co class held on 18/9/11

1. Unit 2

2. von Neumann/Turing
• Stored Program concept
• Main memory storing programs and data
• ALU operating on binary data
• Control unit interpreting instructions from
memory and executing
• Input and output equipment operated by
control unit
• Princeton Institute for Advanced Studies
—IAS
• Completed 1952

3. Structure of von Neumann machine

4. • Structure of IAS –
detail

5. • MemoryAddressRegister(MAR)-This register holds the
address of the memory location to be accessed. The
desired address is loaded into MAR through a common Bus.
• ProgramCounter(PC)-This register keeps track of the
program flow. Its count is incremented after every fetch of
Op Code or Operand from the Program Memory.
• StackPointer(SP)-This is an up down counter which holds
the address of the top of the Stack. Its count is
incremented or decremented after every Push or Pop
operation of the Stack.
• InstructionRegister(IR)- The OpCode fetched from the
program memory is loaded here, and its output is fed to
the Instruction Decoder for generating the Control Signals.
• TemporaryRegisters-These registers are accessible only to
the system,and cannot be accessed by the user.

8. Arithmetic & Logic Unit
• Does the calculations
• Everything else in the computer is there
to service this unit
• Handles integers
• May handle floating point (real) numbers
• May be separate FPU (maths co-
processor)
• May be on chip separate FPU (486DX +)

10. ALU Inputs and Outputs

11. Integer Representation
• Only have 0 & 1 to represent everything
• Positive numbers stored in binary
—e.g. 41=00101001
• No minus sign
• No period
• Sign-Magnitude
• Two’s compliment

12. Sign-Magnitude
• Left most bit is sign bit (MSB)
• 0 means positive
• 1 means negative
• +18 = 00010010
• -18 = 10010010
• Problems
—Need to consider both sign and magnitude in
arithmetic
—Two representations of zero (+0 and -0)

13. Two’s Compliment
• +3 = 00000011
• +2 = 00000010
• +1 = 00000001
• +0 = 00000000
• -1 = 11111111
• -2 = 11111110
• -3 = 11111101

14. Benefits
• One representation of zero
• Arithmetic works easily (see later)
• Negating is fairly easy
—3 = 00000011
—Boolean complement gives 11111100
—Add 1 to LSB 11111101

15. Geometric Depiction of Twos
Complement Integers

16. Negation Special Case 1
• 0= 00000000
• Bitwise not 11111111
• Add 1 to LSB +1
• Result 1 00000000
• There is carry out of MSB which is ignored
, so:
• -0=0

17. Negation Special Case 2
• -128 = 10000000
• bitwise not 01111111
• Add 1 to LSB +1
• Result 10000000
• So:
• -(-128) = -128 X
• Monitor MSB (sign bit)
• It should change during negation

18. Range of Numbers
• 8 bit 2s compliment
—+127 = 01111111 = 27 -1
— -128 = 10000000 = -27
• 16 bit 2s compliment
—+32767 = 011111111 11111111 = 215 - 1
— -32768 = 100000000 00000000 = -215

19. Conversion Between Lengths
• Positive number pack with leading zeros
• +18 = 00010010
• +18 = 00000000 00010010
• Negative numbers pack with leading ones
• -18 = 10010010
• -18 = 11111111 10010010
• i.e. pack with MSB (sign bit)

20. Addition and Subtraction
• Normal binary addition
• Monitor sign bit for overflow
• Take twos compliment of subtrahend and
add to minuend
—i.e. a - b = a + (-b)
• So we only need addition and complement
circuits

23. • If the two numbers are added and they
are both positive or both negative
number, then overflow occurs if the result
has the opposite sign.

24. Hardware for Addition and Subtraction

25. Multiplication
• Complex
• Work out partial product for each digit
• Take care with place value (column)
• Add partial products

26. Multiplication Example
• 1011 Multiplicand (11 dec)
• x 1101 Multiplier (13 dec)
• 1011 Partial products
• 0000 Note: if multiplier bit is 1 copy
• 1011 multiplicand (place value)
• 1011 otherwise zero
• 10001111 Product (143 dec)
• Note: need double length result

27. Unsigned Binary Multiplication

28. Execution of Example

29. Flowchart for Unsigned Binary
Multiplication

30. Multiplying Negative Numbers
• This does not work!
• Solution 1
—Convert to positive if required
—Multiply as above
—If signs were different, negate answer
• Solution 2
—Booth’s algorithm

31. Booth’s Algorithm

32. Example of Booth’s Algorithm

33. Division
• More complex than multiplication
• Negative numbers are really bad!
• Based on long division

35. Division of Unsigned Binary Integers
00001101 Quotient
Divisor 1011 10010011 Dividend
1011
001110
Partial 1011
Remainders
001111
1011
100 Remainder

36. Flowchart for Unsigned Binary Division

38. Real Numbers
• Numbers with fractions
• Could be done in pure binary
—1001.1010 = 24 + 20 +2-1 + 2-3 =9.625
• Where is the binary point?
• Fixed?
—Very limited
• Moving?
—How do you show where it is?

39. Floating Point
• +/- .significand x 2exponent
• Misnomer
• Point is actually fixed between sign bit and body
of mantissa
• Exponent indicates place value (point position)

40. Floating Point Examples

41. Signs for Floating Point
• Mantissa is stored in 2s compliment
• Exponent is in excess or biased notation
—e.g. Excess (bias) 128 means
—8 bit exponent field
—Pure value range 0-255
—Subtract 128 to get correct value
—Range -128 to +127

42. Normalization
• FP numbers are usually normalized
• i.e. exponent is adjusted so that leading
bit (MSB) of mantissa is 1
• Since it is always 1 there is no need to
store it
• (c.f. Scientific notation where numbers
are normalized to give a single digit
before the decimal point
• e.g. 3.123 x 103)

43. FP Ranges
• For a 32 bit number
—8 bit exponent
—+/- 2256 1.5 x 1077
• Accuracy
—The effect of changing lsb of mantissa
—23 bit mantissa 2-23 1.2 x 10-7
—About 6 decimal places

44. Expressible Numbers

45. IEEE 754
• Standard for floating point storage
• 32 and 64 bit standards
• 8 and 11 bit exponent respectively
• Extended formats (both mantissa and
exponent) for intermediate results

46. IEEE 754 Formats

47. • 1. Convert the decimal number to binary.
• 2. Write the binary number in scientific notation using 20
• 3. Normalise the number by moving the binary point and
changing the exponent.
• 4. Write f by taking the fractional part of the normalised
number and adding trailing zeroes to get 23 bits.
• 5. Determine sign bit.
• 6. Add 127 to the exponent (from step 3) to get e.
• 7. Convert e to an 8 bit binary number (add leading
• zeroes if needed).
• 8. Write in IEEE format by concatenating s, e, and f.

48. • (24.5)10
• 11000.1
• Write in scientific notation
• 11000.1 x 20
• Normalise the number by moving the binary
point and changing the exponent
• 1.10001 x 24
• 100 0100 0000 0000 0000 0000
• Sign bit = 0
• Add 127 to get e
• e=127+4=131
• Convert 131 to binary
• 10000011
• 0 1000 0011 100 0100 0000 0000 0000 0000

49. • 1. Regroup the binary digits into groups of 1, 8,
and 23 digits.
• 2. Convert e to a decimal number.
• 3. Subtract 127 to get the exponent .
• 4. Delete the trailing zeroes from f and write 1.f
x 2exponent where the exponent is the value from
step 3 and f is the original f with the trailing
zeroes removed.
• 5. Un-normalise the number by moving the
binary point until the exponent is 0.
• 6. Drop the 20. It is no longer needed.
• 7. Convert the binary number to decimal.
• 8. If s is 1, put a minus sign in front of the
decimal number.

50. • 1100 0001 1100 0100 0000 0000 0000 0000
• 1 1000 0011 100 0100 0000 0000 0000 0000
• Convert e to decimal
• 1000 0011 = 131
• Subtract 127 to get true exponent
• 131-127=4
• Delete the trailing zeroes from f and write
1.f * 2exponent.
• 100 0100 0000 0000 0000 0000
• 10001
• 1.10001 x 24

51. • Un-normalise the number by moving and
binary point until the exponent is 0.
• 1.10001 x 24 = 11000.1 x 20
• 11000.1
• 24.5
• s=1
• Answer is -24.5