phy 201 (blum)1 binary numbers. phy 201 (blum)2 why binary? maximal distinction among values ...
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PHY 201 (Blum) 1
Binary Numbers
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PHY 201 (Blum) 2
Why Binary? Maximal distinction among values
minimal corruption from noise Imagine taking the same physical
attribute of a circuit, e.g. a voltage lying between 0 and 5 volts, to represent a number
The overall range can be divided into any number of regions
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PHY 201 (Blum) 3
Don’t sweat the small stuff For decimal numbers, fluctuations must
be less than 0.25 volts For binary numbers, fluctuations must
be less than 1.25 volts5 volts
0 voltsDecimal Binary
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PHY 201 (Blum) 4
Range actually split in three
High
Low
Forbidden range
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PHY 201 (Blum) 5
It doesn’t matter …. Some of the standard voltages
coming from a computer’s power are ideally supposed to be 3.30 volts, 5.00 volts and 12.00 volts
Typically they are 3.28 volts, 5.14 volts or 12.22 volts or some such value
So what, who cares
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PHY 201 (Blum) 6
How to represent big integers
Use positional weighting, same as with decimal numbers
205 = 2102 + 0101 + 5100
Decimal – powers of ten 11001101 = 127 + 126 + 025 +
024 + 123 + 122 + 021
+ 120 = 128 + 64 + 8 + 4 + 1 = 205 Binary – powers of two
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PHY 201 (Blum) 7
Converting 205 to Binary
205/2 = 102 with a remainder of 1, place the 1 in the least significant digit position
Repeat 102/2 = 51, remainder 0
1
0 1
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PHY 201 (Blum) 8
Iterate 51/2 = 25, remainder 1
25/2 = 12, remainder 1
12/2 = 6, remainder 0
1 0 1
1 1 0 1
0 1 1 0 1
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PHY 201 (Blum) 9
Iterate 6/2 = 3, remainder 0
3/2 = 1, remainder 1
1/2 = 0, remainder 1
0 0 1 1 0 1
1 0 0 1 1 0 1
1 1 0 0 1 1 0 1
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PHY 201 (Blum) 10
Recap
1 1 0 0 1 1 0 1
127 + 126 + 025 + 024
+ 123 + 122 + 021 + 120
205
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PHY 201 (Blum) 11
Finite representation Typically we just think computers do binary
math. But an important distinction between binary
math in the abstract and what computers do is that computers are finite.
There are only so many flip-flops or logic gates in the computer.
When we declare a variable, we set aside a certain number of flip-flops (bits of memory) to hold the value of the variable. And this limits the values the variable can have.
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PHY 201 (Blum) 12
Same number, different representation 5 using 8 bits 0000 0101 5 using 16 bits 0000 0000 0000 0101 5 using 32 bits 0000 0000 0000 0000 0000 0000
0000 0101
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PHY 201 (Blum) 13
Adding Binary Numbers Same as decimal; if the sum of
digits in a given position exceeds the base (10 for decimal, 2 for binary) then there is a carry into the next higher position
1
3 9
+ 3 5
7 4
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PHY 201 (Blum) 14
Adding Binary Numbers
1 1 1 1
0 1 0 0 1 1 1
+ 0 1 0 0 0 1 1
1 0 0 1 0 1 0
carries
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PHY 201 (Blum) 15
Uh oh, overflow*
What if you use a byte (8 bits) to represent an integer
A byte may not be enough to represent the sum of two such numbers.
*The End of the World as We Know It
1 1
1 0 1 0 1 0 1 0
+ 1 1 0 0 1 1 0 0
1 0 1 1 1 0 1 1 0
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PHY 201 (Blum) 16
Biggest unsigned integers 4 bit: 1111 15 = 24 - 1 8 bit: 11111111 255 = 28 – 1 16 bit: 1111111111111111
65535= 216 – 1 32 bit:
11111111111111111111111111111111 4294967295= 232 – 1
Etc.
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PHY 201 (Blum) 17
Bigger Numbers You can represent larger numbers
by using more words You just have to keep track of the
overflows to know how the lower numbers (less significant words) are affecting the larger numbers (more significant words)
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PHY 201 (Blum) 18
Negative numbers Negative x is the number that when added
to x gives zero
Ignoring overflow the two eight-bit numbers above sum to zero
1 1 1 1 1 1 1
0 0 1 0 1 0 1 0
1 1 0 1 0 1 1 0
1 0 0 0 0 0 0 0 0
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PHY 201 (Blum) 19
Two’s Complement
Step 1: exchange 1’s and 0’s
Step 2: add 1 (to the lowest bit only)
0 0 1 0 1 0 1 0
1 1 0 1 0 1 0 1
1 1 0 1 0 1 1 0
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PHY 201 (Blum) 20
Sign bit With the two’s complement approach,
all positive numbers start with a 0 in the left-most, most-significant bit and all negative numbers start with 1.
So the first bit is called the sign bit. But note you have to work harder than
just strip away the first bit. 10000001 IS NOT the 8-bit version of –
1
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PHY 201 (Blum) 21
Add 1’s to the left to get the same negative number using more bits
-5 using 8 bits 11111011 -5 using 16 bits 1111111111111011 -5 using 32 bits 11111111111111111111111111111011 When the numbers represented are whole
numbers (positive or negative), they are called integers.
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PHY 201 (Blum) 22
Biggest signed integers 4 bit: 0111 7 = 23 - 1 8 bit: 01111111 127 = 27 – 1 16 bit: 0111111111111111 32767=
215 – 1 32 bit:
01111111111111111111111111111111 2147483647= 231 – 1
Etc.
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PHY 201 (Blum) 23
Most negative signed integers 4 bit: 1000 -8 = - 23
8 bit: 10000000 - 128 = - 27
16 bit: 1000000000000000 -32768= - 215
32 bit: 10000000000000000000000000000000 -2147483648= - 231
Etc.
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PHY 201 (Blum) 24
Riddle
Is it 214? Or is it – 42? Or is it Ö? Or is it …? It’s a matter of interpretation
How was it declared?
1 1 0 1 0 1 1 0
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PHY 201 (Blum) 25
3-bit unsigned and signed
7 1 1 1
6 1 1 0
5 1 0 1
4 1 0 0
3 0 1 1
2 0 1 0
1 0 0 1
0 0 0 0
3 0 1 1
2 0 1 0
1 0 0 1
0 0 0 0
-1 1 1 1
-2 1 1 0
-3 1 0 1
-4 1 0 0
Think of an odometer reading 999999 and the car travels one more mile.
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PHY 201 (Blum) 26
Fractions Similar to what we’re used to with
decimal numbers
3.14159 =
3 · 100 + 1 · 10-1 + 4 · 10-2 + 1 · 10-3 + 5 · 10-4 + 9 · 10-5
11.001001 =
1 · 21 + 1 · 20 + 0 · 2-1 + 0 · 2-2 + 1 · 2-3 + 0 · 2-4 + 0 · 2-5
+ 1 · 2-6
(11.001001
3.140625)
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Places 11.001001
PHY 201 (Blum) 27
Two’s placeOne’s place
Half’s place
Fourth’s place Eighth’s
place Sixteenth’s place
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PHY 201 (Blum) 28
Decimal to binary 98.61
Integer part 98 / 2 = 49 remainder 0 49 / 2 = 24 remainder 1 24 / 2 = 12 remainder 0 12 / 2 = 6 remainder 0 6 / 2 = 3 remainder 0 3 / 2 = 1 remainder 1 1 / 2 = 0 remainder 1
1100010
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PHY 201 (Blum) 29
Decimal to binary 98.61
Fractional part 0.61 2 = 1.22 0.22 2 = 0.44 0.44 2 = 0.88 0.88 2 = 1.76 0.76 2 = 1.52 0.52 2 = 1.04
.100111
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PHY 201 (Blum) 30
Decimal to binary Put together the integral and
fractional parts 98.61 1100010.100111
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PHY 201 (Blum) 31
Another Example (Whole number part)
123.456 Integer part
123 / 2 = 61 remainder 1 61 / 2 = 30 remainder 1 30 / 2 = 15 remainder 0 15 / 2 = 7 remainder 1 7 / 2 = 3 remainder 1 3 / 2 = 1 remainder 1 1 / 2 = 0 remainder 1
1111011
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PHY 201 (Blum) 32
Checking: Go to All Programs/Accessories/Calculator
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PHY 201 (Blum) 33
Put the calculator in Programmer view
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PHY 201 (Blum) 34
Enter number, put into binary mode
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PHY 201 (Blum) 35
Another Example (fractional part)
123.456 Fractional part
0.456 2 = 0.912 0.912 2 = 1.824 0.824 2 = 1.648 0.648 2 = 1.296 0.296 2 = 0.592 0.592 2 = 1.184 0.184 2 = 0.368 …
.0111010…
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PHY 201 (Blum) 36
Checking fractional part: Enter digits found in binary mode
Note that the leading zero does not display.
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PHY 201 (Blum) 37
Convert to decimal mode, then
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Edit/Copy result. Switch to Scientific View. Edit/Paste
PHY 201 (Blum) 38
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PHY 201 (Blum) 39
Divide by 2 raised to the number of digits (in this case 7, including leading zero)
1 2
3 4
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PHY 201 (Blum) 40
Finally hit the equal sign. In most cases it will not be exact
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PHY 201 (Blum) 41
Other way around Multiply fraction by 2 raised to the desired
number of digits in the fractional part. For example .456 27 = 58.368
Throw away the fractional part and represent the whole number 58 111010
But note that we specified 7 digits and the result above uses only 6. Therefore we need to put in the leading 0. (Also the fraction is less than .5 so there’s a zero in the ½’s place.) 0111010
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PHY 201 (Blum) 42
Limits of the fixed point approach
Suppose you use 4 bits for the whole number part and 4 bits for the fractional part (ignoring sign for now).
The largest number would be 1111.1111 = 15.9375
The smallest, non-zero number would be 0000.0001 = .0625
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PHY 201 (Blum) 43
Floating point representation Floating point representation
allows one to represent a wider range of numbers using the same number of bits.
It is like scientific notation. We’ll do this later in the semester.
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PHY 201 (Blum) 44
Hexadecimal Numbers Even moderately sized decimal
numbers end up as long strings in binary
Hexadecimal numbers (base 16) are often used because the strings are shorter and the conversion to binary is easier
There are 16 digits: 0-9 and A-F
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PHY 201 (Blum) 45
Decimal Binary Hex 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
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PHY 201 (Blum) 46
Binary to Hex Break a binary string into groups of
four bits (nibbles) Convert each nibble separately
1 1 1 0 1 1 0 0 1 0 0 1
E C 9
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PHY 201 (Blum) 47
Addresses With user friendly computers, one rarely
encounters binary, but we sometimes see hex, especially with addresses
To enable the computer to distinguish various parts, each is assigned an address, a number Distinguish among computers on a network Distinguish keyboard and mouse Distinguish among files Distinguish among statements in a program Distinguish among characters in a string
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PHY 201 (Blum) 48
How many? One bit can have two states and thus
distinguish between two things Two bits can be in four states and … Three bits can be in eight states, … N bits can be in 2N states
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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PHY 201 (Blum) 49
IP(v4) Addresses An IP(v4) address is used to
identify a network and a host on the Internet
It is 32 bits long How many distinct IP addresses
are there?
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PHY 201 (Blum) 50
Characters We need to represent characters using
numbers ASCII (American Standard Code for
Information Interchange) is a common way A string of eight bits (a byte) is used to
correspond to a character Thus 28=256 possible characters can be
represented Actually ASCII only uses 7 bits, which is 128
characters; the other 128 characters are not “standard”
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PHY 201 (Blum) 51
Unicode Unicode uses 16 bits, how many
characters can be represented? Enough for English, Chinese,
Arabic and then some.
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PHY 201 (Blum) 52
ASCII 0 00110000 (48) 1 00110001 (49) … A 01000001 (65) B 01000010 (66) … a 01100001 (97) b 01100010 (98) …
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PHY 201 (Blum) 53
Booleans A Boolean variable is something
that is true or false Booleans have two states and
could be represented by a single bit (1 for true and 0 for false)
Booleans appearing in a program will take up a whole word in memory
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Boolean Operators A.k.a. logical operators Have Boolean input and Boolean output Standard:
AND OR NOT XOR (either or but not both) NOR = NOT(OR) NAND = NOT(AND)
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Truth Tables AND
INPUT OUTPUT
A B A AND B
0 0 0
0 1 0
1 0 0
1 1 1
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Truth Tables (Cont.) OR
INPUT OUTPUT
A B A OR B
0 0 0
0 1 1
1 0 1
1 1 1
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Truth Tables (Cont.) XOR (Excluded OR)
INPUT OUTPUT
A B A XOR B
0 0 0
0 1 1
1 0 1
1 1 0
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Numbers from Logic All of the numerical operations we have
talked about are really just combinations of logical operations
E.g. the adding operation is just a particular combination of logic operations
Possibilities for adding two bits 0+0=0 (with no carry) 0+1=1 (with no carry) 1+0=1 (with no carry) 1+1=0 (with a carry)
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Addition Truth TableINPUT OUTPUT
A BSum
A XOR B
CarryA AND
B
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
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All is NAND Actually you can use one logic gate
(the NAND) and a few tricks (like De Morgan’s theorem) to build all of the “combinatorial” circuitry (the circuitry that doesn’t involve memory)
NORs work too But we tend to think in ANDs, ORs
and NOTs
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Bit manipulation You can use an AND to select out
part of a word (where s is a 1 or 0, etc)s t u v w x y z
1 1 1 1 0 0 0 0
s t u v 0 0 0 0
AND
gives
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IP Addresses Revisted LaSalle’s IP address is what’s called a
Class B IP address Of the 32 bits the first two bits are 10
(this identifies us as Class B) The remaining 14 bits of the first two
bytes identify us as LaSalle The remaining 2 bytes are for our
internals use (to assign computers within LaSalle)
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In or Out To see if an address is local to
LaSalle, you would restrict your attention to the first two bytes.
HOW? AND it with FFFF0000
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Subnets A network (like LaSalle’s) can be
divided further into sub-networks Then subnet masks are used to
determine whether or not another computer is on the same subnet