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Information System

Information System. Instructor: Adil Waheed Lecturer University of Sargodha Lahore E-mail:- adil_pu@yahoo.com. Digital Systems. Digital vs. Analog Waveforms. Digital: only assumes discrete values. Analog: values vary over a broad range continuously. Analogue Quantities.

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Information System

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  1. Information System Instructor: Adil Waheed Lecturer University of Sargodha Lahore E-mail:- adil_pu@yahoo.com

  2. Digital Systems Digital vs. Analog Waveforms Digital: only assumes discrete values Analog: values vary over a broad range continuously

  3. Analogue Quantities Continuous Quantity • Intensity of Light • Temperature • Velocity

  4. Digital Values • Discrete set of values

  5. Continuous Signal

  6. Continuous Signal

  7. Digital Representation

  8. Under Sampling

  9. Electronic Processing • Analogue Systems • Digital Systems • Representing quantities in Digital Systems

  10. Digital Systems • Two Voltage Levels • Two States • On/Off • Black/White • Hot/Cold • Stationary/Moving

  11. Binary Number System • Binary Numbers • Representing Multiple Values • Combination of 0v & 5v

  12. Merits of Digital Systems • Efficient Processing & Data Storage • Efficient & Reliable Transmission • Detection and Correction of Errors • Precise & Accurate Reproduction • Easy Design and Implementation • Occupy minimum space

  13. Logic Gates • Building Blocks • AND, OR and NOT Gates • NAND, NOR, XOR and XNOR Gates • Integrated Circuits (ICs)

  14. c 3 2 1 0 c 9 8 1 1 1 1 V 7400 D N 4 5 6 1 2 3 G Logic Gate Symbol and ICs

  15. 1 1 x 0 x 1 0 y 1 0 F 1 0 F 0 time time NOT/OR/AND Logic Gate Timing Diagrams

  16. Operator Precedence • Evaluate the following Boolean equations: a=1, b=1, c=0, d=1. • Q1. F = a * b + c. • Answer: * has precedence over +, so we evaluate the equation as F = (1 *1) + 0 = (1) + 0 = 1 + 0 = 1. • Q2. F = ab + c. • Answer: the problem is identical to the previous problem, using the shorthand notation for *. • Q3. F = ab’. • Answer: we first evaluate b’ because NOT has precedence over AND, resulting in F = 1 * (1’) = 1 * (0) = 1 * 0 = 0. • Q4. F = (ac)’. • Answer: we first evaluate what is inside the parentheses, then we NOT the result, yielding (1*0)’ = (0)’ = 0’ = 1. • Q5. F = (a + b’) * c + d’. • Answer: Inside left parentheses: (1 + (1’)) = (1 + (0)) = (1 + 0) = 1. Next, * has precedence over +, yielding (1 * 0) + 1’ = (0) + 1’. The NOT has precedence over the OR, giving (0) + (1’) = (0) + (0) = 0 + 0 = 0. a

  17. Commutative a + b = b + a a * b = b * a Distributive a * (b + c) = a * b + a * c a + (b * c) = (a + b) * (a + c) (this one is tricky!) Associative (a + b) + c = a + (b + c) (a * b) * c = a * (b * c) Identity 0 + a = a + 0 = a 1 * a = a * 1 = a Complement a + a’ = 1 a * a’ = 0 To prove, just evaluate all possibilities Boolean Algebra Properties

  18. a b c F 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 1 0 1 1 1 1 1 Truth Table Representation a b F a b c F a b c d F 0 0 0 0 0 0 0 0 0 • Define value of F for each possible combination of input values • 2-input function: 4 rows • 3-input function: 8 rows • 4-input function: 16 rows • n-input function: 2n rows • Q: Use truth table to define function F(a,b,c) that is 1 when abc is 5 or greater in binary 0 1 0 0 1 0 0 0 1 1 0 0 1 0 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 1 0 0 ( a ) 1 0 1 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 0 ( b ) 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 a 1 1 1 0 1 1 1 1 ( c )

  19. Number Systems and Codes • Decimal Number System • Caveman Number System • Binary Number System • Hexadecimal Number System • Octal Number System

  20. Decimal Number System • Ten unique numbers 0,1..9 • Combination of digits • Positional Number System • 275 = 2 x 102 + 7 x 101 + 5 x 100 • Base or Radix 10 • Weight 1, 10, 100, 1000 ….

  21. Representing Fractions • Fractions can be represented in decimal number system in a manner = 3 x 102 + 8 x 101 + 2 x 100 + 9 x 10-1 + 1 x 10-2 = 300 + 80 + 2 + 0.9 + 0.01 = 382.91

  22. Binary Number System • Two unique numbers 0 and 1 • Base – 2 • A binary digit is a bit • Combination of bits to represent larger values

  23. Binary Number System

  24. Combination of Binary Bits • Combination of Bits • 100112 = 1910 = (1 x 24) + (0 x 23) + (0 x 22) + (1 x 21) + (1 x 20) = (1 x 16) + (0 x 8) + (0 x 4) + (1 x 2) + (1 x 1) = 16 + 0 + 0 + 2 + 1 = 19

  25. Fractions in Binary • Fractions in Binary • 1011.1012 = 11.625 = (1 x 23) + (0 x 22) + (1 x 21) + (1 x 20) + (1 x 2-1) + (0 x 2-2) + (1 x 2-3) = (1 x 8) + (0 x 4) + (1 x 2) + (1 x 1) + (1 x 1/2) + (0 x 1/4) + (1 x 1/8) = 8 + 0 + 2 + 1 + 0.5 + 0 + 0.125 = 11.625 • Floating Point Notations

  26. Decimal-Binary Conversion • Binary to Decimal Conversion • Sum-of-Weights • Adding weights of non-zero terms • Decimal to Binary Conversion • Sum-of-Weights (in reverse) • Repeated Division by 2

  27. Decimal-Binary Conversion • Binary to Decimal Conversion • Sum-of-Weights • Adding weights of non-zero terms Terms 16,0,0.2 and 1 19

  28. Decimal-Binary Conversion • Binary to Decimal Conversion • Sum-of-Weights • Adding weights of non-zero terms

  29. Decimal-Binary Conversion • Binary to Decimal Conversion • Sum-of-Weights • Adding weights of non-zero terms

  30. Lecture No. 1 Number Systems A Summary

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