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Data Representation and Number system,Numeric 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. 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.
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,A F Kana Digital Logic Design Page 2. Quotient Remainder,1234 10 123 4,123 10 12 3,1 10 0 1 Stop when the quotient is 0.
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 1,1 2 0 1 Stop when the,quotient is 0,1 0 1 0 0 1 1 1 1 0 1 BIN Base 2. Let s express the same decimal number 1341 in octal notation. Quotient Remainder,1341 8 167 5,167 8 20 7,2 8 0 2 Stop when the quotient is 0. 2 4 7 5 OCT Base 8, Let s express the same decimal number 1341 in hexadecimal notation.
Quotient Remainder,1341 16 83 13,5 16 0 5 Stop when the quotient is 0. 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,A F Kana Digital Logic Design Page 3. Quotient Remainder,3315 16 207 3,207 16 12 15,12 16 0 12 Stop when the quotient is 0. C F 3 HEX Base 16,HEX Base 16,Product Integer Part 0 4 C C C.
0 3 16 4 8 4,0 8 16 12 8 12,0 8 16 12 8 12,0 8 16 12 8 12. 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. 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, In summary the conversion from any base to base 10 can be obtained from the formulae.
Where b is the base di the digit at position i m the number of digit after the decimal point n the number. of digits of the integer part and X10 is the obtained number in decimal This form the basic of the polynomial method of. converting numbers from any base to decimal, Example Convert 234 14 expressed in an octal notation to decimal. 2 82 3 81 4 80 1 8 1 4 8 2 2 64 3 8 4 1 1 8 4 64 156 1875. A F Kana Digital Logic Design Page 4, Example Convert the hexadecimal number 4B3 to decimal notation What about the decimal equivalent of the. hexadecimal number 4B3 3,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. 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 bellow 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. A F Kana Digital Logic Design Page 5,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. 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 bn 1 y. Since bn 1 is the digit b 1 repeated n times because bn 1 bn 1n b 1 bn 1 bn 2 b 1 b 1 bn. 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. 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. 781 complement of y,1 to get the ten s complement of y. 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,A F Kana Digital Logic Design Page 6. becomes the sum,99608 nines complement of y,1 to get the ten s complement.
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, 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. Decimal Unsigned Sign Mag 1 s Comp 2 s Comp Excess 4. 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,Table1 3 bit number representation. A F Kana Digital Logic Design Page 7,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 28 different bit. 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. 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. A F Kana Digital Logic Design Page 8, Now add 1 producing 11110100 2 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 01111111,128 10 0000001 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. A F Kana Digital Logic Design Page 9,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 1 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,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,A F Kana Digital Logic Design Page 10. 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 significand without the most significant bit. Exponent biasing, The exponent is biased by 2e 1 1 where e is the number of bits used for the exponent field e g if e 8 then 28 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 28 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. 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 0001001001x24. 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. A F Kana Digital Logic Design Page 11,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. 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. 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. 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. A F Kana Digital Logic Design Page 12,8421 8 4 2 1 2421 Excess 3. Code code code code,0 0000 0000 0000 0011,1 0001 0111 0001 0100. 2 0010 0110 0010 0101,3 0011 0101 0011 0110,4 0100 0100 0100 0111.
5 0101 1011 1011 1000,6 0110 1010 1100 1001,7 0111 1001 1101 1010. 8 1000 1000 1110 1011,9 1001 1111 1111 1100, 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, 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.
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. A F Kana Digital Logic Design Page 13,Gray Code Decimal Equivalent. 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 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 bitstring 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 9710 through 12210 The upper case letters are also. contiguous A through Z maps to 6510 through 9010 Finally the digits are contiguous 0 through 9 maps to 4810. through 5710,A F Kana Digital Logic Design Page 14. 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,ASCII Code Decimal,0 nul 16 dle 32 sp 48 0 64 80 P 96 112 p. 1 soh 17 dc1 33 49 1 65 A 81 Q 97 a 113 q,2 stx 18 dc2 34 50 2 66 B 82 R 98 b 114 r. 3 etx 19 dc3 35 51 3 67 C 83 S 99 c 115 s,4 eot 20 dc4 36 52 4 68 D 84 T 100 d 116 t.
5 enq 21 nak 37 53 5 69 E 85 U 101 e 117 u,6 ack 22 syn 38 54 6 70 F 86 V 102 f 118 v. 7 bel 23 etb 39 55 7 71 G 87 W 103 g 119 w,8 bs 24 can 40 56 8 72 H 88 X 104 h 120 x. 9 ht 25 em 41 57 9 73 I 89 Y 105 i 121 y,10 nl 26 sub 42 58 74 J 90 Z 106 j 122 z. 11 vt 27 esc 43 59 75 K 91 107 k 123,12 np 28 fs 44 60 76 L 92 108 l 124. 13 cr 29 gs 45 61 77 M 93 109 m 125,14 so 30 rs 46 62 78 N 94 110 n 126.
15 si 31 us 47 63 79 O 95 111 o 127 del, 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. ASCII Code Hex,00 nul 10 dle 20 sp 30 0 40 50 P 60 70 p. 01 soh 11 dc1 21 31 1 41 A 51 Q 61 a 71 q,02 stx 12 dc2 22 32 2 42 B 52 R 62 b 72 r. 03 etx 13 dc3 23 33 3 43 C 53 S 63 c 73 s,04 eot 14 dc4 24 34 4 44 D 54 T 64 d 74 t. 05 enq 15 nak 25 35 5 45 E 55 U 65 e 75 u,06 ack 16 syn 26 36 6 46 F 56 V 66 f 76 v.
07 bel 17 etb 27 37 7 47 G 57 W 67 g 77 w,08 bs 18 can 28 38 8 48 H 58 X 68 h 78 x. 09 ht 19 em 29 39 9 49 I 59 Y 69 i 79 y,0a nl 1a sub 2a 3a 4a J 5a Z 6a j 7a z. 0b vt 1b esc 2b 3b 4b K 5b 6b k 7b,0c np 1c fs 2c 3c 4c L 5c 6c l 7c. 0d cr 1d gs 2d 3d 4d M 5d 6d m 7d,0e so 1e rs 2e 3e 4e N 5e 6e n 7e. 0f si 1f us 2f 3f 4f O 5f 6f o 7f del, The difference in the ASCII code between an uppercase letter and its corresponding lowercase letter is 2016 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,A F Kana Digital Logic Design Page 15.

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Tanaman Penutup Tanah Kacang-kacangan dan Mulsa Jerami Sumarni, N., E. Sumiati, dan R. Rosliani Balai Penelitian Tanaman Sayuran, Jl. Tangkuban Parahu 517, Lembang, Bandung 40391 Naskah diterima tanggal 8 Mei 2008 dan disetujui untuk diterbitkan tanggal 31 Oktober 2008 ABSTRAK. Penelitian dilakukan di Kebun Percobaan Balai Penelitian Tanaman Sayuran Lembang, dari bulan Juli-Oktober 2004 ...



Penelitian ini bertujuan untuk mengetahui konsentrasi toleran larutan atonik dan PEG 6000 yang resisten terhadap cekaman kekeringan serta mengetahui interaksi antara larutan atonik dan PEG 6000 pada planlet kacang hijau secara in vitro. Penelitian ini dilaksanakan pada bulan November 2016 sampai Januari 2017 di Laboratorium Botani (ruang In Vitro), Jurusan Biologi, Fakultas Matematika dan Ilmu ...