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# CS220 : Digital Design

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1. CS220 : Digital Design

2. Basic Information • Title: Digital Design • Code: CS220 • Lecture: 3 • Tutorial: 1 • Pre-Requisite: Computer Introduction (CS201)

3. Overall Aims of Course By the end of the course the students will be able to: • Grasp basic principles of combinational and sequential logic design. • Determine the behavior of a digital logic circuit (analysis) andtranslate description of logical problems to efficient digital logic circuits (synthesis). • Understanding of how to design a general-purpose computer, starting with simple logic gates.

4. Contents

5. Assessment schedule

6. List of References Essential Books • “DIGITAL DESIGN”, by Mano M. Morris, 4th edition, Prentice- Hall. Recommended Books • “FUNDAMENTALS OF LOGIC DESIGN”, by Charles H. Roth, Brooks/Cole Thomson Learning. • “INTRODUCTION TO DIGITAL SYSTEMS”, by M.D. ERCEGOVAC, T. Lang, and J.H. Moreno, Wiley and Sons. 1998. • “DIGITAL DESIGN, PRINCIPLES AND PRACTICES”, by John F.Wakely, Latest Edition, Prentice Hall, Eaglewood Cliffs, NJ. • “FUNDMENTALS OF DIGITAL LOGIC WITH VHDL DESIGN”, by Stephen Brown and Zvonko Vranesic, McGraw Hill. • “INTRODUCTION TO DIGITAL LOGIC DESIGN”, by John Hayes, Addison Wesley, Reading, MA.

7. 1. Digital Systems and Binary Numbers 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

8. 1.1 Digital Systems

9. 1.2 Binary Numbers In general, a number expressed in a base-r system has coefficients multiplied by powers of r: r is called baseor radix.

10. Binary-to-Decimal Conversion Example: Example:

11. 1.4 Octal and Hexadecimal Numbers

12. Decimal-to-Octal Conversion Example:

13. Octal-to-Decimal Conversion Example: Example:

15. Binary–Octal and Octal–Binary Conversions Example: Example:

16. Hex–Binary and Binary–Hex Conversions Example: Example:

17. Hex–Octal and Octal–Hex Conversions • For Hexadecimal–Octal conversion, the given hex number is firstly converted into its binary equivalent which is further converted into its octal equivalent. • An alternative approach is firstly to convert the given hexadecimal number into its decimal equivalent and then convert the decimal number into an equivalent octal number. The former method is definitely more convenient and straightforward. • For Octal–Hexadecimal conversion, the octal number may first be converted into an equivalent binary number and then the binary number transformed into its hex equivalent. • The other option is firstly to convert the given octal number into its decimal equivalent and then convert the decimal number into its hex equivalent. The former approach is definitely the preferred one.

18. Example

19. Arithmetic Operation • Addition augend 101101 Added: + 100111 ----------Sum: 1010100

20. Subtraction minuend: 101101 subtrahend: - 100111 ------------- difference: 000110

21. Multiplication

22. 1.5 Complements • Diminished Radix Complement ((r-1)‘s complement) Given a number N in base r having n digits, the (r - 1)’s Complement of N is defined as (rn- 1) -N. • the 9’s complement of 546700 is 999999 – 46700=453299 • the 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

23. Radix Complement Given a number N in base r having n digit, the r’s complement of Nis defined as (rn -N) for N ≠0 and as 0 for N =0 . • The 10’s complement of 012398 is 987602 • The 10’s complement of 246700 is 753300 • The 2’s complement of 1011000 is 0101000

24. Subtraction with Complement • The subtraction of two n-digit unsigned numbers 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 carry x, 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.

25. Example: • Using 10’s complement subtract 72532 – 3250 M = 72532 10’s complement of N = 96750 sum = 169282 Discarded end carry 105 = -100000 answer: 69282

26. Example: • Using 10’s complement subtract 3250 - 72532 M = 03250 10’s complement of N = 27468 sum = 30718 Discarded end carry 105 = -100000 answer: -(100000 - 30718) = -69282 • The answer is –(10’s complement of 30718) = -69282

27. Example Using 2’s complement subtract (a) 1010100 – 1000011 M = 1010100 N = 1000011, 2’s complement of N = 0111101 1010100  0111101 sum = 10010001 Discarded end carry 27=-10000000 answer: 0010001

28. Example (b) 1000011 – 1010100 M = 1000011 N = 1010100, 2’s complement of N = 0101100 1000011  0101100 sum = 1101111 answer: - (10000000 - 1101111) = -0010001 • The answer is –(2’s complement of 1101111) = - 0010001

29. Example Using 1’s complement, subtract1010100 - 1000011 M = 1010100 N = 1000011, 1’s complement of N = 0111100 1010100  0111100 10010000 end-around carry = + 1 answer: 0010001 33

30. Example Using 1’s complement, subtract1000011 - 1010100 M = 1000011 N = 1010100, 1’s complement of N = 0101011 1000011 0101011 1101110 Answer: -0010001 34

31. 1.6 Signed Binary Numbers

32. 1.6 Signed Binary Numbers 1 - Sign and Magnitude representation 2 - 1’s Complement Representation 3 - 2’s Complement Representation Notes 1 - The previous representation are the same for positive numbers and different for negative numbers 2 - For a signed binary number the most significant bit is used for representing the sign of the number We use 0 for positive numbers and 1 for negative numbers Example : represent +76

33. Representing negative numbers in the previous three systems 1’s Complement of a negative number can be obtained by flipping all bits of the positive binary number 2’s Complement of a negative number can be obtained by adding 1 to the 1’s Complement or by flipping bits of the positive binary number after the first one from the right Example : represent -76

34. Arithmetic Addition with Comparison Arithmetic Addition The addition of two numbers in the signed mgnitude system follow the rules of ordinary arithmetic. If the signed are the same, we add the two magnitudes and give the sum the common sign. If the signed are different, we subtract the smaller magnitude from the larger and give the difference the sign of the larger magnitude. EX. (+25) + (-38) = -(38 - 25) = -13

35. Arithmetic Addition without Comparison The addition of two signed binary number with negative numbers represented in signed 2’s complement form is obtained from the addition of the two numbers, including their signed bits. A carry out of the signed bit position is discarded (note that the 4th case).

36. Arithmetic Addition without Comparison     06 00000110 06 11111010     13 00001101 13 00001101     19 00010011 07 00000111     06 00000110 06 11111010     13 11110011 13 11110011     07 11111001 19 11101101 40

37. Arithmetic Subtraction (+/-) A – (+B)= (+/-) A + (-B) (+/-) A – (-B)= (+/-) A + (+B) Example (-6) – (-13)= +7 In binary: (1111010 – 11110011)= (1111010 + 00001101) =100000111 after removing the carry out the result will be : 00000111

38. 1.7 Binary Codes

39. Binary Coded Decimal (BCD)

40. Binary Coded Decimal (BCD) in this system each digit is represented in 4 bits For example : to represent in BCD

41. BCD Addition Example : Evaluate the following operations in BCD System 1 – 3 + 4 2 – 4 + 8 3 - 148 + 576

42. BCD Addition Example : Evaluate the following operations in BCD System 1 – 3 + 4 2 – 4 + 8 3 - 148 + 576 Error We must add 6 (0110) to the result

43. BCD Addition Example : Evaluate the following operations in BCD System 1 – 3 + 4 2 – 4 + 8 3 - 184 + 576

44. Notes 1 – In BCD Addition , we add (0110)=(6) if the result value was greater than (1001)=(9) or if the result was more than 4 digits In previous Example we added 0110 when the result was 1 - greater than 9 (1001) 2 - more than 4 digits (10000) Note : result more than 4 digit is greater than 9(1001) 

45. Decimal Arithmetic Addition for signed numbers Example: (+375) + (- 240) = + 135 in BCD • Apply 10‘s complement to the negative number only. • Addition is done by summing all digits,including the sign digit,and discarding the end carry 0 375 +9 760 ------------ 0 135

46. Decimal Arithmetic • Subtraction for signed and unsigned numbers • Apply 10‘s complement to the subtrahend and apply addition (same as binary case)