The numeric system we use daily is the decimal system, but this system is not convenient for machines since the information is handled codified in the shape of on or off bits; this way of codifying takes us to the necessity of knowing the positional calculation which will allow us to express a number in any base where we need it.
2.1 Radix number systems
The numeric system we use daily is the decimal system, but this system is not convenient for machines since the information is handled codified in the shape of on or off bits; this way of codifying takes us to the necessity of knowing the positional calculation which will allow us to express a number in any base where we need it.
A base of a number system or radix defines the range of values that a digit may have.
In the binary system or base 2, there can be only two values for each digit of a number, either a “0” or a “1”.
In the octal system or base 8, there can be eight choices for each digit of a number:
“0”, “1”, “2”, “3”, “4”, “5”, “6”, “7”.
In the decimal system or base 10, there are ten different values for each digit of a number:
“0”, “1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”.
In the hexadecimal system, we allow 16 values for each digit of a number:
“0”, “1”, “2”, “3”, “4”, “5”, “6”, “7”, “8”, “9”, “A”, “B”, “C”, “D”, “E”, and “F”.
Where “A” stands for 10, “B” for 11 and so on.
2.2 Conversion among radices
· Convert from Decimal to Any Base
Let`s think about what you do to obtain each digit. As an example, let`s start with a decimal number 1234 and convert it to decimal notation. To extract the last digit, you move the decimal point left by one digit, which means that you divide the given number by its base 10.
1234/10 = 123 + 4/10
The remainder of 4 is the last digit. To extract the next last digit, you again move the decimal point left by one digit and see what drops out.
123/10 = 12 + 3/10
The remainder of 3 is the next last digit. You repeat this process until there is nothing left. Then you stop. In summary, you do the following:
Quotient | Remainder
1234/10 = 123 4
123/10 = 12 3
12/10 = 1 2
1/10 = 0 1 (Stop when the quotient is 0)
(Write in reverse order) 1 2 3 4 (Base 10)
=>Now, let`s try a nontrivial example. Let`s express a decimal number 1341 in binary notation. Note that the desired base is 2, so we repeatedly divide the given decimal number by 2.
Quotient | Remainder
1341/2 = 670 1
670/2 = 335 0
335/2 = 167 1
167/2 = 83 1
83/2 = 41 1
41/2 = 20 1
20/2 = 10 0
10/2 = 5 0
5/2 = 2 1
2/2 = 1 0
1/2 = 0 1 (Stop when the quotient is 0)
(Write in reverse order) 1 0 1 0 0 1 1 1 1 0 1 (BIN; Base 2)
ð Let`s express the same decimal number 1341 in octal notation.
Quotient | Remainder
1341/8 = 167 5
167/8 = 20 7
20/8 = 2 4
2/8 = 0 2 (Stop when the quotient is 0)
(Write in reverse order) 2 4 7 5 (OCT; Base 8)
ð Let`s express the same decimal number 1341 in hexadecimal notation.
Quotient | Remainder
1341/16 = 83 13
83/16 = 5 3
5/16 = 0 5 (Stop when the quotient is 0)
(Write in reverse order) 5 3 D (HEX; Base 16)
In conclusion, the easiest way to convert fixed point numbers to any base is to convert each part separately. We begin by separating the number into its integer and fractional part. The integer part is converted using the remainder method, by using a successive division of the number by the base until a zero is obtained. At each division, the reminder is kept and then the new number in the base r is obtained by reading the remainder from the lat remainder upwards.
The conversion of the fractional part can be obtained by successively multiplying the fraction with the base. If we iterate this process on the remaining fraction, then we will obtain successive significant digit. This methods form the basis of the multiplication methods of converting fractions between bases.
||Example|| Convert the decimal number 3315 to hexadecimal notation. What about the hexadecimal equivalent of the decimal number 3315.3?
Quotient | Remainder
3315/16 = 207 3
207/16 = 12 15
12/16 = 0 12 (Stop when the quotient is 0)
(Write in reverse order) C F 3 (HEX; Base 16)
(HEX; Base 16)
Product | Integer Part
0.3*16 = 4.8 4
0.8*16 = 12.8 12 —–C
0.8*16 = 12.8 12 —–C
0.8*16 = 12.8 12——C
:
:
0.4 C C C … (Note: No reverse)
Thus, 3315.3 (DEC) –> CF3.4CCC… (HEX)
· Convert From Any Base to Decimal
Let`s think more carefully what a decimal number means. For example, 1234 means that there are four boxes (digits); and there are 4 one`s in the right-most box (least significant digit), 3 ten`s in the next box, 2 hundred`s in the next box, and finally 1 thousand`s in the left-most box (most significant digit). The total is 1234:
Original Number: 1 2 3 4
How Many Tokens: 1 2 3 4
Digit/Token Value: 1000 100 10 1
Value: 1000 + 200 + 30 + 4 = 1234
or simply, 1*1000 + 2*100 + 3*10 + 4*1 = 1234
Thus, each digit has a value: 10^0=1 for the least significant digit, increasing to 10^1=10, 10^2=100, 10^3=1000, and so forth.
Likewise, the least significant digit in a hexadecimal number has a value of 16^0=1 for the least significant digit, increasing to 16^1=16 for the next digit, 16^2=256 for the next, 16^3=4096 for the next, and so forth. Thus, 1234 means that there are four boxes (digits); and there are 4 one`s in the right-most box (least significant digit), 3 sixteen`s in the next box, 2 256`s in the next, and 1 4096`s in the left-most box (most significant digit). The total is:
1*4096 + 2*256 + 3*16 + 4*1 = 4660
||Example|| Convert 234.14 expressed in an octal notation to decimal.
Solution: 2*82 + 3*81 + 4*80+1*8-1 + 4*8-2
= 2*64 +3*8 +4*1 +1/8 +4/64
=156.1875
||Example|| Convert the hexadecimal number 4B3 to decimal notation. What about the decimal equivalent of the hexadecimal number 4B3.3?
Solution:
Original Number: 4 B 3 . 3
How Many Tokens: 4 11 3 3
Digit/Token Value: 256 16 1 0.0625
Value: 1024 + 176 + 3 + 0.1875 = 1203.1875
||Example|| Convert 234.14 expressed in an octal notation to decimal.
Solution:
Original Number: 2 3 4 . 1 4
How Many Tokens: 2 3 4 1 4
Digit/Token Value: 64 8 1 0.125 0.015625
Value: 128 + 24 + 4 + 0.125 + 0.0625 = 156.1875
· Relationship between Binary – Octal and Binary-hexadecimal
As demonstrated by the table below, there is a direct correspondence between the binary system and the octal system, with three binary digits corresponding to one octal digit. Likewise, four binary digits translate directly into one hexadecimal digit.
|
BIN |
OCT |
HEX |
DEC |
|
0000 |
00 |
0 |
0 |
|
0001 |
01 |
1 |
1 |
|
0010 |
02 |
2 |
2 |
|
0011 |
03 |
3 |
3 |
|
0100 |
04 |
4 |
4 |
|
0101 |
05 |
5 |
5 |
|
0110 |
06 |
6 |
6 |
|
0111 |
07 |
7 |
7 |
|
1000 |
10 |
8 |
8 |
|
1001 |
11 |
9 |
9 |
|
1010 |
12 |
A |
10 |
|
1011 |
13 |
B |
11 |
|
1100 |
14 |
C |
12 |
|
1101 |
15 |
D |
13 |
|
1110 |
16 |
E |
14 |
|
1111 |
17 |
F |
15 |
With such relationship, In order to convert a binary number to octal, we partition the base 2 number into groups of three starting from the radix point, and pad the outermost groups with 0`s as needed to form triples. Then, we convert each triple to the octal equivalent.
For conversion from base 2 to base 16, we use groups of four.
Consider converting 101102 to base 8:
101102 = 0102 1102 = 28 68 = 268
Notice that the leftmost two bits are padded with a 0 on the left in order to create a full triplet.
Now consider converting 101101102 to base 16:
101101102 = 10112 01102 = B16 616 = B616
(Note that `B` is a base 16 digit corresponding to 1110. B is not a variable.)
The conversion methods can be used to convert a number from any base to any other base, but it may not be very intuitive to convert something like 513.03 to base 7. As an aid in performing an unnatural conversion, we can convert to the more familiar base 10 form as an intermediate step, and then continue the conversion from base 10 to the target base. As a general rule, we use the polynomial method when converting into base 10, and we use the remainder and multiplication methods when converting out of base 10.
2.3 Numeric complements
The radix complement of an n digit number y in radix b is, by definition, bn – y. Adding this to x results in the value x + bn – y or x – y + bn. Assuming y <= x, the result will always be greater than bn and dropping the initial `1` is the same as subtracting bn, making the result x – y + bn -bn or just x – y, the desired result.
The radix complement is most easily obtained by adding 1 to the diminished radix complement, which is (b^n – 1) – y. Since (b^n – 1) is the digit b – 1 repeated n times (because bn – 1 = bn -1n = (b – 1)(b(n-1) + b(n-2) + … + b + 1) = (b – 1)b(n-1) + … + (b – 1), see also binomial numbers), the diminished radix complement of a number is found by complementing each digit with respect to b – 1 (that is, subtracting each digit in y from b – 1). Adding 1 to obtain the radix complement can be done separately, but is most often combined with the addition of x and the complement of y.
In the decimal numbering system, the radix complement is called the ten`s complement and the diminished radix complement the nines` complement.
In binary, the radix complement is called the two`s complement and the diminished radix complement the ones` complement. The naming of complements in other bases is similar.
· Decimal example
To subtract a decimal number y from another number x using the method of complements, the ten`s complement of y (nines` complement plus 1) is added to x. Typically, the nines` complement of y is first obtained by determining the complement of each digit. The complement of a decimal digit in the nines` complement system is the number that must be added to it to produce 9. The complement of 3 is 6, the complement of 7 is 2, and so on. Given a subtraction problem:
873 (x)
– 218 (y)
The nines` complement of y (218) is 781. In this case, because y is three digits long, this is the same as subtracting y from 999. (The number of 9`s is equal to the number of digits of y.)
Next, the sum of x, the nines` complement of y, and 1 is taken:
873 (x)
+ 781 (complement of y)
+ 1 (to get the ten`s complement of y)
=====
1655
The first “1” digit is then dropped, giving 655, the correct answer.
If the subtrahend has fewer digits than the minuend, leading zeros must be added which will become leading nines when the nines` complement is taken. For example:
48032 (x)
– 391 (y)
becomes the sum:
48032 (x)
+ 99608 (nines` complement of y)
+ 1 (to get the ten`s complement)
=======
147641
Dropping the “1” gives the answer: 47641
· Binary example
The method of complements is especially useful in binary (radix 2) since the ones` complement is very easily obtained by inverting each bit (changing `0` to `1` and vice versa). And adding 1 to get the two`s complement can be done by simulating a carry into the least significant bit. For example:
01100100 (x, equals decimal 100)
– 00010110 (y, equals decimal 22)
becomes the sum:
01100100 (x)
+ 11101001 (ones` complement of y)
+ 1 (to get the two`s complement)
==========
101001110
Dropping the initial “1” gives the answer: 01001110 (equals decimal 78)
· Signed fixed point numbers
Up to this point we have considered only the representation of unsigned fixed point numbers. The situation is quite different in representing signed fixed point numbers. There are four different ways of representing signed numbers that are commonly used: sign-magnitude, one`s complement, two`s complement, and excess notation. We will cover each in turn, using integers for our examples. The Table below shows for a 3-bit number how the various representations appear.
|
Decimal |
Unsigned |
Sign-Mag |
1’s comp |
2’s comp |
Excess 4 |
|
7 |
111 |
– |
– |
– |
– |
|
6 |
110 |
– |
– |
– |
– |
|
5 |
101 |
– |
– |
– |
– |
|
4 |
100 |
– |
– |
– |
– |
|
3 |
011 |
011 |
011 |
011 |
111 |
|
2 |
010 |
010 |
010 |
010 |
110 |
|
1 |
001 |
001 |
001 |
001 |
101 |
|
+0 |
000 |
000 |
000 |
000 |
100 |
|
-0 |
– |
100 |
111 |
000 |
100 |
|
-1 |
– |
101 |
110 |
111 |
011 |
|
-2 |
– |
110 |
101 |
110 |
010 |
|
-3 |
– |
111 |
100 |
101 |
001 |
|
-4 |
– |
– |
– |
100 |
000 |
Table1. 3 bit number representation
· Signed Magnitude Representation
The signed magnitude (also referred to as sign and magnitude) representation is most familiar to us as the base 10 number system. A plus or minus sign to the left of a number indicates whether the number is positive or negative as in +1210 or -1210. In the binary signed magnitude representation, the leftmost bit is used for the sign, which takes on a value of 0 or 1 for `+` or `-`, respectively. The remaining bits contain the absolute magnitude.
Consider representing (+12)10 and (-12)10 in an eight-bit format:
(+12)10 = (00001100)2
(-12)10 = (10001100)2
The negative number is formed by simply changing the sign bit in the positive number from 0 to 1. Notice that there are both positive and negative representations for zero: +0= 00000000 and -0= 10000000.
· One`s Complement Representation
The one`s complement operation is trivial to perform: convert all of the 1`s in the number to 0`s, and all of the 0`s to 1`s. See the fourth column in Table1 for examples. We can observe from the table that in the one`s complement representation the leftmost bit is 0 for positive numbers and 1 for negative numbers, as it is for the signed magnitude representation. This negation, changing 1`s to 0`s and changing 0`s to 1`s, is known as complementing the bits. Consider again representing (+12)10 and (-12)10 in an eight-bit format, now using the one`s complement representation:
(+12)10 = (00001100)2
(-12)10 = (11110011)2
Note again that there are representations for both +0 and -0, which are 00000000 and 11111111, respectively. As a result, there are only 28 – 1 = 255 different numbers that can be represented even though there are 28different bit patterns.
The one`s complement representation is not commonly used. This is at least partly due to the difficulty in making comparisons when there are two representations for 0. There is also additional complexity involved in adding numbers.
· Two`s Complement Representation
The two`s complement is formed in a way similar to forming the one`s complement: complement all of the bits in the number, but then add 1, and if that addition results in a carry-out from the most significant bit of the number, discard the carry-out.
Examination of the fifth column of Table above shows that in the two`s complement representation, the leftmost bit is again 0 for positive numbers and is 1 for negative numbers. However, this number format does not have the unfortunate characteristic of signed-magnitude and one`s complement representations: it has only one representation for zero. To see that this is true, consider forming the negative of (+0)10, which has the bit pattern: (+0)10 = (00000000)2
Forming the one`s complement of (00000000)2 produces (11111111)2 and adding 1 to it yields (00000000)2, thus (-0)10= (00000000)2. The carry out of the leftmost position is discarded in two`s complement addition (except when detecting an overflow condition). Since there is only one representation for 0, and since all bit patterns are valid, there are 28 = 256 different numbers that can be represented.
Consider again representing (+12)10 and (-12)10 in an eight-bit format, this time using the two`s complement representation. Starting with (+12)10 =(00001100)2, complement, or negate the number, producing (11110011)2.
Now add 1, producing (11110100)(BIN), and thus (-12)10 = (11110100)2 :
(+12)10 = (00001100)2
(-12)10 = (11110100)2
There is an equal number of positive and negative numbers provided zero is considered to be a positive number, which is reasonable because its sign bit is 0. The positive numbers start at 0, but the negative numbers start at -1, and so the magnitude of the most negative number is one greater than the magnitude of the most positive number. The positive number with the largest magnitude is +127, and the negative number with the largest magnitude is -128. There is thus no positive number that can be represented that corresponds to the negative of -128. If we try to form the two`s complement negative of -128, then we will arrive at a negative number, as shown below:
(-128)10 = (10000000)2
(-128)10 = (01111111)2
(-128)10 + (+0000001)2
(-128)10 ——)2
(-128)10 = (10000000)2
The two`s complement representation is the representation most commonly used in conventional computers.
· Excess Representation
In the excess or biased representation, the number is treated as unsigned, but is “shifted” in value by subtracting the bias from it. The concept is to assign the smallest numerical bit pattern, all zeros, to the negative of the bias, and assign the remaining numbers in sequence as the bit patterns increase in magnitude. A convenient way to think of an excess representation is that a number is represented as the sum of its two`s complement form and another number, which is known as the “excess” or “bias” Once again, refer to Table 2.1, the rightmost column, for examples.
Consider again representing (+12)10 and (-12)10 in an eight-bit format but now using an excess 128 representation. An excess 128 number is formed by adding 128 to the original number, and then creating the unsigned binary version. For (+12)10, we compute (128 + 12 = 140)10 and produce the bit pattern (10001100)2. For (-12)10, we compute (128 + -12 = 116)10 and produce the bit pattern (01110100)2
(+12)10 = (10001100)2
(-12)10 = (01110100)2
Note that there is no numerical significance to the excess value: it simply has the effect of shifting the representation of the two`s complement numbers.
There is only one excess representation for 0, since the excess representation is simply a shifted version of the two`s complement representation. For the previous case, the excess value is chosen to have the same bit pattern as the largest negative number, which has the effect of making the numbers appear in numerically sorted order if the numbers are viewed in an unsigned binary representation.
Thus, the most negative number is (-128)10 = (00000000)2 and the most positive number is (+127)10 = (11111111)2. This representation simplifies making comparisons between numbers, since the bit patterns for negative numbers have numerically smaller values than the bit patterns for positive numbers. This is important for representing the exponents of floating point numbers, in which exponents of two numbers are compared in order to make them equal for addition and subtraction.
· Choosing a bias:
The bias chosen is most often based on the number of bits (n) available for representing an integer. To get an approximate equal distribution of true values above and below 0, the bias should be 2(n-1) or 2((n-1)-1)
2.4 Floating point representation
Floating point is a numerical representation system in which a string of digits represent a real number. The name floating point refers to the fact that the radix point (decimal point or more commonly in computers, binary point) can be placed anywhere relative to the digits within the string. A fixed point is of the form a * bn where a is the fixed point part often referred to as the mantissa, or significand of the number b represents the base and n the exponent. Thus a floating point number can be characterized by a triple of numbers: sign, exponent, and significand.
· Normalization
A potential problem with representing floating point numbers is that the same number can be represented in different ways, which makes comparisons and arithmetic operations difficult. For example, consider the numerically equivalent forms shown below:
3584.1 * 100 = 3.5841 * 103 = .35841 * 104.
In order to avoid multiple representations for the same number, floating point numbers are maintained in normalized form. That is, the radix point is shifted to the left or to the right and the exponent is adjusted accordingly until the radix point is to the left of the leftmost nonzero digit. So the rightmost number above is the normalized one. Unfortunately, the number zero cannot be represented in this scheme, so to represent zero an exception is made. The exception to this rule is that zero is represented as all 0`s in the mantissa.
If the mantissa is represented as a binary, that is, base 2, number, and if the normalization condition is that there is a leading “1” in the normalized mantissa, then there is no need to store that “0” and in fact, most floating point formats do not store it. Rather, it is “chopped off” before packing up the number for storage, and it is restored when unpacking the number into exponent and mantissa. This results in having an additional bit of precision on the right of the number, due to removing the bit on the left. This missing bit is referred to as the hidden bit, also known as a hidden 1.
For example, if the mantissa in a given format is 1.1010 after normalization, then the bit pattern that is stored is 1010- the left-most bit is truncated, or hidden.
2.5 Possible floating point format
In order to choose a possible floating point format for a given computer, the programmer must take into consideration the following:
The number of words used (i.e. the total number of bits used.)
The representation of the mantissa (2s complement etc.)
The representation of the exponent (biased etc.)
The total number of bits devoted for the mantissa and the exponent
The location of the mantissa (exponent first or mantissa first)
Because of the five points above, the number of ways in which a floating point number may be represented is legion. Many representation use the format of sign bit to represent a floating point where the leading bit represents the sign
|
Sign |
Exponent |
Mantissa |
· The IEEE standard for floating point
The IEEE (Institute of Electrical and Electronics Engineers) has produced a standard for floating point format arithmetic in mini and micro computers. (i.e. ANSI/IEEE 754-1985). This standard specifies how single precision (32 bit) , double precision (64 bit) and Quadruple (128 bit) floating point numbers are to be represented, as well as how arithmetic should be carried out on them.
-General layout
The three fields in an IEEE 754 float
|
Sign |
Exponent |
Fraction |
Binary floating-point numbers are stored in a sign-magnitude form where the most significant bit is the sign bit, exponent is the biased exponent, and “fraction” is the significant without the most significant bit.
-Exponent biasing
The exponent is biased by (2^(e-1)) – 1, where e is the number of bits used for the exponent field (e.g. if e=8, then (2^(8 – 1) – 1 = 128 – 1 = 127 ). Biasing is done because exponents have to be signed values in order to be able to represent both tiny and huge values, but two`s complement, the usual representation for signed values, would make comparison harder. To solve this, the exponent is biased before being stored by adjusting its value to put it within an unsigned range suitable for comparison.
For example, to represent a number which has exponent of 17 in an exponent field 8 bits wide:
exponentfield = 17 + (2^(8 – 1) – 1 )= 17 + 128 – 1 = 144.
-Single Precision
The IEEE single precision floating point standard representation requires a 32 bit word, which may be represented as numbered from 0 to 31, left to right. The first bit is the sign bit, S, the next eight bits are the exponent bits, `E`, and the final 23 bits are the fraction `F`:
|
S EEEEEEEE FFFFFFFFFFFFFFFFFFFFFFF |
|
0 1 8 9 31 |
To convert decimal 17.15 to IEEE Single format:
Convert decimal 17 to binary 10001. Convert decimal 0.15 to the repeating binary fraction 0.001001 Combine integer and fraction to obtain binary 10001.001001 Normalize the binary number to obtain 1.0001001001×24
Thus, M = m-1 =0001001001 and E = e+127 = 131 = 10000011.
The number is positive, so S=0. Align the values for M, E, and S in the correct fields.
0 10000011 00010010011001100110011
Note that if the exponent does not use all the field allocated to it, there will be leading 0`s while for the mantissa, the zero`s will be filled at the end.
-Double Precision
The IEEE double precision floating point standard representation requires a 64 bit word, which may be represented as numbered from 0 to 63, left to right. The first bit is the sign bit, S, the next eleven bits are the exponent bits, `E`, and the final 52 bits are the fraction `F`:
|
S EEEEEEEEEEE FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF |
|
0 1 11 12 |
-Quad Precision
The IEEE Quad precision floating point standard representation requires a 128 bit word, which may be represented as numbered from 0 to 127, left to right. The first bit is the sign bit, S, the next fifteen bits are the exponent bits, `E`, and the final 128 bits are the fraction `F`:
|
S EEEEEEEEEEEEEEE FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF |
|
0 1 15 16 |
|
|
Single |
Double |
Quadruple |
|
No. of sign bit |
1 |
1 |
1 |
|
No. of exponent bit |
8 |
11 |
15 |
|
No. of fraction |
23 |
52 |
111 |
|
Total bits used |
32 |
64 |
128 |
|
Bias |
127 |
1023 |
16383 |
Table2 Basic IEEE floating point format
2.6 Binary code
Internally, digital computers operate on binary numbers. When interfacing to humans, digital processors, e.g. pocket calculators, communication is decimal-based. Input is done in decimal then converted to binary for internal processing. For output, the result has to be converted from its internal binary representation to a decimal form. Digital system represents and manipulates not only binary number but also many other discrete elements of information.
· Binary coded Decimal
In computing and electronic systems, binary-coded decimal (BCD) is an encoding for decimal numbers in which each digit is represented by its own binary sequence. Its main virtue is that it allows easy conversion to decimal digits for printing or display and faster decimal calculations. Its drawbacks are the increased complexity of circuits needed to implement mathematical operations and a relatively inefficient encoding. It occupies more space than a pure binary representation. In BCD, a digit is usually represented by four bits which, in general, represent the values/digits/characters 0-9
To BCD-encode a decimal number using the common encoding, each decimal digit is stored in a four-bit nibble.
|
Decimal |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
BCD |
0000 |
0001 |
0010 |
0011 |
0100 |
0101 |
0110 |
0111 |
1000 |
1001 |
Thus, the BCD encoding for the number 127 would be:
0001 0010 0111
The position weights of the BCD code are 8, 4, 2, 1. Other codes (shown in the table) use position weights of 8, 4, -2, -1 and 2, 4, 2, 1.
An example of a non-weighted code is the excess-3 code where digit codes is obtained from their binary equivalent after adding 3. Thus the code of a decimal 0 is 0011, that of 6 is 1001, etc
|
Decimal Digit |
8 4 2 1 Code |
8 4 -2 -1 Code |
2 4 2 1 Code |
Excess-3 Code |
|
0 |
0 0 0 0 |
0 0 0 0 |
0 0 0 0 |
0 0 1 1 |
|
1 |
0 0 0 1 |
0 1 1 1 |
0 0 0 1 |
0 1 0 0 |
|
2 |
0 0 1 0 |
0 1 1 0 |
0 0 1 0 |
0 1 0 1 |
|
3 |
0 0 1 1 |
0 1 0 1 |
0 0 1 1 |
0 1 1 0 |
|
4 |
0 1 0 0 |
0 1 0 0 |
0 1 0 0 |
0 1 1 1 |
|
5 |
0 1 0 1 |
1 0 1 1 |
1 0 1 1 |
1 0 0 0 |
|
6 |
0 1 1 0 |
1 0 1 0 |
1 1 0 0 |
1 0 0 1 |
|
7 |
0 1 1 1 |
1 0 0 1 |
1 1 0 1 |
1 0 1 0 |
|
8 |
1 0 0 0 |
1 0 0 0 |
1 1 1 0 |
1 0 1 1 |
|
9 |
1 0 0 1 |
1 1 1 1 |
1 1 1 1 |
1 1 0 0 |
it is very important to understand the difference between the conversion of a decimal number to binary and the binary coding of a decimal number. In each case, the final result is a series of bits. The bits obtained from conversion are binary digit. Bits obtained from coding are combinations of 1`s and 0`s arranged according to the rule of the code used. e.g. the binary conversion of 13 is 1101; the BCD coding of 13 is 00010011
· Error-Detection Codes
Binary information may be transmitted through some communication medium, e.g. using wires or wireless media. A corrupted bit will have its value changed from 0 to 1 or vice versa. To be able to detect errors at the receiver end, the sender sends an extra bit (parity bit) with the original binary message.
A parity bit is an extra bit included with the n-bit binary message to make the total number of 1`s in this message (including the parity bit) either odd or even. If the parity bit makes the total number of 1`s an odd (even) number, it is called odd (even) parity. The table shows the required odd (even) parity for a 3-bit message
|
Three-Bit Message |
Odd Parity Bit |
Even Parity Bit |
||
|
X |
Y |
Z |
P |
P |
|
0 |
0 |
0 |
1 |
0 |
|
0 |
0 |
1 |
0 |
1 |
|
0 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
1 |
0 |
|
1 |
0 |
0 |
0 |
1 |
|
1 |
0 |
1 |
1 |
0 |
|
1 |
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
1 |
No error is detectable if the transmitted message has 2 bits in error since the total number of 1`s will remain even (or odd) as in the original message.
In general, a transmitted message with even number of errors cannot be detected by the parity bit.
– Gray code
The Gray code consist of 16 4-bit code words to represent the decimal Numbers 0 to 15. For Gray code, successive code words differ by only one bit from one to the next
|
Gray Code |
Decimal Equivalent |
|
0 0 0 0 |
0 |
|
0 0 0 1 |
1 |
|
0 0 1 1 |
2 |
|
0 0 1 0 |
3 |
|
0 1 1 0 |
4 |
|
0 1 1 1 |
5 |
|
0 1 0 1 |
6 |
|
0 1 0 0 |
7 |
|
1 1 0 0 |
8 |
|
1 1 0 1 |
9 |
|
1 1 1 1 |
10 |
|
1 1 1 0 |
11 |
|
1 0 1 0 |
12 |
|
1 0 1 1 |
13 |
|
1 0 0 1 |
14 |
|
1 0 0 0 |
15 |
2.7 Character Representation
Even though many people used to think of computers as “number crunchers”, people figured out long ago that it`s just as important to handle character data.
Character data isn`t just alphabetic characters, but also numeric characters, punctuation, spaces, etc. Most keys on the central part of the keyboard (except shift, caps lock) are characters. Characters need to be represented. In particular, they need to be represented in binary. After all, computers store and manipulate 0`s and 1`s (and even those 0`s and 1`s are just abstractions. The implementation is typically voltages).
Unsigned binary and two`s complement are used to represent unsigned and signed integer respectively, because they have nice mathematical properties, in particular, you can add and subtract as you`d expect.
However, there aren`t such properties for character data, so assigning binary codes for characters is somewhat arbitrary. The most common character representation is ASCII, which stands for American Standard Code for Information Interchange.
There are two reasons to use ASCII. First, we need some way to represent characters as binary numbers (or, equivalently, as bit-string patterns). There`s not much choice about this since computers represent everything in binary.
If you`ve noticed a common theme, it`s that we need representation schemes for everything. However, most importantly, we need representations for numbers and characters. Once you have that (and perhaps pointers), you can build up everything you need.
The other reason we use ASCII is because of the letter “S” in ASCII, which stands for “standard”. Standards are good because they allow for common formats that everyone can agree on.
Unfortunately, there`s also the letter “A”, which stands for American. ASCII is clearly biased for the English language character set. Other languages may have their own character set, even though English dominates most of the computing world (at least, programming and software).
Even though character sets don`t have mathematical properties, there are some nice aspects about ASCII. In particular, the lowercase letters are contiguous (`a` through `z` maps to 97(DEC) through 122(DEC)). The upper case letters are also contiguous (`A` through `Z` maps to 65(DEC) through 90(DEC)). Finally, the digits are contiguous (`0` through `9` maps to 48(DEC) through 57(DEC)).
Since they are contiguous, it`s usually easy to determine whether a character is lowercase or uppercase (by checking if the ASCII code lies in the range of lower or uppercase ASCII codes), or to determine if it`s a digit, or to convert a digit in ASCII to an integer value.
The characters between 0 and 31 are generally not printable (control characters, etc). 32 is the space character. Also note that there are only 128 ASCII characters. This means only 7 bits are required to represent an ASCII character. However, since the smallest size representation on most computers is a byte, a byte is used to store an ASCII character. The Most Significant bit(MSB) of an ASCII character is 0.
The difference in the ASCII code between an uppercase letter and its corresponding lowercase letter is 20(HEX). This makes it easy to convert lower to uppercase (and back) in hex (or binary).
· Other Character Codes
While ASCII is still popularly used, another character representation that was used (especially at IBM) was EBCDIC, which stands for Extended Binary Coded Decimal Interchange Code (yes, the word “code” appears twice). This character set has mostly disappeared. EBCDIC does not store characters contiguously, so this can create problems alphabetizing “words”.
One problem with ASCII is that it`s biased to the English language. This generally creates some problems. One common solution is for people in other countries to write programs in ASCII.
Other countries have used different solutions, in particular, using 8 bits to represent their alphabets, giving up to 256 letters, which is plenty for most alphabet based languages (recall you also need to represent digits, punctuation, etc).
However, Asian languages, which are word-based, rather than character-based, often have more words than 8 bits can represent. In particular, 8 bits can only represent 256 words, which is far smaller than the number of words in natural languages.
Thus, a new character set called Unicode is now becoming more prevalent. This is a 16 bit code, which allows for about 65,000 different representations. This is enough to encode the popular Asian languages (Chinese, Korean, Japanese, etc.). It also turns out that ASCII codes are preserved. What does this mean? To convert ASCII to Unicode, take all one byte ASCII codes, and zero-extend them to 16 bits. That should be the Unicode version of the ASCII characters.
The biggest consequence of using Unicode from ASCII is that text files double in size. The second consequence is that endianness begins to matter again. Endianness is the ordering of individually addressable sub-units (words, bytes, or even bits) within a longer data word stored in external memory. The most typical cases are the ordering of bytes within a 16-, 32-, or 64-bit word, where endianness is often simply referred to as byte order. The usual contrast is between most versus least significant byte first, called big-endian and little-endian respectively. Big-endian places the most significant bit, digit, or byte in the first, or leftmost, position. Little-endian places the most significant bit, digit, or byte in the last, or rightmost, position. Motorola processors employ the big-endian approach, whereas Intel processors take the little-endian approach. Table below illustrates how the decimal value 47,572 would be expressed in hexadecimal and binary notation (two octets) and how it would be stored using these two methods.
|
Number Hexadecimal |
Big-Endian |
Little-Endian |
|
B9D4 |
B9D4 |
4D9B |
|
Binary |
||
|
10111001 |
10111001 |
11010100 |
|
11010100 |
11010100 |
10111001 |
With single bytes, there`s no need to worry about endianness. However, you have to consider that with two byte quantities.
While C and C++ still primarily use ASCII, Java has already used Unicode. This means that Java must create a byte type, because char in Java is no longer a single byte. Instead, it`s a 2 byte Unicode representation.
"ASPIRANTS" 