1.Number Systems

7/27/2019 1.Number Systems

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Number Systems


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HET202 Digital Electronics Design 2

Analogue vs Digital

Analogue Continuous range of values

Precision limited by noise

Digital Discrete range of values

(usually 2) but Precision limited by number ofbits

t

t


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Analogue vs Digital The real world is analogue (if you ignore

Quantum effects!).

It is common to translate data from analogue todigital for processing and back to analogue for

output. Why bother?

Advances in integrated circuits (ICs) has made thecomplex processing of digital data very inexpensive.

Digital circuits are inherently more noise resistant.

AnalogueWorld

AnalogueWorld

Digital

Processing

AtoD DtoA


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Why Boolean? It is convenient in electrical systems to use a

two-value system to represent values true/false, on/off, yes/no, 1/0. Two voltage or current levels can be used.

Easier to process and distribute reliably.

Dont think of them as numbers (even though weoften represent them as 0/1 for brevity).

The Need for Binary Numbers.

Multi-value quantities need to be represented inthe digital system. We therefore need numbersmade up from the simpler two value system torepresent a rangeof discrete values.


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Combinational vs. Sequential Two categories of digital circuits:

Combinational: The output is purely a function of the

current input.

Sequential:

The output is dependent upon what has

happened previously and possibly thecurrent inputs.


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Combinational Circuits Output values only depend upon the

current input values (ignoringpropagation effects speed)

System has no state (memory of

what has happened before)

Z=f(W,X,Y)

W

X

Y

Z


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Sequential Circuits Output depends on present and past

inputs. Circuit has state (held in flip-flops).

Example: circuit for elevator call button.

Has to remember the brief button pressuntil the elevator arrives.

SequentialCircuit

(internal state)

Call elevator(push button)

Momentary value

Elevator calledPersistent value


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Positional Number Systems

1024.339

3 x 10-1

3 x 10-2

9 x 10-3

4 x 100

2 x 101

0 x 102

1 x 103

decimal point

Base 10 weightings are powers of 10


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Unsigned Binary Numbers

1001.101

1 x 2-1 = 0.500

0 x 2-2 = 0.000

1 x 2-3 = 0.125

1 x 20 = 1.000

0 x 21 = 0.000

0 x 22 = 0.000

1 x 23 = 8.000

binary point

9.625

Binary weightings are powers of 2

Each bit of thenumber may be

represented bya Boolean value.


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Decimal to BinaryGenerate each digit bysuccessive division or

multiplication.

There is no guaranteethe fraction will be finite!

Number = 36.375

Base= 2

Decimal

Number

Binary

Digits

Converted

Number

0 0 0100100.0110

0.5 1 0100100.011

0.75 1 0100100.01

0.375 0 0100100.0

36 0 0100100

18 0 0100109 1 01001

4 0 0100

2 0 010

1 1 01

0 0 0

Fractional part Multiplication by base (2)

Whole number part Division by base (2)

Fractionpart

Whole

part


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Decimal to Base-7Number = 45.75

Base= 7

Decimal

Number

Binary

Digits

Converted

Number

0.25 1 063.5151

0.75 5 063.515

0.25 1 063.51

0.75 5 063.545 3 063

6 6 06

0 0 0

Fractional part -

Multiplication by base (7)

Whole number part -Division by base (7)


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Binary Addition

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0 carry 1

Easy huh?


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Binary Addition

11 1 1

1 1 1 01 0 1 1

1 1 0 11 0 0 0

1 0 1 10 1 0 0

Carry out of8-bit number

Carry out ofeach column

190+141=331

1

1


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Binary Subtraction

2 2 22 2

0 1 0 11 1 1 01 1 1 00 0 1 0

0 1 1 11 0 1 1

Borrow in fromleft column

11111

Borrow outBoth rowssubtracted

229-46=183

A borrow-out of 1 fromthis column becomes a borrow-in

of 2 in this column


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Exercise Convert to 8-bit binary and do the

arithmetic operation: 120 + 54

224 134

112 89

Convert back to decimal and check the

results.


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Exercise Working120 + 54 224 134

112 89


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Exercise Solution120 + 54

120 = 01111000254 = 001101102

101011102

= 174

224 134

224 = 111000002134 = 100001102

010110102

= 90

112 89

112 = 01110000289 = 010110012000101112

= 23

*


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Exercises Some exercises to highlight overflow

(carry/borrow)


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Binary Number Circle

4-bitbinary

number circle

0

4

8

12

2

6

14

10

1

5

9

13 3

7

15

11

11 - 1 = 10

In real hardware there isa fixed number of bitsavailable. We oftenignore leading zeroes butthey are still there!

Example:

If we only use 4 bits thenthe binary counting

sequence wraps aroundat 150.


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Binary Number Circle

4-bitbinary

number circle

0

4

8

12

2

6

14

10

1

5

9

13 3

7

15

11

8 - 14 = 10

Subtracting across theboundary still works ifyou think of the result asthe distance on thenumber circle.

(Modulo arithmetic ignore the borrow/carry)

8 1000

- 14 - 1110

10 (1)1010


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Representing ve Numbers Several choices for notation

Sign + Magnitude notation (discussed) 1s Complement (not discussed)

2s Complement notation (discussed)

Various excess codes (not discussed)


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Signed Numbers

Sign+Magnitude Notation

Sign Bit Magnitude

Simple binary number0+ve

1 -ve

Problems?

+0

-0

0000

1000


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Signed Numbers

Sign+MagnitudeArithmetic

Difficult to do have to work out what operation toperform

5 + -6 actually calculate -(6 - 5) i.e. exchange the

operands and do subtraction!-5 + -6 actually calculate -(5 + 6) i.e. negate the

addition of the negated numbers!

Required action depends the signs of the numbers andon which has the larger magnitude. Natural for us a

bit hard for the computer since the only way it can

work out the bigger number is to do a subtraction!


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Sign+Magnitude ExamplesValue 4-bit sign + magnitude 8-bit sign + magnitude

+7 0111 00000111

+6 0110 00000110

... ... ...

+1 0001 00000001+0 0000 00000000

-0 1000 10000000

-1 1001 10000001

-2 1010 10000010

... ... ...

-7 1111 10000111


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Signed Numbers 2s ComplementAs for straight binary numbers but with

the weighting of the most significant bitbeing negative.

Example:

4 bit weights are -8,4,2,1 8 bit weights are -128,64,32,16,8,4,2,1

Need to know how many bits are being

used to work out the value of thenumber dont omit leading zeroes.


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Signed Numbers 2s Complement

1001.101

1 x 2-1

= 0.5000 x 2-2 = 0.000

1 x 2-3 = 0.125

1 x 20 = 1.000

0 x 21 = 0.000

0 x 22 = 0.000-1 x 23 =-8.000

binary point

-6.375

Sign

Bit


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2s Complement ExamplesNumber

4-bit 2'scomplement

8-bit 2'scomplement

+7 0111 00000111+6 0110 00000110

... ... ...

+1 0001 00000001

0 0000 00000000

-1 1111 11111111

-2 1110 11111110

... ... ...

-6 1010 11111010

-7 1001 11111001

-8 1000 11111000


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What is OVERFLOW? When the word length is n bits, we say

that an overflowhas occurred if thecorrect representation of the sum(including sign) requires more that n

bits.


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Interpreting valuesas unsigned binary

Interpreting values

as 2s complement

Binary

1011

0011

(0)1110

V=0,C=0

0110

0110(0)1100

V=1,C=0

Binary

1011

1101

(1)1000

V=0,C=1

1011

1010(1)0101

V=1,C=1

Signed

Value

-5

3

-2

6

6-4

Wrong

Unsigned

Value

11

3

14

6

612

Signed

Value

-5

-3

-8

-5

-65

Wrong

Unsigned

Value

11

13

8

Wrong

11

105

Wrong

Examples


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Why Use 2s Complement? Addition and subtraction use the same rules as

unsigned binary.

Same hardware may be used for both. Hardware may be shared.

Carry (C) is used for unsigned, Overflow (V) for signed

Carry

Overflow

Carry

Overflow

OP

SignedNumbers

SignedNumber

OP

UnsignedNumbers

UnsignedNumber

The samehardware

Condition code

register (CCR)


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2s Complement To negate a number in 2s

complement it is necessary tosubtract it from zero.

A short-hand for this is:

Complement the number (flip each bit).

Add 1 to the result (discard any carry).


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2s Complement

Generate (-7) in 8 bits

7 => 000001112

Subtract from zero

0000 00002 0

-0000 01112 7

1111 10012 -7

Generate (-7) in 8 bits

7 => 000001112

Complement & add 1

(discard any carry)

0000 01112 7

1111 10002 ~7

+0000 00012 +1

1111 10012 -7


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Widening 2s Complement Numbers Need to sign-extend the number to

maintain the signed value.

1 0 0 1

1 1 1 1 1 0 0 1

0 1 0 1

0 0 0 0 0 1 0 1

(-8)+1= -7

(-128)+64+32+16+8+1= -7

4+1= 5

4+1= 5


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Number LinesNumber Lines for 8-bit Signed & Unsigned Numbers

0 +127 +128 +255

0 +127-128 -1

Unsigned Range

Signed Range (2s complement)

10000000 11111111 00000000 01111111 10000000 11111111

Which number is larger?00011101 0111010011010111 0001110111010111 10000100

AAA

BBB


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Number Ranges8-bit 2s complement

-128 +127

16-bit 2s complement

-32,768 +32,767

8-bit unsigned

0 +255

16-bit unsigned

0 +65535


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Exercises Convert to 8-bit 2s complement and do

the operation 120 + 54

112 89 straight subtraction

112 + (89) do complement & add


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Exercise Working120 + 54 112 89


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Exercise Solution120 + 54

120 = 01111000254 = 001101102

101011102

= 82

(overflow!)

*112 89

112 = 01110000289 = 010110012

000101112

= 23


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Exercise Working

OR

112 89

= 112 + (~89) + 1

112 89 = 112 + (89)


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Exercise Solution

OR

112 89

112 = 01110000289 = 010110012~89 = 101001102

*

= 112 + (~89) + 1

112 = 011100002

~89 = 101001102+1 = 000000012000101112

= 23

112 89

112 = 011100002

89 = 010110012~89 = 101001102-89 = 101001112

= 112 + (89)

112 = 011100002-89 = 101001112

000101112= 23


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Binary Coded Decimal (BCD) BCD uses decimal arithmetic.

We are simply using a 4-bit code torepresent each decimal digit.


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Binary Coded Decimal (BCD)

0011,1001.0110

decimal point

6 x 10-1 = .6

9 x 10 0 = 9.0

3 x 10 1 = 30.0

39.6

Each group of 4-digits represents a decimal digit


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Binary Coded Decimal Disadvantages (relative to binary):

Less efficient in storage e.g. 4 bytes allows

99,999,999 versus 4,292,967,296 in binary.Arithmetic is more complex (in particular

multiplication and division).

Advantages (relative to binary): Input/Output conversion is much easier i.e.computer human readable.

Less conversion errors i.e. decimal fractions are

converted exactly (within precision). This isimportant in finance. Example - try $6.30 in BCD& binary.


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Other Codes Other codes may be used that are attractive

in a particular application. Example:

What problems might the following binary shaftencoder have when the detector is on thetransition 100 011?

BinaryEncoderdisk

000001010011

100101110111

ClockwiseSequence

Detector3 Sensors

S2 S1 S0

S2S1S0


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Shaft Encoder Timing

1002=4 0112=31102=6

100 011

1112=7

Close-up of100->011 transition

Skewed transitionsdue to tolerancesbetween sensors

(position,threshold).

S2

S1

S0

Shaft appears to jump aroundin position although it is onlymoving a very small amount.

4 6 7 3


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A Unit-distance Code

Binary encoderdisk

Unit distanceencoder disk

Only 1-bit changesat each transition

000

001

010

011100

101

110

111

000

001

101

100110

111

011

010


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Character Codes -ASCII0 1 2 3 4 5 6 7 8 9 A B C D E F

0 nul soh stx etx eot enq ack bel bs ht lf vt ff cr so si

1 dle dc1 dc2 dc3 dc4 nak syn etb can em sub esc fs gs rs us

2 sp ! " # $ % & ' ( ) * + , - . /

3 0 1 2 3 4 5 6 7 8 9 : ; < = > ?

4 @ A B C D E F G H I J K L M N O

5 P Q R S T U V W X Y Z [ \ ] ^ _

6 ` a b c d e f g h i j k l m n o7 p q r s t u v w x y z { | } ~ del

Second Digit

FirstDi

git

ASCII 7-bit code used to represent common English

characters (subset of international set).


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Other Codes

See text 8-4-2-1 (BCD)

6-3-1-1

Excess-3 2-out-of-5

Gray code (similar to unit distance example)

Unicode (character code able to represent thecharacters of many languages)

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1.Number Systems

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