![Two’s Complement Shortcuts
• Algorithm 1 – Simply complement each bit and then
add 1 to the result.
Finding the 2’s complement of (01100101)2 and of its
2’s complement…
N = 01100101 [N] = 10011011
10011010 01100100
+ 1 + 1
--------------- ---------------
10011011 01100101
• Algorithm 2 – Starting with the least significant bit,
copy all of the bits up to and including the first 1 bit
and then complementing the remaining bits.
N = 0 1 1 0 0 1 0 1
[N] = 1 0 0 1 1 0 1 1](https://image.slidesharecdn.com/digitalelectronics-231015153837-d4327054/85/digitalelectronics-ppt-55-320.jpg)
digitalelectronics.ppt
- 1. DIGITAL ELECTRONICS Dr.M.MANIKANDAN Associate Professor Department of Electronics and Engg. MIT- Campus Anna University
- 5. Digital Computer Systems • Digital systems consider discrete amounts of data. • Examples 26 letters in the alphabet 10 decimal digits • Larger quantities can be built from discrete values: Words made of letters Numbers made of decimal digits (e.g. 239875.32) • Computers operate on binary values (0 and 1) • Easy to represent binary values electrically Voltages and currents. Can be implemented using circuits Create the building blocks of modern computers
- 6. A basic organization of a digital computer
- 7. Types of Systems • With no state present – Combinational logic system – Output = Function (Input) • With state present – State updated at discrete times (e.g. once per clock tick) Synchronous sequential system – State updated at any time Asynchronous sequential system
- 8. Example : Digital Counter (e.g., Odometer) • Inputs : Count Up, Reset • Outputs : Visual Display • State : “Value of stored digits • Is this system synchronous or asynchronous? 0 0 1 3 5 6 4 UP RESET
- 13. Understanding Decimal Numbers • Decimal numbers are made of decimal digits: (0,1,2,3,4,5,6,7,8,9) • But how many items does a decimal number represent? 8653 = 8x103 + 6x102 + 5x101 + 3x100 • What about fractions? 97654.35 = 9x104 + 7x103 + 6x102 + 5x101 + 4x100 + 3x10-1 + 5x10-2 In formal notation -> (97654.35)10 • Why do we use 10 digits, anyway?
- 14. Understanding Octal Numbers • Octal numbers are made of octal digits: (0,1,2,3,4,5,6,7) • How many items does an octal number represent? (4536)8 = 4x83 + 5x82 + 3x81 + 6x80 = (1362)10 • What about fractions? (465.27)8 = 4x82 + 6x81 + 5x80 + 2x8-1 + 7x8-2 • Octal numbers don’t use digits 8 or 9 • Who would use octal number, anyway?
- 15. Understanding Binary Numbers • Binary numbers are made of binary digits (bits): 0 and 1 • How many items does an binary number represent? (1011)2 = 1x23 + 0x22 + 1x21 + 1x20 = (11)10 • What about fractions? (110.10)2 = 1x22 + 1x21 + 0x20 + 1x2-1 + 0x2-2 • Groups of eight bits are called a byte (11001001) 2 • Groups of four bits are called a nibble. (1101) 2
- 16. Why Use Binary Numbers? • Easy to represent 0 and 1 using electrical values. • Possible to tolerate noise. • Easy to transmit data • Easy to build binary circuits. AND Gate 1 0 0
- 17. Conversion Between Number Bases Decimal(base 10) Octal(base 8) Binary(base 2) Hexadecimal (base16) • Learn to convert between bases. • Already demonstrated how to convert from binary to decimal. • Hexadecimal described in next lecture.
- 20. Converting Binary to Decimal • To Convert to decimal, use decimal arithmetic to sum the weighted powers of two: • Converting 110102 to N10: N10 = 1 x 24 x 1x 23 + 0 x 22 + 21 + 0 + 20 = 26
- 33. Gray Code • Gray code is not a number system. • It is an alternate way to represent four bit data • Only one bit changes from one decimal digit to the next • Useful for reducing errors in communication. • Can be scaled to larger numbers. Digit Binary Gray Code 0 0000 0000 1 0001 0001 2 0010 0011 3 0011 0010 4 0100 0110 5 0101 0111 6 0110 0101 7 0111 0100 8 1000 1100 9 1001 1101 10 1010 1111 11 1011 1110 12 1100 1010 13 1101 1011 14 1110 1001 15 1111 1000
- 35. Binary Arithmetic • Single Bit Addition with Carry • Multiple Bit Addition • Single Bit Subtraction with Borrow • Multiple Bit Subtraction • Multiplication • BCD Addition
- 36. Binary Addition • Binary addition is very simple. • This is best shown in an example of adding two binary numbers… 1 1 1 1 0 1 + 1 0 1 1 1 - - - - - - - - - - - - - - - - - - 0 1 0 1 1 1 1 1 1 1 1 0 0 carries
- 37. Binary Subtraction • We can also perform subtraction (with borrows in place of carries). • Let’s subtract (10111)2 from (1001101)2… 1 10 0 10 10 0 0 10 1 0 0 1 1 0 1 - 1 0 1 1 1 ------------------------ 1 1 0 1 1 0 borrows
- 38. Binary Multiplication • Binary multiplication is much the same as decimal multiplication, except that the multiplication operations are much simpler… 1 0 1 1 1 X 1 0 1 0 ----------------------- 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 ----------------------- 1 1 1 0 0 1 1 0
- 41. Parity Codes • Parity codes are formed by concatenating a parity bit, P to each code word of C. • In an odd-parity code, the parity bit is specified so that the total number of ones is odd. • In an even-parity code, the parity bit is specified so that the total number of ones is even. Information Bits P 1 1 0 0 0 0 1 1 Added even parity bit 0 1 0 0 0 0 1 1 Added odd parity bit
- 42. Parity Code Example • Concatenate a parity bit to the ASCII code for the characters 0, X, and = to produce both odd-parity and even-parity codes. Character ASCII Odd-Parity ASCII Even-Parity ASCII 0 0110000 10110000 00110000 X 1011000 01011000 11011000 = 0111100 10111100 00111100
- 43. ASCII Code • American Standard Code for Information Interchange • ASCII is a 7-bit code, frequently used with an 8th bit for error detection (more about that in a bit). Character ASCII (bin) ASCII (hex) Decimal Octal A 1000001 41 65 101 B 1000010 42 66 102 C 1000011 43 67 103 … Z a … 1 ‘
- 45. ASCII Codes and Data Transmission • ASCII Codes • A – Z (26 codes), a – z (26 codes) • 0-9 (10 codes), others (@#$%^&*….) • Complete listing in Mano text • Transmission susceptible to noise • Typical transmission rates (1500 Kbps, 56.6 Kbps) • How to keep data transmission accurate?
- 46. Overview • Hexadecimal numbers • Related to binary and octal numbers • Conversion between hexadecimal, octal and binary • Value ranges of numbers • Representing positive and negative numbers • Creating the complement of a number • Make a positive number negative (and vice versa) • Why binary?
- 47. Understanding Binary Numbers • Binary numbers are made of binary digits (bits): 0 and 1 • How many items does an binary number represent? (1011)2 = 1x23 + 0x22 + 1x21 + 1x20 = (11)10 • What about fractions? (110.10)2 = 1x22 + 1x21 + 0x20 + 1x2-1 + 0x2-2 • Groups of eight bits are called a byte (11001001) 2 • Groups of four bits are called a nibble. (1101) 2
- 48. Understanding Hexadecimal Numbers • Hexadecimal numbers are made of 16 digits: (0,1,2,3,4,5,6,7,8,9,A, B, C, D, E, F) • How many items does an hex number represent? (3A9F)16 = 3x163 + 10x162 + 9x161 + 15x160 = 1499910 • What about fractions? (2D3.5)16 = 2x162 + 13x161 + 3x160 + 5x16-1 = 723.312510 • Note that each hexadecimal digit can be represented with four bits. (1110) 2 = (E)16 • Groups of four bits are called a nibble. (1110) 2
- 49. Putting It All Together • Binary, octal, and hexadecimal similar • Easy to build circuits to operate on these representations • Possible to convert between the three formats
- 50. Converting Between Base 16 and Base 2 • Conversion is easy! • Determine 4-bit value for each hex digit • Note that there are 24 = 16 different values of four bits • Easier to read and write in hexadecimal. • Representations are equivalent! 3A9F16 = 0011 1010 1001 11112 3 A 9 F
- 51. Converting Between Base 16 and Base 8 1. Convert from Base 16 to Base 2 2. Regroup bits into groups of three starting from right 3. Ignore leading zeros 4. Each group of three bits forms an octal digit. 3A9F16 = 0011 1010 1001 11112 3 A 9 F 352378 = 011 101 010 011 1112 5 2 3 7 3
- 52. How To Represent Signed Numbers • Plus and minus sign used for decimal numbers: 25 (or +25), -16, etc. • For computers, desirable to represent everything as bits. • Three types of signed binary number representations : signed magnitude, 1’s complement, 2’s complement. • In each case: left-most bit indicates sign: positive (0) or negative (1). 000011002 = 1210 Sign bit Magnitude 100011002 = -1210 Sign bit Magnitude
- 53. One’s Complement Representation • The one’s complement of a binary number involves inverting all bits. • 1’s comp of 00110011 is 11001100 • 1’s comp of 10101010 is 01010101 • For an n bit number N the 1’s complement is (2n-1) – N. • Called diminished radix complement by Mano since 1’s complement for base (radix 2). • To find negative of 1’s complement number take the 1’s complement. 000011002 = 1210 Sign bit Magnitude 111100112 = -1210 Sign bit Magnitude
- 54. Two’s Complement Representation • The two’s complement of a binary number involves inverting all bits and adding 1. • 2’s comp of 00110011 is 11001101 • 2’s comp of 10101010 is 01010110 • For an n bit number N the 2’s complement is (2n-1) – N + 1. • Called radix complement by Mano since 2’s complement for base (radix 2). • To find negative of 2’s complement number take the 2’s complement. 000011002 = 1210 Sign bit Magnitude 111101002 = -1210 Sign bit Magnitude
- 55. Two’s Complement Shortcuts • Algorithm 1 – Simply complement each bit and then add 1 to the result. Finding the 2’s complement of (01100101)2 and of its 2’s complement… N = 01100101 [N] = 10011011 10011010 01100100 + 1 + 1 --------------- --------------- 10011011 01100101 • Algorithm 2 – Starting with the least significant bit, copy all of the bits up to and including the first 1 bit and then complementing the remaining bits. N = 0 1 1 0 0 1 0 1 [N] = 1 0 0 1 1 0 1 1
- 56. Finite Number Representation • Machines that use 2’s complement arithmetic can represent integers in the range -2n-1 <= N <= 2n-1-1 where n is the number of bits available for representing N. Note that 2n-1-1 = (011..11)2 and –2n-1 = (100..00)2 • For 2’s complement more negative numbers than positive. • For 1’s complement two representations for zero. • For an n bit number in base (radix) z there are zn different unsigned values. (0, 1, …zn-1)
- 57. 1’s Complement Addition • Using 1’s complement numbers, adding numbers is easy. • For example, suppose we wish to add +(1100)2 and +(0001)2. • Let’s compute (12)10 + (1)10. (12)10 = +(1100)2 = 011002 in 1’s comp. (1)10 = +(0001)2 = 000012 in 1’s comp. 0 1 1 0 0 + 0 0 0 0 1 -------------- 0 0 1 1 0 1 0 -------------- 0 1 1 0 1 Add carry Final Result Step 1: Add binary numbers Step 2: Add carry to low-order bit
- 58. 1’s Complement Subtraction • Using 1’s complement numbers, subtracting numbers is also easy. • For example, suppose we wish to subtract +(0001)2 from +(1100)2. • Let’s compute (12)10 - (1)10. (12)10 = +(1100)2 = 011002 in 1’s comp. (-1)10 = -(0001)2 = 111102 in 1’s comp. Step 1: Take 1’s complement of 2nd operand Step 2: Add binary numbers Step 3: Add carry to low order bit Add carry Final Result 1’s comp Add 0 1 1 0 0 - 0 0 0 0 1 -------------- 0 1 1 0 0 + 1 1 1 1 0 -------------- 1 0 1 0 1 0 1 -------------- 0 1 0 1 1
- 59. 2’s Complement Addition • Using 2’s complement numbers, adding numbers is easy. • For example, suppose we wish to add +(1100)2 and +(0001)2. • Let’s compute (12)10 + (1)10. (12)10 = +(1100)2 = 011002 in 2’s comp. (1)10 = +(0001)2 = 000012 in 2’s comp. 0 1 1 0 0 + 0 0 0 0 1 ---------------- 0 0 1 1 0 1 Final Result Step 1: Add binary numbers Step 2: Ignore carry bit Add Ignore
- 60. 2’s Complement Subtraction • Using 2’s complement numbers, follow steps for subtraction • For example, suppose we wish to subtract +(0001)2 from +(1100)2. • Let’s compute (12)10 - (1)10. (12)10 = +(1100)2 = 011002 in 2’s comp. (-1)10 = -(0001)2 = 111112 in 2’s comp. 0 1 1 0 0 - 0 0 0 0 1 ----------------------- 0 1 1 0 0 + 1 1 1 1 1 ------------------------- 1 0 1 0 1 1 Final Result Step 1: Take 2’s complement of 2nd operand Step 2: Add binary numbers Step 3: Ignore carry bit 2’s comp Add Ignore Carry
- 61. 2’s Complement Subtraction: Example # 2 • Let’s compute (13)10 – (5)10. (13)10 = +(1101)2 = (01101)2 (-5)10 = -(0101)2 = (11011)2 • Adding these two 5-bit codes… • Discarding the carry bit, the sign bit is seen to be zero, indicating a correct result. Indeed, (01000)2 = +(1000)2 = +(8)10. 0 1 1 0 1 + 1 1 0 1 1 -------------- 1 0 1 0 0 0 carry
- 62. 2’s Complement Subtraction: Example #3 • Let’s compute (5)10 – (12)10. (-12)10 = -(1100)2 = (10100)2 (5)10 = +(0101)2 = (00101)2 • Adding these two 5-bit codes… • Here, there is no carry bit and the sign bit is 1. This indicates a negative result, which is what we expect. (11001)2 = -(7)10. 0 0 1 0 1 + 1 0 1 0 0 ---------------- 1 1 0 0 1