class-exercises161、1011100010112=

8=2、

(

156

)10

=

(

)23、convert

0.3910

to

anumber

in

radix2.Theprecision

must

achieve10%。ten

minutesclass-exercises161、1011100010112=

5613

8=

B8B2、

(

156

)10

=

(

10011100

)23、將0.3910轉換為二進制數,要求精度達到10%。0.01102Review

of

the

last

lessonPositional

Number

SystemsMSD，LSD，MSB，LSBGeneral

Positional-Number-System

ConversionsA

decimal

fraction

A

number

in

Binary

radixBinary

addition

and

subtraction2.5

Representation

of

NegativeNumbers(負數的表示)Signed-Magnitude

Representation

P34[符號–數值表示法（原碼）]Complement

Number

Systems

P35(補碼數制)Radix-Complement

Representation

P36(基數補碼表示法)Two’s-Complement

Representation

P37(二進制補碼表示法)2.6

Two’s-Complement

Addition

and

Subtraction

P39(二進制補碼的加減法)Table

2-6

P40(十進制數與4位二進制數對照表)2.6.3

Overflow

(溢出)

P412.5.1

Signed-Magnitude

Representationa

number

consists

of

a

magnitude

and

a

symbol indicating

whether

the

magnitudeis

positive

or negative.MSB

as

the

Sign

bit

(0

=

plus,

1

=

minus)[最高有效位表示符號位（

0=正，1=負）]01111111＝＋12700101110＝＋4600000000＝＋011111111＝－12710101110＝－4610000000＝－02.5.1

Signed-MagnitudeRepresentation [符號–數值表示法（原碼）] Two

possible

representations

of

Zero[零有兩種表示（+0、–0）]An

n-bit

signed-magnitude

integer

range

is (n位二進制整數表示范圍)：–

(

2n-1

–

1)

~

+

(

2n-1

–

1)– The

signed-magnitude

system

has

an

equalnumber

of

positive

and

negative

integers.P35represent

the

result

with8-bitsigned-magnitude

integer!110-110=?the

signed-magnitude

system

negatesa

number

by

changing

its

sign

.Complement

Number

Systemsitnegates

a

number

bytaking

its

complement

as

definedby the

system.2.5.2 Complement

Number

Systems(補碼數制)radix

–

Complement(基數補碼)Diminished

Radix

–

Complement [基數減1補碼(反碼)]2.5.22.5.2Complement

Number

Systems(補碼數制)an

n-bit

numberDD

=

dn–1dn–2·

· ·d1d0

.The

radix

point

is

on

the

right

and

sothe

number

is

an

integer.

If

anoperation

produces

a

result

that

requiresmore

than

n

digits,

we

throw

away

theextra

highorder

digit(s).

If

a

number

Dis

complemented

twice,

the

result

isD.

(P35)2.5.3

Radix-Complement

RepresentationThe

complement

of

an

n-digit

numberis

obtained

by

subtracting

it

from

r

n

.r’s

complement= r

n

-D(n位數D的基數補碼等于從

r

n

中減去該數)Example

:

Table

2-4

P362.5.3

Radix-ComplementRepresentation2.5.3

Radix-Complement

Representationr’s

complement= r

n

-DIf

D

is

between

1

and

rn

–

1,

thissubtraction

produces another

numberbetween

1

and

r

n

-

1.what

is

the

result

of

the

subtraction,IfD

is

0?2.5.3

Radix-Complement

Representationr’s

complement= r

n

-DIf

D

isbetween1

and

rn

–

1,

thissubtraction

produces

anothernumberbetween1and

r

n

-1.what

is

the

result

ofthesubtraction,

IfD

is

0?

0000000…0

(n-bit)(positive)Diminished

Radix–

ComplementRepresentation[基數減1補碼表示法（反碼）]The

Diminished

Radix

–

Complement

ofan

n-digit

number

is

obtained

bysubtracting

it

from

r

n

-1[

n位數的反碼等于從

rn

–

1

中減去該數]Example:

Table

2-4

,2-5

P.36(r-1)’s

Complement

=

r

n

–

1-D*2.5.6

(P38)there

are

two

representations

of

zero,positive

zero

(00

×

×

×

00)

andnegative

zero

(11

×

×

×

11).2.5.3

Radix-ComplementRepresentation2.5.3

Radix-ComplementRepresentationAdvantager

n

–D=

[(r

n-1)-D]+1This

can

becomplementingplished

bythe

individual

digits

of

D,2.5.4

Two’s

–

Complement

Representation(二進制補碼表示法)Two’s-Complement(二進制補碼的求取)：MSB

(the

signbit): 1=minus;

0=plusWeight

of

the

MSB:-2n-13.

The

range

of

representablenumbers

is-(2

n-1)

through

+(2

n-1

-1).Two’s-Complement(二進制補碼的求取)Example

1. Write

the

8-bit

two’

plementrepresentation

for

the

decimal

number:

-119.(若約定字長是一個字節，試求－119的補碼表示。)?＝011101112，as

formula(公式):2n-D=

(2n-1-D)+128-1：

1

1

1

1

1

1

1

1subtract(減去)+119; -0

1

1

1

0

1

1

11

0

0

01

0

0

0+

1?plus(加)1：－11910：1

0

0

0

1

0

0

12Two’s-Complement(二進制補碼的求取)Example

1. Write

the

8-bit

two’

plementrepresentation

for

the

decimal

number:

-119.(若約定字長是一個字節，試求－119的補碼表示。)＝011101112，?Two’s-Complement(二進制補碼的求取)100010002

+

1

100010012=-11910Example

1. Write

the

8-bit

two’

plementrepresentation

for

the

decimal

number:

-119.(若約定字長是一個字節，試求－119的補碼表示。)＝011101112，帶符號位一起按位取反再+1,得到相反數的補碼.表2-61位十進制數與4位二進制數(P40)十進制二進制原碼二進制反碼二進制補碼-8————1000-7111110001001-6111010011010-5110110101011-4110010111100-3101111001101-2101011011110-110011110111101000或00001111或000000001000100010001200100010001030011001100114010001000100501010101010160110011001107011101110111Sumupfor theComplement(總結)1. Positive

number

has

the

same：Sign-Magnitude,Ones’

–

Complement,and

Two’s-

Complement(正數的原碼、反碼、補碼相同)Sumupfor theComplement(總結)Complement

Number

Systemssigned-magnitude

system010001101111(010001)-1710(110001)Complement(總結)(010001)不變-1710(101111)(010001)符號位改-1710變(110001)符號位不變其余按位取反加1.Complement

Number

Systems連同符號位一起按位取反加1.signed-magnitude

system符號位改變Sign

extension(符號位擴展)We

can

convert

an

n-bit

two’

plementnumber

X

into

an

m-bit

one,

but

some

careis

needed.If

m

>

n,

we

must

append

m

-

n

copies

of X’s

sign

bit

to

the

left

of

X

.

That

is,

we pad

a

positive

number

with

0s

and

a negative

one

with

1s;

this

is

called

signextension.If

m

<

n,

we

discard

X’s

n

–

m

leftmost

bits;however,

the

result

is

valid

only

if

allof

the

discarded

bits

are

the

same

asthe sign

bit

of

the

result

.2.6

Two’s

–Complement

Additionand

Subtraction

(二進制補碼的加法和減法)we

define

[x]to

be

the

two’

plementrepresentation

of

x[x+y]=

([x]+[y]

)[x-

y]=

(

[x]+

[-y]

)2.6

Two’s

–Complement

Additionand

Subtrac