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# The first microprocessor - PowerPoint PPT Presentation

It could be pedagogical to study a small 4 bit nibble computer in order to understand how real numbers are stored and handled! Normaly 32 bit registers are used, but the principles from the nibble computer will remain the same. The first microprocessor.

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It could be pedagogical to study a small 4 bit nibble computer in order to understand how real numbers are stored and handled!

Normaly 32 bit registers are used, but the principles from the nibble computer will remain the same.

William Sandqvist [email protected]

Intel 4004 1971 was a 4 bit (nibble) processor. The customer Busicom wanted a dedicated calculator chip, but the disobedient engineers created a general purpose processor.

William Sandqvist [email protected]

4.3Imagine a ”Nibble” computer

• Computer registers are ”rings”.

A four bit register can hold 24 = 16 numbers.

Either 8 positive (+0…+7) and 8 negative (-1…-8) ”signed numbers”, or 16 (0…F) ”unsignednumbers”.

If a result is bigger than 16, the register overflows. (In some processors indicated with a Carry-flag).

William Sandqvist [email protected]

• Imagine a ”Nibble” fixed point format, Q3!

• Imagine a ”Nibble” floating point format!

William Sandqvist [email protected]

Q3 [b3b2b1b0]

Fixed point format

The Q-format is the same as the signed numbers but downscaled:

-1 < Q3 < 1-b0

The step between numbers is the weight of b0 0.125

Fixed point number line!

Fixed point circle!

William Sandqvist [email protected]

Floating point format

Sign b3

Significand 1.b0

1.02 = 1.0101.12 = 1.510

Exponent b2b1-1 (exess1)

200-1 = 0.5201-1 = 1210-1 = 2211-1 = 4

+ 1.0  0.5 = +0.5+ 1.0  1 = +1+ 1.0  2 = +2+ 1.0  4 = +4

+ 1.5  0.5 = +0.75+ 1.5  1 = +1.5+ 1.5  2 = +3+ 1.5  4 = +6

- 1.0  0.5 = -0.5- 1.0  1 = -1- 1.0  2 = -2- 1.0  4 = -4

- 1.5  0.5 = -0.75- 1.5  1 = -1.5- 1.5  2 = -3- 1.5  4 = -6

Floating point number line!

William Sandqvist [email protected]

Maximum Quantisation Error MQE:

Fixed point: (2-3)/2 = 2-4 = 0.125/2 = 0.0625

Float: (6-4)/2 = 1

William Sandqvist [email protected]

In fixed point formats the number zero is represented as [0000]

In floating point format, there is no representation for zero!

We could use, 0000 as +0 and 1000 as –0, as in the IEEE standard format!

Floating point circle!

The IEEE 32 and 64 bit float standard

William Sandqvist [email protected]

Converting Real numbers to Binary

William Sandqvist [email protected]

Significand 1.b2b1b01.0002 = 1.000101.0012 = 1.125101.0102 = 1.25101.0112 = 1.375101.1002 = 1.5101.1012 = 1.625101.1102 = 1.75101.1112 = 1.87510

Exponent b4b3-1 (exess1)00-1 = -1 200-1 = 0.501-1 = 0 201-1 = 110-1 = 1 210-1 = 211-1 = 2 211-1 = 4

• 0.2510 = 0.012 = 1.02  2-2 exponent -2 not representable

• 0.812510 = 0.11012 = 1.10122-1 representable

• -1,37510 = -1.0112 = -1.0112  20 representable

• 4.2510 = 100.012 = 1.00012  22 significand 1.0001 not representable

• 7.510 = 111.12 = 1.1112  22 representable

William Sandqvist [email protected]

Algorithm:1. Check for zeroes2. Align significands3. Add/Sub significands4. Normalize result

a = 0011112 0 01 111  + 011.111b = 0100102 0 10 010  + 10 1.010

Align significands. 10 > 01 shift smaller number a to right to get same exponent:Significand 0.1111 exponent 10

Normalize result:

10.0011 exp 10 1.00011 exp 11Rounding: 1.00011 ~ 1.001

Result: 0 11 001

a = 1.875 b = 2.5 a+b = 4.5 (4.375)

William Sandqvist [email protected]

Algorithm: 1. Check for zeroes2. Add exponents and subtract Bias3. Multiply significands4. Normalize

Exponents: 01 10 Bias = 1exp = 01+10 -1 = 10

Multiply significands

Normalize result:

10.010110 exp 10 1.0010110 exp 11

Rounding: 1.0010110 ~ 1.001

Result: 0 11 001

a = 1.875 b = 2.5 a*b = 4.5 (4.6875)

William Sandqvist [email protected]

By using the exess–127 format for the exponent, we can compare floats as if they were sign and magnitude integers!

Dec  IEEE-754

William Sandqvist [email protected]

5-4 Cache conflict misses

Data cache 1 k Word (Word = 32 bit). Cache line 4 Words.

Conflict misses depending on memory placement of x and a?

for (i = 0; i < 50; i++)for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

William Sandqvist [email protected]

Memorys are often Byte-organized, but we could draw it as if it was organized in Memory-lines with the same size as the Cache-line.

If the CPU needs a data-item in a line, the whole line is transfered fom memory to cache.

Memory-line: i Cache-line: j = i % K

William Sandqvist [email protected]

a and x always conflict miss …

for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

Conflict miss if the distans between the startaddress in memory of a and of x is a multiple of 1 k Words, 4096 Bytes.

William Sandqvist [email protected]

a and x conflict miss 1:4 …

for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

Conflict miss 1:4 if the distans between the startaddress in memory of a and of x is a multiple of 1 k Words  3 Words, 4096 Bytes  12 Bytes.

William Sandqvist [email protected]

a and xno conflict miss …

for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

No conflict miss if the distans between the startaddresses in memory of a and of x is not a multiple of 1 k Words, and not a multiple of 1 k Words 3, 2, 1 Words!

William Sandqvist [email protected]

8-1 Arbitration

3 Masters and 4 Slaves are connected to a classical bus.

Sequential execution, 7 cycles needed.

1. M1  Rd S12. M2  Rd S23. M3  Rd S44. M2  Rd S35. M3  Rd S36. M2  Rd S27. M1  Rd S1

Many other alternatives are possible!

William Sandqvist [email protected]

Weighted round robin slave-side arbitration.

Connections and weights:

• Best case execution time (3 cycles)

• M1  Rd S1 ; M2  Rd S2 ; M3  Rd S4

• M1  Rd S1 ; M2  Rd S3 ;

• ; M2  Rd S2 ; M3  Rd S3

• Worst case execution time (4 cycles)

• M1  Rd S1 ; M2  Rd S2 ; M3  Rd S4

• M1  Rd S1 ; ; M3  Rd S3

• ; M2  Rd S3 ;

• ; M2  Rd S2 ;

William Sandqvist [email protected]