CS 105T Digital Logic Design

1. Sem. 2 -1438-1439H Chapter 1 CS 105TDigital Logic Design Dr. Mona Jamjoom

2. Communication Office# 2.501.39 Office hours : • Monday (9-12),(12:30-1:30). • Tuesday (9-11),(12:30-1:30). E-mail: • mmjamjoom@pnu.edu.sa

3. Textbook “DIGITAL DESIGN”, by Mano M. Morris, 4th edition, Prentice- Hall.

5. Conduct in Class • Don’t distract the students. I don’t insist that you pay attention, but you must allow others to participate. This ‎means: ‎ • Do not distract others with conversation.‎ • Do not distract others with your phone.‎ • Do not distract others by using your laptop in front of the class.‎ • If you snore, I will wake you up. ‎ • Do not attend the class late; you have only 10 minutes to attend it. Otherwise, you will be marked absent on ‎that day.‎ • There are no makeup quizzes or exams (if you miss a quiz or exam, you will get zero for that).‎ Computer Science Department

6. Rules of e-mail messages: • Write your message in paschal Arabic (NOT in slang Arabic). • In the email Subject, you have to write: • Course#- Section# - full Name – brief description for the email , for example: CS105D-1C10 – Nora AlAhmad – about sheet 1 CS105D-11C12– Sara AlAli – about ............. • In general, subject should summarize the message. • We will not reply to any email that contains something like: ????? !!!!! , or هاااااااااااام or أرجو الرد,or any impolite words. • If you violate any one of the above rules, your email will not be opened. • NOTE: Send the email from your @pnu.edu.sa address only. Computer Science Department

7. Computer Science Department

8. Topics

9. Outline 1.1 Digital Systems 1.2 Binary Numbers 1.3 Number-Base Conversions 1.4 Octal and Hexadecimal Numbers 1.5 Complements 1.6 Signed Binary Numbers 1.7 Binary Codes

10. 1.1 Digital Systems (1-2) • Digital system is a system that uses discrete values such as numbers and letters. • The signal in most digital systems use two values : • 0 and 1 which called a bit. • Discrete elements of information are represented with a group of bits called binary codes. • Thus, Digital system is a system that manipulates discrete elements of information represented internally in binary form.

11. Arithmetic & data processing operations 1.1 Digital Systems (2-2) Central Processing Unit Stores programs, inputs, outputs & intermediate data Receives results of computations and prints them to user Memory Unit Data and program prepared by user Output Unit Input Unit Block diagram of a digital Computer

12. 1.2 Binary Numbers (1-6) In general, a number expressed in a base-r system has coefficients multiplied by powers of r: an.rn + an-1.rn-1 + ....... + a2.r2 + a1.r + a0 + a-1.r-1 + a-2.r-2 + .......... + a-m.r-m • ris called base or radix. • ajranges in thr value from 0 to r-1

13. 1.2 Binary Numbers (3-6)

14. 1.2 Binary Numbers (2-6) Example: (127.4)8 = 1 x 82 + 2 x 81 + 7 x 80 + 4 x 8-1 = (87.5)10 (1101)2 = 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = (13)10

15. 1.2 Binary Numbers (2-6) Example: Is: (13)10 =(13)8 ?

16. 1.2 Binary Numbers (4-6) Arithmetic Operations Follow the same rules of as for decimal numbers Addition

17. 1.2 Binary Numbers (5-6) Arithmetic Operations Subtraction

18. 1.2 Binary Numbers (6-6) Arithmetic Operations Multiplication

19. 1.3 Number-Base Conversions (1-6) Base r – to – Decimal Conversion Rule: an.rn + an-1.rn-1 + ....... + a2.r2 + a1.r + a0 + a-1.r-1 + a-2.r-2 + .......... + a-m.r-m an.rn + an-1.rn-1 + ....... + a2.r2 + a1.r + a0 Integral part a-1.r-1 + a-2.r-2 + ....... + a-m.r-m Fractional part Integral part Fractional part Result

20. 1.3 Number-Base Conversions (2-6) Base r – to – Decimal Conversion Example: Find the decimal equivalent ofthe binary number (1001.0101)2. 1001 = 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20 = 8+0+0+1=9 Integral part .0101 = 0 x 2-1 + 1 x 2-2 + 0 x 2-3 + 1 x 2-4 = 0+.25+0+0.0625=0.3125 Fractional part (9.3125)10 Result

21. 1.3 Number-Base Conversions (3-6) Base r – to – Decimal Conversion Example: (1010.011)2= 23 + 21 + 2-2 + 2-3 =(10.375)10 (127.4)8 = 1 x 82 + 2 x 81 + 7 x 80 + 4 x 8-1 = (87.5)10 (1101)2 = 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 = (13)10

22. 1.3 Number-Base Conversions (4-6) Decimal – to – Base r Conversion Rule: Convert each part differently. • Divide number & its quotients by r. • Accumulate reminders. Integral part • Multiply number & its quotients by r. • Accumulate integers. Fractional part Integral part Fractional part Result

23. 1.3 Number-Base Conversions (5-6) Decimal – to – Base r Conversion Example: Find the binary equivalent of the decimal number (41)10. Divide by 2 Result (101001)2

24. 1.3 Number-Base Conversions (6-6) Decimal – to – Base r Conversion Example: Find the binary equivalent of the decimal number (0.6875)10. Result Multiply by 2 (0.1011)2

25. 1.4 Octal and Hexadecimal Numbers (1-17)

26. 1.4 Octal and Hexadecimal Numbers (2-17) Decimal-to-Octal Conversion Example: Find the octal equivalent of the decimal number (153.513)10. Divide by 8 Multiply by 8 Result (231.406517)8

27. 1.4 Octal and Hexadecimal Numbers (3-17) Octal-to-Decimal Conversion Example: Find the decimal equivalent ofthe octal number (137.21)8. 137= 1 x 82 + 3 x 81 + 7 x 80 = 64+24+7=95 Integral part .21 = 2 x 8-1 + 1 x 8-2 =0.265 Fractional part Result (95.265)10

28. 1.4 Octal and Hexadecimal Numbers (4-17) Decimal-to-Hexadecimal Conversion Example: Find the hexadecimal equivalent of the decimal number (82.25)10. Divide by 16 Multiply by 16 Result (52.4)16

29. 1.4 Octal and Hexadecimal Numbers (5-17) Hexadecimal-to-Decimal Conversion Example: Find the decimal equivalent ofthe hexadecimal number (1E0.2A)16. 1E0= 1 x 162 + 14 x 161 + 0 x 160 = 256+224+0=480 Integral part .2A = 2 x 16-1 + 10 x 16-2 =0.164 Fractional part Result (480.164)10

30. 1.4 Octal and Hexadecimal Numbers (6-17) Binary–Octal and Octal–Binary Conversions Binary – Octal Rule: • Divide integral part into three bits starting from right integral bit. • Divide fractional part into three bits starting from left fractional bit. • Adding zero‘s if necessary.

31. 1.4 Octal and Hexadecimal Numbers (7-17) Binary–Octal and Octal–Binary Conversions Example: Find the octal equivalent ofthe binary number (1110100.0100111)2. 001 110 100 . 010011 100 (1 6 4 . 23 4)8

32. 1.4 Octal and Hexadecimal Numbers (8-17) Binary–Octal and Octal–Binary Conversions Octal – Binary Rule: BOTH integral and fractional parts: • Convert each digit seperately into binary of three bits

33. 1.4 Octal and Hexadecimal Numbers (9-17) Binary–Octal and Octal–Binary Conversions Example: Find the binary equivalent ofthe octal number (374.26)8. 3 7 4 . 26 011 111 100 . 010110 (11 111 100 . 01011)2

34. 1.4 Octal and Hexadecimal Numbers (10-17) Binary–Hex and Hex–Binary Conversions Binary – Hex Rule: • Divide integral part into four bits starting from right integral bit. • Divide fractional part into four bits starting from left fractional bit. • Adding zero‘s if necessary.

35. 1.4 Octal and Hexadecimal Numbers (11-17) Binary–Hex and Hex–Binary Conversions Example: Find the hexdecimal equivalent ofthe binary number (1011001110.011011101)2. 0010 1100 1110 . 01101110 1000 (2 C E . 6E 8)16

36. 1.4 Octal and Hexadecimal Numbers (12-17) Binary–Hex and Hex–Binary Conversions Hex – Binary Rule: BOTH integral and fractional parts: • Convert each digit seperately into binary of four bits

37. 1.4 Octal and Hexadecimal Numbers (13-17) Binary–Hex and Hex–Binary Conversions Example: Find the binary equivalent ofthe hexadecimal number (17E.F6)16. 1 7 E . F6 0001 0111 1110 . 11110110 (1 0111 1110 . 1111011)2

38. 1.4 Octal and Hexadecimal Numbers (14-17) Hex–Octal and Octal–Hex Conversions Hex – Octal Rule: • Convert Hex number into binary. • Convert the result binary number into octal. OR • Convert Hex number into decimal. • Convert the result decimal number into octal.

39. 1.4 Octal and Hexadecimal Numbers (15-17) Hex–Octal and Octal–Hex Conversions Example: Find the Hex equivalent ofthe octal number (762.013)8. 7 6 2 . 01 3 Binary (0001 1111 0010 . 0000 0101 1000)2 Hex (1 F 2 . 05 8)16

40. 1.4 Octal and Hexadecimal Numbers (16-17) Hex–Octal and Octal–Hex Conversions Octal – Hex Rule: • Convert Octal number into binary. • Convert the result binary number into Hex. OR • Convert Octal number into decimal. • Convert the result decimal number into Hex.

41. 1.4 Octal and Hexadecimal Numbers (17-17) Hex–Octal and Octal–Hex Conversions Example: Find the octal equivalent ofthe Hex number (2F.C4)16. 2 F . C4 Binary (000 101 111. 110 001 000)2 Octal (5 7 . 61)8 (0 5 7 . 61 0)8

42. 1.5 Complements (1-9) Complement‘s Types RadixComplement (r’s complement) Diminished radix Complement ((r-1)’s complement)

43. 1.5 Complements (1-9) Diminished Radix Complement ((r-1)‘s complement) Rule • Given a number N in base r having n digits, the (r- 1)’scomplement of N is defined as (rn- 1) -N. Examples • 9’s complement of 546700 is 999999 – 546700= 453299 • 1’s complement of 1011000 is 0100111. Note • The (r-1)’s complement of octal or hexadecimal numbers is obtained by subtracting each digit from 7 or F (decimal 15), respectively.

44. 1.5 Complements (2-9) Radix Complement Rule • The 10’s complement of 012398 is 987602 • The 10’s complement of 246700 is 753300 • The2’s complement of 1011000 is 0101000 • Given a number N in base r having n digit, the r’s complement of N is defined as (rn -N) for N ≠0 and as 0 for N =0 . OR • Adding 1 to (r-1)‘s complement. Examples

45. 1.5 Complements (2-9) Note: • The complement of the complement restores the number to its original value.

46. 1.5 Complements (3-9) Subtraction with Complement The subtraction of two n-digit unsignednumbers M – N in base r can be done as follows: • M + (rn - N), note that (rn - N) is r’s complement of N. • If (M  N), the sum will produce an end carryx, which can be discarded; what is left is the result M- N. • If (M < N), the sum does not produce an end carry and is (N - M). Take the r’x complement of the sum and place a negative sign in front.

47. 1.5 Complements (4-9) Example: Using 10’s complement subtract 72532 – 3250. M = 72532 10’s complement of N = 96750 sum = 169282 Discard end carry. answer: 69282

48. 1.5 Complements (5-9) Example: Using 10’s complement subtract 3250 - 72532 . M = 03250 10’s complement of N = 27468 sum = 30718 The answer is –(10’s complement of 30718) = -69282

49. 1.5 Complements (6-9) Example: Using 2’s complement subtract 1010100 – 1000011 . M = 1010100 N = 1000011, 2’s complement of N = 0111101 10101000111101 = 10010001 Discard end carry. answer: 0010001

50. 1.5 Complements (7-9) Example: Using 2’s complement subtract 1000011 – 1010100. M= 1000011 N= 1010100, 2’s complement of N= 0101100 1000011 + 0101100 = 1101111 The answer is –(2’s complement of 1101111) = -0010001