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Number Representation (Part 2)

Number Representation (Part 2). Computer Architecture (Fall 2006). Unsigned Representations. Unsigned number representations Binary (base-2) Octal (base-8) Hexadecimal (base-16) Given n bits Range: 0 to 2 n -1 decimal numbers. Need for Signed Representations.

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Number Representation (Part 2)

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  1. Number Representation(Part 2) Computer Architecture (Fall 2006)

  2. Unsigned Representations • Unsigned number representations • Binary (base-2) • Octal (base-8) • Hexadecimal (base-16) • Given n bits • Range: 0 to 2n-1 decimal numbers

  3. Need for Signed Representations • Unsigned representations cannot represent negative numbers • Such as: -2, -51 etc. • However negative numbers are frequently used • Need a representation for positive & negative numbers – that is, signed numbers

  4. Standard Signed Representations • Signed numbers are represented using 3 different standards • Sign-bit Magnitude (SBM) representation • 1’s Complement Representation • 2’s Complement Representation • This is the representation that is used by computers today!

  5. Strategy for Signed Representation • General strategy is as follows: • All representations assume fixed size • 8, 16, 32, or 64 bits operated on as a single unit-word • Given n bits (unsigned range: 0 to 2n-1) • Break the range into two halves • One half represents positive numbers • 0 to 2n-1-1 corresponds to positive numbers • Decimal value: 0 to 2n-1-1 • Another half represents negative numbers • 2n-1 to 2n-1 corresponds to negative numbers • Decimal range: -2n-1 to -1

  6. Sign-Bit Magnitude (SBM) Representation • Uses left-most bit called sign bit to indicate sign • Sign bit = 0 implies positive number • Sign bit = 1 implies negative number • Example: 27 26 25 24 23 22 21 20 1 0 0 0 1 0 0 1 Sign-bit Decimal value = -910

  7. SBM Examples (1) • Represent 1010 using 8-bit SBM • Converting to binary: 1010 = 10102 • 8-bit SBM representation: 000010102 • Sign bit is 0 to indicate positive value. • Represent -1510 using 8-bit SBM • Converting to binary: 1510 = 11112 • 8-bit SBM representation: 100011112 • Sign bit is 1 to indicate negative value

  8. SBM Examples (2) • Convert 8-bit SBM 000001012 to Decimal • The sign bit (left most bit) is 0 indicating positive value • Converting 00001012 to decimal we get 510 • 000001012 in SBM = 510 • Convert 8-bit SBM 100011012 to Decimal • The sign bit (left most bit) is 1 indicating negative value! • Converting 00011012 to decimal we get 1310 • Result: 100011012 = -1310

  9. 1’s Complement Representation • There are a few drawbacks with SBM • There are 2 different representations for zero • +0 (00000000) and -0 (10000000) • Logic circuits for addition and subtraction of binary numbers are complicated as they have to handle sign bit separately. • 1’s Complement Representation • Fixed size representation • Most significant bit is reserved as sign-bit • Positive numbers are represented using standard binary notation • Negative numbers are represented by inverting all bits in the standard binary notation

  10. Example of 1’s Complement • Represent 1210 using 8-bit 1’s Complement • Converting to binary: 1210 = 11002 • 8-bit 1’s complement representation: 000011002 • Sign bit is 0 to indicate positive value. • Represent -1310 using 8-bit SBM • Converting to binary: 1310 = 11012 • 8-bit representation: 000011012 • Since original decimal number was negative each bit in 8-bit representation is inverted! • 1s Complement representation = 11110010 • Note: Sign bit becomes 1 to indicate negative value

  11. Example of 1’s Complement • Convert 1’s complement 000011112 to decimal • Sign bit is 0 to indicating positive value. • Converting 000011112 to decimal we get 1510 • Final result: +1510 • Convert 1’s complement 111101012 to decimal • Sign bit is 1 indicating negative value. • First invert all the bits to get: 000010102 • Convert above binary to decimal to get 1010 • Final result: -1010

  12. 1’s Complement • Still has 2 different representations for 0 • +0 (00000000) and -0 (11111111) • However, A – A operations can be easily represented • A – A = A + (-A) = 11111111 (which is effectively 0 in 1’s complement) • General subtraction does not work in all cases!

  13. 2’s Complement • Overcomes the limitations of SBM and 1’s complement representation! • Fixed size representation • Enables subtraction of numbers via addition! • Also reserves a special sign bit (left most bit) • 0 (sign bit) indicates positive numbers • 1 (sign bit) indicates negative numbers • Conversion to and from 2’s complement requires similar steps • Several of them are identical to 1’s complement

  14. Convert Decimal to 2’s Complement Signed Decimal Number Positive Negative Invert all the n bits Decimal to n-bit Binary Decimal to n-bit Binary Add 12

  15. 2’s Complement Example • Represent +2010 using 8-bit 2’s Complement • Since number is positive simply represent +2010 as a 8-bit binary number! • +2010 = 000101002 • Represent -1810 using 8-bit 2’s complement • Since number is negative we need to do 1’s complement and add 12 • Step 1: Convert 1810 to 8-bit binary = 000100102 • Step 2: Invert bits = 111011012 • Step 3: Add 12 = 111011102 • Final result: -1810 = 111011102 in 8-bit 2’s complement

  16. 2’s Complement to Decimal Binary number in n-bit 2’s complement Value of Sign-bit (Left most bit) Sign Bit=0 Sign Bit=1 Invert all bits Binary to decimal (positive number) Add 1 Binary to decimal (Negative number)

  17. Example • Convert 8-bit 2’s complement to decimal • Case 1: 000010102 • Sign bit is 0 indicating positive number • Simply convert binary to decimal to get 1010 • Case 2: 111110102 • Sign bit is 1 indicating negative number • Step 1: Invert all bits to get 000001012 • Step 2: Add 12 to get 000001102 • Convert binary to decimal to get -610 • Note the negative sign on decimal number!

  18. Subtraction using 2’s Complement • Perform 510 – 310 using 4-bit 2’s complement 510 – 310 = 510 + (-310) 510 = 01012 -310 = (00112→11002 + 12) = 11012 510 + (-310) = 01012 + 11012 = 00102 00102 = 210 INV

  19. Subtraction using 2’s Complement • Perform 210 – 410 using 4-bit 2’s complement 210 – 410 = 210 + (-410) 210 = 00102 -410 = (01002→10112 + 12) = 11002 210 + (-410) = 00102 + 11002 = 11102 11102 is 2’s complement result and because it is negative (sign bit is 1) it needs to be converted 11102→ 00012 + 12 = 00102 = -210 INV INV

  20. Subtraction Circuit • Circuit to subtract two 3-bit 2’s complement numbers • Recollect that subtraction is performed via addition in 2’s complement representation. • Addition is performed using Full Adder modules • Cascaded together in the form of Ripple Carry Adder • One of the numbers have to be converted to 2’s complement representation • Invert all the bits • Add 12

  21. Logic Circuit to add 3 bits: Full Adder • Refer to truth table shown earlier • Sum = A + B + C • Carry = (A•B) + (B•C) + (C•A) Full Adder A A + B B Sum C AB BC Carry CA

  22. Ripple Carry Adder • Several full adder’s can be cascaded to add multiple-bits! • Circuit reflects the way binary numbers are added • The following circuit adds 2 3-bit binary numbers namely A2A1A0 and B2B1B0 to yield result C3C2C1C0 FA2 A2 FA1 A1 FA0 A0 B2 B1 B0 0 What happens this bit is set to 1 instead of 0? C3 C2 C1 C0

  23. Subtraction Circuit FA2 A2 FA1 A1 FA0 A0 B2 B1 B0 1 C2 C1 C0 Each bit is inverted (Step 1 of Converting to 2’s Complement) Add 1 to inverted Bn bits to generate 2’s complement!

  24. Add/Subtract Circuit • Circuit to Add or Subtract two 3-bit 2’s complement number The XOR gate inverts the bits only when control input is 1 B2 B1 B0 FA2 A2 FA1 A1 FA0 A0 0 C3 C2 C1 Control Input 0 = Add 1 = Subtract C0

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