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IDENTIFICATION
Product Name:

Application Notes

Product Code:

DEC—OB—NAAA-D

Date Created:

July I2,I965

Maintainer:

Special Systems Group

PUP-B
LIBRARY

A??LECATIGN

NOTE

80i

SCALENG FGR FiXED-POINT, 2's COMPLEMENT ARETHMETEC

lNTRODUCTlON
in the programming at arithmetic operations on a Fixedepoint, 2‘s complement arithmetic com~

pater, the position of the scale point; that is, the decimai point in a decimal number or the

binary paint in a binary number, must be kept track at by the programmer“

ane numbers

have been entered in the computer, there is no hardware or movable machine point to repre—
sent

the scale points.

The scale point exists only in the mind at the programmer, and only by

keeping track at its imaginary position is he able to correctly interpret the machine's calculated
results

The tundamentai properties of scaled numbers can be simply explained it we imagine a hypo~
thetical decimai machine that is capable of manipulating numbers consisting of a sign and five
decimal

digits,

in this hypothetical computer,

in real computers, the machine acts as it

there were a scale point between the sign and the leftmost decimal digit.
machine points

It is called the

Thus, every number contained in the computer can be thought of as a signed

decimal traction.

Example:

+Al2345
machine point

However, the programmer is tree to assign a decimal point at any position in the numbers

For

example, the above number could represent +l23e45 it the scale point were thought of as being
three pieces to the right at the machine point.

in that case the number wauid be written

$2345 i323, where D3 indicates that the decimal point is three places to the right of the machine
points

in other wards, D3 is the decimal scaie Factor.

changing the contents of the machine.

Any scale Factor may be chosen without

Examples:

+Al2345 D2
+£2345 D4
+Al 2345 DO

+l2 9345

+l234i-85

+3. i2345

The scale factor need not be restricted by the size of the machine words

Numbets in our

hypothetical computer can be assigned scaie factors that exceed five, or the scale factors can
even

be negative.

Examples:

$2345 D7 +i234500
~5321345 D—A +30000l2345
=

Of course, these are merely programmer‘s representations; the machine number is always re—
stricted to a sign and five digits.

Addition and Subtraction

In addition and subtraction, the scale factors of the numbers to be combined must be identical.

Thus,

#142204 D3 added to —123332 D3 gives

a sum

165536 D3 (422.04

This rule has the same basis as in ordinary arithmetic.

233332

if the scale factors differ,

655.36)

one

number

must be shifted until the scale points are aiigneds

Examples:

The number

+Al427l Dl (+i Q4271)
+28496 D3 (+384.9o)

+Al427l Dl is brought into the accumulator and shifted right two decimal places

before addition or subtractione

The scale point shifts with the numbers

+384.96

(+.OOi42 D3)
(+ .38496 D3)

“+3"8""‘5';‘3" 8”

”(T38638“ 33"";

+OOT e42

Notice that if the number of the higher scale factor had been shifted left instead, its two most

significant digits would have been test and the resulting sum wouid have been seriously in error.
The shifting of numbers off the ieft east at the accumulator in this manner is called overflow.

Multiplication
When two numbers are multiplied together, the scale factor of the product is the algebraic sum
of the scale factors of the multiplier and multiplicand.

(:00200 D3)

Example:

(106000 D2): :OOOTZ DS,

l2.

Normally the most significant part of the product is in the AC and the least significant part is
in another register.

Thus, the product in the above example would appear in the computer
The machine point for

with the machine point between the Sign and leftmost digit in the AC.

the least significant portion of the product is ignored, since this involves double-precision
arithmetic.
+000 l 2 +00000 D5
A

it is important to remember that the two decimal

numbers, :00200 and 106000, when multiplied

in the computer will result in the machine product of

:OOOTZ txOOOOO regardless of the positions

of the scale points in the multiplier and multiplicand.

the programmer.

Examples:

Thus fractions,

well

The scale points must be kept track of by

integers, can be multiplied in exactly the same way.

4220000 05 +90060 D3 +£300i2 +00000 D8 (+20,000 +.6 +12,000)
+00200 DO
+£6000 DO +IPOOl2 +OOOOO D0 (+ .002 +.06 +.000i2)
+POQOO [3—2 160000 0-4 +9090 +00000 D-o (+.00002 +.00006 +.00000000l2)
x

1‘

Division

The remarks above for multiplication apply directly to division, with two exceptions.

First,

the scale point of the result is the algebraic difference of the scale points of the operands.

Second, the scale point of the divisor must be smaller than the scale point of the dividend;
that is, the scale point of the quotient must be positive.

Again, the contents of the registers

in our decimal computer will look the same no matter where scale points of the numbers are

located

SCALlNG ON A BiNARY COMPUTER

The fundamentai atoperties of scaled numbers in a computer as outlined above can now be

angelied to the binary and octal numbering systems as used in a Ts complement, fixedepoint,

binary computer.

The decimal scale factor becomes the binary scale factor and is indicated

by a B in front of the scale factor. The machine point is still between the sign and the leftmost
til", but in this case the sign is a binary digit.

in a l2~bit computer such as PUP—8,

then,

the leftmost bit is the Sign bit and there are eleven bits for the number, with the machine point
between the first and second bits.

Because the number system used by the machine is now different from that customarily used by

the programmer, conversion becomes one of our new considerations.

The programmer may be

dealing with decimal or octal numbers, but because the machine is binary, the scale factors
must

be determined from the binary equivalents.

As will be explained below,

scaling analysis

is performed on each problem so that the binary scale factors chosen result in the most efficient
use

of the lZebit word,

Having selected the appropriate Scale factor for a given number, it is

expressed in decimal or octal form followed by the binary scale factor.

For example, the

combination 975 BlO means that the decimal number 975, when converted to binary form, has a

binary point ten places to the right of the machine points
The decimal number 975 BlO when converted to binary would appear in the machine as:

on noon no
A

machine point

binary point

Grouping these bits into threes, it is more convenient to write this-number in its octal form:
3636 hit)

Notice that in this octal form,

cannot indicate

the point which separates the integral from

the fractional part, because it comes within one of the octal digits,
becomes part of the leading digits.

Also, the sign bit, bit 0,

Negafive numbere 5:52;”: be wr‘ifi'en eifher one of fwo wayse

For exomfle; Momwier fhe ocfo1

number:

~32 86

AS a pos‘éfive octo!

number, 3.2 86 wou1d be sfored 1n fhe compufer as:
000 001 101000
A.

binary poinf
As (1 negofive number in 2‘3 eomplemeni‘, it would be stored:

111110010111
A

111110011000
A

binary poinf
1n the oc’ro1 form; 11 would. be wrifien:

7630 86

Again we conno’r separate 1he integro1 from ’rhe Frocfionoi pom and fhe Sign is Encorporofed
into ‘rhe

leading ocfoi £11911“.

The summary of the above FU185 of binary scoiing Es given in Appendix 1

OVERFLOW
1n ocidifion fo shiffing digits off 1he 1ef’r end of fhe occumulai’or, overflow can e130 occug" 1n

orifhmefic operations,

Suppose we are working with signed quanfifies and we odd fhe nembees:

Decimal Vo1ue

Binary Represenfefion

0.6.2131

18 85

0/10 010 000 000

2200 85

5 BS

([100 101 000 000

500 85

23 BS

010 111 000 000

2700 85

EquivaIenf

Notice that there was no carry to the left of the first machine position (i ,e., into the sign bit)

However, if we try to add the numbers:
Decimal Value

Binary Representation

28 B5

OAll lOO 000 000
090 lOl 000 000

3400 B5

lAOO CW 000 000

4100 35

5 B5
33 B5

Octai

Equivalent

500 B5

The result as given in the machine would be erroneous because the magnitude portion of the
AC is not large enough to hold the sum.

This situation is described as overflow.

lf overflow occurs in division, the link is set to i,
to

division takes place, and control returns

the main program.

Overflow occurs'in the PDP—8 when:

In addition, the sign of the addend and augend are the same and the

sign of the sum is different and/or the link is set.
2

in division, the magnitude of the divisor is less than that of the dividend

when the scale factors are the same (iv.e., when the quotient _>_ l «0 BO).
3.

A digit is shifted off the left end of the accumulator.

Overflow is something which must be avoided in all normal circumstances.

To accomplish this,

the programmer must have some knowledge of the magnitude of the numbers with which he is

working and, accordingly, must locate each number at such a scale that overflow cannot occur
even

in the “worst case.”

ln this connection, the concept of ”minimum binary scale” is helpful

the

largest positive integer that can be contained is:
Oil

iiiAOOO 000
binary point

which in decimal is 3i

(Lee, 2 ml).

At a binary scale of 5,

The lorgesr posifive integers which con be conroinecl or orher binory scales ore robulared in

Appendix 2%

ii, For exomple, we have rhe number 75 re piece in rhe computer, the rcble in

Appendix 2 indicates rhor if con be conroined or o binory scole of 7 or higher.

A binary scale

of 6 or lower would nor be sufficienr io hold o number oi fhar magnitude.

ii: o binary scole rector greorer then 7 were used, rhe number would be shifrecl far’rher re the

righr rhon neceseary resulring in underfiow‘. The number of significonr iiguree rhar can be
corried in rhe froc’rionol parr is rhereby reduced”

if ihe number 75 were carried as 75 B7,

rhere is room in the machine word ior irociionol resulrs since four binary birs can follow rhe

binary poinr; if if were carried 75 El l
precision orirhme’ric

there is no provision For (3 fractionol por’r in single-n

The programmer may no? always be successful in his attempts ro arrange numbers so rhor over~
flow will

nor occur.

if overflow does occur, rhe PDP~8 does not holr bur an indicaiion of some

rype is given. The rype oi indicorion depends upon the operorion which produced rhe overflow.
if a programmer suspecrs rhor overflow moy occur as o resulr of on oddirion or division, he

should follow such an operorion by c program sequence rhor would correct rhe error or or leosf
indicore fhaf such on overflow rook place.

The proper locorion of the binary poini‘ oncl rhe avoidance of overflow,

bear, roices some

eiiorf on ihe par? of fire programmer.

PROGRAMMlNG TECHNlQUES FOR SCALENG

General
A complete ser oi shifr subrourines in borh single and double precision is ovailable from rhe

PDPwB Library (DigirolmgwgmUwSy/m) for use in scaling numbers borh before and after ori’rhmeric

operorions. When using rhe rourinesr

ii is

imporroni‘ to keep in mind rhor o lefr shifr of one

posirion ciecreoses rhe binory scole iocror by one., Similoriy a righr shift“ of one position 1r:
creases

rhe binory scole Focror by oner

One technique used in scaling is to express numbers in a symbolic form that would clearly

imply the position of the binary point.

The general form is:

x2"q

x“

X is the absolute value of the number.

where:

2q is the factor such that q is the smallest integer that makes 2Q
greater than the maximum value of X.
q corresponds to the minimum binary scale factor which was pre~

viously discussed

X' is the scaled form of X (i.e.’, X is X' with the binary point
q places to the right of the machine point)

A scaling analysis should be performed on each problem to insure maximum accuracy (i .e., the
most efficient
same

use

of the binary word so that there are no leading

At the

time, the programmer must insure that there will be no loss of the most significant bits by

overflow at any step in the calculation.
must

insignificant bits)

These are the two bounds within which the programmer

keep the numbers as they are stored and manipulated in the machine.

Analysis
in the programming of any given problem or equation, there are three steps prior to the actual

coding which should be taken to insure maximum accuracy and to prevent error due to over—
flow.

Step l:

Determine the limits of the values of all numbers to be used in the

problem (maximum and minimum).
Step 2:
true

Determine the scale factors and set up the relationships between the

numbers and the scaled fractions.

Step 3: Substitute the scaled quantities into the original equation and cancel where possible“

The scale factors that do not cancel specify the number

it the scole factor of a term is negotivef the

of required shift operations.

if the scale

number must be shifted right before manipulation is performedi

factor is positive, the number must be shifted left if it is to be stored at
minimum binary scale

Addition Scaling
As emphasized before,

quantities to be added (or subtracted) must have the same scale factors.

However, in order to prevent an overflow in the summing process,
final sum according to its limit.

it is not enough to scale the

Generally the program must be scaled by the largest limit

which applies to any element in the sum or partial sum generated during the summing process.

Example 'l
Program the operation specified by:
n

i=i

where
mum

ail< K for i

l, 2, 3,

The maximum vaiue of A is i K

‘

that is, the maxi-

value of the sum is obtained by multiplying the Upper iimit of any element in the list to

be summed by the number of elements in the list.

Statement

Solve the above problem for n

10 and K

iO0.0.

Analysis
Step l:

Therefore, A i K, ’n iOOiO
Step 2:

of: iO0.0 tori: i, 2, 3,

A‘

‘

i0: i000

Step 3:

Machine Instruction Coding

Assume that the ten values of the numbers are stored in consecutive locations starting «at location AT

ADDUP,

B7 and that the sum is stored in A.

/lN|T|ALlZE

CLA
DCA

LOOP,

TAD

M12

DCA

COUNTR

TAD

ADR]

DCA

POlNTR

TAD

JMS

SPSR

/SUMMATION LOOP
/ENTER SHIFT RIGHT SUBROUTINE

POINTR
TAD

DCA

ISZ

POINTR

lSZ

COUNTR

JMP

LOOP

/EXIT
POTNTR,
M3,
COUNTR,
M12,
ADRT,
A,

/CONSTANTS

Ow3
0

0-42
A]
O

Multiplication Scaling
When multiplication is performed in digitai computers,
numbers is one ”2n“ hit number,

note

that the product of two ”n” bit

UsuoHy the high—order portion is left in the AC and the

Eow~order portion is stored 3r; e epechéed register.

The fundamental rule, again, is:

Scale focfor of mulfipHer + scale {Cider of: mumpiicend

producf

scale Facing? of:

Sfcfemenf

Program ’rhe mulfipiicaf‘ion operation:
Z

cub

Anaiysis
Step I:

04400.0

b<IOO0.0
Therefore

x<400,000.

Step 2:

29
bzflob
217
c1

217x 2290
=22a

Step 3:

210E)
b

Machine Insfrucfion Coding

Assume Hm? fhe values of c: and b are sfored En iocations A and B.

fhaf‘ ’rhey are scaled B9 and BIO,

SETMUL:

Assume

respecfively.

CLA
TAD

JMS

MULT

DCA

SVA

TAD

MP"!

DCA

SVA +1

TAD

JMS

DPSL

/MOVE PRODUCT TO TEMP STORE

/SHEFT PRGDUCT LEFT 2 PLACES

SVA

SVA,

0 0*

M2,

drifter the multiplication (22-bit signed product), the shitt brings two more significant bits into

the high~order portion of the product.

Knowing the maximum value at y more definitely (i .e.,

if a and b could never be maximum at the some time) would allow tor even more accuracyg

this example, the iimit oi y was not known so it was assumed to be 400, 000 as calculated in

Step l of the analysis.

Division Scaling

When division is performed in digital computers, the dividend is a ”2n” bit word and the
divisor is an ”n“ bit word.

Remember that in division the divisor
without overtiow.
occur

must be greater than the dividend for division to occur

Therefore, the programmer should scale the values so that division will

with maximum dividend and minimum divisor.

For example, it both dividend and div-

isor are stored at minimum binary scale, the dividend should be shifted one position to the

right by a doublewshitt subroutine before division to insure that overflow does not take place.
i

Statement

Program the division operation:

2:533

Analysis
Step i :

m<24, GOO
60<n<l , 000

Therefore,

24<y<400

m:2

Step2:

n~210n‘
<5.

y”2>/
2

Step3:

‘215m‘w25

y‘-—-

i0I
n

y222”i

‘
.

in this exompie, the

2“] impiies that the quotient wouid have

of the machine point

(i.e.,

in the sign

one

significant bit to the ieft

bit). Thus, the division wouid resuit in overfiow.

This probiem couid be averted by shifting the dividend right one position,

previously

mentioned, before division takes piece.
3.

Machine instructions

Assume that the singie~precision division subroutine
is used.

(Digital 8~i2~F)

The dividend is stored in iocotions numi and numi + i at c:

scoie factor of Bi5 and the divisor is stored in num2 at o scoie factor
of Bio.

SETDIV,

CLA

CMA

3M3

DPSR

/LOAD --i [NTO AC
/SHiFT RIGHT SUBROUTINE

NUMi

CALDIV,

/SAVE MSB
/MOVE LSB

DCA

SVA

TAD

L3H

DCA

CALDIV + i

TAD

SVA

JMS

DiV

/DiViDE Bié/BiO

ANS

/SAVE QUOTiENT

NUM2,

DCA

APPENDIX I

RULES

BINARY

SCALING

ADDITION
The binary scale Factor at the addend,

augend, and sum are alike.

23.9 B8 + I69.7 88:146I2 B8

SUBTRACTION
The binary scale Factor of the minuend,

subtrahend, and difference are alike,

107.8 B7

23.2 B7: 84,.6 B7

MULTIPLICATION
The binary scale factor of the product is the sum at the binary scale factor of the multiplier
and multiplicand.

12.2 B6 x 3 87: 36.6 BI3
24.9 85x I35.5 88

3373.95 BI3

(minimum binary scale)

DIVISION

The binary scale Factor of the quotient is the binary scale Factor of the dividend minus the

binary scale Factor of the divisor.
88 BIB—3H 85:8 BIO

APPENDIX 2

MINIMUM

BINARY

Binary ScaIe

SCALE

Maximum DecimaI Capacity

I27

255

I 023

2 047

4 095

8 WI

I6 383

32 767

65 535

I3I O7I

262 I43
524 287
I 048 575

I9
20
2I

2 097 ISI

4 I94 303

8 388 607

(2n wI)

APPHCATTON

NOTE

802

The question of matrix inversion comes up occasionally concerning the PDP—B,
There is na general answer ta the matrix-«inversion problem. The approach depends upon the
”behaviar“ at the given matrix.

Basically, the problem is:

Given a matrix A! tind a matrix B, such that
AB

l where l is the unit matrix.

There are three basic approaches.
are

All numbers below are decimal, all operations

Floating-point without EAE.
GAUSS

-—

Time

JQRDAN METHOD
%

Storage

(2.5) (l .46 mils) (M3) where the matrix is M x M
$15

3M2

630 t about 550

(This does not include input/output which would require about
450 locations.)
(well behaved) matrix:

For a ll) x TO

Time

i"*e"*"3.é>o seconds

Storage a: T480 words + 450 tor 1/0
For a 20 x 20 (well

Time

”A“!

Storage

behaved) matrix:

29.28 seconds
%

2380 words "i“ 450 “For

l/G

RANK ANNlHlLATlON

5M3 (l .96 mils) where the matrix is M
Storage 5:36M2 l2M 630 about 400 words
Time

(This does not include l/O which would require about 450
locations.)
For a TC x l0 well behaved

Time

nonsymmetric matrix:

539.80 seconds

Storage

’3‘“? T750 t 4:50

tor

l/O

For a 20 x 20 well behaved eoneymetric matrix:
Time

”$78.4 seconds

Storage

$323670 t 450 tor

l/O

@AUSS

Time

SElDEL (lTERATlVE METHOD)
is:

Storage

(.88 mils) per Iteration (M x M matrix)
(it would be impossible to estimate the number at iterations re-—
quired. The method may not work in some cases.)
M

3M2

6M + 630 + about 200 (not including

l/O)

For a l0 x l0 matrix:
Time

2388 mils / iteration

Storage

x l

WO + 450 for l/O

For 20 x 20 matrix:
Time

2:352 mils/ iteration

Storage 7/2l50 + 450 for [/0

Generally, a matrix is well behaved it the elements along the maiar diagonal
As nandiagonal elements become
larger than the diagonal elements, the matrix becomes ill behaved and it becomes increasingly

dominate are greater than the other elements at the matrix.

difficult to invert,

NOTE

APPLlCATlON

804

THROUGHPUT TO lBMwCOMPATlBLE MAGNETTC TAPE
A common data—acquistion situation is the realwtime conversion of an

signal(s)

analog

form together with the recording of the resulting digital information on mag—-

digital
lBM~compatible format (lBM Binary);

netic tape in

A typical system consists of a l388 Converter,

PDP-8, 57A and 570 Tape Trans-

port. The use of the 57A permits the controlling program to load one core buffer while the
57A writes (via the data break) the data in a second core buffer on tape and produces the in—
terrecord gap.

The function of the buffers is then reversed by the program.

Within a record, the character density must conform to lBM standards; 200, 556,
IBM specifications govern the longitudinal tolerance as each
or 800 characters to the inch.
Each interrecord gap must be regardless of density (within allow-=to lBM standardss
conform
These comments apply for tape that will
able tolerance) 3/4 inch to
character may be recorded.

be processed on TBM Magnetic

Tape Units lBM 729 ll, lV, V, and Vi.

In a given file on tape (disregarding the end~of-file gap and character), the
effective density of recorded information is a function of the relationship between record
as

compared

to gap

length.

For example a file consisting of a single

long

length

record with no record

gaps would consist entirely of‘data or would be TOGO/o efficiento On the other hand, a file consisting of many 3/4—inch records would be only 50% efficient since half of the available tape

would be used for the 3/4vinch interrecord gaps required by each recordo

While the 57A is writing the contents of the one buffer on tape and then writing
the blank interrecord gap, the second buffer is being filled by information coming from the
A-to—D converters

Since the conversion rate must be constant, this process continues both dur—-

ing the emptying of the first buffer and during gapping.

There is,

therefore, a difference in

the rate at which data enters one bLTffe—r from the Auto—D converter and the rate at which data
as distinguished from gap is written on tape from the second buffera
Consider a 570 Transport with a speed of 75

ips recording at 200 cpi

The number

of characters that may be recorded per second is equal to:

75 x 200

15000 cps

Suppose, though, that the situation discussed above (3/4—inch records separated
3/4~inch
gaps) existed. The actual conversion rate (assuming a l2—bit conversion) would
by
be;

l5,000 x l/2 x 1/2

3750 conversions per second

The second factor of l/2 arises because only half of the tape is available for
actual data (0.50 duty factor) due to the interrecord gaps.

Figure l illustrates the duty factor as a function of characters per record for all
three standard lBM densities.

A 3/4—inch interrecord gap is assumed.

DUTY FAC‘E’OR {EFFEC‘FE‘JENESS}
253

5l2

$025

204%

$000

200 bpt

,63t

.773

.872

.932

.335

555

bpi

.3“

.55t

,‘i’lé

.834

$66

800 bpf

.299

‘53

3'73

.372

LOO

200
4

o
g.

.80

......

o
.

u”

‘50

‘

W
556

‘

”cw—W‘BOV.
a

/
/

800

/9
°/

256

{Hz

l024

2048

4096

CHAR] RECORD

Note that these rates could never be realized in practice with a double—buffer

type of throughput since the assumption of the necessity of a double buffer implies that at least
two records-an

Figure 2 lists the throughput rates which could be achieved if no interrecord gaps
required. Next, the duty factor is used to calculate actual (or effective) character rates

were
as a

ideal case and therefore one interrecord gapwwill be written.

function of characters per record and this information is plotted.

Consider the top curve of

in this case the nominal (or ideal) character rate is 90, 000 cps. The vertical bars
the two "ends” of the curve illustrate how far below the nominal character rate the actual

Figure 3.
at

character rate is at any point.

Note that for small records the actual character rate falls off

very rapidly.

Writing Rates (c ps)
200 bpi
45 ips
75 ips

il2.5 ips

Effect of Record

ips

200 bpi
556 hpi

800 bpi
75

ips

200 bpi
556 bpi
800 hpi

ii2.5 ips

200 lopi
556 bpi

800 bpi

9, 000
l5, 000
22,500

556 bpi

800 bpi

25, 000

36, 000
60, 000
90, 000

4i , 700

62,500

Length

256

Si 2

i024

2048

4096

5, 670
9,520
l0,770
9, 450
i5, 900
i7, 950
l4,l80
23, 800
26, 900

6, 960
l3,800
l6,600
ii, 600
23, 000
27, 640
l7,400
34, 500

7, 850
l7,8i0
22,700
13, iOO
29, 530
37, 800
l9,620
44, 500
56, 700

8, 380
20,800
27,850
i3, 990
34, 670
46, 400
20,970
52, 000
69, 600

8, 680
22,700
3i,400

4i , 500

i4, 490
37, 870
52, 300
2l,730
56, 800
78, 400

One interesting aspect oi Figure 3 Es that by proper choice of record

length (or
oelow), tape speed! and density? a desired AQC conversion rate may be
selecteda Note the horizontai line at 40,000 cpso Assuming Habit conversions, the conver—
sion rate here wooid be 20,000 conversions per second and this couid be achieved by using a
double butter, each one at which was capobie oi holding the number of characters indicated on
the lower scale below the intersection of the horizontai iine (at 40578063 cps) with the respective
equipment curves.
core

buffer size,

see

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80,000 1

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22251300

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CHARACTERS / RECORD

Figure 4 defines certain details of EBM-«cornpotibie tapess Note that the longitudinal check character of the end of a record is separated from the other characters in a record
by a gap that is about four times as large as the intercharacter spacing within a record (regardless at density) and that the record gap proper starts following the check character. (No longi—
tudinal check bit is written it the iongitudinai coent in the associated trock is even).
This fact is of consequence only it records composed at an extremely few characters
are

of concern and would effect the numerical data used to construct

vicinity (irew just

Figure 2 in the immediate

the right) of the vertical scales

Figure 5 illustrates the time availabie between onalog~to-digital conversions as
a function of
tape speed, density and butter sizes, Figure 5 may be used to determine how many
machine cycles are available for programming in throughput situations of this nature.

Longitudina Cheek Chemo?"
+ .p—OIZS 003

200 CPI K .0207

Gay

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0029::

(Note 33

+ 40046
556 CW:
800 Cpl: 005‘ + .0033 .HOOZE

Notes 5 0 §

Record Gap

$0025

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Top. Marian

NOiES:
i.

Tape is shown with oxide side up, Read/Write head on same side as oxide.

Tape shown representing i b” in all tracks, NRZl recording; 1 bit produced by reversal of flux polarity, tape
fully saturated in each direction.

3. Variation permitted in the location of the Check Character assuming nominal values For tape speed and all ascillator timings in the Tape Control. No longitudinal check bit is written it longitudinal count in the track is even.

Myiar Tape: 3/4”, +5/32", ~l/i6".

Acetate Tape.

3/4", 46/32“, ~l/8”.

Zero Backward creep.

Forward

creep less than 0.2“ per cycle.

Dimensions of tape measured at 50% relative humidity and 70°F.

Tape thickness (Mylar or iBM HD) is 0.0022“,

+.0003", -.0004”.

interchangeability, skew of each tape unit is adiusted to 0.25 user: or less at the read bus of
Maximum skew for any reel of tape, read by any tape
unit connected to any tape control, must be equal to or less than the read character gate for the bit density and tape
speed at which the tape was written.
a.

To insure complete

the tape unit when readingwhile—writing continuous l bits.

Time from First Bit

Write Skew

Gate. i5%
729 ll or V, 550 (:92
729 ll or V, 200 cpi
729lV or Vl, 556 cpi
729EVor Vi, 200 cpl

R358

Fall

6,3 Psec

l6,l riser:
44.0 usec

l6.9 usec

29.5 usec

729 V, 800 Cpi

4.3psec
H.4usec
6.3usec

729 Vl, 800 cpl

4.0 usec

6.8 usec

When reading, while writing coded

l0.8usec

l0.4usec

information, all bits within a character must be received before the rise at the

write skew gate.

8..

Read or Write Character

729 H

l\/, 556 cpi
729 ll or V, 200 cpi
72‘? lV or Vl, 556 cpi
729 lV or Vl, 200 cpi
72‘? V, 800 cpi
729 Vl, 800 cpi
or

Gate, 6%
iO.5 usec

29.2 usec
7.5 usec
2i .0 usec
7.9 psec

5.4usec

Time Between Characters: Writing~shallnotbeless than tall at the skew gate timing
pius i usec, including vori~
ations due to tape speed, skew and bit configuration. Reading—eshall not be less than read character
gate timing plus
i usec, including variations due to tape
speed, skew, and bit configuration.

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2048

BUFFER StZE
WORDS

It must be emphasized that the data graphed as continuous lines in

Figures l, 3,

and 5 actually consists for each curve of a series of discrete points, which are shown as continuous lines for convenience only.

For example there can never be a record consisting of

356.l35 characters, and no point associated with this number of characters is actually present
intended to be present in the curves of Figure l.