Introduction to Arduino Machine Language Programming

Numbers and Their Computer Representation

Introduction

Base 10 result of ten fingers

Arabic symbols 0-9, India created Zero and Positional Notation

Other Systems: Roman Numerals: essentially additive, Importance of Roman Numeral lies in whether a symbol precedes or follows another symbol. Ex. IV = 4 versus VI = 6. This was a very clumsy system for arithmetic operations.

Positional Notation (Positive Real Integers)

Fractional numbers will not be considered but it should be noted that the addition of said would be a simple and logical addition to the theory presented.

The value of each digit is determined by its position. Note pronunciation of 256 “Two Hundred and Fifty Six?

Ex. 256 = 2*10² + 5*10¹ + 6*10⁰

Generalization to any base or radix

Base or Radix = Number of different digit which can occur in each position in the number system.

N = A_nrⁿ + A_n-1r^n-1 + … + A₁r¹ + A₀r⁰   (or simple A₁r + A₀)

Introduction to Binary System

The operation of most digital devices is binary by nature, either they are on or off.

Examples: Switch, Relay, Tube, Transistor, and Transistor-Transisor-Logic Integrated circuit (TTL IC)

Thus it is only logical for a digital computer to in base 2.

Note: Future devices may not have this characteristic, and this is one of the reasons the basics and theory are important. For they add flexibility to the system.

In the Binary system there are only 2 states allowed; 0 and 1 (FALSE or TRUE, OFF or ON)

Example:      Most Significant Bit

Bit = One Binary Digit (0 or 1)

This positional related equation also gives us a tool for converting from a given radix to base 10 – in this example Binary to Decimal.

Base Eight and Base Sixteen

Early in the development of the digital computer Von Neuman realized the usefulness of operating in intermediate base systems such as base 8 (or Octal)

By grouping 3 binary digits or bits one octal digit is formed. Note that 2³ = 8

Binary to Octal Conversion Table

2²2¹2⁰

0 0 0   =  0

0 0 1   =  1

0 1 0   =  2

0 1 1   =  3

1 0 0   =  4

1 0 1   =  5

1 1 0   =  6

1 1 1   =  7                 Symbols (not numbers) 8 and 9 are not used in octal.

Example:    100 001 010 110

4     1      2     6 ₈  = 4*8³ + 1*8² + 2*8¹ + 6*8⁰ = 2134

This is another effective way of going from base 2 to base 10

Summary: Base 8 allows you to work in the language of the computer without dealing with large numbers of ones and zeros. This is made possible through the simplicity of conversion from base 8 to base 2 and back again.

In microcomputers groupings of 4 bits (as opposed to 3 bits) or base 16 (2⁴) is used. Originally pronounced Sexadecimal, base 16 was quickly renamed Hexadecimal (this really should be base 6).

Binary to Hex Conversion Table

2³2²2¹2⁰

0 0 0 0   =  0

0 0 0 1   =  1

0 0 1 0   =  2

0 0 1 1   =  3

0 1 0 0   =  4

0 1 0 1   =  5

0 1 1 0   =  6

0 1 1 1   =  7

1 0 0 0   =  8

1 0 0 1   =  9

1 0 1 0   =  A

1 0 1 1   =  B

1 1 0 0   =  C

1 1 0 1   =  D

1 1 1 0   =  E

1 1 1 1   =  F

In Hex Symbols for 10 to 15 are borrowed from the alphabet. This shows how relative numbers really are or in other words, they truly are just symbols.

Example:    1000 0101 0110

8        5       6 ₁₆  = 8*16² + 5*16¹ + 6*16⁰ = 2134

It is not as hard to work in base 16 as you might think, although it does take a little practice.

Conversion from Base 10 to a Given Radix (or Base)

Successive Division            is best demonstrated by an example

To get the digits in the right order let them fall to the right.

For this example:  43₁₀ = 101011₂                 Quick Check (Octal)  101  011 = 5*8 + 3 = 43₁₀

Another example: Convert 43₁₀ from decimal to Octal

For this example: 43₁₀ = 53₈             Quick Check (Octal) 5*8 + 3 = 43₁₀

Generalization of the procedure OR Why It Works

Where r = radix, N = number, A = remainder, and n = the number of digits in radix r for number N. Division is normally done in base 10.

Another way of expressing the above table is:

N   = r*N₁ + A₀

N₁ = r*N₂ + A₁

N₂ = r*N₃ + A₂

:

N_n-1 = r*N_n + A_n-1

N_n   =  r*0   + A_n

or (now for the slight of hand)

N   = r*( r*N₂ + A₁)+ A₀                                               substitute N₁

N  = r²N₂ + rA₁+ A₀                                                    multiply r through equation

N  = r²(r*N₃ + A₂) + rA₁+ A₀                            substitute N₂

:

N = A_nrⁿ + A_n-1r^n-1 + … + A₁r¹ + A₀r⁰    

Nomenclature

Bit              =          1 binary digit

Byte           =          8 bits

Nibble         =          one half byte = 4 bits

Word          =          Computer Dependent

Binary Arithmetic

Binary Addition

Binary addition is performed similar to decimal addition using the following binary addition rules:

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 10   (0 with a carry of 1)

Examples:

Problem	2110 + 1010 = 3110	4510 + 5410 = 9910	310 + 710 = 1010
	101012 +       010102 _______________ 111112	1011012 +       1101102 _______________ 11000112	0112 +       1112 _______________ 10102
Check			1*23 + 0*22 + 1*21 + 0*20 = 1*8 + 0*4 + 1*2 + 0*1 = 1010

Octal Addition

Octal addition is also performed similar to decimal addition except that each digit has a range of 0 to 7 instead of 0 to 9.

Problem	2110 + 1010 = 3110	4510 + 5410 = 9910	310 + 710 = 1010
	258 +       128 _______________ 378	558 +       668 _______________ 1438	38 +       78 _______________ 128
Check	3*81 + 7*80 3*8  + 7*1  = 3110	1*82 + 4*81 + 3*80 64 + 32 + 3  = 9910	1*81 + 2*80 8 + 2 = 1010

Hexadecimal Addition

Hex addition is also performed similar to decimal addition except that each digit has a range of 0 to 15 instead of 0 to 9.

Problem	2110 + 1010 = 3110	4510 + 5410 = 9910	310 + 710 = 1010
	1516 +       0A16 _______________ 1F16	2D16 +       3616 _______________ 6316	316 +       716 _______________ A16    (not 10)
Check	1*161 + 15*160 16 + 15 = 3110	6*161 + 3*160 96 + 3 = 9910	10*160 1010

Binary Multiplication

Decimal	Binary
1110 x        1310 _______________ 3310 1110– _______________ 14310	10112 x   11012 _______________ 10112 00002- 10112– 10112— _______________ 100011112
Check	8*161 + 15*160 128 + 15 = 14310

Binary Division

Check: 1*16¹ +  5*16⁰ = 16 + 5 = 21₁₀

Practice arithmetic operations by making problems up and then checking your answers by converting them back to base 10 via different bases (i.e., 2, 8, and 16).

How a computer performs arithmetic operations is a much more involved subject and has not been dealt with in this section.

Complements and Negative Numbers OR Adding a Sign Bit

Addition, Multiplication, and Division is nice but what about subtraction and negative numbers? From grade school you have learned that subtraction is simply the addition of a negative number. Mathematicians along with engineers have exploited this principle along with modulo arithmetic — a natural outgrowth of adders of finite width — to allow computers to operate on negative numbers without adding any new hardware elements to the arithmetic logic unit (ALU).

Sign Magnitude

Here is a simple solution, just add a sign bit. To implement this solution in hardware you will need to create a subtractor; which means more money.

sign                magnitude

Example:    – 2        =          1                      0010₂

Ones Complement

Here is a solution that is a little more complex. Add the sign bit and invert each bit making up the magnitude — simply change the 1’s to 0’s and the 0’s to 1’s.

sign                magnitude

Example:    – 2        =          1                      1101₂

To subtract in 1’s complement you simply add the sign and magnitude bits letting the last carry bit (from the sign) fall into the bit bucket, and then add 1 to the answer. Once again let the last carry bit fall into the bit bucket. The bit bucket is possible due to the physical size of the adder.

0 1010₂               10

+   _ 1 1101₂             +(-2)

0 1000₂                 8

+______1₂                       Adjustment

0 1001₂

Although you can now use your hardware adder to subtract numbers, you now need to add 1 to the answer. This again means adding hardware. Compounding this problem, ones complement allows two numbers to equal 0 (schizophrenic zero).

Twos Complement

Here is a solution that is a little more complex to set up, but needs no adjustments at the end of the addition. There are two ways to take the twos complement of a number.

Method 1 = Take the 1’s complement and add 1

__0 0010₂             2          <- start

+    1 1101₂             1’s complement (i.e. invert)

+              1₂                       add 1

1 1110₂

Method 2 = Move from right to left until a 1 is encountered then invert.

0 00102	start  = 210
02	no change
102	no change but one is encountered
1102	invert = change 0 to 1
11102	invert = change 0 to 1
1 11102	invert = change 0 to 1

Subtraction in twos complement is the same as addition. No adjustment is needed, and twos complement has no schizophrenic zero although it does have an additional negative number (see How It Works).

0 1010₂               10

+    1 1110₂             +(-2)

0 1001₂                 8

Examples:

Problem	3310 – 1910 = 1410	6910 – 8410 = -1510
	0 1000012 +       1 1011012 _______________ 0 0011102	0 10001012 +       1 01011002 _______________ 1 11100012
Check	convert to intermediate base E16 = 1410	convert back to sign magnitude – 00011112 convert to intermediate base (16) – F16 = – 1510

Why It Works

Real adders have a finite number of bits, which leads naturally to modulo arithmetic — the bit bucket.

Overflow

With arithmetic now reduced to going around in circles, positive numbers can add up to negative and vice-versa. Two tests provide a quick check on whether or not an “Overflow” condition exists.

Test 1 =  If the two numbers are negative and the answer is positive, an overflow has occurred.

Test 2 =  If the two number are positive and the answer is negative, an overflow has occurred.

If computers were calculators and the world was a perfect place, we would be done. But they are not and so we continue by looking at a few real world problems and their solutions.

Character Codes OR Non-Numeric Information

Decimal Number Problem

Represent a Decimal Numbers in a Binary Computer. A binary representation of a decimal number, a few years ago, might have been “hard wired” into the arithmetic logic unit (ALU) of the computer. Today it, more likely than not, is simply representing some information that is naturally represented in base 10, for example your student ID.

Solution

In this problem, ten different digits need to be represented. Using 4 bits 2⁴ or 16 combinations can be created. Using 3 bits 2³ or 8 combinations can be created. Thus 4 bits will be required to represent one Decimal Digit. It should here be pointed out how 16 combinations can be created from 4 bits (0000 – 1111) while the largest numeric value that can be represented is 15. The reason that the highest numeric value and the number of combinations are different, is due to zero (0) being one of the combinations. This difference points up the need to always keep track of wetter or not you are working zero or one relative and what exactly you are after — a binary number or combinations.

The most common way of representing a decimal number is named Binary Coded Decimal (BCD). Here each binary number corresponds to its decimal equivalent, with numbers larger than 9 simply not allowed. BCD is also known as an 8-4-2-1 code since each number represents the respective weights of the binary digits. In contrast the Excess-3 code is an unweighted code used in earlier computers. Its code assignment comes from the corresponding BCD code plus 3. The Excess-3 code had the advantage that by complementing each digit of the binary code representation of a decimal digit (1’s complement), the 9’s complement of that digit would be formed. The following table lists each decimal digit and its BCD and Excess-3 code equivalent representation. I have also included the negative equivalent of each decimal digit encoded using the Excess-3 code. For instance, the complement of 0100 (1 decimal) is 1011, which is 8 decimal. You can find more decimal codes on page 18 of “Digital Design” by M. Morris Mano (course text).

Binary Coded Decimal (BCD)		Excess-3
Decimal Digit	Binary Code 8‑4-2-1	Decimal Digit	Binary Code	9’s Compliment
0 1 2 3 4 5 6 7 8 9 N/A N/A N/A N/A N/A N/A	0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111	N/A N/A N/A 0 1 2 3 4 5 6 7 8 9 N/A N/A N/A	0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111	1111 1110 1101 1100 1011 1010 1001 1000 0111 0110 0101 0100 0011 0010 0001 0000

Alphanumeric Character Problem

Represent Alphanumeric data (lower and upper case letters of the alphabet (a-z, A-Z), digital numbers (0-9), and special symbols (carriage return, line feed, period, etc.).

Solution

To represent the upper and lower case letters of the alphabet, plus ten numbers, you need at least 62 (2×26+10) unique combinations. Although a code using only six binary digits providing 2⁶ or 64 unique combinations would work, only 2 combinations would be left for special symbols. On the other hand a code using 7 bits provides 2⁷ or 128 combinations, which provides more than enough room for the alphabet, numbers, and special symbols. So who decides which binary combinations correspond to what character. Here there is no “best way.” About thirty years ago IBM came out with a new series of computers which used 8 bits to store one character (2⁸ = 256 combinations), and devised the Extended Binary-Coded Decimal Interchange Code (EBCDIC pronounced ep-su-dec) for this purpose. Since IBM had a near monopoly on the computer field, at that time, the other computer makers refused to adopt EBCDIC, and that is how the 7bit American Standard Code for Information Interchange (ASCII) came into existence. ASCII has now been adopted by virtually all micro-computer and mini-computer manufacturers. The table below shows a partial list of the ASCII code. Page 23 of the text lists all 128 codes with explanations of the control characters.

DEC	HEX	CHAR	DEC	HEX	CHAR
32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63	20 21 22 23 24 25 26 27 28 29 2A 2B 2C 2D 2E 2F 30 31 32 33 34 35 36 37 38 39 3A 3B 3C 3D 3E 3F	! “ # $ % & ‘ ( ) * + , – * / 0 1 2 3 4 5 6 7 8 9 : ; < = > ?	64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95	40 41 42 43 44 45 46 47 48 49 4A 4B 4C 4D 4E 4F 50 51 52 53 54 55 56 57 58 59 5A 5B 5C 5D 5E 5F	@ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z [ ] ^ _

The word “string” is commonly used to describe a sequence of characters stored via their numeric codes — like ASCII).

Although ASCII requires only 7 bits, the standard in computers is to use 8 bits, where the leftmost bit is set to 0. This allows you to code another 128 characters (including such things as Greek letters), giving you an extended character set, simply by letting the leftmost bit be a 1. This can also lead to a computer version of the tower of Babel. Alternatively, the leftmost bit can be used for detecting errors when transmitting characters over a telephone line. Which brings us to our next problem.

Synthesis

Although ASCII solves the communication problem between English speaking computers, what about Japanese, Chinese, or Russian computers which have different, and in all these examples, larger alphabets?

Communication Problem

Binary information may be transmitted serially (one bit at a time) through some form of communication medium such as a telephone line or a radio wave. Any external noise introduced into the medium can change bit values from 1 to 0 or visa versa.

Solution

The simplest and most common solution to the communication problem involves adding a parity bit to the information being sent. The function of the parity bit is to make the total number of 1’s being sent either odd (odd parity) or even (even parity). Thus, if any odd number of 1’s were sent but an even number of 1’s received, you know an error has occurred. The table below illustrates the appropriate parity bit (odd and even) that would be appended to a 4-bit chunk of data.

Synthesis

What happens if two binary digits change bit values? Can a system be devised to not only detect errors but to identify and correct the bit(s) that have changed? One of the most common error-correcting codes was developed by R.W. Hamming. His solution, known as a Hamming code, can be found in a very diverse set of places from a Random Access Memory (RAM) circuit to a Spacecraft telecommunications link. For more of error correcting codes read pages 299 to 302 of the text.

Although detecting errors is nice, preventing them from occurring is even better. Which of course brings us to our next problem.

Shaft Encoder Problem

As a shaft turns, you need to convert its radial position into a binary coded digital number.

Solution

The type of coder which will be briefly described below converts a shaft position to a binary-coded digital number. A number of different types of devices will perform this conversion; the type described is representative of the devices now in use, and it should be realized that more complicated coders may yield additional accuracy. Also, it is generally possible to convert a physical position into an electric analog-type signal and then convert this signal to a digital system. In general, though, more direct and accurate coders can be constructed by eliminating the intermediate step of converting a physical position to an analog electric signal. The Figure below illustrates a coded-segment disk which is coupled to the shaft.

The shaft encoder can be physically realized using electro-mechanical (brush) or electro-optical technology. Assuming an electro-optical solution, the coder disk is constructed with bands divided into transparent segments (the shaded areas) and opaque segments (the unshaded areas). A light source is put on one side of the disk, and a set of four photoelectric cells on the other side, arranged so that one cell is behind each band of the coder disk. If a transparent segment is between the light source and a light-sensitive cell, a 1 output will result; and if an opaque area is in front of the photoelectric cell, there will be a O output.

There is one basic difficulty with the coder illustrated: if the disk is in a position where the output number is changing from 011 to 100, or in any position where several bits are changing value, the output signal may become ambiguous. As with any physically realized device, no matter how carefully it is made, the coder will have erroneous outputs in several positions. If this occurs when 011 is changing to 100, several errors are possible; the value may be read as 111 or 000, either of which is a value with considerable errors. To circumvent this difficulty, engineers use a “Gray,” or “unit distance,” code to form the coder disk (see previous Figure). In this code, 2 bits never change value in successive coded binary numbers. Using a Gray coded disk, a 6 may be read as 7, or a 4 as 5, but larger errors will not be made. The Table below shows a listing of a 4-bit Gray code.

Decimal	Gray Code
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15	0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000

Synthesis

Gray code is used in a multitude of application other than shaft encoders. For example, CMOS circuits draw the most current when they are switching. If a large number of circuits switch at the same time unwelcome phenomena such as “Ground Bounce” and “EMI Noise” can result. If the transistors are switching due to some sequential phenomena (like counting), then these unwelcome visitors can be minimized by replacing a weighted binary code by a Gray code.

If the inputs to a binary machine are from an encoder using a Gray code, each word must be converted to conventional binary or binary-coded decimal bit equivalent. How can this be done? Before you can answer this question, you will need to learn about Boolean Algebra — what a coincidence, that’s the topic of the  next Section.

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