PPT - Electrical and Computer Engineering

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EE2174: Digital Logic and
Lab
Professor Shiyan Hu
Department of Electrical and Computer Engineering
Michigan Technological University
CHAPTER 2
Number System
Overview


Digital Computers
Number Systems
Representations
 Conversions




Arithmetic Operations
Decimal Codes
Alphanumeric Codes
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Chapter 1: Digital Computers and Information
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Digital Systems

Major characteristic
manipulation of discrete elements of
information (any set restricted to a finite
# of elements)
 e.g. 10 decimal digits, 26 letters


Discrete elements (in digital systems)
can be represented by signals (physical
quantities).

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Most common signals: electrical (voltage,
current)
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Voltage Ranges
The 2 binary values (HIGH, LOW) of a digital signal
are represented by ranges of voltage values
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Information Representation

Binary signals (2 discrete values)
0 and 1 (LOW and HIGH, FALSE and TRUE)
 Binary quantity: binary digit/bit
 Information: group of bits/words (size: 8,
16, 32, 64, …)


Digital hardware computes binary
functions of binary numbers:
Combinational (memoryless)
 Sequential (using memory)

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Number Systems

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Representation of numbers
Radix: “base”, the primitive unit for
group of numbers, e.g. for decimal
arithmetic radix=10 (“base” 10)
For every system, we need arithmetic
operations (addition, subtraction,
multiplication)
Also, conversion from one base to the
other
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Number Systems - Decimal




“base” 10 (radix is 10)
10 digits: 0..9
(251.3)10 = 2102 + 5101 + 1100 + 310-1
Note: ‘.’ is called the radix point
(decimal point for base 10)
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Number Systems – Decimal (cont.)

In general, a decimal number with n digits to
the left of the decimal point, and m digits to
the right of the decimal point is written as:
An-1 An-2 … A1 A0 . A-1 A-2 … A-m+1 A-m
where Ai is a coefficient between 0..9,
and i denotes the weight (=10i) of Ai.
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Number Systems – Decimal (cont.)
The value of
An-1 An-2 … A1 A0 . A-1 A-2 … A-m+1 A-m
is calculated by
i=n-1..0 (Ai  10i ) + i=-m..-1 (Ai  10i )
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Number Systems – General



“base” r (radix r)
r digits
N = An-1 r n-1 + An-2r n-2 +… + A1r + A0 +
A-1 r -1 + A-2r -2 +… + A-m r -m
Most
Significant
Bit (MSB)
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Least
Significant
Bit (LSB)
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Number Systems – General (cont.)


e.g. let r = 8
(312.5)8 = 382 + 181 + 280 + 58-1
= (202.625)10
Conversion from n-ary (any system
with radix n) to decimal follows
similar process as above
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Number Systems (cont.)

Most common number systems for
computers:
Binary (r = 2)
 Octal (r = 8)
 Hexadecimal (r = 16)

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Binary Numbers-Base 2



Computers represent all data as “strings
of bits”, each bit being either 0 or 1
“base” 2, with 2 digits: 0 and 1
e.g.
(101101.10)2 = 125 + 024 + 123 + 122 + 021 +
120 + 12-1 + 02-2
(in decimal) = 32 + 0 + 8 + 4 + 0 + 1 + ½ + 0
= (45.5)10
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Binary Numbers-Base 2 (cont.)

e.g.
(1001.011)2 = 123 + 022 + 021 + 120 +
02-1 + 12-2 + 12-3
(in decimal) = 8 + 1 + 0.25 + 0.125
= (9.375)10
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Powers of two
Memorize at least through 216
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Octal Numbers-Base 8


“base” 8, with 8 digits: 0..7
e.g.
(762)8 = 782 + 681 + 280
(in decimal) = 448 + 48 + 2
= (498)10
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Hexadecimal Numbers-Base 16




r = 16
Digits (convention): 0..9, A, B, C, D, E, F
A=10, B=11, … , F = 15
e.g.
(3FB)16 = 3162 + 15161 + 11160
(in decimal)
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= 768 + 240 + 11
= (1019)10
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Base Conversions





Any base r  decimal – Easy! (already
covered, see slides 13-14, 16-17, 19-20)
Decimal  Binary
Octal  Binary
Hex  Binary
Decimal  Any base r
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Decimal to Binary
Let N be a decimal number.
1.
2.
3.
Find the greatest number that is a power of 2 and
when subtracted from N it produces a positive
difference N1
Put a 1 in the MSB
Repeat Step 1, starting from N1 and finding
difference N2. Put a 1 in the corresponding bit.
Stop when the difference is zero.
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Decimal to Binary (cont.)
e.g. N = (717)10
717 – 512 = 205 = N1
512 = 29
205 –128 = 77 = N2
128 = 27
77 – 64 = 13 = N3
64 = 26
13 – 8 = 5 = N4
8 = 23
5 – 4 = 1 = N5
4 = 22
1 – 1 = 0 = N6
1 = 20
 (717)10 = 29 +
27 + 26
+ 23 + 22 +
20
= ( 1 0 1 1 0 0 1 1 0 1)2
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Binary to Octal and Hex

Octal:
8 = 23  every 3 binary bits convert 1 octal

Hex:
16 = 24  every 4 binary bits convert 1 hex
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Binary  Octal
(011 010 101 000 . 111 101 011 100)2
(3 2
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5
0
. 7
5
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3
4 )8
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Binary  Hex
( 0110 1010 1000 . 1111 0101 1100 )2
( 6
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A
8
. F
5
Chapter 1: Digital Computers and Information
C )16
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Octal  Hex
Go through Binary!
Hex  Binary  Octal
Octal  Binary  Hex
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Convert Decimal to any base r


Integer part: Divide by the base, keep
track of remainder, and read-up
e.g. (153)10 = ( ? )8 , r = 8
153 / 8 = 19 + 1/8
rem = 1
LSB
19 / 8 = 2 + 3/8
rem = 3
2 / 8 = 0 + 2/8
rem = 2
MSB
stop

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(153)10 = ( 231)8
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Convert Decimal to any base r


Fractional part: Multiply by the base, keep
track of integer part, and read-down
e.g. (0.78125)10 = ( ? )16 , r = 16
0.7812516 = 12.5 integer = 12 = C MSB
0.5  16 = 8.0
integer = 8 = 8 LSB
stop

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(0.78125)10 = (0.C8)16
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Binary Arithmetic Operations:
Addition


Follow same rules as in decimal addition, with
the difference that when sum is 2 indicates a
carry (not a 10)
Learn new carry rules




0+0 = 0c0 (sum 0 with carry 0)
0+1 = 1+0 = 1c0
Carry
1 1 1 1 1 0
1+1 = 0c1
Augend 0 0 1 0 0 1
1+1+1 = 1c1
Addend 0 1 1 1 1 1
Result
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1
0 1 0 0 0
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Binary Arithmetic Operations:
Addition (cont.)



“Half addition” (rightmost bit position, aka
LSB): only 2 bits are added, yielding a sum
and a carry
“Full addition” (remaining positions): three
bits are added, yielding a sum and a carry
In Chapter 3, we’ll see many different
hardware implementations of half-adders
and full-adders
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Binary Arithmetic Operations:
Subtraction

Learn new borrow rules




0-0 = 1-1 = 0b0 (result 0 with borrow 0)
1-0 = 1b0
0-1 = 1b1
Borrow
1 1 0 0
…
Minuend
1 1 0 1 1
Subtrahend 0 1 1 0 1
Result
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0 1 1 1 0
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Keys to Success



Recall and use the “algorithms” you use
to perform base-10 arithmetic.
Generalize them to the base in use
(carry, borrow rules change)
Preserve the base! In binary, 1+1=10
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Two’s complement

Two’s complement of a binary number is
computed as complementing each bit and
add 1
It is used for subtrahend in subtraction
 For example, (3)10=(011)2, so (-3)10=(100+1)2
 (111)-(011)=(111)+(101)=(1100) removing the
leading carry = (100)


Why?

9-Apr-15
x-(011)=x-[111-100]=x-[1000-1-100]=x(100+1)-1000
Arithmetic
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Binary Arithmetic Operations:
Multiplication


Shift-and-add algorithm, as in base 10
M’cand
M’plier
0
0
0
0
0
0
1
0
1
1
0
1
1
0
(1)
(2)
(3)
Sum
0
0
0
1
0
0
1
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
0
1
0
0
0
0
Check: 13 * 6 = 78
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CODES


Representations of info (set) obtained by
associating one or more codewords (a binary
pattern/string) with each element in the set
n-bit binary code: a group of n bits that can
n
encode up to 2 distinct elements
e.g. A set of 4 distinct numbers can be
represented by 2-bit codes s.t. each number
in the set is assigned exactly one of the
combinations/codes in {00,01,10,11}
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CODES (cont.)



To encode m distinct elements with an
n
n-bit code: 2
>= m
Note: The codeword associated with
each number is obtained by coding the
number, not converting the number to
binary.
We will see: BCD, ASCII, Unicode
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Binary-Coded Decimal (BCD)


A decimal code:
Decimal numbers (0..9)
are coded using 4-bit
distinct binary words
Observe that the
codes 1010 .. 1111
(decimal 10..15) are
NOT represented
(invalid BCD codes)
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Binary-Coded Decimal (cont.)



To code a number with n decimal digits, we
need 4n bits in BCD
e.g. (365)10 = (0011 0110 0101)BCD
This is different to converting to binary,
which is (365)10 = (101101101)2
Clearly, BCD requires more bits. BUT, it is
easier to understand/interpret
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BCD Addition

When 2 BCD codes are added:
If the binary sum is less than 1010 (9 in
decimal), the corresponding BCD sum digit
is correct
 If the binary sum is equal or more than
1010, must add 0110 (6 in decimal) to the
corresponding BCD sum digit in order to
produce the correct carry into the digit to
the left

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BCD Addition (cont.)

Example: Add 448 and 489 in BCD.
0100 0100 1000 (448 in BCD)
0100 1000 1001 (489 in BCD)
10001 (greater than 9, add 6)
1 0111 (carry 1 into middle digit)
1101
(greater than 9, add 6)
1001 1 0011
(carry 1 into leftmost digit)
1001 0011 0111 (BCD coding of 93710)
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ASCII character code



We also need to represent letters and other
symbols  alphanumeric codes
ASCII = American Standard Code for
Information Interchange. Also know as
Western European
It contains 128 characters:



94 printable ( 26 upper case and 26 lower case
letters, 10 digits, 32 special symbols)
34 non-printable (for control functions)
Uses 7-bit binary codes to represent each of
the 128 characters
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ASCII Table
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Unicode




Established standard (16-bit
alphanumeric code) for international
character sets
Since is 16-bit, it has 65,536 codes
Represented by 4 Hex digits
ASCII is between 000016 .. 007B16
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Unicode Table (first 191 char.)
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ASCII Parity Bit



Parity coding is used to detect errors in
data communication and processing
An 8th bit is added to the 7-bit ASCII
code
Even (Odd) parity: set the parity bit so
as to make the # of 1’s in the 8-bit code
even (odd)
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ASCII Parity Bit (cont.)

For example:
Make the 7-bit code 1011011 an 8-bit even
parity code  11011011
 Make the 7-bit code 1011011 an 8-bit odd
parity code  01011011


Both even and odd parity codes can
detect an odd number of error. An even
number of errors goes undetected.
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Summary


Conversion between different number
system
Binary Arithmetic
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Arithmetic
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