ECE456: Number Systems (review) Instructor: Dr. Honggang Wang II-209B, [email protected] ECE Dept.,...
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ECE456: Number Systems (review) Instructor: Dr. Honggang Wang II-209B, [email protected] ECE Dept., Fall 2013
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Transcript of ECE456: Number Systems (review) Instructor: Dr. Honggang Wang II-209B, [email protected] ECE Dept.,...
ECE456: Number Systems (review)
Instructor: Dr. Honggang Wang
II-209B, [email protected]
ECE Dept., Fall 2013
Dr. Wang
Administrative Issues (9/16/13)
• Project team set-up due Wednesday, Sept. 25
• If you missed the first class, go to the course website for syllabus and 1st lecture.
http://www.faculty.umassd.edu/honggang.wang/teaching.html
Dr. Wang
1 Mega bytes = 2? bytes
a) 26 bytes (x)
b) 217 bytes (x)
c) 1024 bytes (x)
d) 216 bytes (x)
e) 1E7 bytes (x)
f) 210 bytes (x)
g) 232 bytes (x)
h) No answer (x)
From Background Survey
Number Systems
• Do you know the equivalent hexadecimal, octal, and decimal values of the binary number 11001010?
• What is the equivalent binary number of the decimal number 63?
Dr. Wang
From Background Survey
Dr. Wang
ConventionsTerm Normal Usage Usage as a Power of 2
Kilo (K) 103 210 =1,024
Mega (M) 106 220 =1,048,576
Giga (G) 109 230 =1,073,741,824
Tera (T) 1012 240 =1,099,511,627,776
Mili (m) 10-3
Micro (m) 10-6
Nano (n) 10-9
Pico (p ) 10-12
• Powers of 2 are most often used in describing memory capacity.– Ex: 1Kilobyte (KB) =1024 bytes= 210 bytes
• Powers of 10 are used to describe the CPU clock frequencies: cycles per second (Hz)– Ex: Pentium 4 --1.8GHz = 1.8x109 Hz
Dr. Wang
DefinitionsTerm Definition
bit 0 or 1
byte (B) a group of 8 bits
nibble (nybble) half a byte (4 bits)
word (w) a group of bits that is processed simultaneously.
a word may consist of 8/16/32/other number of bits
machine dependent
(ex: 8086 – 16 bits; 80386/80486/Pentium – 32 bits)
double word 2 words
msb (most significant bit) the leftmost bit in a word
lsb (least significant bit) the rightmost bit in a word
Hz (hertz) reciprocal of second
Dr. Wang
Review of Number Systems
• Overview
• Number systems conversions
Chapter 19 (online chapter)Or Appendix A in 7th edition
Dr. Wang
Number SystemsNumber Systems
• Two basic types of number systems:
– Non-positional• Ex: Roman numerals: I, II, III, IV, V … X, XI … C
• Normally only useful for small numbers
– Positional• Ex: the decimal systems
• Each position in which a digit/symbol is written has a different
positional value, which is a power of the base
Dr. Wang
Decimal number systems 1. a base of 10 (determines the magnitude of a place).
2. is restricted to 10 re-usable digits/symbols (0,1,2,3,4,5,6,7,8,9)
3. the value of a digit depends on its position
(digit x positional value = digit x baseposition)
4. sum of the value of all digits gives the value of the number.
Positional Number Systems (Example)
58710 = 5 x 102 + 8 x 101 + 7 x 100
= 5 x 100 + 8 x 10 + 7 x 1
= 500 + 80 + 7
= 587
Dr. Wang
Positional Number Systems
In general (base is b),
N = ...P3P2P1P0 . P-1P-2P-3...
= ... + P3b3 + P2b2 + P1b1 + P0b0 + P-1b-1 + P-2b-2 + P-3b-3 + ...
Decimal (base is 10):
375.1710 = 3 x 102 + 7 x 101 + 5 x 100 + 1 x 10-1 + 7 x 10-2
= 3 x 100 + 7 x 10 + 5 x 1 + 1 x 0.1 + 7 x 0.01
= 300 + 70 + 5 + 0.1 + 0.07
= 375.17
0Increase by 1 Decrease by 1
Dr. Wang
• Specify the value of the digit 5 in the following decimal
numbers:
Exercise (1)
25
51
4538
Dr. Wang
• Binary– Base 2– 2 symbols:0,1
• Octal– Base 8– 8 symbols: 0,1,2,3,4,5,6,7
– ,3,4,5,6,7,8,9
• Hexadecimal– Base 16– 16 symbols: 0,1,2,3,4,5,6,7,8,9,
A,B,C,D,E,F– More compact representation
of the binary system
• Decimal– Base 10– 10 symbols: 0,1,2,3,4,5,6,7,8,9
Decimal
(base 10)
Binary
(base 2)
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
16 10000
17 10001
Octal
(base 8)
Hexadecimal
(base 16)
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
10 8
11 9
12 A
13 B
14 C
15 D
16 E
17 F
20 10
21 11
Dr. Wang
Example of Equivalent Numbers
Binary: 1 0 1 0 0 0 0 1 0 1 0 0 1 1 12
Octal: 502478
Decimal: 2064710
Hexadecimal: 50A716
Notice how the number of digits gets
smaller as the base increases.
Dr. Wang
Agenda
• Overview of number systems
– Positional and non-positional
– Base, positional value, symbol value
– Binary, decimal, octal, hexadecimal
• Number systems conversions
Dr. Wang
Number Systems Conversions
• Binary, Octal, and Hex to Decimal• Decimal to Hex, Octal, and Binary• Binary Hex• Binary Octal• Hex Octal
Dr. Wang
1. Binary, Octal, Hex Decimal
In general (base is b: 2 for binary, 8 for Octal, 16 for Hex),
N = ...P3P2P1P0 . P-1P-2P-3...
= ... + P3b3 + P2b2 + P1b1 + P0b0 + P-1b-1 + P-2b-2 + P-3b-3 + ...
Multiply the decimal equivalent of each digit by its positional/place value (a power of the base) and sum these products
Dr. Wang
Exercise (2)
• Convert the following numbers to their decimal equivalents
10011012
1101.112
1AB.616
173.258
Dr. Wang
2. Decimal Binary, Octal, or Hex
To convert decimal numbers to any base we divide with the corresponding base until the quotient is zero and write the remainders in reverse order.
Dr. Wang
Decimal Octal, Binary, Hex
• Divide the decimal number successively by 8 (for Octal), 2 (for Binary), 16 (for Hex)
• After each division record the remainder– Octal: 0,1,…, or, 7
– Binary: either a 1 or 0
– Hex: 1, 2,…, or,9, or A, B, …, or F
• Continue until the result of the division (quotient) is 0
• Write the remainders in reverse order
Dr. Wang
Exercise (3)
• Convert 123|10 to Base 8
• Convert 59|10 to Base 2
• Convert 42|10 to Base 16
Dr. Wang
Number Systems Conversions(Agenda)
Binary, Octal, and Hex to DecimalDecimal to Hex, Octal, and Binary• Binary Hex• Binary Octal• Hex Octal
Dr. Wang
Binary Hex
Dr. Wang
Binary to Hexadecimal Conversion
10100010111001|2 = ?|16
Work from right to left
Divide into 4-bit groups
##10 1000 1011 1001
2 8 B 9
Decimal Binary Hexadecimal
0 0000 0
1 0001 1
2 0010 2
3 0011 3
4 0100 4
5 0101 5
6 0110 6
7 0111 7
8 1000 8
9 1001 9
10 1010 A
11 1011 B
12 1100 C
13 1101 D
14 1110 E
15 1111 F
NOTE: # is a place holder for zero!
Dr. Wang
Hexadecimal to Binary Conversion
FACE|16 = ?|2
F A C E
1111 1010 1100 1110
Decimal Binary Hexadecimal
0 0000 0
1 0001 1
2 0010 2
3 0011 3
4 0100 4
5 0101 5
6 0110 6
7 0111 7
8 1000 8
9 1001 9
10 1010 A
11 1011 B
12 1100 C
13 1101 D
14 1110 E
15 1111 F
FACE|16=1111101011001110|2
Dr. Wang
Binary Octal
Dr. Wang
10101110001101|2=?|8
#10 101 110 001 101
2 5 6 1 5
10101110001101|2=25615|8
Binary to Octal Conversion
Binary Octal
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
Work from right to left Divide into 3 bit groups
Dr. Wang
1247|8=?|2
1 2 4 7
001 010 100 111
Octal to Binary Conversion
Note: one need not write the leading zeros
1247|8=001010100111 |2
=1010100111|2
Binary Octal
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
Dr. Wang
Hexadecimal Octal
Dr. Wang
How do we convert fromhexadecimal to octal and
vice versa?
Convert to binary first!
Dr. Wang
Exercise (4)
• Do you know the equivalent hexadecimal, octal, and decimal values of the binary number 11001010? ______Yes _______No If you answered Yes, please indicate them below:
– Equivalent hexadecimal number:________________– Equivalent octal number: __________________– Equivalent decimal number: __________________
• What is the equivalent binary number of the decimal number 63? _____________________________
From Background Survey
Dr. Wang
Exercise (5)
• Convert 18110 to binary and hex
• Convert 121F16 to decimal
• Convert 010101011002 to hex
• Convert A17F16 to octal
Dr. Wang
Summary
1. Basic number systems concepts (base, positional/place value, symbol value)
2. Convert back and forth between decimal numbers and their binary, octal, and hexadecimal equivalents
3. Abbreviate binary numbers as octal or hexadecimal numbers
4. Convert octal and hexadecimal numbers to binary numbers
Dr. Wang
• Specify the value of the digit 5 in the following decimal
numbers:
Solution (1)
the 5 in 25 = 5 x 100 = 5
the 5 in 51 = 5 x 101 = 50
the 5 in 4538 = 5 x 102 = 500
Dr. Wang
Solution (2)
10011012 = 1 x 26 + 0 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21+1 x 20 = 64 + 0 + 0 + 8 + 4 + 0 + 1 = 7710
1101.112 = 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 + 1 x 2-1 + 1 x 2-2 = 8 + 4 + 0 + 1 + 1/2 + 1/4 = 13.7510
1AB.616 = 1 x 162 + A x 161 + B x 160 + 6 x 16-1 = 1 x 256 + 10 x 16 + 11 x 1 + 6 x 16 = 256 + 160 + 11 + 0.375 = 427.37510
173.258 = 1 x 82 + 7 x 81 + 3 x 80 + 2 x 8-1 + 5 x 8-2
= 1 x 64 + 7 x 8 + 3 x 1 + 2/8 + 5/64 = 64 + 56 + 3 + 0.25 + 0.078125 = 123.32812510
Dr. Wang
Convert 123|10 to Base 8:
8 )123
8 )15 R 3
8 )1 R 7
0 R 1
Therefore, 123|10 = 173|8
Solution (3-1)
Base you are converting to
Read Up!
Dr. Wang
Solution (3-2)
• Convert 59|10 to Base 2:
59|10 =1110112
• Convert 42|10 to Base 16:
16 )42
16 )2 R A
0 R 2Read Up!
Therefore, 42|10 = 2A|16
Dr. Wang
Solution (4)
• Do you know the equivalent hexadecimal, octal, and decimal values of the binary number 11001010? ______Yes _______No If you answered Yes, please indicate them below:
– Equivalent hexadecimal number:__CA___________– Equivalent octal number: _______312___________– Equivalent decimal number: _____202_____________
• What is the equivalent binary number of the decimal number 63? ________111111_____________
From Background Survey
Dr. Wang
Solution (5)
• Convert 18110 to binary (10110101) and hex (B5)
• Convert 121F16 to decimal (4639 10 )
• Convert 010101011002 to hex (2AC16)
• Convert A17F16 to octal (1205778)
