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CHAPTER 2Logic Circuits
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Chapter ObjectivesIn this chapter you will be introduced
to:Logic functions and circuitsBoolean algebra for dealing with
logic circuitsLogic gates and synthesis of simple circuitsCAD tools
and the VHDL hardware description language
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x 1 = x 0 = (a) Two states of a switchS x (b) Symbol for a
switch Binary Switch
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(a) Simple connection to a batteryS (b) Using a ground
connection as the return pathBattery Light Power supplyS Light
xxLight Controlled by a Switch
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(a) The logical AND function (series connection)S Power supplyS
S Power supplyS (b) The logical OR function (parallel connection)
Light Light x1x2x1x2AND and OR Logic Functions
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Series-Parallel Connection
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Inverting Circuit
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Truth Table for Two Input Variables
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Truth Table for Three Input Variables
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x 1 x 2 x 1 x 2 + x 1 x 2 x 1 x 2 (a) AND gate (b) OR gatex x
(c) NOT gateThe Basic Gates
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Inverter (NOT Circuit)Performs a basic logic function called
inversion or complementationChanges one logic level (HIGH / LOW) to
the opposite logic levelIn terms of bits, it changes a 1 to a 0 and
vice versa
INPUTOUTPUTAA0110
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The AND GateComposed of two or more inputs and a single
outputPerforms logical multiplication.The logical operation of the
AND gate is such that the output is HIGH (1) when all the inputs
are HIGH, otherwise it is LOW (0)
INPUT
OUTPUT
A
B
A.B
0
0
0
0
1
0
1
0
0
1
1
1
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The OR GateComposed of two or more inputs and a single
outputPerforms logical additionThe logical operation of the OR gate
is such that the output is HIGH (1) when any of the inputs are
HIGH, otherwise it is LOW (0)
INPUT
OUTPUT
A
B
A + B
0
0
0
0
1
1
1
0
1
1
1
1
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X = AB + A
INPUT
OUTPUT
A
B
X = AB + A
0
0
0
0
1
0
1
0
1
1
1
1
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x 1 x 2 x 3 f x 1 x 2 + ( ) x 3 . = What is the Truth Table?
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x 1 x 2 1 1 00 f 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 1 (a) Network
that implementsf x 1 x 1 x 2 + = A B 1 0 1 0 1 0 1 0 1 0 x 1 x 2 A
B f Time(c) Timing diagramLogic Network Example
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Boolean AlgebraA mathematical system for formulating logical
statements with symbols so that problems can be solved in a manner
similar to ordinary algebra.Boolean algebra is the mathematics of
digital systems.
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Axioms of Boolean Algebra
Multiplication Rules
Addition Rules
1a. 0 . 0 = 0
1b. 1 + 1 = 1
2a. 1 . 1 = 1
2 b. 0 + 0 = 0
3a. 0 . 1 = 1 . 0 = 0
3b. 0 + 1 = 1 + 0= 1
4a. If x=0, then x' = 1
4a. If x=1, then x' = 0
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Commutative Laws of Boolean AlgebraThe commutative law of
addition for two variables is algebraically expressed as:x + y = y
+ xThe commutative law of multiplication for two variables is
expressed as:xy = yxIn summary, the order in which the variables
are ORed or ANDed make no difference.
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Associative Laws of Boolean AlgebraThe associative law of
addition of three variables is expressed as:x + (y + z) = (x + y) +
zThe associative law of multiplication of three variables is
expressed as:x(yz) = (xy)zIn summary, ORing or ANDing a grouping of
variables produces the same result regardless of the grouping of
the variables.
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Distributive Law of Boolean AlgebraThe distributive law of three
variables is expressed as follows:x (y+z) = xy + xzThis law states
that ORing several variables and ANDing the result is equivalent of
ANDing the single variable with each of the variables in the
grouping, then ORing the result.
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Single Variable Theorems
Rule NumberBoolean Expression5 a.x * 0 = 0 5 b. x + 1 = 1 6 a.x
* 1= x 6 b.x + 0 = x 7 a.x * x = x 7 b.x + x = x 8 a. x * x =0 8
b.x + x =1 9.(x) = x
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DualityTo reflect the principle of duality, the axioms and
single-variable theorems are listed in pairs.For example, see 5a
and 3a. When x =0, by 5a, the result is 0.When x =1, by 5a, the
result is 0, which is also proved by 3a.
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Two- and Three-Variable Properties
Property Boolean Expression10 a. Commutative10 b. Commutativex *
y = y *x x + y = y + x 11 a. Associative11 b. Associativex * (y *
z) = (x * y) * zx + (y + z) = (x + y) + z12 a. Distributive12 b.
Distributivex * (y + z) = x * y + x * zx + (y * z) = (x + y) * (x +
z)13 a. Absorption13 b. Absorption x + x * y = x x * ( x + y) =
x
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Two- and Three-Variable Properties
Property Boolean Expression14 a. Combining14 b. Combiningx * y +
x * y= x(x + y ) * ( x + y) = x15 a. DeMorgans Theorem15 b.
DeMorgans Theorem(x * y)= (x+ y) (x + y)= x * y 16 a. 16 b. x + x *
y = x + yx * (x + y) = x * y17 a. Consensus17 b. Consensus x * y +
x * z + y * z = x * y + x * z(x + y ) * (x + z) * (y + z ) = (x +
y) * (x+ z)
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Boolean SimplificationExample 1:F = AB + CD + AB + CD = AB + CD
(by identity 5)Example 2:F = ABC + ABC + AC = AB(C + C) + AC (by
identity 13) = AB(1) + AC (by identity 7) = AB +AC (by identity
4)
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Logic Circuit Implementation
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DeMorgans Theorem(A B) = A + B(1)That is, the complement of the
product is equivalent to the sum of the complements.This is true
for any number of variables.(A B C Z) = A + B + C + + Z(A + B) = A
B(2)Similarly, the complement of the sum is equivalent to the
product of the complements.Similarly, (A + B + C + + Z) = A * B * C
* * Z
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Proof of DeMorgans Theorem
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Methods to Complement a FunctionInterchange 1s and 0s for the
values of F in the truth table.Use DeMorgans theorem on algebraic
functionChange F to F, and F to FChange OR to ANDChange AND to
ORComplement each individual variableExample 1:F = AB+ CD +
BDApplying DeMorgans theorem,F = (A + B)(C + D)(B + D)
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Methods to Complement a FunctionExample 2:Simplify F = (x1 + x3)
. (x1+ x3)F = x1 x1+ x1 x3 + x3 x1+ x3 x3 (Distributive Property)x1
x1 and x3 x3 = 0 ( Identity 8)F = x1 x3 + x1 x3Example 3:F = xyz +
xyz +xz= xy(z + z ) + xz (Factoring out)= xy .1 + xz ( By Identity
6)= xy + xz ( By Identity 4)
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Practice ProblemsFind the complement: (xyz)Expand: x +
yzSimplify:xy +xy + xyxy + xz + xy + yzwy + wyz + wxy + wxy
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Venn Diagram Representationx x x (a) Constant 1(b) Constant 0(c)
Variablex (d) x x
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Venn Diagram Representationx y z x x y x y (e) (f) (g) (h) x y
*x y + x y z + * x y * y
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Verification of the Distributive Propertyx y z x y z x y z x y z
x y z x y z x x y * x y * x + z * x y z + ( ) * (a) (d) (c) (f) x z
* y z + (b) (e)
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Verify xy + xz + yz = xy + xzx y z y x z x y z x y * y z * x z *
x y z x y *
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Boolean FunctionsExample 1:Prove: (A + B) (A + B) = AB + ABLHS =
AA + AB + BA + BB (by distributive property)= O + AB + BA + O = AB
+ AB = RHSExample 2:Prove: AC + B C + AC + BC = AB + AB + ABLHS =
A(C+C) + B(C+C) = A*1 + B*1 = A + BRHS = AB + AB + AB = AB + A(B +
B) = A B + A = A + B (by identity 11)LHS = RHS
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Precedence of OperationsNOT, AND, and then ORExample: A*B + A*B
1. Complements2. AND operations3. OR operation
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f = m0 * 1 + m1 * 1 + m2 * 0 + m3 *1 Synthesis Using Gates
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f (a) Canonical sum-of-products f (b) Minimal-cost realizationx
2 x 1 x 1 x 2 Two Implementations
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Terms and DefinitionsSynthesisThe designing of a new system that
implements a desired functional behavior.AnalysisThe task of
determining the function performed by a system.
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Terms and DefinitionsSum-of-Products (SOP)Product-of-Sums
(POS)Canonical SOPCanonical POSMintermA product term with all n
variables in asserted or negated form.MaxtermThe complement of a
minterm
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Three-Variable Minterms and Maxterms
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A Three-Variable Function
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A Three-Variable FunctionCanonical Sum-of-Productsf(x1, x2, x3)
= x1x2x3 + x1x2x3 + x1x2x3 + x1x2x3= (m1, m4, m5, m6)= m(1, 4, 5,
6)Canonical Product-of-Sumsf(x1, x2, x3) = x1x2x3 + x1x2x3 + x1x2x3
+ x1x2x3f(x1, x2, x3) = (x1+x2+x3)(x1+x2+x3)(x1+x2+x3)(x1+x2+x3)=
(M0, M2, M3, M7)= M(0, 2, 3, 7)
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Practice ProblemsShow that the minimal Sum-of-Products is:f(x1,
x2, x3) = x2x3 + x1x3Show that the minimal Product of Sums is:f(x1,
x2, x3) = (x1 + x3)(x2 + x3)
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(b) A minimal product-of-sums realizationMinimal
Realizations
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More Practice ProblemsDetermine the canonical Sum-of-Products
expressions for the following functions:f(x1, x2, x3) = m(2, 3, 4,
6, 7)f(x1, x2, x3, x4) = m(3, 7, 9, 12, 13, 14, 15)Now determine
the minimized SOPs for these two functions.Find the canonical
Product of Sums expression for:f(x1, x2, x) = M(0, 1, 5)
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Four More Logic GatesNANDNORExclusive OR (XOR)Exclusive NOR
(XNOR)
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The NAND GateThe NAND, which is composed of two or more inputs
and a single output, is a very popular logic element because it may
be used as a universal function.It may be employed to construct an
inverter, an AND gate, an OR gate, or any combination of these
functions.The term NAND is formed by the concatenation NOT-AND and
implies an AND function with an inverted output.The logical
operation of the NAND gate is such that the output is LOW (0) only
when all the inputs are HIGH (1).
INPUT
OUTPUT
A
B
X = (AB)
0
0
1
0
1
1
1
0
1
1
1
0
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The NOR gateThe NOR gate, which is composed of two or more
inputs and a single output, also has a universal property.The term
NOR is formed by the concatenation NOT-OR and implies an OR
function with an inverted output.The logical operation of the NOR
gate is such that the output is HIGH (1) only when all the inputs
are LOW.
INPUT
OUTPUT
A
B
(A + B)
0
0
1
0
1
0
1
0
0
1
1
0
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The Exclusive-OR (XOR) and Exclusive NOR (XNOR) GatesThese gates
are usually formed from the combination of the other logic gates
already discussed.
Because of their functional importance, these gates are treated
as basic gates with their own unique symbols.
The Exclusive-OR is an "inequality" function and the output is
HIGH (1) when the inputs are not equal to each other.
The Exclusive-NOR is an "equality" function and the output is
HIGH (0) when the inputs are equal to each other.
INPUT
XOR OUTPUT
XNOR OUTPUT
A
B
A(B
(A(B)
0
0
0
1
0
1
1
0
1
0
1
0
1
1
0
1
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NAND and NOR Gates
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DeMorgans Theorem in Terms of Logic Gates
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Using NAND Gates to Implement a Sum-of-Products
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Using NOR Gates to Implement a Product-of-Sums
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x1 f (a) POS implementation(b) NOR implementationf x3 x2 x1 x3
x2 NOR Gate Realization of an Example Function
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f f (a) SOP implementation(b) NAND implementation x1 x3 x2 x3 x2
x1 NAND Gate Realization of an Example Function
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Truth Table for a Three-Way Light Control
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SOP and POS Realizations
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Implementation of a Multiplexer0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1
0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 (a) Truth table sx1x2f (s, x1, x2)f x
1 x 2 s (b) Circuit f s x 1 x 2 0 1 (c) Graphical symbol0 1 (d)
Morecompact truth-table representationf (s, x1, x2)sx1x2
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A Typical CAD SystemSee Figure 2.29, page 61, in your
textbook.
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fx3x1x2A Simple Logic Function
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VHDLVHSIC Hardware Description LanguageVery High Speed
Integrated CircuitEntity DeclarationDescribes
PortsInputOutputArchitecture DeclarationDescribes Functions
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ENTITY example1 IS PORT ( x1, x2, x3 : IN BIT ; f : OUT BIT ) ;
END example1 ;
VHDL Entity Declaration
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ARCHITECTURE LogicFunc OF example1 IS BEGIN f
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Complete VHDL Code for the Circuit
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VHDL Code for a Four-Input Function
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Logic Circuit for the VHDL Code
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Example 2.10A circuit that controls a given digital system has
three inputs: x1, x2, and x3.It has to recognize three different
conditions:Condition A is true if x3 is true and either x1 is true
or x2 is false.Condition B is true if x1 is true and either x2 or
x3 is false.Condition C is true if x2 is true and either x1 is true
or x3 is false.The control circuit must produce an output of 1 if
at least two of the conditions A, B, and C are true.Design the
simplest circuit that can be used for this purpose.
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Solution to Example 2.10Using 1 for true and 0 for false,
express the three conditions A, B, and C as:A = x3(x1 + x2) = x3x1
+ x3x2B = x1(x2 + x3) = x1x2 + x1x3C = x2(x1 + x3) = x2x1 + x2x3The
desired output can be expressed as:f(x1, x2, x3) = AB + AC + BC
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Solution to Example 2.10Determine the product term AB:= (x3x1 +
x3x2)(x1x2 + x1x3)= x3x1x1x2 + x3x1x1x3 + x3x2x1x2 + x3x2x1x3=
x3x1x2 + O + x3x2x1 + O= x1x2x3Determine the product term AC =
x1x2x3Determine the product term BC = x1x2x3
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Solution to Example 2.10f(x1, x2, x3 ) = AB + AC + BC= x1x2x3 +
x1x2x3 + x1x2x3= x1(x2 + x2)x3 + x1x2(x3 + x3)= x1x3 + x1x2=x1(x3 +
x2)
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Example 2.11Solve Example 2.10 using Venn Diagrams
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(a) FunctionA (b) FunctionB (c) FunctionC (d) Functionf x1 x3 x2
x1 x2 x1 x2 x1 x2 x3 x3 x3 Solution to Example 2.11
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