1. 1. xii Prefacecircuits except in Chapters 11 and 12. The practical application of digital design will he enhancedby doing the suggested experiments in Chapter 11 while studying the theory presentedin the text.Each chapter has a list of refereuces and a set of problems. Answers to selected problemsappear iu at the end of the book to aid the student and to help the independent reader. A solutionsmanual is available for the instructor from the publisher.I would like to thauk Charles Kime for iutroducing me to Verilog. My greatest thanks go toJack Levine for guiding me arid checking the sections, examples, and problem solutions to allVerilog HDL materiaL Thanks go to Tom Robbins for helping me decide to write the third editionand my editor Eric Frauk for his patience throughout the revision. Appreciation goes to GaryCovington and Donna Mitchell for providing the CD-ROM from SynaptiCad. Thanks also tothose who reviewed the third edition: Thomas G. Johnson, California State University; UmitUyar, City University ofNew York; Thomas L. Drake, Clemson University; and Richard Molyet,University of Toledo. Finally, I am grateful to my wife Sandra for encouraging me to pursuethis project.M. MORRIS MANaBinary Systems1-1 DIGITAL SYSTEMSDigital systems have such a prominent role in everyday life that we refer to the present technologicalperiod as the digital age. Digital systems are used in communication, business transactions,traffic control, space guidance, medical treatment, weather monitoring, the Internet, andmany other commercial, industrial, and scientific enterprises. We have digital telephones, digitaltelevision, digital versatile discs, digital cameras, and of course, digital computers. Themost striking property of the digital computer is its generality. It can follow a sequence of instructions,called a program, that operates on given data. The user can specify and change theprogram or the data according to the specific need. Because of this flexibility, general-purposedigital computers can perform a variety of information processing tasks that range over a widespectrum of applications.One characteristic of digital systems is their ability to manipulate discrete elements of information.Any s'etthat is restricted to a finite number of elements contains discrete infonnation.Examples of discrete sets are the 10 decimal digits, the 26 letters of the alphabet, the 52playing cards, and the 64 squares of a chessboard. Early digital computers were used for numericcomputations. In this case, the discrete elements used were the digits. From this application,the term digital computer emerged. Discrete elements of information are representedin a digital system by physical quantities called signals. Electrical siguals such as voltages andcurrents are the most common. Electronic devices called transistors predominate in the circuitrythat implements these signals. The signals in most present-day electronic digital systemsuse just two discrete values and are therefore said to be binary. A binary digit, called abit, has two values: 0 and 1. Discrete elements of information are represented with groups ofbits called binary codes. For example, the decimal digits 0 through 9 are represented in a digitalsystem with a code of four bits. By p,sing various techniques, groups of bits can be made1
2. 2. 2 Chapter 1 Binary Systems Section 1-2 Binary Numbers 3to represent discrete symbols, which are then used to develop the system in a digital format.Thus, a digital system is a system that manipulates discrete elements of information that isrepresented internally in binary form.Discrete quantities of information either emerge from the nature of the data being processedor may be quantized from a continuous process. For example, a payroll schedule is an inherentlycliscrete process that contains employee names, social security numbers, weekly salaries,income taxes, and so on. An employee's paycheck is processed using discrete data values suchas letters of the alphabet (names), digits (salary), and special symbols (such as $). On the otherhand, a research scientist may observe a continuous process, but record only specific quantitiesin tabular form. The scientist is thus quantizing his continuous data, making each numberin his table a discrete quantity. In many cases, the quantization of a process can be performedautomatically by an analog-to-digital converter.The general-purpose digital computer is the best-known example of a digital system. Themajor parts of a computer are a memory unit, a central processing unit, and input-output units.The memory unit stores programs as well as input, output, and intermediate data. The centralprocessing unit perlonns arithmetic and other data processing operations as specified by the program.The program and data prepared by a user are transferred into memory by means of aninput device such as a keyboard~ An output device, such as a printer, receives the results of thecomputations and the printed results are presented to the user. A digital computer can accommodatemany input and output devices. One very useful device is a communication unit thatprovides interaction with other users through the Internet. A digital computer is a powerful instrumentandean penorm not only arithmetic computations, but also logical operations. In addition,it can be programmed to make decisions based on internal and external conditions.There are fundamental reasons why commercial products are made with digital circuits.Like a digital computer, most digital devices are programmable. By changing the program ina programmable device, the same underlying hardware can be' used for many different applications;Dramatic cost reductions in digital devices have come about because of the advancesin digital integrated circuit technology. As the number of transistors that can be put on a pieceof silicon increases to produce complex functions, the cost per unit decreases and digital devicescan be bought at an increasingly reduced price. Equipment built with digital integratedcircuits can perform at a speed of hundreds of millions of operations per second. Digital systemscan be made to operate with extreme reliability by using error-correcting codes. An exampleof this is the digital versatile disk (DVD) in which digital information representing video,audio, and other data is recorded without a loss of a single item. Digital information on a DVDis recorded in such a way that by examining the code in each digital sample before it is playedback, any error can be automatically identified and corrected.A digital system is an interconnection of digital modules. To understand the operation of eachdigital module, it is necessary to have a basic knowledge of digital circuits and their logical function.The first seven chapters of this book present the basic tools of digital design such as logicgate structures, combinational and sequential circuits, and programmable logic devices. Chapter8 introduces digital design at the register transfer level (RTL). Chapters 9 and 10 deal withasynchronous sequential circuits and the various integrated digital logic families. Chapters IIand 12 introduce commercial integrated circuits and show how they can be connected in thelaboratory to perform experiments with digital circuits.1-2An important trend in digital design is the use of hardware description language (HDL).HDL resembles a programming language and is suitable for describing digital circuits in textualform. It is used to simulate a digital system to verify its operation before hardware is builtin. It is also used in conjunction with logic synthesis tools to automate the design. HDL descriptionsof digital circuits are presented throughout the book.As previously stated, digital systems manipulate discrete quantities of information that arerepresented in binary form. Operands used for calculations may be expressed in the binarynumber system. Other discrete elements, including the decimal digits, are represented in binarycodes. Data processing is carried out by means of binary logic elements using binary signals.Quantities are stored in binary storage elements. The purpose of this chapter is to introduce thevarious binary concepts as a frame of reference for further study in the succeeding chapters.BINARY NUMBERSA decimal number such as 7,392 represents a quantity equal to 7 thousands plus 3 hundreds,plus 9 tens, plus 2 units. The thousands, hundreds, etc. are powers of 10 implied by the positionof the coefficients. To be more exact, 7,392 should be written as7 X 103 + 3 X 10' + 9 X 10' + 2 X 100However, the convention is to write only the coefficients and from their position deduce thenecessary powers of 10. In general, a number with a decimal point is represented by a seriesof coefficients as follows:The aj coefficients are any of the 10 digits (0, I, 2, ... ,9), and the subscript valuejgives theplace value and, hence, the power of 10 by which the coefficient must be multiplied. This canbe expressed as10'a, + lO'a, + 103a3 + lO'a, + lO'a, + 100aD + lO-l a_1 + lO-'a_, + 1O-3a_3The decimal number system is said to be of base, or radix, 10 because it uses 10 digits andthe coefficients are multiplied by powers of 10. The binary system is a different number system.The coefficients of the binary numbers system have only two possible values: 0 or I. Eachcoefficient aj is multiplied by 2J For example, the decimal equivalent of the binary numberIIOlO.11 is 26.75, as shown from the multiplication of the coefficients by powers of2:I X 24 + I X 23 + 0 X 2' + I X 21 + 0 X 2 + I X T 1 + I X T 2 = 26.75In general, a number expressed in a base-r system has coefficients multiplied by powers of r:all rll + all-I rll - 1 + ... + a2 . r 2 + al . r + ao + a~l . r-l + a-2 . r~2 + ... + a_m r-mThe coefficients aj range in value from ato r - 1. To distinguish between numbers of differentbases, we enclose the coefficients in parentheses and write a subscript equal to the base used(except sometimes for decimal numbers, where the content makes it obvious that it is decimal).An example of a base-5 number is
3. 3. Chapter 1 Binary Systems Section 1-3 Number Base Conversions 5r-allowable digits. Examples of addition, subtraction, and multiplication of two binary numbersare as follows:augend:addend:sum:The sum oftwo binary numbers is calculated by the same rules as in decimal, except thatfhe digits of fhe sum in any significant position can be only 0 or 1. Any carry obtained in a givensignificant position is used by the pair of digits one significant position higher. The subtractionis slightly more complicated. The rules are still the same as in decimal, except that the borrowin a given significant position adds 2 to a minuend digit. (A borrow in the decimal system adds10 to a minuend digit.) Multiplication is very simple. The multiplier digits are always 1 or O.Therefore, tbe partial products are equal either to the multiplicand or to O.1-3 NUMBER BASE CONVERSIONSproduct:The conversion of a number in base r to decimal is done by expanding the number in a powerseries and adding an the terms as shown previously. We now present a general procedure forthe reverse operation of converting a decimal number to a number in base r. If the number includesa radix point, it is necessary to separate the number into an integer part and a fractionpart, since each part must be converted differently. The conversion of a decimal integer to anumber in base r is done by dividing the number and all successive quotients by r and accumulatingfhe remainders. This procedure is best illustrated by example.Convert decimal 41 to binary. First, 41 is divided by 2 to give an integer quotient of 20 and aremainder of!. The quotient is again divided by 2 to give a new quotient and remainder. Thisprocess is continued until the integer quotient becomes o. The coefficients of the desired binarynumber are obtained from the remainders as follows:1011X 101101100001011110111multiplicand:multiplier:101101-100111000110minuend:subtrahend:difference:101101+1001111010100IntegerQuotient Remainder Coefficient41/2 = 20 + ,ao = 1 20/2 = 10 + 0 al = 010/2 = 5 + 0 Q2 = 05/2 = 2 + ,'2 a3 = 12/2 = 1 + 0 a4 = 01/2 = 0 + 12 as = 1Therefore, the answer is (41) 10 = (a,a4a3aZa,aO)2 = (101001)2(4021.2), = 4 X 53 + 0 X 52 + 2 X 51 + 1 X 5 + 2 X 5-' = (511.4)10The coefficient values for base 5 can be only 0, 1, 2, 3, and 4. The octal nnmber system is abase-8 system fhat has eight digits: 0, 1,2,3,4, 5, 6, 7. An example of an octal number is127.4. To determine its equivalent decimal value, we expand the number in a power series witha base of 8:(127.4)8 = 1 X 82 + 2 X 8' + 7 X 8 + 4 X 8-' = (87.5)10Note that the digits 8 and 9 cannot appear in an octal number.It is customary to borrow the needed r digits for the coefficients from the decimal systemwhen the base of the number is less than 10. The letters of the alphabet are used to supplementthe 10 decimal digits when the base of the number is greater than 10. For example, in the hexadecimal(base 16) number system, the first ten digits are borrowed from fhe decimal system.The letters A, B, C, D, E, and F are used for digits 10, 11, 12, 13, 14, and 15, respectively. Anexample of a hexadecimal number is(B65F)16 = 11 X 163 + 6 X 162 + 5 X 16' + 15 X 16 = (46,687)111As noted before, the digits in a binary number are called bits. Wheu a bit is equal to 0, it doesnot contribute to the sum during the conversion. Therefore, the conversion from binary to decimalcan be obtained by adding the numbers with powers of two corresponding to the bits thatare equal to 1. For example,(110101)2 = 32 + 16 + 4 + 1 = (53)10There are four 1'8 in the binary number. The corresponding decimal number is the sum of thefour powers of two numbers. The first 24 numbers obtained from 2 to the power of n are listedin Table 1-1. In computer work, 210 is referred to as K(kilo), 220 as M(mega), 230 as G(giga),and 240 as T(tera). Thus 4K = 212 = 4096 and 16M, = 224 = 16,777,216. Computer capacityis usually given in bytes. A byte is equal to eight bits and can accommodate one keyboardcharacter. A computer hard disk with 4 gigabytes of storage has a capacity of 4G = 232 bytes(approximately 4 billion bytes).Arithmetic operations with numbers in base r follow the same rules as for decimal numbers.When a base other than the familiar base lOis used, one must be careful to use only theTable 1-1Powers of Twon 2" n 2" n 2"0 1 8 256 16 65,5361 2 9 512 17 131,0722 4 to 1,024 18 262,1443 8 11 2,048 19 524,2884 16 12 4,096 20 1,048,5765 32 13 8,192 21 2,097,1526 64 14 16,384 22 4,194,3047 128 15 32,768 23 8,388,608
4. 4. Chapter 1 Binary Systems Section 1-4 Octal arid Hexadecimal Numbers 7The conversion from decimal integers to any base-r system is similar to the example, exceptthat division is done by r instead of 2.Convert decimal 153 to octal. The required base r is 8. First, 153 is divided by 8 to give an integerquotient of 19 and a remainder of I. Then 19 is divided by 8 to give an integer quotientof 2 and a remainder of 3. Finally, 2 is divided by 8 to give a quotient of 0 and a remainder of2. This process can be conveniently manipulated as follows:To convert a decimal fraction to a number expressed in base r, a similar procedure is used.Multiplication is by r iustead of 2, and the coefficients found from the integers may range invalue from 0 to r - 1 instead of 0 and 1.The answer, to seven significant figures, is obtained from the integer part of the products(41.6875)10 = (101001.1011)21-4 OCTAL AND HEXADECIMAL NUMBERSIII 100 000 110h = (26153.7406)87 4 0 6From Examples 1-2 and 1-4, we have101 0115 3Convert (0.513)10 to octal.( 10 110 0012 6 10.513 X 8 = 4.1040.104 X 8 = 0.8320.832 X 8 = 6.6560.656 X 8 = 5.2480.248 X 8 = 1.9840.984 X 8 = 7.872(0.513) 10 = (0.406517 ... )8The conversion of decimal numbers with both integer and fraction parts is done by convertingthe integer and the fraction separately and then combining the two answers. Using theresults of Examples 1-1 and 1-3, we obtain(153.513)10 = (231.406517)8The conversion from and to binary, octal, aud hexadecimal plays an important role in digitalcomputers. Since 23 = 8 and 24 = 16, each octal digit corresponds to three binary digits andeach hexadecimal digit corresponds to four binary digits. The fIrst 16 numbers in the decimal,binary, octal, and hexadecimal number systems are listed in Table 1-2.The conversion from binary to octal is easily accomplished by partitioning the binary numberinto groups of three digits each, starting from the binary point and proceeding to the leftand to the right. The corresponding octal digit is then assigned to each group. The followingexample illustrates the procedure:The arithmetic process can be manipulated more conveniently as follows:Integer Remainder4120 I10 05 02 II 0o 1 101001 = answer1~ L= (231)8153192oThe conversion of a decimal fraction to binary is accomplished by a method similar to thatused for integers. However, multiplication is used instead of division, and integers are accumulatedinstead of remainders. Again, the method is best explained by example.Convert (0.6875)10 to binary. First, 0.6875 is multiplied by 2 to give aninteger and a fraction.The new fraction is multiplied by 2 to give a new integer and a new fraction. This process iscontinued until the fraction becomes 0 or until the number of digits have sufficient accuracy.The coefficients of the binary number are obtained from the integers as follows:Integer Fraction Coefficient0.6875 X 2 = 1 + 0.3750 a_1 = 10.3750 X 2 = 0 + 0.7500 a-2 = 00.7500 X 2 = I + 0.5000 a-3 = 10.5000 X 2 = 1 + 0.0000 a_4 = ITherefore, the answer is (0.6875)10 = (O.a -1 a _2a -3a -4)z = (0.1011)2
5. 5. 8 Chapter 1 Binary SystemsTable 1-2Numbers with Different BasesDecimal Binary Odal Hexadecimal(base 10) (base 2) (base 8) (base 16)00 0000 00 001 0001 01 102 0010 02 203 0011 03 304 0100 04 405 0101 05 506 0110 06 607 0111 07 708 1000 10 809 1001 11 910 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E15 1111 17 FConversion from binary to hexadecimal is similar, except that the binary number is divided(into groups offour digits:The corresponding hexadecimal (or octal) digit for each group of binary digits is easily rememberedafter studying the values listed in Table 1-2.Conversion from octal or hexadecimal to binary is doneby reversing the preceding procedure.Each octal digit is converted to its three-digit binary equivalent. Similarly, each hexadecimaldigit is converted to its four-digit binary equivalent. This is illustrated in the following examples:(673.124)8 ~ (110 III 011 001 010 100 ),6 7 3 I 2 4and(306.D h6 = ( 0011 0000 0110 1101 ),3 0 6 DBinary numbers are difficult to work with because they require three or four times as manydigits as their decimal equivalent. For example, the binary number Illlllllllll is equivalentto decimal 4095. However, digital computers use binary numbers and it is sometimes necessaryfor the human operator or user to communicate directly with the machine by means ofbinary numbers. One scheme that retains the binary system in the computer, but reduces thenumber of digits the human must consider, utilizes the relationship between the binary numbersystem and the octal or hexadecimal system. By this method, the human thinks in terms ofoctal or hexadecimal numbers and performs the required conversion by inspection when directcommunication with the machine is neCessary. Thus the binary number 1lll1l1llll1 has 12digits and is expressed in octal as 7777 (four digits) or in hexadecimal as FFF (three digits).During communication between people (about binary numbers in the computer), the octal orhexadecimal representation is more desirable because it can be expressed more compactly witha third or a quarter of the number of digits required for the equivalent binary number. Thus, mostcomputer manuals use either octal or hexadecimal numbers to specify binary quantities. Thechoice between them is arbitrary, although hexadecimal tends to win out, since it can representa byte with two digits.1-5 COM Pl EMEN TSComplements are used in digital computers for simplifying the subtraction operation and forlogical manipulation. There are two types of complements for each base-r system: the radixcomplement and the diminished radix complement. The first is referred to as the r's complementand the second as the (r - I)'s complement. When the value of the base r is substitutedin the name, the two types are referred to as the 2's complement and l's complement forbinary numbers, and the 10's complement and 9's complement for decimal numbers.Diminished Radix ComplementGiven a number N in base r having n digits, the (r - I)'s complement ofN is defined as(r" - I) - N. For decimal numbers, r = 10 and r - I = 9, so the 9's complement of N is(10" - I) - N. In this case, 10" represents a number that consists of a single I followed bynOs. 10" - I is a number represented by n 9s. For example, if n ~ 4, we have 104 = 10,000and 104- I = 9999. It follows that the 9's complement of a decimal number is obtained bysubtracting each digit from 9. Some numerical examples follow:The 9's complement of 546700 is 999999 - 546700 = 453299.The 9's complement of 012398 is 999999 - 012398 = 987601.Forbinarynumbers,r = 2andr -1 = I, sathe I's complementofN is (2" -I) - N.Again, 2" is represented by a binary number that consists of a I followed by nOs. 2" - I is abinary number represented by n I's. For example, if n = 4, we have 24 ~ (10000), and24- I = (llll),. Thus the l's complement of a binary number is obtained by subtractingeach digit from 1. However, when subtracting binary digits from 1, we can have eitherI - 0 = I or I - I = 0, which causes the bit to change from 0 to I or from I to O. Therefore,the I's complement of a binary number is formed by changing I's to O's and O's to I 's. The followingare some numerical examples:Section 1-5 Complements 9The I's complement of 1011000 is 0100111.The I's complement of 0101101 is 1010010.0010), = (2C6B.F2)162IlllF1011B011061100C(102
6. 6. Chapter 1 Binary Systems Section 1-5 Complements 11The (r - I)'s complement of octal or hexadecimal numbers is obtained by subtracting eachdigit from 7 or F (decimal 15), respectively.adix ComplementThe r's complement of an n-digit number N in base r is defined as rll - N, for N i: 0 and 0for N = O. Comparing with the (r - I)'s complement, we note that the r's complement isobtained by adding I to the (r - I)'s complement since r" - N ~ [(r" - I) - N] + 1.Thus, the lO's complement of decimal 2389 is 7610 + I = 7611 and is obtained by adding Ito the 9's-complement value. The 2's complement of binary 101100 is 010011 + I = 010100and is obtained by adding I to the I 's-complement value.Since 10" is a number represented by a I followed by nO's, 10" - N, which is the 10'scomplement of N, can be formed also by leaving all least significant O's unchanged, subtractingthe first nonzero least significant digit from 10, and subtracting all higher significant digitsfrom 9.The lO's complement of 012398 is 987602.The lO's complement of 246700 is 753300.The lO's complement of the first number is obtained by snbtracting 8 from 10 in the least significantposition and subtracting all other digits from 9. The lO's complement of the second numberis obtained by leaving the two least significant O's unchanged, subtracting 7 from 10, andsnbtracting the other three digits from 9.Similarly, the 2's complement can be formed by leaving all least significant O's and the firstI unchanged, and replacing I's with O's and O's with I's in all other higher significant digits.The 2's complement of 1101100 is 0010100.The 2's complement of 0110111 is 1001001.The 2's complement of the first nnmber is obtained by leaving the two least significant O's andthe first I unchanged, and then replacing I's with O's and O's with I's in the other four mostsignificantdigits. The 2's complement of the second nnmber is obtained by leaving the leastsignificant I unchanged and complementing all other digits.In the previous definitions, it was assumed that the numbers did not have a radix point. If theoriginal nnmber N contains a radix point, the point should be removed temporarily in orderto form the r's or (r - I)'s complement. The radix point is then restored to the complementednumber in the same-relative position. It is also worth mentioning that the complementof the complement restores the number to its original value. The r's complement of N isrn- N. The complement of the complement is Til ~ (rn- N) = N, and is equal to theoriginal number.~btraction with ComplementsThe direct method of snbtraction taught in elementary schools uses the borrow concept. In thismethod, we borrow a 1 from a higher significant position when the minuend digit is smallerthan the subtrahend digit. The method works well when people perform subtraction with paperand pencil. However, when subtraction is implemented with digital hardware, the method is lessefficient than the method that uses complements.The subtraction of two n-digit unsigned numbers M - N in base r can be done as follows:1. Add the minuend, M, to the r's complement of the subtrahend, N. This performsM + (rn- N) = M - N + r".2. IfM ;0> N, the sum will produce an end cany, r", which can be discarded; what is left isthe result M - N.3. IfM < N, the sum does not produce an end carry and is equal to r" - (N - M), whichis the r's complement of (N - M). To obtain the answer in a familiar form, take the r'scomplement of the sum and place a negative sign in front.The following examples illustrate the procedure:Using 10's complement, subtract 72532 - 3250.M = 72532lO's complement of N = + 96750Sum = 169282Discard end carry 105 = -100000Answer ~ 69282 Note that M has 5 digits and N has only 4 digits. Both numbers must have the same numberof digits, so we write N as 03250. Taking the 10's complement ofN produces a 9 in the mostsignificant position. The occurrence of the end carry signifies that M 2 N and that the resultis positive.Using 10's complement, subtract 3250 - 72532.M ~ 03250lO's complement of N ~ + 27468Sum ~ 30718There is no end carry.Therefore, the answer is -(10's complement of 30718) = -69282 Note that since 3250 < 72532, the result is negative. Since we are dealing with unsignednumbers, there is really no way to get an unsigned result for this case. When subtracting withcomplements, the negative answer is recognized from the absence of the end carry and thecomplemented result. When working with paper and pencil, we can change the answer to asigned negative number in order-to put itin a familiar form.
7. 7. . ~13Section 1-6 Signed Binary NumbersNote that the negative result is obtained by taking the I's complement of the sum since this isthe type of complement used. The procedure with end-around carry is also applicable for subtractingunsigned decimal numbers with 9's complement.1-6 SIGNED BINARY NUMBERSPositive integers (including zero) can be represented as unsigned numbers. However, to representnegative integers, we need a notation for negative values. In ordinary arithmetic, a negativenumber is indicated by a minus sign and a positive number by a plus sign. Because ofhardware limitations, computers must represent everything with binary digits. It is customaryto represent the sign with a bit placed in the leftmost position of the number. The conventionis to make the sign bit 0 for positive and I for negative.It is important to realize that both signed and unsigned binary numbers consist of a string ofbits when represented in a computer. The user determines whether the number is signed or unsigned.If the binary number is signed, then the leftmost bit represents the sign and the rest ofthe bits represent the number. If the binary number is assumed to be unsigned, then the leftmostbit is the most significant bit of the number. For example, the string of bits 01001 can be consideredas 9 (unsigned binary) or as +9 (signed binary) because the leftmost bit is O. The stringof bit' 1100 I represent the binary equivalent of 25 when considered as an unsigned number oras -9 when considered as a signed number. This is because the I that is in the leftmost positiondesignates a negative and the other four bits represent binary 9. Usually, there is no confusionin identifying the bits if the type of representation for the number is known in advance.The representation of the signed numbers in the last example is referred to as the signedmagnitudeconvention. In this notation, the number consists of a magnitude and a symbol (+or -) or a bit (0 or I) indicating the sign. This is the representation of signed numbers nsed inordinary arithmetic. When arithmetic operations are implemented in a computer, .it is moreconvenient to use a different system for representing negative numbers, referred to as the signedcomplementsystem. In this system, a negative number is indicated by its complement. Whereasthe signed-magnitude system negates a number by changing its sign, the signed-complementsystem negates a number by taking its complement. Since positive numbers always start witho(plus) in the left-most position, the complement will always start with aI, indicating a negativenumber. The signed-complement system can use either the l's or the 2's complement, butAs an example, consider the number 9 represented in binary with eight bits. +9 is representedwith a sign bit of 0 in the leftmost position, followed by the binary equivalent of9, which gives00001001. Note that all eight bits must have a value and, therefore, Os are inserted followingthe sign bit up to the first 1. Although there is only one way to represent +9, there are three different100010011111011011110111the 2's complement is the most common.ways to represent -9 with eight bits:signed-magnitude representation:signed-l 's-complement representation:signed-2's-complement representation:In signed-magnitude, -9 is obtained from +9 by changing the sign bit in the leftmost positionfrom 0 to 1. In signed-I's complement, -9 is obtained by complementing all the bits ofChapter 1 Binary SystemsSubtraction with complements is done with binary numbers in a similar manner using theprocedure outlined previously,Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction(a) X - Y and (b) Y - X using 2's complements.(a) X = 1010100Subtraction of unsigned numbers can also be done by means of the (r - I)'s complement.Remember that the (r - l)'s complement is one less than the r's complement. Because ofthis, the result of adding the minuend to the complement of the subtrahend produces a sum thatis 1 less than the correct difference when an end carry occurs. Removing the end carry andadding 1 to the sum is referred to as an end~around carry.~2's complement ofY = + 0111101Sum = 10010001Discard end carry 2' = -10000000Answer: X - Y = 0010001(b) Y = 1O000112's complement of X = + 0101100Sum = 1101111There is no end carry.Therefore, the answer is Y - X = -(2's complement of 1101111) = -0010001100001101010111101110X=Y=Repeat Example 1-7 using I's complement.(a) X - Y = 1010100 - 1000011I's complement of X = +Sum =1010100I's complement of Y = + 0111100Sum = 10010000End-around carry = +__~IAnswer: X - Y = 0010001(b) Y - X = IQQQQll - 1010100There is no end carry.Therefore, the answer is Y - X = -(1's complement of 1101110) = -0010001
8. 8. 14 Chapter 1 Binary SystemsTable 1-3Signed Binary NumbersSigned-2'. Signed-1'. SignedDecimal complement complement magnitude+7 0111 0111 0111+6 0110 0110 0110+5 0101 0101 0101+4 0100 0100 0100+3 0011 0011 0011+2 0010 0010 0010+1 0001 0001 0001+0 0000 0000 0000-0 1111 1000-1 1111 1110 1001-2 1110 1101 1010-3 1101 1100 1011-4 1100 lOll 1100-5 1011 1010 1101-6 1010 1001 1110-7 1001 1000 1111-8 1000+9, including the sign bit. The signed-2's-complement representation of -9 is obtained by takingthe 2's complement of the positive number, including the sign bit.Table 1-3 lists all possible 4-bit signed binary numbers in 1he three representations. Theequivalent decimal number is also shown for reference. Note that the positive numbers in allthree representations are identical and have 0 in the leftmost position. The signed-2's complementsystem has only one representation for 0, which is always positive. The other two systemshave either a positive 0 or a negative 0, which is something not encountered in ordinary arithmetic.Note that all negative numbers have a I in the leftmost bit position; this is the way wedistinguish them from the positive numbers. With fOUf bits, we can represent 16 binary numbers.In the signed-magnitude and the 1's complement representations, there are eight positivenumbers and eight negative numbers, including two zeros. In the 2's complement representation,there are eight positive numbers, including one zero and eight negative numbers.The signed-magnitude system is used in ordinary arithmetic, but is awkward when employedin computer arithmetic because of the separate handling of the sign and the magnitude.Therefore, the signed-complement is normally used. The l's complement imposes some difficultiesand is seldom used for arithmetic operations. It is useful as a logical operation since thechange of I to 0 or 0 to I is equivalent to a logical complement operation, as will be shown inthe next chapter. The following discussion of signed binary arithmetic deals exclusively withthe signed-2's-complement representation of negative numbers. The same procedures can beapplied to the signed-I 's-complement system by including the end-around carry as done withunsigned numbers.Section 1-6 Signed Binary Numbers 15Arithmetic AdditionThe addition of two numbers in the signed-magnitude system follows the rules of ordinaryarithmetic. If the signs are the same, we add the two magnitudes and give the sum the commonsign. If the signs are different, we subtract the smaller magnitude from the larger and give theresult the sign of the larger magnitude. For example, (+25) + (-37) = -(37 - 25) = -12and is done by subtracting the smaller magnitude 25 from the larger magnitude 37 and usingthe sign of 37 for the sign of the result. This is a process that requires the comparison of the signsand the magnitudes and then performing either addition or subtraction. The same procedure appliesto binary numbers in signed-magnitude representation. In contrast, the rule for addingnumbers in the signed-complement system does not require a comparison or subtraction, butonly addition. The procedure is very simple and can be stated as follows for binary nnmbers:The addition of two signed binary numbers with negative numbers represented in signed2's-complement form is obtained from the addition of the two numbers, including their sign bits.A carry out of the sign-bit position is discarded.Numerical examples for addition follow:+ 6 00000110 - 6 11111010+13 00001101 +13 00001101 - -+19 00010011 + 7 00000111+ 6 00000110 -6 11111010-13 - llllOOl1 -13 llllOOl1 -- 7 11111001 -19 11101101Note that negative numbers must be initially in 2's complement and thatifthe sum obtainedafter the addition is negative, it is in 2's-complement form.In each of the four cases, the operation performed is addition with the sign bit included.Any carry out of the sign-bit position is discarded, and negative results are automatically in 2'scomplementform.In order to obtain a correct answer, we must ensure that the result has a sufficient numberof bits to accommodate the sum. If we start with two n-bit numbers and the sum occupiesn + I bits, we say that an overflow occurs. When one performs the addition with paper and pencil,an overflow is not a problem, becanse we are not limited by the width of the page. Wejustadd another 0 to a positive number or another 1 to a negative number in the most-significantposition to extend them to n + I bits and then perform the addition. Overflow is a problem incomputers because the number of bits that hold a number is finite, and a result that exceeds thefinite value by I cannot be accommodated.The complement form of representing negative numbers is unfamiliar to those used to thesigned-magnitude system. To determine the value of a negative number when in signed-2'scomplement, it is necessary to convert it to a positive number to place it in a more familiarform. For example, the signed binary number 11111001 is negative because the leftmost bit is1. Its 2's complement is 00000111, which is the binary equivalent of +7. We therefore recognizethe original negative number to be equal to -7.
9. 9. 16 Chapter 1 Binary Systems Section 1-7 Binary Codes 17Arithmetic SubtractionSubtraction of two signed binary numbers when negative numbers are in 2's-complement formis very simple and can be stated as follows: . .Take the 2's complement of the snhtrahend (including the sign bit) and add It to the mmu-end(including the sign bit). A cany out of the sign-bit position is discarded. . .This procedure occurs because a subtraction operation can be changed to a~ additIo~ ope~ationif the sign of the subtrahend is changed. This is demonstrated by the followmg relatIOnshIp:(A) - (+B) = (A) + (-B);(A) - (-B) = (A) + (+B).Bnt chauging a positive number to a negative number is easily done by taking its 2's complement.The reverse is also true because the complement of a negative number in complement formproduces the equivalent positive nnmber. Consider the snbtraction of (-6) - (-:-13) = +7. Inbinary with eight bits, this is written as (11111010 - 11110011). The subtractIOn IS changedto addition by taking the 2's complement of the subtrahend (-13) to gIve (+13). In bmary,this is 11111010 + 00001101 = 100000111. Removing the end carry, we obtam the correctanswer: 00000111( +7).It is worth noting that binary numbers in the signed-complement system are added and subtractedby the same basic addition and subtraction rules as unsigned numbers. Therefore, computersneed only one common hardware circnit to haudle both types of arithmetic. The user orprogrammer must interpret the results of such addition or subtraction dIfferently, dependmg onwhether it is assumed that the numbers are signed or unsigned.1-7 BINARY CODESDigital systems use signals that have two distinct values and circuit elements that have tw~stable states. There is a direct analogy among binary signals, binary circuit elements, and bInarydigits. A binary number of n digits, for example, may be represented by n binary circuitelements, each having au output signal equivalent to a or I. Digital systems represent aud manipulatenot only binary numbers, but also many other discrete elements of information. A~ydiscrete element of information distinct among a group of quantities can be represented WItha binary code. The codes must be in binary because computers can only hold I's aud a's. It mustbe realized that binary codes merely change the symbols, not the meauing of the elements ofinformation that they represent. Ifwe inspect the bits of a computer at random, we will find thatmost of the time they represent some type of coded information rather than binary numbers.An n-bit binary code is a group of n bits that assume up to 21ldistinct combinations of 1'sand a's with each combination representing one element of the set that is being coded. A setof four 'elements cau be coded with two bits, with each element assigned one of the followingbit combinations: 00, 01, 10, II. A set of eight elements requires a 3-bit code aud a set of 16elements requires a 4-bit code. The bit combination of an n-bit code is determined from the countin binary from a to 2" ~ 1. Each element must be assigned a unique binary bit combination audno two elements can have the same value; otherwise, the code assignment will be ambiguous.BCD CodeAlthough the minimum number of bits required to code 2/1 distinct quantities is n, there isno maximum number of bits that may be used for a binary code. For example, the 10 decimaldigits cau be coded with 10 bits, and each decimal digit cau be assigned a bit combination ofnine a's and a 1. In this particular binary code, the digit 6 is assigned the bit combination0001000000.Although the binary number system is the most natural system for a computer, most people aremore accustomed to the decimal system. One way to resolve this difference is to convert thedecimal numbers to binary, perform all arithmetic calculations in binary, and then convert thebinary results back to decimal. This method requires that we store the decimal numbers in thecomputer so they can be converted to binary. Since the computer can accept only binary values,we must represent the decimal digits by means of a code that contains.1 's and O's. It is alsopossible to perform the arithmetic operations directly with decimal numbers when they arestored in the computer in coded form.A binary code will have some unassigned bit combinations if the number of elements inthe set is not a multiple power of 2. The 10 decimal digits form such a set. A binary code thatdistinguishes among 10 elements must contain at least four bits, but 6 out of the 16 possiblecombinations remain unassigned. Different binary codes can be obtained by arranging fourbits in 10 distinct combinations. The code most commonly used for the decimal digits is thestraight binary assignment as listed in Table 1-4. This is called binary coded decimal aud is commonlyreferred to as BCD. Other decimal codes are possible and a few of them are presentedlater in this section.Table 1-4 gives the 4-bit code for one decimal digit. A number with k decimal digits will require4k bits in BCD. Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110,with each group of 4 bits representing one decimal digit. A decimal number in BCD is the sameas its equivalent binary number only when the number is between 0 and 9. A BCD numberTable 1-4Binary Caded Decimal (BCD)Decimal BCDsymbol digit0 00001 00012 00103 00114 01005 01016 01107 01118 10009 1001i
10. 10. 18 Chapter 1 Binary Systems Section 1-7 Binary Codes 19greater than 10 looks different than its equivalent hinary number, even though both contain l'sand D's. Moreover, the binary combinations 1010 through 1111 are not used and have no meaningin the BCD code. Consider decimal 185 and its corresponding value in BCD and binary:(185ho = (00011000 OlO1hco = (1011100l)2This condition occurs when the sum is greater than or equal to 16. Although the other four bitsare less than 1001, the binary sum requires a correction because of the carry. Adding 0110, weobtam the required BCD digit sum 0111 (7) and a BCD carry.The addition of two n-digit unsigned BCD numbers follows the sarne procedure. Considerthe addition of 184 + 576 = 760 in BCD:BCD carryThe first, least significant pair of BCD digits produces a BCD digit sum of 0000 and a carryfor the next palr of dIgIts. The second pair ofBCD digits plus a previous carry produces a digitsum of 0110 and a carry for the next pair of digits. The third pair of digits plus a carry producesa bmary sum of 0111 and does not require a correction.Decimal Arithmetic1 10001 1000 0100+0101 -- -Ol-ll OllO --Olll 10000 1010OllO OllO -- --DIll OllO 0000The repr~sen~ation ofsigned decimal numbers in BCD is similar to the representation of signednumbers m bmary. We can use either the familiar sign and magnitude system or the signed-complementsystem. The sign of a decimal number is usually represented with four bits to conformto the 4-bit code of the decimal digits. It is customary to designate a plus with four D's and aminus with the BCD equivalent of 9, which is 1001.The s~gned-magnitude system is seldom used in computers. The signed-complement systemcan be e!lher the 9's or the 10's complement, but the lO's complement is the one most oftenused. To obtain the lO's complement of a BCD number, we first take the 9's complement andthen add one to the least significant digit. The 9's complement is calculated from the subtractionof each digit from 9.The procedures developed for the signed-2's complement system in the previous sectionapply also to the signed-lO's complement system for decimal numbers. Addition is done byadding all digits, including the sign digit, and discarding the end carry. This assumes that allnegative numbers are in 1O's complement form. Consider the addition (+375) + (-240) =+135 done in the signed-complement system.o 375+9 760o 135184+576760Binary sumAdd 6BCD sumThe 9 in the leftmost position of the second number represents a minus and 9760 is the lO'scomplement of 0240. The two numbers are added and the end carry is discarded to obtainThe BCD value has 12 bits, but the equivalent binary nnmber needs only 8 bits. It is obvionsthat a BCD number needs more bits than its equivalent binary value. However, there is an advantagein the use of decimal numbers because computer input and output data are generatedby people that use the decimal system.It is important to realize that BCD numbers are decimal numbers and not binary numbers,although they use bits in their representation. The only difference between a decimal numberand BCD is that decimals are written with the symbols 0, 1,2, ... , 9 and BCD numbers use thebinary code 0000, 0001, 0010, ... , 1001. The decimal value is exactly the same. Decimal 10is represented in BCD with eight bits as 0001 0000 and decimal 15 as 0001 0101. The correspondingbinary values are 1010 and 1111 and have only four bits.Consider the addition of two decimal digits in BCD, together with a possible carry from a previousless significant pair of digits. Since each digit does not exceed 9, the sum cannot begreater than 9 + 9 + 1 = 19, with the 1 in the sum being a previous carry. Suppose we addthe BCD digits as if they were binary numbers. The binary sum will produce a result in the rangefrom 0 to 19. In binary, this will befrom 0000 to 10011, but in BCD, it is from 0000 to 1 1001;the first 1 being a carry and the next four bits being the BCD digit sum. When the binary sumis equal to or less than 1001 (without a carry), the corresponding BCD digit is correct. However,when the binary sum is greater than or equal to 1010, the result is an invalid BCD digit.The addition of 6 ~ (OllO)2 to the binary sum converts itlo the correct digit and also producesa carry as required. This is because the difference between a carry in the most significant bitposition of the biuary sum and a decimal carry differ by 16 - 10 ~ 6. Considerthe followiugthree BCD additions:4 0100 4 0100 8 1000+5 +0101 +8 +1000 +9 1001- -- - -- - --9 1001 12 llOO 17 10001+OllO +OllO -- --10010 101llIn each case, the two BCD digits are added as if they were two binary numbers. If the biuarysum is greater or equal to 1010, we add 0110 to obtain the correct BCD digit sum and a carry.In the first example, the sum is equal to 9 and is the correct BCD digit sum. In the second example,the binary sum produces an invalid BCD digit. The addition of 0110 produces the correctBCD digit sum 0010 (2) and a carry. In the third example, the binary sum produces a carry.BCD Addition
11. 11. 20 Chapter 1 Binary Systems Section 1-7 Binary Codes 21+135. Of course, the decimal numbers inside the computer must be in BCD, including the signdigits. The addition is done with BCD digits as described previously.The subtraction of decimal numbers either unsigned or in the signed-l O's complement systemis the same as in the binary case. Take the lO's complement of the subtrahend and add itto the minuend. Many computers have special hardware to perform alithmetic calculations directlywith decimal numbers in BCD. The user of the computer can specify by programmedinstructionsto perform the arithmetic operation with decimal numbers dIrectly WIthout havmgto convert them to binary.Other Decimal CodesBinary codes for decimal digits require a minimum of four bits per digit. Many different codescau be formulated by arranging four bits in 10 distinct possible combinations. The BCD andthree other representative codes are shown in Table 1-5. Each code uses only 10 bit combinationsout of possible 16 combinations that can be alTanged with four bits. The other six unusedcombinations in each case have no meaning and should be avoided.The BCD and the 2421 codes are examples of weighted codes. In a weighted code. each bitposition is assigned a weighting factor in such a way that each digit can be ~valuatedby addingthe weights of all the I's in the coded combination. The BCD code has weights of 8, 4, 2, andI, which correspond to the power of two values of each bit. The bit assignmeut 0110 for example,is interpreted by the weights to represent decimal 6 because 8 X a + 4 X I + 2 XI + 1 X a = 6. The bit combination 1101 wheu weighted by the respective digits 2421 givesGray Codethe decimal equivalent of 2 X I + 4 X I + 2 X a + 1 X I ~ 7. Note that some digits canbe coded in two possible ways in the 2421 code. Decimal 4 can be assigned to bit combinations0100 or 1010 since both combinations add up to a total weight offour.The 2421 and the excess-3 codes are examples of self-complementing codes. Such codeshave the property that the 9's complement of a decimal number is obtained directly by changingI's to a's and a's to I's in the code. For example, decimal 395 is represented in the excess3code as 0110 1100 1000. The 9's complement 604 is represented as 1001 0011 0111, whichis simply obtained by complementing each bit of the code (as with the I's complement of binarynumbers).The excess-3 code has been used in some older computers because of its self-complementingproperty. This is an unweighted code where each coded combination is obtained from thecorresponding binary value plus 3. Note that the BCD code is not self-complementing.The 8, 4, -2. -I code is an example of assigning both positive and negative weights to a decimalcode. In this case, the bit combination 0110 is interpreted as a decimal 2 and is calculatedfrom 8 X a + 4 X I + (-2) X I + (-1) X a = 2.The output data of many physical systems produce quantities that are continuous. These datamust be converted into digital form before they are applied to a digital system. Continuous oranalog information is converted into digital form by means of an analog-to-digital converter.It is sometimes convenient to use the Gray code shown in Table 1-6 to represent the digitaldata when it is converted from analog data. The advantage of the Gray code over the straightTable 1-5Four Different Binary Codes for the Decimal Digits Table 1-6Gray CodeDecimal BCDdigit 8421 2421 Excess-3 B 4-2-1 Gray Decimalcode equivalenta 0000 0000 0011 a a a aI 0001 0001 0100 a I I I 0000 a2 0010 0010 0101 a I I a 0001 I3 0011 0011 0110 a 101 0011 24 0100 0100 0111 a 1 a a 0010 35 0101 1011 1000 101 I 0110 46 0110 1100 1001 I a I a 0111 57 0111 1101 1010 100 I 0101 68 1000 1110 1011 100 a 0100 79 1001 1111 1100 I I 1 1 1100 81101 91010 0101 0000 a a a I 1111 10Unused 1011 0110 0001 a a I a 1110 11bit 1100 0111 0010 a a I 1 1010 12combi- 1101 1000 1101 I I a a 1011 13nations 1110 1001 1110 I 101 1001 141111 1010 1111 I I I a 1000 15
12. 12. 22 Chapter 1 Binary Systemsbinary number sequence is that only one bit in the code group changes when going from onenumber to the next. For example, in going from 7 to 8, the Gray code changes from 01 00 to1100. Only the first bit changes from 0 to I; the other three bits remain the same. When comparingthis with binary numbers, the change from 7 to 8 will be from 0111 to 1000, whichcauses all four bits to change values.The Gray code is used in applications where the normal sequence of binary numbers mayproduce an error or ambiguity during the transition from one number to the next. If binarynumbers are used, a change from 0111 to 1000 may produce an intermediate erroneous number1001 if the rightmost bit takes longer to change in value than the other three bits. The Graycode eliminates this problem since only one bit changes in value during any transition betweentwo numbers.A typical application of the Gray code occurs when analog data are represented by continuouschange of a shaft position. The shaft is partitioned into segments, and each segment is assigneda number. If adjacent segments are made to correspond with the Gray-code sequence,ambiguity is eliminated when detection is sensed in the line that separates any two segments.ASCII Character CodeMany applications of digital computers require the handling of data not only of numbers, butalso of letters. For instance, an insurance company with thousands of policy holders will usea computer to process its files. To represent the names and other pertinent information, it is necessaryto formulate a binary code for the letters of the alphabet. In addition, the same binarycode must represent numerals and special characters (such as $). An alphanumeric characterset is a set of elements that includes the 10 decimal digits, the 26 letters of the alphabet, and anumber of special characters. Such a set contains between 36 and 64 elements if only capitalletters are included, or between 64 and 128 elements if both. uppercase and lowercase lettersare included. In the first case, we need a binary code of six bit..., and in the second, we need abinary code of seven bits.The standard binary code for the alphanumeric characters is ASCII (American StandardCode for Information Interchange). It uses seven bits to code 128 characters, as shown inTable 1-7. The seven bits of the code are designated by bl through b" with b, being the mostsignificantbit. The letter A, for example, is represented in ASCII as 1000001 (column 100,row 0001). The ASCII code contains 94 graphic characters that can be printed and 34 nonprintingcharacters used for various control functions. The graphic characters consist of the 26uppercase letters (A through Z), the 26 lowercase letters (a through z), the 10 numerals (0through 9), and 32 special printable characters such as %, *, and $.The 34 control characters are designated in the ASCII table with abbreviated names. Theyare listed again below the table with their functional names. The control characters are used forrouting data and arranging the printed text into a prescribed format. There are three types of controlcharacters: format effectors, information separators, and communication-control characters.Format effectors are characters that control the layout of printing. They include the familiartypewriter controls such as backspace (BS), horizontal tabulation (HT), and carriage return(CR). Information separators are used to separate the data into divisions such as paragraphs andpages. They include characters such as record separator (RS) and file separator (FS). The communication-control characters are useful during the transmission of text between remote ter-Section 1-7 Binary Codes 23Table 1-7American Standard Code for Information Interchange (ASCII),b7~jJ5b.b,b2 b, 000 001 010 011 100 101 110 1110000 NUL DLE SP 0 @ P P0001 SOH DCI I A Q a q0010 STX DC2 2 B R b r0011 ETX DC3 # 3 C S c s0100 EOT DC4 $ 4 D T d0101 ENQ NAK % 5 E U e u0110 ACK SYN & 6 F V f v0111 BEL ETB 7 0 W g w1000 BS CAN 8 QL) X h x1001 lIT EM 9 I Y Y1010 LF SUB * J Z j z1011 VT ESC + K [ k1100 FF FS < LI1101 CR OS M ] m1Il0 SO RS > N II n1111 SI US / ? 0 0 DELControl charactersNUL Null DLE Data-link escapeSOH Start of heading DCI Device control 1STX Start of text DC2 Device control 2ETX End of text DC3 Device control 3EOT End of transmission DC4 Device control 4ENQ Enquiry NAK Negative acknowledgeACK Acknowledge SYN Synchronous idleBEL Bell ETB End-of-transmission blockBS Backspace CAN CancelHT Horizontal tab EM End of mediumLF Line feed SUB SubstituteVT Vertical tab ESC EscapeFF Form feed FS File separatorCR Carriage return OS Group separatorSO Shift out RS Record separatorSI Shift in US Unit separatorSP Space DEL Deleteminals. Examples of communication-control characters are STX (start of text) and ETX (endof text), which are used to frame a text message when transmitted through telephone wires.ASCII is a 7-bit code, but most computers manipulate an 8-bit quantity as a single unitcalled a byte. Therefore, ASCII characters most oftep are stored one per byte. The extra bit is
13. 13. 24 Chapter 1 Binary Systems Section 1'8 Binary Storage and Registers 25sometimes used for other purposes, depending on the application. For example, some printersrecognize 8-bitASCII characters with the most-significant bit set to O. Additional 128 8-bit characterswith the most-significant bit set to I are used for other symbols such as the Greek alphabetor italic type font.Error-Detecting CodeRegistersA register is a group of binary cells. A register with n cells can store any discrete quantity ofinformation that contains n bits. The state of a register is an n-tup1e number of 1's and O's, witheach bit designating the state of one cell in the register. The content of a register is a functionof the interpretation given to the information stored in it. Consider, for example, a 16-bit registerwith the following content:To detect errors in data communication and processing, an eighth bit is sometimes added to theASCII character to indicate its parity. A parity bit is an extra bit included with a message to makethe total number of 1's either even or odd. Consider the following two characters and theireven and odd parity:BINARY STORAGE AND REGISTERSThe binary information in a digital computer must have a physical existence in some information-storage medium for storing individual bits. A binary cell is a device that possesses twostable states and is capable of storing one bit of information. The inputto the cell receives excitationsignals that set it to one of the two states. The output of the cell is a physical quantitythat distinguishes between theCtwo states. The infonnation stored in a cell is 1 when it is in onestable state and 0 when in the 9ther stable state.1100001111001001A register with 16 cells can be in one of 216 possible states. If one assumes that the content ofthe register represents a binary integer, then the register can store any binary number from 0to 216- I. For the particular example shown, the content of the register is the binary equivalentof the decimal number 50121. If it is assumed that the register stores alphanumeric charactersof an eight-bit code, the content of the register is any two meaningful characters. For theASCll code with an even parity placed in the eighth most-significant bit position, the registercontains the two characters C (left eight bits) and I (right eight bits). On the other hand, if oneinterprets the content of the register to be four decimal digits represented by a four-bit code,the content of the register is a four-digit decimal number. In the excess-3 code, the registerholds the decimal number 9096. The content of the register is meaningless in BCD because thebit combination 1100 is not assigned to any decimal digit. From this example, it is clear that aregister can store discrete elements of information and that the same bit configuration may beinterpreted differently for different types of data.Register TransferA digital system is characterized by its registers and the components that perform data processing.A register tranifer operation is a basic operation in digital systems. It consists.of a transferof binary information from one set of registers into another set of registers. The transfer maybe direct from one register to another. or may pass through data processing circuits to performan operation. Figure 1-1 illustrates the transfer of the information among registers and demonstratespictorially the transfer of binary information from a keyboard into a register in the memoryunit. The input unit is assumed to have a keyboard, a control circuit, and an input register.Each time a key is struck, the control enters an equivalent eight-bit alphanumeric charactercode into the input register. We shall assume that the code used is the ASCll code with an oddparitybit. The information from the input register is transferred into the eight least significantcells of a processor register. After every transfer, the input register is cleared to enable the controlto insert a new eight-bit code when the keyboard is struck again. Each eight-bit charactertransferred to the processor register is preceded by a shift of the previous character to the nexteight cells on its left. When a transfer of four characters is completed, the processor register isfull, and its contents are transferred into a memory register. The content stored in the memoryregister shown in Fig. 1-1 came from the transfer of the characters "J," "0," "H," and "N" afterthe four appropriate keys were struck.To process discrete quantities of information in binary form, a computer must be providedwith devices that hold the data to be processed and circuit elements that manipulate individualbits of information. The device most commonly used for holding data is a register.With odd parity1100000101010100With even parity0100000111010100ASCII A = 1000001ASCII T = 1010100In each case, we insert an extra bit in the leftmost position of the code to produce an even numberof I's in the character for even parity or an odd number of I's in the character for oddparity. In general, one or the other parity is adopted, with even parity being more common.The parity bit is helpful in detecting errors during the transmission of information from onelocation to another. This is handled by generating an even parity bit in the sending end for eachcharacter. The 8-bit characters that include parity bits are transmitted to their destination. Theparity of each character is then checked in the receiving end. If the parity of the received characteris not even, it means that at least one bit has changed value during the transmission. Thismethod detects one, three, or any odd combination of errors in each character that is transmitted.An even combination of errors is undetected. Additionar error detection codes may beneeded to take care of an even combination of errors.What is done after an error is detected depends on the particular application. One possibilityis to request retransmission of the message on the assumption that the errorwas random andwill not occur again. Thus, if the receiver detects a parity error, it sends back the ASCll NAK(negative acknowledge) control character consisting of an even parity eight bits 10010101. Ifno error is detected, the receiver sends back anACK (acknowledge) control character, 00000110.The sending end will respond to an NAK by transmitting the message again nntil the correctparity is received. If, after a number of attempts, the transmission is still in error, a message canbe sent to the operator to check for malfunctions in the transmission path.1-8
14. 14. Binary logic deals with variables that take on two discrete values and operations that assumelogical meaning. The two values the variables take may be called by different names (true andfalse, yes and no, etc.), bnt for our purpose, it is convenient to think in terms of bits and assignthe values of I and O. The binary logic introduced in this section is equivalent to an algebra calledBoolean algebra. The formal presentation of Boolean algebra is covered in more detail in Chapter2. The purpose of this section is to introduce Boolean algebra in a heuristic manner and relateit to digital logic circnits and binary signals.Definition of Binary LogicBinary logic consists of binary variables and logical operations. The variables are designated byletters of the alphabet snch as A, B, C, x, y, z, etc., with each variable having two and only twodistinct possible values: I and O. There are three basic logical operations: AND, OR, and NOT.26 Chapter 1 Binary SystemsMEMORY UNITJ 0 H N I' :1 Memory01001010010011111100100011001110 RegisterPROCESSOR UNITI 8 cells 8 cells IProcessor8 cells 8 cells RegisterINPUT UNIT InputI 8 cells I Registert~IKeyboard I , CONTROLIFIGURE 1-1Transfer of information with registersManipulation of binary variables is done by means of digital logic circuits. Figure 1-2 illustratesthe process of adding two lO-bit binary numbers. The memory unit, which normally consistsof millions of registers, is shown in the diagram with only three of its registers. The part of theprocessor unit shown consists of three registers-RI, R2, and R3-together with digital logiccircuits that manipulate the bits of RI and R2 and transfer into R3 a binary number equal totheir arithmetic sum. Memory registers store information and are incapable of processing thetwo operands. However, the information stored in memory can be transferred to processor registers.Results obtained in processor registers can be transferred back into a memory registerfor storage until needed again. The diagram shows the contents of two operands transferred fromtwo memory registers into Rl and R2. The digital logic circuits produce the sum, which istransferred to register R3. The contents of R3 can now be transferred back to one of the memoryregisters.The last two examples demonstrated the information-flow capabilities of a digital system ina very simple manner. The registers of the system are the basic elements for storing and holdingthe binary information. Digital logic circuits process the binary information stored in theregisters. Digital logic circuits and registers are covered in Chapters 2 throngh 6. The memoryunit is explained in Chapter 7. The register transfer level for describing and designing digitalsystems is covered in Chapter 8.Section 1-9 Binary LogicMEMORY UNIT. 10000000000SumOperand 1100111000011Ope~d ~IOOOIOOOOIO IL-~ 10 0 0 1 0000101 R11Digital Logiccircuits for ----+- 0 I 0 0 I 0 0 0 I Il R3binary addition100 III 0 0 0 0 Il R2PROCESSOR UNITFIGURE 1-2Example of binary information processing1-9 BINARY LOGIC27, j
15. 15. 28 Chapter 1 Binary Systems Section 1-9 Binary Logic 29the same as those llsed for multiplication and addition. However, binary logic should not be confusedbinary arithmetic, we have 1 + I = 10 (read: "one plus one is equal to 2"), whereas in binarybetween the values that the variables may take and tbe result of the operation. The truth tablesfor the operations AND and OR with variables x and yare obtained by listing all possible valuescombination is then listed in a separate row. The truth tables for AND, OR, and NOT are listedLogic Gates1. AND: This operation is represented by a dot or by the absence of an operator. For example,x y ~ z or xy = z is read "x AND y is equal to z." The logical operation AND is interpretedto mean that z = I if and only if x = I and y ~ I; otherwise z = O. (Rememberthat x, y, and z are binary variables and can be equal either to I or 0, and nothing else.)2. OR: This operation is represented by a plus sign. For example, x + y = z is read "xOR y is equal to z," meaning that z ~ I if x = lor if y = lor if both x ~ 1 and y = 1.If both x ~ 0 and y = 0, then z = O.3. NOT: This operation is represented by a prime (sometimes by an overbar). For example,x' ~ z (or ex = z) is read "not x is equal to z," meaning that z is what x is not. In otherwords, if x = 1, then z = 0; but if x = 0, then z = 1. The NOT operation is also referredto as the complement operation, since it changes a 1 to 0 and a 0 to 1.Binary logic resembles binary arithmetic, and the operations AND and OR have similaritiesto multiplication and addition, respectively. In fact, the symbols used for AND and OR arewith binary arithmetic. One should realize that an arithmetic variable designates a numberthat may consist of many digits. A logic variable is always eitber 1 or O. For example, inlogic, we have I + I ~ I (read: "one OR one is equal to one").For each combination of the values of x and y, there is a value of z specified by tbe definitionof the logical operation. These definitions may be listed in a compact form using truth tables.A truth table is a table of all possible combinations of the variables sbowing tbe relationthat the variables may have when combined in pairs. The result of the operation for eachin Table 1-8. These tables clearly demonstrate the definition of the operations.Logic gates are electronic circuits that operate on one or more input signals to produce an outputsignal. Electrical signals such as voltages or currents exist throughout a digital system ineither of two recognizable values. Voltage-operated circuits respond to two separate voltage levelsthat represent a binary variable equal to logic I or logic O. For example, a particular digitalsystem may define logic 0 as a signal equal to 0 volt and logic I as a signal equal to 4 volts.In practice, each voltage tevel has an acceptable range as shown in Fig. 1-3. The input terminalsof digital circuits accept binary signals within the allowable range and respond at the outputterminals with binary signals that fall within the specified range. The intermediate regionbetween the allowed regions is crossed only during state transition. Any desired informationfor computing or control can be operated on by passing binary signals through various combinationsof logic gates with each signal representing a particular binary variable.The graphic symbols used to designate the three types of gates are shown in Fig. 1-4. Thegates are blocks of hardware that produce the equivalent of logic 1 or logic 0 output signals ifinput logic requirements are satisfied. The input signals x and y in theAND and OR gates mayexist in one offom possible states: 00, 10, 11, or 01. These input signals are shown in Fig. 1-5together with the corresponding output signal for each gate. The timing diagrams illustrate theresponse of each gate to the four input signal combinations. The horizontal axis of the timingRangefor logic-lRangefor logic-OVolts431oFIGURE 1-3Example of binary signalsTransition occursbetween these limits;~ ;~Y X---[>---X'(a) Two-input AND gate (b) Two-input OR gate (c) NOT gate or inverterFIGURE 1-4Symbols for digital logic circuitsTable 1-8Truth Tables of Logical OperationsAND OR NOTx y X'y x Y x+y 0 0 0 0 0 0 iott10 1 0 0 1 1 1 01 0 0 1 0 11 I 1 1 1 1
16. 16. 30 Chapter 1 Binary Systems Problems 311 01---",o,-------,o,,---11l yOR:x+y~NOT: x' ~ I 1 1FIGURE 1-5Input-output signals for gatesdiagram represents time and the vertical axis shows the signal as it changes between the twopossible voltage levels. The low level represents logic 0 and the high level represents logic 1.The AND gate responds with a logic I output signal when both input signals are logic 1. TheOR gate responds with logic I output signal if any input signal is logic 1. The NOT gate iscommonly referred to. as an inverter. The reason for this name is apparent from the signal responsein the timing diagram, where it is shown that the output signal inverts the logic senseof the input signal.AND and OR gates may have more than two inputs. AnAND gate with three inpnts and anOR gate with four inputs are shown in Fig. 1-6. The three-inputAND gate responds with logicI outpnt if all three inpnts are logic I. The output produces logic 0 if any input is logic O. Thefour-input OR gate responds with logic I if any input is logic I; its output becomes logic 0 onlywhen all inputs are logic O.List the octal and hexadecimal numbers from 16 to 32. Using A and B for the last two digits, listthe numbers from 10 to 26 in base 12.What is the exact number of bytes in a system that contains (a) 32K byte, (b) 64M bytes, and(c) 6.4G byte?What is the largest binary number that can be expressed with 12 bits? What is the equivalent decimaland hexadecimal?Convert the following numbers with the indicated bases to decimal: (4310)5' and (198)12'q 1-5 Determine the base of the numbers in each case for the following operations to be correct:---"'- (a) 14/2 ~ 5; (b) 54/4 ~ 13, (c) 24 + 17 ~ 40.1-6 The solution to the quadratic equation x2 ~ llx + 22 = 0 is x = 3 and x = 6. What is the base1-91-101-11of the numbers?Express tbe following numbers in decimal: (10110.0101)', (16.5)16' and (26.24)g.Convert the following binary numbers to hexadecimal and to decimal: (a) 1.11010, (b) 1110.10.Explain why the decimal answer in (b) is 8 times that of (a).Convert the hexadecimal number 68BE to binary and then from binary convert it to octal.Convert the decimal number 345 to binary in two ways: (a) convert directly to binary; (b) convertfirst to hexadecimal, then from hexadecimal to binary. Which method is faster?Do the following conversion problems:(a) Convert decimal 34.4375 to binary.(b) Calculate the binary equivalent of 1/3 out to 8 places. Then convert from binary to decimal.How close is the result to 1/3?(c) Convert the binary result in (b) into hexadecimal. Then convert the result to decimal. Is theanswer the same?1-12 Add and multiply the following numbers without converting them to decimal.(a) Binary numbers 1011 and IOJ.(b) Hexadecimal numbers 2E and 34.1-13 Perform tbe following division in binary: 1011111 + IOJ.1-14 Find the 9's- and the 10's-complement of the following decimal numbers:(a) 98127634 (b) 72049900 (c) 10000000 (d) 00000000.1-15 (a) Find the 16's-complement ofAF3B.(b) Convert AF3B to binary.(c) Find the 2's-complement of the result in (b).(d) Convert the answer in (c) to hexadecimal and compare with the answer in (a).~ 1-16 Obtain the l's and 2's complements of the following binary numbers:(a) 11101010 (b) 01111110 (c) 00000001 (d) 10000000 (e) 00000000.1-71-81-17 Perform subtraction on the following unsigned numbers using the 10's-complement of the subtrahend.Where the result should be negative, lO's complement it and affix a minus sign. Verifyyour answers.(a) 7188 - 3049 (b) 150 - 2100 (c) 2997 - 7992 (d) 1321 - 375~ 1-18 Perform subtraction on the following unsigned binary numbers using the 2's-complement of thesubtrahend. Where the result should be negative, 2's complement it and affix a minus sign.(a) 11011 - 11001 (b) 110100 - 10101 (c) 1011 - 110000 (d) 101010 - 1010111-19 The following decimal numbers are shown in sign-magnitude fonn: +9826 and +801. Convertthem to signed 1O's-complement form and perfOITIl the following operations: (Note that the sumis + 10627 and requires six digits).(a) (+9826) + (+801) (b) (+9826) + (-801)(c) (-9826) + (+801) (d) (-9826) + (-801)1-20 Convert decimal +61 and +27 to binary using the signed-2's complement representation andenough digits to accomodate the numbers. Then perform the binary equivalent of (+27) + (-61),(-27) + (+61) and (-27) + (-61). Convert tbe answers back to decimal and verify that tbeyare correct.1 ~1f~+B+C+D(b) Four-input OR gatexAND: x . y~~BC(a) Three-input AND gateFIGURE 1-6Gates with multiple inputsPROBLEMS1-11-21-4
17. 17. Boolean Algebraand Logic Gates332-1 BAS leD EFIN I T ION SBoolean algebra, like any other deductive mathematical system, may be defined with a set ofelements, a set of operators, and a number of unproved axioms or postulates. A set of elementsis any collection of objects having a common property. If S is a set, and x and yare certain objects,then XES denotes that x is a member of the set, S, and y
18. 18. erties of bistable electlical switching circuits can be represented by this algebra. For the formaldefinition of Booleau algebra, we shall employ the postulates formulated by E. V. Huntingtonin 1904.Boolean algebra is an algebraic structure defined by a set of elements, B, together with twobinary operators, +and, provided that the following (Huntington) postulates are satisfied:1. (a) Closure with respect to the operator +.(b) Closure with respect to the operator'.2. (a) An identity element with respect to +, designated by 0: x + 0 = 0 + x = x.(b) An identity element with respect to " designated by I: x . I = I x ~ x.3. (a) Commutative with respect to +: x + y ~ y + x.(b) Commutative with respect to : x . y = y x.4. (a) . is distributive over +: x . (y + z) = (x' y) + (x' z).(b) + is distributive over': x + (Y' z) = (x + y) . (x + z).5. For every element x E B, there exists an element Xl E B (called the complement of x)such that (a) x + x' = I and (b) x . x' = O.6. There exists at least two elements x, y E B such that x '" y.Comparing Boolean algebra with arithmetic and ordinary algebra (the field of real numbers),we note the following differences:1. Huntington postulates do not include the associative law. However, this law holds forBoolean algebra and can be derived (for both operators) from the other postulates.2. The distributive law of + over', i.e., x + (Y' z) = (x + y) (x + z), is valid forBoolean algebra, but not for ordinary algebra.3. Boolean algebra does not have additive or multiplicative inverses; therefore, there areno subtraction or division operations.4. Postulate 5 defines an operator called complement that is not available iu ordinary algebra.5. Ordinary algebra deals with the real numbers, whicb consitute an infinite set of elements.Boolean algebra deals with the as yet undefined set of elements, B, but in the two-valuedBoolean algebra defined next (and of interest in our subsequent use of this algebra),B is defined as a set with only two elements, 0 and 1.Boolean algebra resembles ordinary algebra in some respects. The choice of symbols + and is intentional to facilitate Boolean algebraic manipulations by persons already familiar withordinary algebra. Although one can use some knowledge from ordinary algebra to deal withBoolean algebra, the beginner must be careful not to substitute the rules of ordinary algebrawhere they are not applicable.It is important to distinguish between the elements of the set of an algebraic structure andthe variables of an algebraic system. For example, the elements of the field of real numbers arenumbers, whereas variables such as a, b, c, etc., used in ordinary algebra, are symbols thatstand for real numbers. Similarly, in Boolean algebra, one defines the elemeuts of the set B, andvariables such as x, y, and z are merely symbols that represent the elements. At this point, it isimportant to realize that in order to have a Boolean algebra, one must show:342-2Chapter 2 Boolean Algebra and Logic Gates3. Commutative law. A binary operator * on a set S is said to be commutative wheneverx * y = y * x for all x, yES4. Identity element. A set S is said to have an identity element with respect to a binary operation* on S if there exists an element e E S with the propertye*x = x*e = x for every XESExample: The element 0 is an identity element with respect to the operation + on the setof integers I = { ... ,-3,-2,-1,0, 1,2, 3, ... }, sincex + 0 = 0 + x = x for any x E IThe set of natural numbers, N, has no identity element since 0 is excluded from the set.5. Inverse. A set S having the identity element e with respect to a binary operator * is saidto have an inverse whenever, for every XES, there exists an element yES such thatx * y = eExample: In the set of integers, I, with e = 0, the inverse of an element a is (-a)since a + (-a) = O.6. Distributive law. If * and are two binary operators on a set S, * is said to be distributiveover' wheneverx* (Y' z) = (x*y) (x*z)An example of an algebraic structure is afield. A field is a set of elements, together with twobinary operators, each having properties I throngh 5, and both operators combined to giveproperty 6. The set of real numbers, together with the binary operators + and " form the fieldof real numbers. The field of real numbers is the basis for arithmetic and ordinary algebra. Theoperators and postulates have the following meanings:The binary operator +defines addition.The additive identity is O.The additive inverse defines subtraction.The binary operator' defines multiplication.The multiplicative identity is 1.The multiplicative inverse of a = I/a defines division, i.e., a 1/a ~ 1.The only distributive law applicable is that of . over +:a' (b + c) = (a' b) + (a' c)AXIOMATIC DEFINITIONOF BOOLEAN ALGEBRAIn 1854 George Boole introduced a systematic treatment of logic and developed for this purposean algebraic system now called Boolean algebra. In 1938 C. E. Shannon introduced atwo-valued Boolean algebra called switching algebra, iU which he demonstrated that the prop-Section 2-2 Axiomatic Definition of Boolean Algebra 35
19. 19. 36 Chapter 2 Boolean Algebra and Logic Gates1. the elements of the set B,2. the rules of operation for the two binary operators, and3. that the set of elements, B, together with the two operators, satisfies the six Huntingtonpostulates.One can formulate many Boolean algebras, depending on the choice of elements of Bandthe mles of operation. In our subsequent work, we deal only with a two-valuedBoolean algebra,i.e., one with only two elements. Two-valued Boolean algebra has appl~catlOns III ~et t~eory(the algebra of classes) and in propositional logic. Our interest here is wlth the apphcatlOnof Boolean algebra to gate-type circuits.Two-Valued Boolean AlgebraA two-valued Boolean algebra is defined on a set of two elements, B = {O, I}, with rules forthe two binary operators + and, as shown in the following operator tables (the mle for thecomplement operator is for verification of postulate 5):Section 2-3 Basic Theorems and Properties of Boolean Algebra2-3 BAS leT H EO REM SAN D PRO PERTIE SOF BOOLEAN ALGEBRADuality37x'Y x,z (x . y) + (x . z)0 0 00 0 00 0 00 0 00 0 00 1 11 0 11 1 1x y z y+z x' (y + z)0 0 0 0 00 0 1 I 00 I 0 1 00 1 I 1 01 0 0 0 0I 0 1 I 11 1 0 1 11 1 1 1 1(b) The distributive law of + over' can be shown to hold true by means of a truth tablesimilar to the one above.S. From the complement table, it is easily shown that(a) x + x' ~ I, since 0 + 0' ~ 0 + I = I and I + I' ~ I + 0 = 1.(b) x . x' = 0, since 0 . 0' = O' I = 0 and I . I' = I . 0 .= 0, which verifies postulate5.6. Postulate 6 is satisfied because the two-valued Boolean algebra has two distinct elements,I and 0, with I '" O.We have just established a two-valued Boolean algebra having a set of two elements, I and0, two binary operators with operation rules equivalent to the AND and OR operations, and acomplement operator equivalent to the NOT operator. Thus, Boolean algebra has been definedin a formal mathematical manner and has been shown to be equivalent to the binary logic presentedheuristically in Section 1-9. The heuristic presentation is helpful in understauding theapplication of Boolean algebra to gate-type circuits. The formal preseutation is necessary fordeveloping the theorems and properties of the algebraic system. The two-valued Boolean algebradefined in this section is also called "switching algebra" by engineers. To emphasize thesimilarities between two-valued Boolean algebra and other binary systems, this algebra wascalled "binary logic" in Section 1-9. From here on, we shall drop the adjective "two-valued"from Boolean algebra in subsequent discussions.The Huntington postulates have been listed in pairs and designated by part (a) and part (b). Onepart may be obtained from the other if the bimuy operators and the identity elements are interchanged.This important property of Boolean algebra is called the duality principle. It statesthat every algebraic expression deducible from the postulates of Boolean algebra remains validif the operators and identity elements are interchanged. In a two-valued Boolean' algebra, theidentity elements and the elemeuts of the set, B, are the same: I and O. The duality principlex y x'Y x Y x+y x x'0 0 0 0 0 0 0 I0 I 0 0 1 1 t 01 0 0 1 0 11 1 1 1 1 1These rules are exactly the same as the AND, OR, and NOT operations, respectively, definedin Table 1-8. We must now show that the Huntington postulates are valid for the set B = {O, I}and the two binary operators defined before.1. Closure is obvious from the tables since the result of each operation is either 1 or 0andl,OEB.2. From the tables, we see that(a) 0 + 0 ~ 0 0 + I = I + 0 = I;(b) I . I = I I . 0 ~ O' I = O.This establishes the two identity elements, 0 for + and I for ., as defined by postulate 2.3. The com~utativelaws are obvious from the symmetry of the binary operator tables.4. (a) Thedistributivelawx'(Y + z) = (xy) + (x'z)canbeshowntoholdtruefromthe operator tables by formiug a tmth table of all possible values of x, y, and z Foreach combination, we derive x . (y + z) and show that the value IS the same as(xy) + (xz).
20. 20. 38 Chapter 2 Boolean Algebra and Logic GatesSection 2-3 Basic Theorems and Properties of Boolean Algebra 39has many applications. If the dual of an algebraic expression is desired, we simply interchangeOR and AND operators and replace I's by O's and O's by I's.. Note that theorem I (b) is the dual of theorem l(a) and that each step of the proof in part (b)IS the dual of part (a). Any dual theorem can be similarly derived from the proof of its carre.sponding pair.Basic TheoremsTable 2-1 lists six theorems of Boolean algebra and four of its postulates. The notation is simplifiedby omitting the binary operator whenever this does not lead to confusion. The theoremsand postulates listed are the most basic relationships in Boolean algebra. The theorems, like thepostulates, are listed in pairs; each relation is the dual of the one paired with it. The posmlatesare basic axioms of the algebraic structure and need no proof. The theorems must be provenfrom the postulates. The proofs of the theorems with one variable are presented below. At theright is listed the number of the postulate that justifies each step of the proof.THEOREM 1(a): x + x = x.=x+O=xby postulate: 2(b)5(a)4(b)2(b)5(a)THEOREM 3: (x')' = x. From postulate 5, we have x + x' = I and x . x' = 0, which definesthe complement ofx. The complement ofx' is x and is also (x')'. Therefore, since the ComplementThe theorems iuvolving two or three variables may be proven algebraically from the postulatesand the theorems that have already been proven. Take, for example, the absorption theorem.THEOREM 6(a): x + xy = x.by postulate: 2(b)4(a)3(a)2(a)2(b)THEOREM 2(a): x + 1 = 1.x + I = I . (x + I)=(x + x')(x + I)= x + x' 1= x + x'= 1THEOREM 2(b): x 0 = 0 by dUality.IS umque, we have that (x')' = x.x+xy=x'l+xy= x(l + y)= x(y + I)~ x Ix V xV x + xV0 0 0 00 1 0 01 0 0 11 1 1 1=xTHEOREM 6(b): x(x + y) = xby duality.The theorems of Boolean algebra can be shown to hold true by means of truth tables. In truthtables, both sides ofthe relation are checked to yield identical results for all possible combinatIOnsof vanables Involved. The following truth table verifies the first absorption theorem.IiIIiIby postulate: 2(b)5(a)4(b)5(b)2(a)by posmlate: 2(a)5(b)4(a)5(a)2(b)Cb) x 1 -xCbl x . x' ~ 0Cb) x x = xCb) x . 0 ~ 0(b) xy = yxCb) x(yz) ~ (xy)zCb) x + yz ~ (x + y)(x + z)(b) (xy)' = x' + y'Cb) x(x + y) ~ xTHEOREM l(b): x x = x.xx=xx+O=xx+xx'(a) x + 0 = xCal x + x' ~ 1Ca) x + x ~ xCa) x + 1 = 1(x')' ~ xCal x + y ~ y + x .Ca) x + (y + z) = (x +y) + zCa) x(y + z) ~ xy + xzCa) (x + y)' = x'y'Ca) x + xy ~ xPostulate 2Postulate 5Theorem 1Theorem 2Theorem 3, involutionPostulate 3, commutativeTheorem 4, associativePostulate 4, distributiveTheorem 5, DeMorganTheorem 6, absorptionx + x = (x + x) 1= (x + x)(x + x')= x + XXi= x(x + x')= x ITable 21Postulates and Theorems of Boolean Algebra
21. 21. .1!I!__]I!!!!"o!!!!!!~!!!I!!!!!!!!!!I!!!!!!!!!!!!!!!!!!!!!I!!!!!!I!I _40 Chapter 2 Boolean Algebra and Logic GatesThe algebraic proofs of the associative law and DeMorgan's theorem are long and will not beshown here. However, their validity is easily shown with truth tables. For example, the truth tablefor the first DeMorgan's theorem (x + y)' = x'y' is shown below.Section 2-4 Boolean Functions 41Table 2-2Truth Tables for F, and F2x y z F, F.0 0 0 0 00 0 1 1 10 1 0 0 00 1 1 0 11 0 0 1 11 0 1 1 11 1 0 1 01 1 1 1 0x-- ---y-----) >0----1z ---_-----''---__-1FIGURE 2-1Gate implementation of F, ~ x + y' ztions. The table shows that the function is equal to I when x = 1 or when yz = 01. It is equalto 0 otherwise.A Boolean function can be transformed from an algebraic expression into a circuit diagramcomposed of logic gates. The logic-circuit diagram for F, is shown in Fig. 2-1. There is an inverterfor input y to generate the complement. There is anAND gate for the term y'z and an ORgate that combines the two terms. In logic-circuit diagrams, the variables of the function aretaken as the inputs of the circnit and the binary variable FJ is taken as the output of the circuit.There is only one way that a Boolean function can be represented in a truth table. However,when the function is in algebraic form, it can be expressed in a variety of ways. The particularexpresslOn used to designate the function will also dictate the interconnection of gatesin the logic circuit diagram. By manipulating a Boolean expression according to Boolean algebrarules, it is sometimes possible to obtain a simpler expression for the same function andthus reduce the number of gates in the circuit and the number of inputs to the gate. Considerfor example the following Boolean function:F2 = x'y'Z + x'YZ + xy'The implementation of this function with logic gates is shown in Fig. 2-2(a). Input variables xand yare complemented with inverters to obtain x' and y'. The three terms in the expressionare implemented with three AND gates. The OR gate forms the logical OR of the three terms.The truth table for F, is listed in Table 2-2. The function is equal to 1 when xyz = 001 or 011x' y' x'y'1 1 11 0 00 1 00 0 0x y x+y (x + y)'0 0 0 10 1 1 01 0 1 01 1 1 0Operator PrecedenceThe operator precedence for evaluating Boolean expressions is (1) parentheses, (2) NOT, (3)AND, and (4) OR. In other words, the expression inside the parentheses must be evaluated beforeall other operations. The next operation that holds precedence is the complement, then followsthe AND, and finally the OR. As an example, consider the truth table for DeMorgan'stheorem. The left side of the expre