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Storage of Bits

Storage of Bits. Computers represent information as patterns of bits A bit ( Bi nary Digi t ) is one of the two digits 0 or 1 Storing a bit in the machine requires a device that can be in one of two states: On or Off , True or False , 0 or 1. GATES AND FLIP-FLOPS.

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Storage of Bits

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  1. Storage of Bits • Computers represent information as patterns of bits • A bit (Binary Digit) is one of the two digits 0 or 1 • Storing a bit in the machine requires a device that can be in one of two states:On or Off , True or False , 0 or 1

  2. GATES AND FLIP-FLOPS • There are several gates that perform logical operations • The basic gates are: • AND • OR • EXLUSIVE-OR(X-OR) • NOT

  3. A Y B Logic Circuit AND Gate Truth Table A B Y 0 0 0 0 1 0 1 0 0 1 1 1 Logic Equation Y= A.B

  4. A Y B Logic Circuit OR Gate Truth Table A B Y 0 0 0 0 1 1 1 0 1 1 1 1 Logic Equation: Y = A + B

  5. A Y B Logic Circuit Exclusive OR (X-OR) Truth Table A B Y 0 0 0 0 1 1 1 0 1 1 1 0 Logic Equation Y = A + B

  6. Y A Logic Circuit NOT Gate Truth Table A Y 0 1 1 0 Logic Equation Y= A

  7. Y A A Y B B • NAND Gate

  8. A A Y Y B B • NOR Gate

  9. 0 A 0 0 X B 0 0 0 C Example(1)Draw the logic circuit and find the truth table for the following logic equation: X=(A+B)C ABCX 0 0 0 0

  10. 0 A 0 0 X B 0 1 0 C Example(1)Draw the logic circuit and find the truth table for the following logic equation: X=(A+B)C ABCX 0 0 0 0 0 0 1 0

  11. 0 A 1 0 X B 1 0 0 C Example(1)Draw the logic circuit and find the truth table for the following logic equation: X=(A+B)C ABCX 0 0 0 0 0 0 1 0 0 1 0 0

  12. 1 A 1 1 X B 1 1 0 C No. of States = 2No. Of Inputs Example(1)Draw the logic circuit and find the truth table for the following logic equation: X=(A+B)C ABCX 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1

  13. FLIP - FLOP • A flip-flop is a circuit that has one of two output values; its output remains fixed until a temporary pulse from another circuit causes it to shift to other value • The significance of a flip-flop is that it stores the information • There are several kinds of flip flops like: • J-K flip-flop , S-R flip-flop, D flip-flop , T flip-flop

  14. Decimal System Binary System Octal System Hexadecimal System Decimal System Consists of 10 digits 0,1,2,3,4,5,6,7,8,9 The base for this system is 10 To analyze any decimal number: 652= 102x6 + 101x5 + 100x2 = 600 + 50 + 2 = 652 Numbering Systems

  15. Consists of two digits 0 and 1 The base for the binary system is 2 It is the system used to represent information inside the computer system It is important to be familiar with conversion methods between the different numbering systems Binary System

  16. Binary System • Decimal -to- Binary conversion: To convert (8)10from decimal to binary: 2 8 (8)10=(1000)2 2 4 0 LSB 2 2 0 2 1 0 2 0 1 MSB

  17. Binary System • Binary -to- Decimal Conversion: Convert (1011)2 into decimal number: 23x1 + 22x0 + 21x1 + 20x1= 8 + 0 + 2 + 1 = (13)10

  18. Octal System • Consists of 8 digits: 0,1,2,3,4,5,6,7 • The base for the octal system is 8 • It has many applications in information coding systems

  19. Octal System • Decimal -to- Octal Conversion: To convert (13)10 into Octal number: (13)10=(15)8 8 13 8 1 5 0 1 • Octal -to- Decimal Conversion: (15)8= 81x1 + 80x5= 8 + 5 = (13)10

  20. Hexadecimal System • It consists of 16 digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F • The base for hexadecimal system is 16 • It is commonly used to describe all the information in the computer system(the information represented as binary numbers and described by hexadecimal system)

  21. Hexadecimal System • Decimal -to- Hexadecimal Conversion: Convert(302)10into Hexadecimal notation: 16 302 (302)10=(12E)16 16 18 14 16 1 2 16 0 1

  22. Hexadecimal System • Hexadecimal -to- Decimal Conversion: (A4B)16= 162x10 + 161x4 + 160x11= 2560 + 64 + 11 =(2575)10

  23. Hexadecimal System • Hexadecimal -to- Binary: (140B)= (0001 0100 0000 1011)2 • Binary -to- Hexadecimal: (10001001011111)2=( ? )16 2 2 5 F

  24. Fractions • Decimal -to-Binary: Result Carry (3.5)10=11.? 0.5x2=1.0 0.0 1 =11.1 1x2-1 (0.6)10= 0.6x2= 0.2 1 =(0.10011)2 0.2x2= 0.4 0 0.4x2= 0.8 0 1x2-1+1x2-4+1x2-5= 0.8x2= 0.6 1 0.5+0.0625+0.3125= 0.6x2= 0.2 1 0.59375 Round-off Problem

  25. Storing Integers: 1. Excess Notation: • It is easy to distinguish the patterns that represent negative values from those do not. • Those that represent negative values have a 0 as their most significant bit, while those do not represent negative values have a 1 as their MSB. • The MSB is often called Sign Bit

  26. 1)Excess Notation: The following is 3-bit , excess-4 notation: Bit PatternValue represented 111 + 3 110 + 2 101 + 1 100 0 011 - 1 010 - 2 001 - 3 000 - 4

  27. 2) Two’s Complement Notation • A notation used to represent both positive and negative integers • It simplifies the arithmetic operations, such as Addition and Subtraction , and then Multiplication and Division • By using this notation, the subtraction operation can be done in similar way the addition performed

  28. Two’s Complement: • To obtain two’s complement of a decimal number, for example ( 5 ): • Get the binary value 101, and to represent that this is a positive number, put a leading 0 as MSB 0101 • Get the 1’s complement of the above value by inverting each digit: 0101 1010 • Add 1 to the result to get + 1 ------------- 2’s Complement 1011

  29. Two’s Complement: Example 0011 1100 1101 Perform 5 - 3. Sol.Can be performed as 5 + (-3) using 2’s complement notation: 0101 1101 2’s complement of 3 ------- + 1 0010 The result(+2) Dropped

  30. 3) Floating- Point Notation: • A notation used to represent very large or very small positive and negative numbers • Consists of byte. The MSB represents sign bit. 0 for positive and 1 for negative • The other bits divided as: 8 7 6 5 4 3 2 1 Sign Exponent Mantissa Bit

  31. Floating-Point Notation: • To convert 3 1/2 into F.P notation: 1) The number is +ve, the sign bit is 0 2) (3 1/2)10 is (11.1)2 , to represent it in F.P the radix should be shift left two places, the value becomes .111(which is the mantissa) and the exponent is 2 3) Convert the exponent into excess-4 form by adding 4 to it (which results 6) and convert it into binary 110 (which is the exponent part) 4) The resulting F.P number is(01101110)

  32. Floating-Point Notation: • Convert (10110101) from F.P notation into decimal number: 1) Sign bit 1 , the No. is negative 2)The exponent is 011 which is 3 ,subtract 4 from it results (-1). This means the radix should be shift left one place. 3)The mantissa is .0101 the radix shifted one place left, the resulted value is (00101) which is 1/8 + 1/32 = 5/32 4)The resulted decimal value is - (5/32)

  33. Memory System • For purpose of storing data, a computer contains a large collection of circuits, each capable of storing a bit. • The computer system contains two main types of memory: • Main Memory • Secondary Memory(Mass Storage)

  34. Main Memory: • The storage circuits in a machine’s main memory are arranged in manageable units called cells(or words) , with a typical cell size being 8 bits • Every 8 bits are called byte • Every 1024 bytes called kilobyte(210 KB), and each 1024 KB is Megabyte … • The main memory is divided to: • Random Access Memory(RAM) • Read Only Memory(ROM)

  35. Mass Storage: • Secondary storage is a cheap and very large (huge amount) storage area • It is a permanent storage • Classified into: • Tapes • Disks • Compact Disks (CDs)

  36. Communication Errors: During the data transmission between two points, we may lose data. To resolve such a problem, a variety of error detection and error correction techniques have been developed such as: • Parity Bit • Cyclic Redundancy Check(CRC) • Hamming Distance

  37. Parity Bit: • It is a bit added as MSB used to detect errors without correcting them. • Is based on the principle that if each bit pattern being manipulated has an odd number of 1’s (Odd Parity)and a pattern found with an even number of 1’s (Even Parity) • This is achieved by adding an extra bit for parity check. Thus an ASCII of 8 bits becomes 9 bits

  38. Parity Bit: • To obtain the odd parity for character(A): 1) From ASCII table,the hex code of A is 41, in binary is : 0100 0001 2) Since the number of 1’s in the code is even. The odd parity bit must be 1 to get odd number of 1’s, and the complete code becomes: 10100 0001 3)If an error happened in 3rd bit from right ,the received character becomes: 1 0100 0101 , the number of 1’s is even, so the receiver detect that an error has been occurred

  39. Parity Bit: 4) This method is used to detect single error and never correct it 5) It fails to detect the error if it happens in the parity bit itself

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