1- Introduction. Bit: The most basic unit of information in a digital computer (On/Off ; 0/1 state)

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2 Objectives: Chapter 2 : Introduction 1- Introduction 2- Positional Numbering System 3- Decimal to binary conversion 4- Signed integer representation 5- Floating-point representation 1- Introduction Bit: The most basic unit of information in a digital computer (On/Off ; 0/1 state) Byte: A set of 8bits Word: two or more adjacent bytes that are manipulated collectively Word size: The size of a word in bits depends on the computer organization (16, 32, 64 bits, ) Nibbles (or nybbles): set of 4 bits Usually a set of 8 bits is divided into two nibbles, a low order nibble and a high order nibble 1

3 2- Positional Numbering System Any numeric value is represented through increasing powers of a radix (or base) The set of valid numerals (digits) is equal in size to the radix of that system The least numeral is 0 and the highest one in 1 smaller than the radix Example: In the decimal system (base 10) The radix is 10 The number of valid numerals is 10 (equal to the radix) The set of valid numerals is: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} The most important radices (bases) in computer science) are: Binary Radix 2 or base 2 Numerals: {0, 1} Octal Radix 8 or Base 8 Numerals: {0, 1, 2, 3, 4, 5, 6, 7} Hexadecimal Radix 16 or base 16 Numerals: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F} (Binary to Decimal conversion) Any numeric value is represented through increasing powers of a radix (or base) = 2x x x3 0 = = 1x x x x x x x2-2 = = 1x x x x x x x5-4 = Questions: Convert the following decimal fractions to binary with a maximum of six places to the right of the binary point: a b c d Ans a b c d

4 3- Decimal to binary conversion Keep in mind the following tables or how to obtain them! A real number can take any value (ex ; ) Whole number: No fractions (ex: 10, 1231, 3543,, -12, , ) Unsigned number: Only positive numbers (ex: , , 12357, ) Unsigned whole numbers: No fraction and only positive numbers : Division-remainder 3

5 A binary number with N bits can represent 2 N unsigned integers from 0 to 2 N -1 Example: Having N=4 bits, we can represent 2 4 = 16 unsigned integers from 0 to 2 4-1=16-1=15 The number 16 CANNOT be represented with only 4 bits!! Example: Convert to base 3 using the division-remainder method remainder of remainder of remainder of 2 (104) 10 = (10212) remainder of remainder of 1 From Down to UP. 0 Questions: 1. Perform the following base conversions using subtraction or division-remainder: a = 3 b = 5 c = 7 d = 9 e = 3 f = 5 g = 7 h = 9 Ans. a b c d a b c d

6 Converting Decimal to Binary fractions Radix points (.) separate the integer part of a number from its fractional part Example of fractions : Base 10 : Base 3 : Base 2 : The radix point is called a decimal point in a decimal system, a binary point in a binary system, and so on To convert fractions from decimal to any other base system we repeatedly multiply by the destination radix Convert to base x 5 = The integer part is 2 From UP to x 5 = The integer part is 0 Down x 5 = The integer part is x 5 = The integer part is 4 the fractional part is zero, we are done (0.4304) 10 = (0.2034) 5 Some fractions in one base could be indeterminate Fractions that contain repeating strings of digits to the right of the radix point Example: (2/3) 10 =(0.666 ) 10 An indeterminate fraction in one base could be determinate in another base (and vice-versa). Example: (2/3) 10 =0.2 3 =(0.666 ) 10-2/3 is indeterminate in base 10 but determinate in base 3. When a fraction is indeterminate, an approximation is needed We fix the number of digits to the right of the radix point Also, approximation is needed due to the limited computing resources (example: limited size of the processor s registers) 5

7 Example: Convert to binary with 4 bits to the right of the binary point x 2= The integer part is x 2= The integer part is 1 Down x 2= The integer part is x 2= The integer part is 1 ( ) 10 = (0.0101) 2 Convert the following decimal fractions to binary with a maximum of six places to the right of the binary point: a b c d Ans. a b c d Converting between Power-of-Two Radices To convert between any base to any other base (different than base 10), it is easier to pass through base 10. Example: 31214= ( )3? First step: = 3x x x x4 0 = Second step: by using the division-remainder method: = So 31214=

8 Converting from Binary to Octal Each octal digit is equivalent to a group of 3 binary digits called octet Each hexadecimal digit is equivalent to a group of 4 binary digits called hextet Example: Convert to octal 1- Make Groups of 3 bits (from right to left): Add zero(s) on the left to complete the last octet Convert each octet to its corresponding octal digit Finally: = Converting from Binary to Hexadecimal Example: Convert to hexadecimal 1- Make Groups of 4 bits (from right to left): Add zero(s) on the left to complete the last hextet Convert each hextet to its corresponding hexadecimal digit C 9 D 4- Finally: = 2C9D16 (MTA 1st Semester 2012/2013) Convert the following binary numbers to the destination base. Show your conversion steps clearly. a. ( ) 2 = (114) 8 b. ( ) 2 = (D5) 16 or (0D5) 16 7

9 4- Signed integer representation An integer is a whole number Signed integers are the set of positive and negative whole numbers How should we encode and deal with the actual sign of the number? Two concepts are used Signed Magnitude concept Complement concept Signed Magnitude Signed magnitude is the most intuitive method The MSB (Most Significant Bit) of a binary number is kept as the sign of the number MSB = 1: negative number MSB = 0: positive number The remaining bits represent the magnitude (or absolute value) of the numeric value Example: In a 8 bit word signed magnitude system give the decimal representation of the following numbers ? - The MSB is 0: The number is positive - The remaining 7 bits are: = The decimal number is ? - The MSB is 1: The number is negative - The remaining 7 bits are: = The decimal number is -1 Example: In a 8 bit word signed magnitude system give the decimal representation of the following numbers ? - The MSB is 1: The number is negative - The remaining 7 bits are: = The decimal number is ? - The MSB is 0: The number is positive - The remaining 7 bits are: = The decimal number is +65 8

10 In a N bit word signed magnitude system 1 bit is used for the sign of the number N-1 bits are used for the magnitude of the number The largest integer is 2 N-1-1 The smallest integer is -(2 N-1-1) Example: in a 8 bit word signed magnitude system The largest integer is = = The smallest integer is = -(2 7-1) = Steps that guides you to solve the Signed-magnitude arithmetic At first we look at the signs of the two operands We arrange the operands in a certain way based on their signs We perform the calculation without regard to the signs Finally, we supply the sign as appropriate 1- Adding operands that have the same sign Example: Add to using signed-magnitude arithmetic. sign carries (79) (35) (114) We find = in signed-magnitude representation. Overflow condition(important) In the last example, adding the seventh bits to the left gives no carry If there is a carry, we say that we have an overflow condition and the carry is discarded, resulting in an incorrect sum. See the Next Example for overflow condition: 9

11 Example: Add to using signed-magnitude arithmetic 1 carries (65) (97) (34) The previous addition overflows The last carry is discarded The sum s result is incorrect 2- Subtracting operands Example1: Subtract from using signedmagnitude arithmetic borrows (99) (79) (20) We find = in signed-magnitude representation. 10

12 Example 2: Subtract from using signedmagnitude arithmetic. Here the subtrahend, , is larger than the minuend, With the result obtained in Example 1, we know that the difference of these two numbers is Because the subtrahend is larger than the minuend, all that we need to do is change the sign of the difference. So we find = in signedmagnitude representation Example 3: Add to using signed-magnitude arithmetic = 13 and = -19 The result is negative We subtract 13 from 19 The result of the binary subtraction is: (-6) Example 4: Subtract (-24) from (-43) using signed-magnitude arithmetic. This is equivalent to adding -43 to 24 The result is negative We subtract 24 from 43 The result of the binary subtraction is: (-19) 11

13 General rules when operands have different signs 1- Determine which operand has the larger magnitude 2- The sign of the result is the same as the sign of the operand with the larger magnitude 3- the magnitude must be obtained by subtracting (not adding) the smaller one from the larger one Questions: Use signed magnitude arithmetic to add to (explain and show your detailed calculations clearly). Do we have an overflow? Why? = +103; = -31 we are adding a positive number to a negative one. So we should subtract 31 from 103 and the result is positive. The result is = There is no overflow because we are subtracting. No overflow occurs when subtracting (Important) -Subtract from ( ) using 8-bit signed magnitude arithmetic (explain and show your detailed calculations clearly). The result should be in signed magnitude format. What can you notice? Sol. The result is = 910 which is wrong! (you must explain) Use signed magnitude arithmetic to Calculate: - add to sum of -40 and 13 - Subtract -2 from Problems related to signed magnitude Too much decisions to make (larger number? ; borrows? ; what signs?). The number 0 could have two representations : and Complicated method Expensive circuits 12

14 Complement system Complement system is used to represent/convert negative numbers only When using complement system the subtraction is converted to an addition Advantages of complement system Simplify computer arithmetic No need to process sign bits separately The sign of a number is easily checked by looking at its high-order bit (MSB). One's Complement We say that we work in one s complement (C1) To convert a negative number to its one s complement then switching all of the 1s with 0s and vice versa!! A positive number is directly converted to its binary representation. Example: Express 2310 and -910 in 8-bit binary one s complement form = + ( ) = C1-910 = - ( ) = C1 Questions: Show the representation (assuming 8-bit registers) using signed-1's complement: a. 77 b. -42 c. 119 d. 107 Sol. One's complement: One's complement: One's complement: One's complement:

15 Subtraction and Addition in one's complement: In one s compliment the subtraction is converted into addition Example: = (-910) Example1: Add 2310 to -910 using 8-bit binary one s complement arithmetic. The result is C1 = +( ) = 1410 Example2 : Add 910 to using 8-bit binary one s complement arithmetic = - ( )2 = C1 910 = + ( ) = C (-2310) = C C1 Result: C1 = -( ) =

16 Disadvantages of one's complement In One s complement, we still have two representations for zero: and Computer engineers long ago stopped using one s complement A more efficient representation for binary numbers is the two s complement Two's Complement We say that we work in two s complement(c2) To convert a negative number to its Two's complement then convert it first to one's complement then add one A positive number is directly converted to its binary representation Example: Express 2310, -2310, and -910 in 8-bit binary two s complement form = + ( ) = c = -( ) = c1 + 1 = c2-910 = -( ) = c1 + 1 = c2 Questions: Show the representation of -16 (assuming 8-bit registers) using signed-2's complement: a. 77 b. -42 c. 119 d. 107 Sol. One's complement: One's complement: One's complement: One's complement:

17 Subtraction and Addition in two's complement: Unlike C1 arithmetic, in C2 the last carry is discarded Example 1: Add 910 to using two s complement arithmetic. The result is C2 = -( ) = Note in the previous example how a negative binary number in C2 is converted to decimal At first all 0 and 1 in the C2 s number are switched: A 1 is then added to the last number: = So C2 = -( ) = Example 2: Find the sum of 2310 and -910 in binary using two s complement arithmetic = +( )2 = C2-910 = -( ) = C (-910) = C C2 Result: C2 = +( ) =

18 Advantages of two s complement It is the most popular choice for representing signed numbers The algorithm for adding and subtracting is quite easy It has the best representation for 0 (all 0 bits) It is self-inverting It is easily extended to larger numbers of bits. Drawback the asymmetry seen in the range of values that can be represented by N bits. Examples: With signed-magnitude, 4 bits allow us to represent the values -7 (11112) through +7 (01112). Using two s complement, we can represent the values: -8 (1000C2) through +7 (0111C2) Overflow in complement systems (C1 and C2) An overflow occurs if two positive numbers are added and the result is negative or if two negative numbers are added and the result is positive. It is not possible to have overflow when if a positive and a negative number are being added together. To Detect Overflow Check the last two carries - If these are different: there is an overflow - If these are equal: there is no overflow Example 1: Find the sum of and 810 in binary using two s complement arithmetic. The result is C2 = -( )2 = !!! Note that the last two carries are different 17

19 Questions : Given the 8-bit binary number: What decimal number does this represent if the computer uses: a. signed-magnitude representation b. signed-1's complement c. signed-2's complement Ans. a b c Assuming 2 s complement 8-bit representation, consider the following: +70 = = Is this correct? Why or why not? Ans. No. The result is a different sign than the two numbers we are adding so overflow has occurred. Express how the number zero is represented in an 8-bit signed magnitude system, 8-bit Complement one system and 8-bit complement two system? (MID-TERM Make Up) Signed magnitude: ; Complement one: ; Complement two: MTA 1st Semester 2012/2013 Consider the decimal numbers -93. a) Represent this number in an 8-bit binary signed magnitude form. b) Represent this number in an 8-bit binary two complement s form. c) Use 8-bit two complement s arithmetic to subtract 93 from (+103). Show your calculations clearly. Show the result in two complement and decimal form. Hint: you should calculate MTA 2nd Semester 2012/2013 Consider the binary number What does this binary number represent in decimal (base 10) when it is used in: a) An 8-bit unsigned whole numbers system b) An 8-bit signed magnitude system? c) An 8-bit one s complement system? d) An 8-bit two s complement system? - Find the sum of -40 and 13 using 8-bit two s complement arithmetic (explain and show your detailed calculations clearly). Give the result in complement 2 and decimal. 18

20 (MTA 2nd Semester 2012/2013) A computer system has two different ways to represent signed numbers: Signed magnitude and two s complement. a) Show how this computer represents the number 7710 in these two different binary systems (an 8-bit signed magnitude and 8-bit two s complement) = ( )10= Since 77 is positive, 7710 = in signed magnitude and two s complement system. b) Show how this computer represents the number (-42)10 in these two different binary systems (an 8-bit signed magnitude and 8-bit two s complement) = ( )10 = In signed magnitude system: = - (101010)2 = In two s complement system: = -( )2 = ( )C2= C2 c) Show how this computer uses two s complement arithmetic to add these numbers in binary (77 + (-42)). Is there an overflow? Why or why not? There is no overflow because the last two carries are identical. 19

21 5- Floating-point representation Computers use a form of scientific notation for floatingpoint representation Numbers written in scientific notation have three components: Scientific notation in base 10: Scientific notation in base 2: In digital computers, floating-point numbers consist of three parts: A sign bit, an exponent part: representing the exponent on a power of 2, a fractional part called a significand: which is a fancy word for a mantissa. 20

22 For simplicity, in all this course, we will use a simplified 14 bits model Sign bit: 1 bit Exponent: 5 bits Significand: 8 bits Exercise 1: Represent the number 17 in a 14 bits floating point representation 17 = 17.0 x 10 0 = 1.7 x 10 1 = 0.17 x 10 2 Analogically in binary: 1710 = x 2 0 = x 2 1 = x 2 2 = x2 3 = x 2 4 = x 2 5 = x 2 6 = x 2 7 =... As a convention, we stop when the MSB of the significant is 1 : x 2 5 The exponent is 510 = The significant is: So:

23 The last floating point representation is not suitable for negative exponents Example: - the number 0.25 = = 0.12 x How to represent the negative exponent -1?! To solve such problems we use an excess-16 bias All negative and positive exponents are added by 16 We say that the real exponent is replaced by a biased exponent All exponents are converted to positive biased exponents With an excess-16 bias Exponent values less than 16 will indicate negative exponent values Exponent values more than 16 will indicate positive exponent values exponents of all zeros or all ones are typically reserved for special numbers (such as zero or infinity). 22

24 Example 1: Represent the number 17 in a 14 bits floating point form with excess-16 bias The number is positive: sign bit is = x 2 5 The exponent is 510 (5+16)10 = 2110 = The significant is: So 17 in floating point form with excess-16 bias is: Example 2: Represent the number in a 14 bits floating point form with excess-16 bias. The number is positive: sign bit is = x 2 0 = 0.12 x 2-1 The exponent is -110 (-1+16)10 = 1510 = The significant is So 0.25 in floating point form with excess-16 bias is: Example 3: Express in normalized floatingpoint form with excess-16 bias. The number is negative: sign bit is = = x2 0 = x2-1 = = 0.1x2-4 The exponent is -410 (-4+16)10 = 1210 = The significant is So in floating point form with excess-16 bias is:

25 Floating-point arithmetic To add/subtract two numbers in floating point form Both numbers should have the same exponent If exponents are different we change one of the numbers so that both of them are expressed in the same power of the base We add the binary numbers We represent the result in a normalized floating point form Example: Add the following binary numbers as represented in a normalized 14-bit format with an excess-16 bias. The second number is x2 0 The first number is x2 2 = x2 0 Now : The result is x 2 0 = x 2 2 In floating point form with excess

26 Question 1: A computer system uses a simplified 14-bit floating point with excess-16 bias to represent binary numbers. a) How does this system represent the number in binary? (MTA 2013) b) Show how this floating point system represents the decimal number in binary. (MTA) c) How does this system represent the number -33 in binary? +33? ( 2012 MOCK MTA) d) Express in the revised 14-bit floating-point model with bias = 16 (TMA ) e) Show how the number -58 is represented in binary (TMA 2013) (1 mark for the exponent, 0.5 mark for mantissa, 0.5 mark for sign bit, 0.5 mark for correct representation of the whole 14-bit binary number). Sol. a) Sign bit = 1; Exponent = = 1310 = (01101)2 ; Mantissa: Binary number in floating point format (with excess 16) is : b) 15.5 = = x 2 4 Exponent = 4 with excess 16 : 4+16 = = Mantissa: The system represents 15.5 in binary as follows: c) 33 = (100001)2 = ( )2 x 2 6 Exponent = = 2210 = (10110)2 ; Mantissa: in floating point format (with excess 16) is : in floating point format (with excess 16) is : d) or e) = -(111010)2 = x 2 6 (1.5 Marks) The sign bit is 1 : The number is negative. (1 mark) The exponent is: = 2210 = ; (2 marks) Significand: (1 Mark) The number in floating point form: (0.5 Mark) 25

27 Question 2: Consider the binary number: It is represented in a normalized 14-bit format (1 sign bit, 5 bits for exponent and 8 bits for the significand/mantissa) with an excess-16 bias. Convert this number to float (in base 10). The binary number in floating point format is: The sign bit is 1 : The number is negative. (1 mark) The exponent is: = 1510 ; After removing the exess-16 bias, the exponent becomes = -1. So the real exponent is -1. (2 marks) So the number in base two is: -( )2 x 2-1 = -(0.0101)2 (1 Mark) In decimal: -(0.0101)2 = (0x x x x2-4)10 = ( )10 = ( )10 = Question 3: Part a: Calculate the bias of the 14-bit floating-point model.[5 marks] Part b: Convert the below model to float: [5 marks] Answer: Part a: bias = 2 power 5 = 2 power 4 = 16 2 Part b: The sign bit is 1, negative number = 21 in base 10, the bias = 16, the exponent = = 5 The significant = The answer = -0.1 * 2 power 5 = 2 power 4 = -16 Question 4: is the binary representation of the unsigned decimal number 93. What is the binary representation of -93 in an 8-bit Complement one system, 8-bit Complement two system and a 14-bit floating point with an excess-16 bias system (1 sign bit, 5 bits exponent and 8 bits significand)? In a complement one system: = -( )2 = C1 (1.5 Mark) In a complement two system: = -( )2 = C1= C2 (1.5 Mark) In a floating point system: -93= -( )2 = -( )2 x 27 = -( )2 x 27 (1 Mark) Sign bit: 1; Exponent = 7+16 = 2310 = ; Mantissa: (3x1 Mark = 3Marks) So, the number in a floating point format is : (0.5 Mark) 26

28 Question 5: Assume a simple model for floating-point representation is used, the representation uses a 14-bit format, 5 bits for the exponent, with a bias of 16, a normalized significand of 8-bits, and a single sign bit for the number: I. Show how the computer would represent the numbers [20.0] and [0.375] using this floating-point format. [4 marks] II. Show how the computer would add the two floating-point numbers in part-i by changing one of the numbers so they are both expressed using the same power. III. Show how the computer would represent the sum in part-ii. [2 mark] IV. What decimal value for the sum is the computer actually storing? [1 mark]