Download Number Systems and Conversion + Boolean Algebra

CS 352
Introduction to Logic Design
Lecture 1
Ahmed Ezzat
Number Systems and Boolean
Algebra,
Ch-1 + Ch-2
1
Outline
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Introduction
Number Systems and Conversion
Binary Arithmetic
Representation of Negative Numbers
Binary Codes
Basic Boolean Operations
Boolean Expressions and Truth Tables
Basic Theorems
Commutative, Associative, and Distribitution laws
Simplification Theorems
Multiplying Out & Factoring
DeMorgan’s Laws
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Introduction
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Digital Systems: works with discrete values, and can be
divided into:
– System Design: Breaks overall system into subsystems
and specify characteristics of each subsystem
– Logic Design: Determine how to interconnect basic
building blocks to perform a specific function
– Circuit Design: Specify interconnection of specific
components, such as resistors and transistors, to form a
gate, flip-flop, or other logic block
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Introduction (2)
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Switching Circuits: A form of digital system, and comes in 2
flavors:
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Combinational Circuits:
Output values depend only
on the present input values
and not on past values
Sequential Circuits:
Output depend on both the
present and past input
values
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Introduction (3)
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Boolean Algebra describes the relationship between input
and output signals mathematically. To be economic, we
need to simplify the boolean algebra  Karnaugh map and
Quine-McCluskey procedures.
Basic memory element used in designing sequential circuit is
called Flip/Flop (F/F). F/Fs are the basic building blocks for
counters, registers, etc.
VHDL is a hardware description language used in the design
of Combinational logic, Sequential logic and digital systems.
This course is devoted to the study of “Logic Design.”
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Number Systems and Conversion (1)
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Decimal numbers are written using positional notations:
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1
0
-1
2
1
0
-1
101.012 = 1 x 2 + 0 x 2 + 1 x 2 + 0 x 2 + 1 x 2
= 4 + 1 + ¼ = 5.25
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123.45 = 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 10
Any positive integer R (R > 1) can be used as a Radix or base
of a number system
For base R, then R digits (0, 1, 2, 3, …R-1) digits are needed to
represent any number in that base
Numbers written in positional notation can be written in a power
series of R.
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Number Systems and Conversion (2)
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1
0
-1
101.018 = 1 x 8 + 0 x 8 + 1 x 8 + 0 x 8 + 1 x 8
= 64 + 1 + 1/64 = 65.015625
1
-2
0
3F16 = 3 x 16 + 15 x 16 = 48 + 15 = 63
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Number Systems and Conversion (3):
Converting Decimal Number N  Base R
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To convert a decimal number N to a base R number:
Notice that N = (a n a n-1 ... a 2 a 1 a 0 )R
N = an Rn + a n-1 R n-1 + ... + a1 R1 + a0
Divide N by R to get Q1 remainder (i.e., a0)
N/R = Q1 = an R n-1 + a n-1 R n-2 + ... + a2 R1 + a 1
Divide Q1 by R to get Q2 remainder (i.e., a1)
Repeat process until an is found.
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Number Systems and Conversion (4):
Converting Decimal Number N  Base R
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Convert 53 decimal to a binary number:
Check your work!  110101  1 + 0 + 4 + 16 + 32 = 53
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Number Systems and Conversion (5):
Converting Decimal Number N  Base R
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To convert a decimal fraction F to base R:
Notice that F = (.a-1 a -2 ... a -m)R
Multiply F by R = F * R = a -1 + F1
Multiply F1 by R = F1 * R = a -2 + F2
Repeat process until am is found or process
repeats.
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Number Systems and Conversion (6):
Converting Decimal Fraction F  Base R
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Convert 0.625 to a binary number:
0.62510 = (0.101) 2
0.710 = 01 0110 0110 0110…2
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Number Systems and Conversion (7):
Converting Between two bases other than Decimal
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Best to convert source number to decimal first and then
convert the decimal number to the new base:
e.g., Convert 231.34 to base 7
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Number Systems and Conversion (8):
Converting Decimal Fraction F  Base R
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Converting between binary and octal and binary and
hexadecimal is easily done by inspection because each
octal digit corresponds to exactly three binary digits and
each hexadecimal digit corresponds to exactly 4 binary
digits.
(1 000 111 . 011 001)2 = (1 0 7 . 3 1)8 = (4 7 . 6 4) 16
(F C 2 . 7) 16 = (1111 1100 0010 . 0111) 2
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Binary Arithmetic (1)
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Arithmetic is usually done in binary in digital systems
because such logic networks are easier to design.
The methods for binary addition, subtraction, multiplication,
and division are the same as for decimal addition, etc. except
that everything is done in powers of 2.
Add 5 + 12 in binary:
11
 carries
00101 = 5
+01100 = 12
--------10001 = 17
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Binary Arithmetic (2)
Subtract 5 from 12 in binary:
111  borrows
01100 = 12
- 00101 = 5
--------00111  7
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Binary Arithmetic (3)
Multiply 13 by 5 in binary:
1101 = 13 (multiplicand)
x 0101 = 5 (multiplier)
--------1101
0000
1101
0000
------------1000001 = 65
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Binary Arithmetic (4)
Divide 14510 (dividend) by 1110 (divisor) in binary:
Quotient is 11012 = 1310, and remainder is 102 = 210
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Representation of Negative Numbers (1)
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Sign and Magnitude Representation: For an n-bits number,
the MSB is the sign bit and the remaining (n-1) bits
represents the magnitude. Thus an n-bit word can represent
+2n-1  -2n-1 integer numbers
2’s Complement and 1’s Complement Representation: is
more common because arithmetic operations are easier.
Thus an n-bit word can represent +(2n-1 -1)  0  -2n-1
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Representation of Negative Numbers (2)
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Representation of Negative Numbers (3)
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Representation of Negative Numbers (4)
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Representation of Negative Numbers (5)
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Representation of Negative Numbers (6)
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1’s complement addition is similar to 2’s complement
addition except that instead of discarding the last carry-bit, it
is added to the n-bit (LSB) of the sum in the position furthest
to the right.
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Binary Codes (1)
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Computers internally use binary, however I/O devices
typically use decimal numbers.
A Binary-Coded-Decimal (BCD – 8-4-2-1) is formed by
replacing each decimal digit by its 4-bit binary equivalent:
342.61 --> 0011 0100 0010 . 0110 0001
Excess-3 code is obtained from BCD by adding 3 (0011).
The 2-out-of-5 code has exactly 2 bits out of the 5 are 1.
Gray code has successive decimal codes differ in only one
digit – typically used when transforming analog values into
digital form
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Binary Codes (2)
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Basic Boolean Operations (1)
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Basic Mathematics for studying logic design of digital
systems is Boolean algebra
Basic Operations are AND, OR, and Complement
(inverse)
Truth Tables list all combinations of inputs and the
corresponding outputs for combinational circuits.
An AND gate produces a 1 iff all inputs are 1
An OR gate produces a 1 iff one or more inputs are 1
A NOT gate (inverter) produces an output opposite to the
input
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Basic Boolean Operations (2)
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Basic Boolean Operations (3)
Switching Algebra
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Boolean Expressions and Truth Tables (1)
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Boolean expression is the application of basic operations
to one or more variables (x, y, etc.) or constants (0 or 1)
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Boolean Expressions and Truth Tables (2)
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A truth table specifies the values of a Boolean expression
for every possible combination of values of the variables
in the expression
Two expressions are equal iff they have the same truth
tables
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Boolean Expressions and Truth Tables (3)
Expressions (AB’ + C) = (A + C)(B’ + C)
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// equivalent
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Basic Theorems (1)
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Basic Theorems (2)
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The above six axioms completely define switching
algebra. All other facts about the system can be proved
starting with the axioms.
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Commutative, Associative and Distributive
Laws
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Commutative law:
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Associative law:
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Distributive law:
XY = YX
X+Y=Y+X
(XY)Z = X(YZ) = XYZ
(X + Y) + Z = X + (Y + Z) = X + Y + Z
XYZ = 1 iff X = Y = Z = 1 X + Y + Z = 0 iff X = Y = Z = 0
X(Y + Z) = XY + XZ
X + YZ = (X + Y)(X + Z)
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Simplification Theorems (1)
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XY + XY’ = X
(X + Y)(X + Y’) = X
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X + XY = X
X(X + Y) = X
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(X + Y’) Y = XY
XY’ + Y = X + Y
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Simplification Theorems (2)
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Multiplying Out and Factoring (1)
Multiply Out (Sum Of Products): An expression is said to be in sum-ofproducts form when all products are the products of single variables only.
The distributive laws are used to multiply out an expression to obtain a sumof-products form. Applying 2nd Distributive law [ (X + YZ) = (X + Y)(X + Z) ]
first, when applicable, can reduce the amount of work required to arrive at a
simplified sum-of-products form.
Example:
AB + AC + CD' + E is a sum of products
A(B+C) + CD + E' is not a sum of products
Multiplying out is a technique that is simpler than traditional multiplication of
the given expression
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Multiplying Out and Factoring (2)
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Multiplying out:
(A + BC)(A + D + E)
Assume X = A,
Y = BC,
Z=D+E
Then (X + Y)(X + Z) = X + YZ  A + BC(D + E)
= A + BCD + BCE
Traditional Multiplication Alternative:
(A + BC)(A + D + E) = A + AD + AE + ABC + BCD + BCE
= A(1 + D + E + BC) + BCD + BCE = A + BCD + BCE
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Multiplying out:
AB'(C + D) + (C + D)'
Assume X = AB’ Y = (C + D)’
and using XY’ + Y = X + Y
Then: AB'(C + D) + (C + D)' = XY’ + Y  X + Y
= AB' + (C + D)'
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Multiplying Out and Factoring (3)
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Multiplying out:
(A + B)(B + C)(B + D')(ACD' + E)
= (A + B)(B + CD')(ACD' + E)
by D2
= (B + ACD')(ACD' + E)
by D2
= ACD' + BE
by D2
Multiplying out:
(A + B')(A + D' + C)(A + D' + B)
= (A + B')(A + D' + BC)
by D2
= A + B'(D' + BC)
by D2
= A + B'D'
by D1
Factoring (Product-of-Sums):
A + B’CD
Assume: X = A, Y = B’, Z = CD
= (X +YZ) = (X + Y)(X + Z) = (A + B’)(A + CD)
= (A + B’)(A + C)(A + D)
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Multiplying Out and Factoring (4)
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Factoring:
ABC + A'B'C + CD'
= C(AB + A'B') + CD'
= C(AB + A'B' + D')
= C((AB + A')(AB + B') + D')
= C((A + A')(B + A')(A + B')(B + B') + D')
= C((B + A')(A + B') + D')
= C(B + A'+ D')(A + B'+ D')
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by D1
by D1
by D2
by D2
by D2
by D2
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Multiplying Out and Factoring (5)
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Factoring:
ABC + A'B'C
= C(AB + A'B')
= C((AB + A')(AB + B'))
= C( (A' + A) (A' + B) (B' + B)(B' + A) )
by D1
by D2
by D2
= C(A' + B)(B' + A)
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Multiplying Out and Factoring (6)
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Factoring:
C’D + C’E’ + G’H
= C’(D + E’) + G’H
= (C’ + G’H)(D + E’ + G’H)
= (C’ + G’)(C’ + H)(D + E’ + G’)(D + E’ + H)
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by D1
by D2
by D2
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DeMorgan’s Laws (1)
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DeMorgan's laws: the inverse or complement of any
Boolean expression can be found by successively
applying the following theorems:
(X + Y)' = X'Y'
(XY)' = X'+ Y'
DeMorgan's laws are generalized as follows:
(X 1 + X 2 + X 3 + ... + X n )' = X1'X2'X3' ... Xn'
(X 1 X 2 X 3 ... X n )' = X1'+ X2'+ X3'+ ... + Xn'
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DeMorgan’s Laws (2)
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Method:
Replace each variable with its complement.
Replace 0 with 1 and 1 with 0.
Replace OR with AND and AND with OR.
Note: ANDing takes precedence over ORing.
Use parentheses to preserve operation
hierarchy in the complemented expression.
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DeMorgan’s Laws (3)
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Example:
[(a'b + 1)(cd + e') + f(g'+ 0) + h]'
= ([a'b + 1]' + [cd + e']') * ([f(g'+ 0)]') * (h')
= ([a'b]'0 + [cd]'e) (f'+ [g'+ 0]') h'
= ((a + b')0 + (c'+ d')e)(f' + g1)h'
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DeMorgan’s Laws (4)
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[(A’ + B)C’]’
= [(A’ + B)’ + C]
= AB’ + C
[(AB’ + C)D’ + E]’
= [(AB’ + C)’ + D] E’
= [(AB’)’ C’ + D]E’
= [(A’ + B)C’ + D]E’
[A’B + AB’]’
= (A’B)’ (AB’)’
= (A + B’)(A’ + B)
= AA’ + AB + A’B’ + BB’
= AB + A’B’
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