
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Implementation F=AB+CD+E by NAND gate only.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 26
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Rules for obtaining the NAND logic diagram from a Boolean function
1. Simplify the function and express it in sum of products.
2. Draw a NAND gate for each product term of the function that has at least two literals. The inputs
to each NAND gate are the literals of the term. This constitutes a group of first-level gates.
3. Draw a single NAND gate (using the AND-invert or invert-OR graphic symbol) in the second
level, with inputs coming from outputs of the 1st level.
4. A term with a single literal requires an inverter in the first level or may be complemented and
applied as an input to the second-level NAND gate.
Note: If we simplify the function combining 0’s in a map, we obtain the simplified expression of the
complement of the function in sum of product. The complement of the function can then be
implemented with two levels of NAND gates using the rules stated above. If the normal output is
desired, it would be necessary to insert a one-input NAND gate.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 27
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Example 3-9: Implement the following function with NAND gates
F(x,y,z) = ∑ (0, 6)
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 28
NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Implement the function F= (A+B) (C+D) E with NOR gates
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 29
NOR Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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Example 3-9: Implement the following function with NAND gates
F(x,y,z) = ∑ (0, 6)
Sum of Product
Product of sums
Sum of Product
Product of sums
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 30
Don’t-Care Conditions 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
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You don’t always need all 2𝑛 input combinations in an n-variable function.
–If you can guarantee that certain input combinations never occur.
–If some outputs aren’t used in the rest of the circuit.
A four bit decimal code, for example, has six combinations which are not used. Any
digital circuit using this code operates under the assumption that these unused
combinations will never occur as long as the system is working properly.
The unused combinations is known as don’t care conditions and can be used on the map
to provide further simplification of the boolean expression.
It should be realized that a don’t care minterm is a combination of variables whose
logical value is not specified. It cannot be marked with a 1 or, 0 in the map as it is not
specified as 0 or 1. To distinguish the don’t care condition from 1’s and 0’s, an X is used.
Thus, an X inside a square in the map indicates that we don’t care whether the value of 0
or 1 is assigned to F for the particular minterms.
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 31
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
Example 3-12: Simplify F(w,x,y,z) = Σ(1,3,7,11,15) which has the don’t care conditions
d(w,x,y,z) = Σ (0,2,5)
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 32
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
Exercise. 3-18 : Implement the following function with either NAND or NOR gates. Use
only four gates.
F = w’xz+w’yz+x’yz’+wxy’z and d= wyz
NAND Implementation
Sum of products:
Combine 1’s and some of X’s
F= x y+xz
wx
yz
x
y
z
F
x (x y)
(x z)
Lecture materials on "Simplification of Boolean Functions" By- Mohammed abdul kader, Assistant Professor, EEE, IIUC 33
NAND Implementation 101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010
NOR Implementation
Products of sums:
Combine 0’s and some of X’s
F = x y +xz
So, F= (x+y) (x +z)
wx
yz
x
z
y
x (x +z)
(x +y )
F