Digital Logic Design Lecture 1 PDF
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جامعة الدمام
2024
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Summary
This document covers lecture 1 of NET 104: Digital Logic Design, focusing on an introduction to various number systems, including decimal, binary, octal, and hexadecimal. It details the representation and conversion between these systems. The lecture example for adding and subtracting using binary numbers is also included.
Full Transcript
NET 104:Digital Logic Design Lecture 1 Introduction Digital Logic Design 10/19/2024 1 Digital Computer Systems Digital systems consider discrete amounts of data Examples 26 letters in the alphabet...
NET 104:Digital Logic Design Lecture 1 Introduction Digital Logic Design 10/19/2024 1 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 Digital Logic Design 10/19/2024 2 Understanding Decimal Numbers Decimal numbers are made of decimal digits: (0,1,2,3,4,5,6,7,8,9) Base = 10 But how many items does a decimal number represent? 8653 = 8 x103 + 6 x102 + 5 x101 + 3 x100 Number = d3 x B3 + d2 x B2 + d1 x B1 + d0 x B0 = Value What about fractions? 97654.35 = 9x104 + 7x103 + 6x102 + 5x101 + 4x100 + 3x10-1 + 5x10-2 In formal notation → (97654.35)10 Digital Logic Design 10/19/2024 3 Understanding Binary Numbers Binary numbers are made of binary digits (bits): 0 and 1 How many items does a binary number represent? 8 4 2 1 = Weights (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 Digital Logic Design 10/19/2024 4 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? 512 64 8 1 = Weights (4536)8 = 4x83 + 5x82 + 3x81 + 6x80 = (2398)10 What about fractions? (465.27)8 = 4x82 + 6x81 + 5x80 + 2x8-1 + 7x8-2 Octal numbers don’t use digits 8 or 9 Digital Logic Design 10/19/2024 5 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 a hex number represent? 4096 256 16 1 = Weights 3 2 1 0 (3A9F)16 = 3x16 + 10x16 + 9x16 + 15x16 = 1500710 What about fractions? 2 1 0 -1 (2D3.5)16 = 2x16 + 13x16 + 3x16 + 5x16 = 723.312510 Note that each hexadecimal digit can be represented with four bits (1110)2 = (E)16 Digital Logic Design 10/19/2024 6 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 1 AND Gate 0 0 10/19/2024 Digital Logic Design 7 Convert an Integer from Decimal to Another Base For each digit position: 1. Divide decimal number by the base (e.g., 2) 2. The remainder is the lowest-order digit 3. Repeat first two steps until no divisor remains Example for (13)10: Quotient Remainder Coefficient 13/2 = 6 , 1 a0 = 1 6/2 = 3 , 0 a1 = 0 3/2 = 1 , 1 a2 = 1 1/2 = 0 , 1 a3 = 1 Answer (13)10 = (a3 a2 a1 a0)2 = (1101)2 MSB LSB 10/19/2024 Digital Logic Design Most and Least 8 significant bit Convert a Fraction from Decimal to Another Base For each digit position: 1. Multiply decimal number by the base (e.g. 2) 2. The integer is the highest-order digit 3. Repeat first two steps until fraction becomes zero Example for (0.625)10: Integer Fraction Coefficient 0.625 x 2 = 1 + 0.250 a-1 = 1 0.250 x 2 = 0 + 0.500 a-2 = 0 0.500 x 2 = 1 + 0 a-3 = 1 Answer (0.625)10 = (0.a-1 a-2 a-3 )2 = (0.101)2 MSB LSB 10/19/2024 Digital Logic Design 9 Conversion Between Base 16 and Base 2 Conversion is easy! Determine the 4-bit value for each hex digit Note that there are 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 10/19/2024 Digital Logic Design 10 Understanding Binary Coded Decimal Binary Coded Decimal (BCD) represents each decimal digit with four bits Ex. 0011 0010 1001 = 32910 3 2 9 Digit BCD Code This is NOT the same as 0 0000 0011001010012 = 80910 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 10/19/2024 Digital Logic Design 11 The Growth of Binary Numbers 10/19/2024 Digital Logic Design 12 Putting It All Together Decimal Binary Octal Hexadecimal BCD 0 0000 0 0 000 1 0001 1 1 0001 2 0010 2 2 0010 3 0011 3 3 0011 4 0100 4 4 0100 5 0101 5 5 0101 6 0110 6 6 0110 7 0111 7 7 0111 8 1000 10 8 1000 9 1001 11 9 1001 10 1010 12 A 0001 0000 11 1011 13 B 0001 0001 12 1100 14 C 0001 0010 13 1101 15 D 0001 0011 14 1110 16 E 0001 0100 15 1111 17 F 0001 0101 10/19/2024 Digital Logic Design 13 Extra Digital Logic Design 10/19/2024 14 How To Represent Signed Numbers Plus and minus signs are used for decimal numbers: 25 (or +25), −16, etc In computers, everything is represented as bits Three types of signed binary number representations: signed magnitude 1’s complement 2’s complement In each case left-most bit indicates the sign: ‘0’ for positive and ‘1’ for negative 10/19/2024 Digital Logic Design 15 Signed Magnitude Representation The left most bit is designated as the sign bit while the remaining bits form the magnitude 000011002 = 1210 Sign bit Magnitude 100011002 = −1210 Sign bit Magnitude 10/19/2024 Digital Logic Design 16 One’s Complement Representation The one’s complement of a binary number is done by complementing (i.e. inverting) all bits 1’s comp of 00110011 is 11001100 1’s comp of 10101010 is 01010101 For a n-bit number N the 1’s complement is n (2 − 1) − N Called “diminished radix complement” by Mano To find the negative of a 1’s complement number take its 1’s complement 000011002 = 1210 111100112 = −1210 Sign bit Magnitude Sign bit Code 10/19/2024 Digital Logic Design 17 One’s Complement Representation 7 0111 4 bits 6 0110 .. 16 combinations.. 1 0001 0 0000 −0 1111 −1 1110.... −6 1001 −7 1000 10/19/2024 Digital Logic Design 18 Two’s Complement Representation The two’s complement of a binary number is done by complementing (inverting) all bits then 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 n (2 −1) − N + 1 Called “radix complement” by Mano To find the negative of a 2’s complement number take its 2’s complement 000011002 = 1210 111101002 = −1210 Sign bit Magnitude Sign bit Code 10/19/2024 Digital Logic Design 19 Two’s Complement Shortcuts Algorithm 1: Complement each bit then add 1 to the result 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, then complement the remaining bits N =01100110 [N] =10011010 10/19/2024 Digital Logic Design 20 Two’s Complement Representation 7 0111 4 bits 6 0110 .. 16 combinations 2 0010 1 0001 0 0000 −1 1111 −2 1110.... −7 1001 −8 1000 10/19/2024 Digital Logic Design 21 2’s Complement Addition Using 2’s complement representation, adding numbers is easy Step 1: Add binary numbers Step 2: Ignore the resulting carry bit For example: (12)10 + (1)10 0 1 1 0 0 (12)10 = +(1100)2 Add + 0 0 0 0 1 = 011002 in 2’s comp. -------------- (1)10 = +(0001)2 Final 0 0 1 1 0 1 Result = 000012 in 2’s comp. Ignore 10/19/2024 Digital Logic Design 22 2’s Complement Subtraction Using 2’s complement representation, subtracting numbers is also easy Step 1: Take 2’s complement of 2nd operand Step 2: Add binary numbers Step 3: Ignore the resulting carry bit 0 1 1 0 0 For example: (12)10 − (1)10 - 0 0 0 0 1 (12)10 = +(1100)2 2’s comp 0 1 1 0 0 = 011002 in 2’s comp. Add + 1 1 1 1 1 (−1)10 = −(0001)2 -------------- Final = 111112 in 2’s comp. Result 1 0 1 0 1 1 Ignore 10/19/2024 Digital Logic Design Carry 23 2’s Complement Subtraction (Cont’d) Example 2: (13)10 − (5)10 (13)10 = +(1101)2 = (01101)2 (−5)10 = −(0101)2 = (11011)2 Adding these two 5-bit codes: 01101 + 11011 Carry 1 01000 Discarding the carry bit, the sign bit is seen to be zero, indicating a positive result Indeed: (01000)2 = +(8)10 10/19/2024 Digital Logic Design 24 2’s Complement Subtraction (Cont’d) Example 3: (5)10 − (12)10 (5)10 = +(0101)2 = (00101)2 (−12)10 = −(1100)2 = (10100)2 Adding these two 5-bit codes: 00101 + 10100 Carry 0 11001 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 10/19/2024 Digital Logic Design 25 Gray Code Gray Gray code is not a number Digit Binary Code system 0 0000 0000 1 0001 0001 It is an alternate way to 2 0010 0011 represent four bit data 3 0011 0010 Only one bit changes from one 4 0100 0110 5 0101 0111 decimal digit to the next 6 0110 0101 Useful for reducing errors in 7 0111 0100 communication 8 1000 1100 9 1001 1101 Can be scaled to larger 10 1010 1111 numbers 11 1011 1110 12 1100 1010 13 1101 1011 14 1110 1001 15 1111 1000 10/19/2024 Digital Logic Design 26 The Gray Code Binary-to-Gray code conversion The MSB in the Gray code is the same as MSB in the binary number. Going from left to right, add each adjacent pair of binary code bits to get the next Gray code bit. Discard carries. Digital Logic Design 10/19/2024 27 Oct-24 27 The Gray Code Gray-to-Binary Conversion The MSB in the binary code is the same as the corresponding bit in the Gray code. Add each binary code bit generated to the Gray code bit in the next adjacent position. Discard carries. Digital Logic Design 28 Text: ASCII Characters ASCII: Maps 128 characters to 7-bit code. both printable and non-printable (ESC, DEL, …) characters http://www.asciitable.com/ 00 nul 10 dle 20 sp 30 0 40 @ 50 P 60 ` 70 p 01 soh 11 dc1 21 ! 31 1 41 A 51 Q 61 a 71 q 02 stx 12 dc2 22 " 32 2 42 B 52 R 62 b 72 r 03 etx 13 dc3 23 # 33 3 43 C 53 S 63 c 73 s 04 eot 14 dc4 24 $ 34 4 44 D 54 T 64 d 74 t 05 enq 15 nak 25 % 35 5 45 E 55 U 65 e 75 u 06 ack 16 syn 26 & 36 6 46 F 56 V 66 f 76 v 07 bel 17 etb 27 ' 37 7 47 G 57 W 67 g 77 w 08 bs 18 can 28 ( 38 8 48 H 58 X 68 h 78 x 09 ht 19 em 29 ) 39 9 49 I 59 Y 69 i 79 y 0a nl 1a sub 2a * 3a : 4a J 5a Z 6a j 7a z 0b vt 1b esc 2b + 3b ; 4b K 5b [ 6b k 7b { 0c np 1c fs 2c , 3c < 4c L 5c \ 6c l 7c | 0d cr 1d gs 2d - 3d = 4d M 5d ] 6d m 7d } 0e so 1e rs 2e. 3e > 4e N 5e ^ 6e n 7e ~ 0f si 1f us 2f / 3f ? 4f O 5f _ 6f o 7f del 10/19/2024 Digital Logic Design 29 Thank You Digital Logic Design 10/19/2024 30