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Boolean Algebra and Digital Circuits
Reading: Chapter 8 (138-162) from the text book
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Defn of a Boolean Algebra
A Boolean algebra consists of:• a set B={0, 1},• 2 binary operations on B (denoted by + & ×),• a unary operation on B (denoted by '), such
that :0 + 0 = 0 0 × 0 = 01 + 0 = 1 0 × 1 = 00 + 1 = 1 1 × 0 = 01 + 1 = 1 1 × 1 = 1
0’=1 and 1’=0. 2
Rules of a Boolean AlgebraThe following axioms (‘rules’) are satisfied for all elements x, y& z of B:
(1) x + y = y + x (commutative axioms) x× y = y × x
(2) x + (y + z) = (x + y) + z (associative axioms) x × (y × z) = (x × y) × z
(3) x × (y + z) = (x × y) + (x × z) x + (y × z) = (x + y) × (x + z) (distributive axioms)
(4) x + 0 = x x × 1 = x (identity axioms)(5) x + x' = 1 x × x' = 0 (inverse axioms) 3
Duality
• To form the dual of an expression, replace all + operators with × operators, all × operators with + operators, all 1’s with 0’s, and all 0’s with 1’s.
• The principle of duality says that if an expression is valid in Boolean algebra, the dual of that expression is also valid.
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Duality
Exercise: Form the dual of the expressiona + (bc) = (a + b)(a + c)
Solution: Following the replacement rules…a(b + c) = ab + ac
• Take care not to alter the location of the parentheses if they are present.
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Laws of Boolean Algebra
• In addition to the laws given by the axioms of Boolean Algebra, we can show the following laws
x'' = x (double complement)x + x = x x× x = x (idempotent )(x + y)' = x' × y' (x × y)' = x' + y' (de Morgan’s
laws)x + 1 = 1 x × 0 = 0 (annihilation)x + (x × y) = x x× (x + y) = x (absorption)0' = 1 1' = 0 (complement)
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Exercise
Simplify the Boolean expression(x' × y) + (x × y)
Solution: (x' × y) + (x × y)= (y × x') + (y × x) (commutative)= y × (x' + x) (distributive)= y × (x + x') (commutative)= y × 1 (inverse)= y (identity)Thus (x' × y) + (x × y) = y 7
Digital Circuits
• The circuitry in a digital computer operates with signals that can take only 2 values ‘on/off’ (i.e. 0/1).
• We’ll use the particular Boolean Algebra where B has just the 2 elts 0 & 1, and where
• Boolean addition corresponds to parallel switch contacts:
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Boolean Addition
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Boolean Multiplication• Boolean multiplication corresponds to series
switch contacts:
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Boolean Notation
• This means that in effect we’ll be employing Boolean Algebra notation.
• The truth tables can be rewritten as
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Notational Short-cuts
We will employ short-cuts in notation:
(1) In ‘multiplication’ we’ll omit the symbol ×, & write xy for x × y (just as in ordinary algebra)
(2) The associative law says that x + (y + z) = (x + y) + z
So we’ll write this as simply x + y + z, because the brackets aren’t necessary.
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Notational Short-cuts
Similarly, write the product of 3 terms as xyz
(3) In ordinary algebra, the expression x + y × z means x + (y × z), because of the convention that multiplication takes precedence over addition.
e.g. x + yz means x + (y × z), and not (x + y) × zSimilarly, ab + cd means (a × b) + (c × d)
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Digital Circuits• A digital circuit (or logic gate circuit) is an
electronic device for carrying out digital computations (e.g. addition of 2 numbers)
• It accepts 1 or more inputs, each of which has 2 possible states (0 for ‘off’ & 1 for ‘on’)
• For each possible combination of inputs,one or more outputs are produced
.
...
LogicCircuitInputs Outputs
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Describing Circuit Functionality: Inverter
• Truth table completely specifies outputs for all input combinations.
• The above circuit is an inverter. – An input of 0 is inverted to a 1.– An input of 1 is inverted to a 0.
AY
01
10
Input Output
A Y
Symbol
Truth Table
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The AND Gate
• This is an AND gate. • So, if the two inputs signals
are asserted (high (ON)) the output will also be asserted (ON).
• Otherwise, the output willbe asserted (low (OFF)).
ABY
000
010
100
111
A
BY
Truth Table
Input 1 input2 output
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The OR Gate
• This is an OR gate. • So, if at least one of the two
input signals is asserted (ON), then output will be asserted (ON).
• Otherwise, the output willbe asserted (low (OFF)).
ABY
000
011
101
111
AB
Y
Input 1 Input 2 Output
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Consider Three-input Gate
3 Input OR Gate
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Boolean Functions
• Boolean algebra deals with binary variables and logic operations.
• Function results in binary 0 or 1
x00001111
y00110011
z01010101
F00001011 F = x(y+z’)
xy
zz’
y+z’ F = x(y+z’)
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Boolean Functions
x00001111
y00110011
z01010101
xy00000011
x
y
z
G = xy +yz
yz
xy
We will learn how to transform between expression and truth table.
yz00010001
G00010011
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Truth Table to ExpressionConverting a truth table to an expression– Each row with output of 1 becomes a product term– Sum product terms together to have the Boolean
function.
x00001111
y00110011
z01010101
G00010011
xyz + xyz’ + x’yz
Any Boolean Expression can be represented in sum of products form!
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Equivalent Representations of Circuits
• Number of 1’s in truth table output column equals ANDterms for Sum-Of-Products (SOP)
x y z
x00001111
y00110011
z01010101
G00010011
G = xyz + xyz’ + x’yz
G
x x xx x
xx
x x
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Reducing Boolean Expressions
• Is this the smallest possible implementation of this expression? No!
• Use Boolean Algebra rules to reduce complexity while preserving functionality.
• Step 1: Use idempotent law (a + a = a). So xyz + xyz’ + x’yz = xyz + xyz + xyz’ + x’yz
G = xyz + xyz’ + x’yz
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Reducing Boolean Expressions
• Step 2: Use distributive law a(b + c) = ab + ac. So xyz + xyz + xyz’ + x’yz = xy(z + z’) + yz(x + x’)
• Step 3: Use Inverse law (a + a’ = 1). So xy(z + z’) + yz(x + x’) = xy.1 + yz.1
• Step 4: Use Identity law (a . 1 = a). So xy + yz = xy.1 + yz.1 = xyz + xyz’ + x’yz
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Reduced Hardware Implementation• Reduced equation requires less hardware!• Same function implemented!
x y z
x00001111
y00110011
z01010101
G00010011
G = xyz + xyz’ + x’yz = xy + yz
G
x x
x x
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x y F
0 0 1
0 1 1
1 0 0
1 1 0
Karnaugh maps
• Alternate way of representing Boolean function–All rows of truth table represented with a
square– Each square represents a minterm
0 1y
x
0
1
1
0 0
1
0 1y
x
0
1
x’y’
xy’ xy
x’y
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Karnaugh maps
• Easy to convert between truth table, K-map, and SOP.
–Unoptimized form: number of 1’s in K-map equals number of minterms (products) in SOP.
–Optimized form: reduced number of minterms
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F(x,y) = x’y + x’y’ = x’
Karnaugh Maps• A Karnaugh map is a graphical tool for assisting in the general
simplification procedure.• Two variable maps.
0A
1 01
B 0 101
F=AB +A’B 0A
1 11
B 0 101
• Three variable maps.
0A
1 11
00 01
01
BC
01 1
111 10
F=AB’C’ +AB C +ABC +ABC + A’B’C + A’BC’
F=AB +AB +AB
A B C F0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 11 0 1 11 1 0 11 1 1 1
+
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Rules for K-Maps We can reduce functions by circling 1’s in
the K-map.
Each circle represents minterm reduction.
Following circling, we can deduce minimizedand-or form.
F(x,y) = x’y + x’y’ = x’
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0 1y
x
0
1
1
0 0
1
Rules for K-Maps
Rules to consider
1. Every cell containing a 1 must be included at least once.
2. The largest possible “power of 2 rectangle” must be enclosed.
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Karnaugh Maps
• A Karnaugh map is a graphical tool for assisting in thegeneral simplification procedure.
• Two variable maps.
0A
1 01
B 0 101
F=AB +A’B 0A
1 11
B 0 101 F=A+B
• Three variable maps.
F=A+B C +BC 0
A
1 11
00 01
01
BC
01 1
111 10
F=AB +AB +AB
F=AB’C’ +AB C +ABC +ABC + A’B’C + A’BC’ 31
More Karnaugh Map Examples
Examples
g = b'
0 101
ab
cab
00 01 11 1001
0 101
ab
cab
00 01 11 1001
0 10 1f = a
0 0 1 00 1 1 1
cout = ab + bc + ac
1 10 0
0 0 1 10 0 1 1
f = a
1. Circle the largest groups possible.2. Group dimensions must be a power of 2. 3. Remember what circling means!
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