Monday, October 11 Assignment(s) due:Assignment #7: IEEE FLOATING POINT FORMATS

Monday, October 11

Assignment(s) due: Assignment #7: IEEE FLOATING POINT FORMATSEMULATOR PROGRESS WORKSHEET 1

Quiz(zes) due: Quiz #7: Debug

Note: Quiz #8: Floating Point Formats is due next Monday

Tonight is the cutoff date for Assignment #5 (Complement Arithmetic)

Next Monday is the cutoff date for all Debug and Floating Point assignmentsand quizzes(these topics will be covered on the midterm)

Any assignments turned in to me by Thursday afternoon will be available in the labon Monday afternoon

Please note: The midterm is next Monday (at the beginning of class)

- an 8.5 in. * 11 in. cheat sheet will be allowed

We will review for the midterm tonight

Conversion to/from floating point formats:

- decimal number to hexadecimal floating point format:

- convert decimal number to binary

- convert to normalized scientific form (varies)

- convert exponent to its excess-N form

- add sign bit

- convert binary number to hexadecimal

- hexadecimal floating point format to decimal value

- convert hexadecimal number to binary

- determine sign

- determine exponent

- determine fraction from its normalized form

- convert to its non-scientific binary form

- convert to decimal

- a problem using the same format as Lab #9

- 32 bits

- 1 bit for the sign bit

- 7 bits for the exponent- using excess-64- using base 16 for the exponent

- 24 bits for the normalized fraction- radix point to the left of the most significant

bit

- round-off because of rounding

- how would 25.12510 be stored in hexadecimal?

Step 1: Convert 25.12510 to binary

Step 1: Convert 25.12510 to binary 25.12510 = 11001.0012

Step 2: Normalize the binary number


Step 2: Normalize the binary number .00011001001 * (162)10



Step 3: Calculate the exponent



Step 3: Calculate the exponent2 + 64 = 6610 = 10000102




Step 4: Put the number together

0 1000010 000110010010000000000000





0 1000010 000110010010000000000000

In hexadecimal:0100 0010 0001 1001 0010 0000 0000 0000

4 2 1 9 2 0 0 0

- the same problem using the same format as Lab #9, part 2 (IEEE format)

- 32 bits



- 24 bits for the normalized fraction- radix point to the right of the most significant bit

- the most significant bit is "hidden"


- how would 25.12510 be stored in hexadecimal?

Step 1: Convert 25.12510 to binary


Step 2: Normalize the binary number


Step 2: Normalize the binary number 1.1001001 * (24)10


Step 3: Calculate the exponent



Step 3: Calculate the exponent4 +127 = 13110 = 100000112




0 10000011 10010010000000000000000





0 10000011 10010010000000000000000

In hexadecimal0100 0001 1100 1001 0000 0000 0000 0000

4 1 C 9 0 0 0 0



- using the IEEE standard short real

- 32 bits



- 23 bits for the normalized fraction- radix point to the right of the most significant bit

- the most significant bit is hidden


- how would 414E000016 be represented as a decimal number?

Step 1: Convert 414E000016 to binary 414E000016 = 01000001010011100000 . . .2

or 0 10000010 10011100000 . . .2


or 0 10000010 10011100000 . . .2

Step 2: The exponent is 13010 or 3 (130 - 127)


or 0 10000010 10011100000 . . .2


Step 3: The fraction is1.10011100000 . . .2


or 0 10000010 10011100000 . . .2



Step 4: The normalized form is1.1001112 * (23)10


or 0 10000010 10011100000 . . .2




Step 5: In binary1100.1112


or 0 10000010 10011100000 . . .2




Step 5: In binary1100.111

Step 6: In decimal12.875

Data Table

N: 18X: 19

20: 40021: 82022: 518

23: 019

24: 119

25: 41826: 31927: 518

28: 730

29: 623

30: 900

Data Table

N: 18X: 19

20: 40021: 82022: 518

23: 019

24: 119

25: 41826: 31927: 518

28: 730

29: 623

30: 900

If the inputcards are:

013 052 070 005 041

What is theoutput?

Data Table

N: 18X: 19

016

005

013

007

6. Given the following information in DEBUG:

-u 100, 10A21F0:0100 BB2001 MOV BX, 12021F0:0103 BE0200 MOV SI, 221F0:0106 0307 MOV AX, [BX]21F0:0108 0300 ADD AX, [BX + SI]21F0:010A 034002 ADD AX, [BX + SI + 02]21F0:010D 894004 MOV [BX + SI +04], AX21F0:0110 CD20 INT 20

-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0120

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0120 0002

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0312 0120 0002

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0312 0120 00020336

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0312 0120 000203360246

12 03 24 00 10 FF



-e 120 12 03 24 00 10 FF 0120 0121 0122 0123 0124 0125 0126 0127

AX BX SI0312 0120 000203360246

12 03 24 00 10 FF 46 02

More practice:

Given the following information in DEBUG:

-u 100

21F0:0100 MOV AX,ABCD

21F0:0103 MOV BX,0120

21F0:0106 MOV CX,0004

21F0:0109 SUB AX,[BX]

21F0:010B ADD BX,0002

21F0:010E LOOP 109

21F0:0110 INT 20

-d 120,12F EB 02 46 00 EB 38 5A 01 - 10 FF 01 02 A3 FF 88 04

Trace through the program above and indicate below the contents of the registers and memory location just before the INT 20 instruction is executed.

AX = BX = CX =

The number 1.27510 = ? 2

a) 1.010001

b) 1.01

c) 1.1001101

d) 1.1

If the number 1.010001 normalized using the IEEE standard (using radix 2 to normalize, radix point to the right of the most significant bit, and the most significant bit being hidden) it will result in whichfraction and exponent?

a) .1010001 and 1

b) .1010001 and 0

c) .010001 and 1

d) .010001 and 0

For the binary number 1 10000101 01010001000000000000000:

Using excess-127, what is the exponent in base 10?

1. -6

2. 0

3. 4

4. 6

For the binary number 1 10000101 01010001000000000000000:

With an exponent of 6 and using radix 2, what is the binary normalized form?

1. -.010100001 * (26)10

2. -.010100000101 * (2-6)10

3. -1.01010001 * (26)10

4. -1.101010001 * (26)10

If the binary normalized form is -1.01010001 * (26)10: The number in binary is:

1. -1010100.01

2. -101010001.1

3. -101010001

4. -.00000101010001

If the binary number is -1010100.01: The number in decimal is:

1. -168.25

2. -84.25

3. -84.5

4. -168.1

Vocabulary:

address DS (data segment)

ASCII code IP (instruction pointer)

bit LOOP instruction

byte memory

compilation/translation register

Noise and Distortion:

-occurs every time a signal is transmittted

- may be random

- white noise (a background hiss)

- impulse noise (amplitude peaks that block out data)

- external, usually from equipment

- internal, signal from another channel

- echoes on the line

- amplitude changes

- circuit failures

Transmission in the presence of noise:

Error Detection

- error-detecting Codes

- add redundant bits to each unit of data

- receiver knows when an error has occurred

- receiver can ask for retransmission

- to detect single-bit errors

- often used with 7-bit ASCII code characters

- add one parity bit so number of 1-bits in the word is always even (even parity) or odd (odd parity)

- works about 70% of the time

- e.g. to transmit the character ‘A’ (6510 => 4116 => 10000012)

1 0 10 0 0 0






- using odd parity:

1 0 10 0 0 0

1 0 10 0 0 01






- using odd parity:

- using even parity:

1 0 10 0 0 0

1 0 10 0 0 0

1 0 10 0 0 0

1

0

Error Detection and Correction

- error-correcting Codes

- add more redundant bits to each unit of data

- receiver can detect and sometimes correct the error

-Hamming Codes

- use a “block parity” method of adding parity bits

- use more than 1 parity bit

- allow for detection and correction of single-bit errors

- Hamming algorithm:

- start numbering the bits at 1 (from the left)

bit 1 bit 2 bit 3 bit 4 bit 5



- all bits whose position is a power of 2 are parity bits

- i.e. bits 1, 2, 4, 8, 16 etc.

parity bit


parity bit

parity bit




- i.e. bits 1, 2, 4, 8, 16 etc.

- then fill in the data bits

parity bit

data bit


data bit

parity bit

parity bit




- i.e. bits 1, 2, 4, 8, 16 etc.

- then fill in the data bits

- so a code with 2 data bits

- would require:

parity bit

data bit


data bit

parity bit

parity bit

parity bit

data bit


data bit

parity bit

parity bit

- Hamming algorithm continued:

- each data bit is checked by those parity bits before it whose bit positions add up to its position

- data bit 3 is checked by parity bit 1 and parity bit 2

parity bit

data bit


data bit

parity bit

parity bit


- each data bit is checked by those parity bits before it whose bit positions add up to its position



parity bit

data bit


data bit

parity bit

parity bit


- or - parity bit 1 checks data bits 3 and 5

- parity bit 2 checks data bit 3

- parity bit 4 checks data bit 5

parity bit

data bit


data bit

parity bit

parity bit

parity bit

data bit


data bit

parity bit

parity bit

parity bit

data bit


data bit

parity bit

parity bit


- each parity bit forms a group of itself and the data bits it checks

- 3 groups are formed

- there will be an overlap of data bits

- each group must add up to an even number of 1’s (or none) for even parity

- or an odd number of 1’s for odd parity

- groups:

- bits 1, 3 and 5

- bits 2 and 3

- bits 4 and 5

- each group contains only 1 parity bit

- groups:

Group 1: bit 1

Group 2: bit 2

Group 3: bit 4

parity bit

data bit


data bit

parity bit

parity bit


- groups:

Group 1: bit 1

Group 2: bit 2

Group 3: bit 4

- data bit 3 is checked by parity bit 1 and parity bit 2- (1 + 2 = 3 or 3 = 1 + 2)

parity bit

data bit


data bit

parity bit

parity bit


- groups:

Group 1: bit 1 bit 3


Group 3: bit 4


parity bit

data bit


data bit

parity bit

parity bit


- groups:



Group 3: bit 4



parity bit

data bit


data bit

parity bit

parity bit


- groups:

Group 1: bit 1 bit 3 bit 5





parity bit

data bit


data bit

parity bit

parity bit

0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1

Legal codewords for the 2-data-bit Hamming code (even parity):

0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1

Parity Bit Data Bits P1 D3, D5 P2 D3 P4 D5


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1



0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0

1


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0

1 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0

1 0 1


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0

1 0 1

1 1 0


0

1

1

0

0

0

P1 P4P2 D3 D5

P1

P1

P1

P2

P2

P2

P4

P4

P4

D3

D3

D3

D5

D5

D5

1 1


0 0 0

1 0 1

1 1 0

110


- to transmit the data number 01

- we use the Hamming code word

- if we transmit

- there is an error in a parity bit

- the receiver checks the groups:

10P1 P2 P4D3 D5

01 1

10P1 P2 P4D3 D5

11 1



- if we transmit


- the receiver checks the groups: p1, d3, d5 ok

10P1 P2 P4D3 D5

01 1

10P1 P2 P4D3 D5

11 1



- if we transmit


- the receiver checks the groups: p1, d3, d5 okp2, d3 no

10P1 P2 P4D3 D5

01 1

10P1 P2 P4D3 D5

11 1



- if we transmit


- the receiver checks the groups: p1, d3, d5 okp2, d3 nop4, d5 ok

10P1 P2 P4D3 D5

01 1

10P1 P2 P4D3 D5

11 1



- if we transmit


- the receiver checks the groups: p1, d3, d5 okp2, d3 nop4, d5 ok

- only one group is wrong- the error is in the parity bit for that group

10P1 P2 P4D3 D5

01 1

10P1 P2 P4D3 D5

11 1



- if we transmit

- there is an error in a data bit

- the receiver checks the groups: p1, d3, d5 nop2, d3 nop4, d5 ok

- two groups are wrong- the error is in the common bit

10P1 P2 P4D3 D5

01 1

11P1 P2 P4D3 D5

01 1

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

7-bit character ‘A’ using the Hamming code and odd parity:

- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1,p2,p4,p8,


- 6510 => 4116 => 10000012


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3p2, 3p4p8


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5p2, 3p4, 5p8


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5p2, 3, 6p4, 5, 6p8


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5, 7p2, 3, 6, 7p4, 5, 6, 7p8


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5, 7, 9p2, 3, 6, 7p4, 5, 6, 7p8, 9


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5, 7, 9p2, 3, 6, 7, 10p4, 5, 6, 7p8, 9, 10


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 011

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 011 1

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 011 1 1

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 011 1 1 0

groups: p1, 3, 5, 7, 9, 11p2, 3, 6, 7, 10, 11p4, 5, 6, 7p8, 9, 10, 11


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

01 0 011 1 1 0

- what if we transmitted:

1 1 1 0

groups: p1, 3, 5, 7, 9, 11 Check 1, skip 1p2, 3, 6, 7, 10, 11 Check 2, skip 2p4, 5, 6, 7 Check 4, skip 4p8, 9, 10, 11 Check 8, skip 8

Or . . . Starting at the parity bit:


- 6510 => 4116 => 10000012

10P1 P2 P4 P8 3 5 6 7 9 10 11

00 0 01

10P1 P2 P4 P8 3 5 6 7 9 10 11

01 0 011 1 1 0

groups: p1, 3, 5, 7, 9, 11 Check 1, skip 1p2, 3, 6, 7, 10, 11 Check 2, skip 2p4, 5, 6, 7 Check 4, skip 4p8, 9, 10, 11 Check 8, skip 8

- what if we transmitted:

1 1 1 0

Or . . . Starting at the parity bit:

Reconstitute the groups,and check

From Lab #10:

Identify the parity bits and the data bits:

Complete the table for the 11-bit Hamming codeword:

Group Parity Bit Data Bits1 P12 P23 P44 P8

Using even parity, which bit (if any) is in error:

Bit# 1 2 3 4 5 6 7 8 9 10 11

a) 0 1 1 1 0 0 0 0 1 1 0

b) 0 1 1 1 0 0 1 1 0 1 0

c) 1 1 0 1 1 1 1 0 0 1 1

Monday, October 11

Assignment(s) due: Assignment #7: IEEE FLOATING POINT FORMATSEMULATOR PROGRESS WORKSHEET 1

Quiz(zes) due: Quiz #7: Debug

Note: Quiz #8: Floating Point Formats is due next Monday

Tonight is the cutoff date for Assignment #5 (Complement Arithmetic)

Next Monday is the cutoff date for all Debug and Floating Point assignments(these topics will be covered on the midterm)

Any assignments turned in to me by Thursday afternoon will be available in the labon Monday afternoon

Please note: The midterm is next Monday (at the beginning of class)

- an 8.5 in. * 11 in. cheat sheet will be allowed

We will review for the midterm tonight

Documents

Monday, October 11 Assignment(s) due:Assignment #7: IEEE FLOATING POINT FORMATS