WEEK 2 Number Systems & Boolean Algebra

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Number Representation (Recall)

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Addition (Unsigned Numbers)

Binary addition of unsigned numbers is done just like in decimal. Add digits and generate carries

Important: We can get a non-zero

carry out; if we are limited to n-bits,

this becomes an overflow condition

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Subtraction (Unsigned Numbers) Binary subtraction is also similar to decimal

subtraction. Use borrow when needed

Important: Doesn’t seem to work if

second term is larger!! A negative result

is often considered an underflow

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Subtraction (Unsigned Numbers) Using r’s Complement

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Subtraction …Examples

10’s complement subtraction

A = 72532, A’ = 27468

B = 03250, B’ = 96750

A – B = 72532 B – A = 03250

+ 96750 + 27468

1 69282 30718 - 69282

Minuend is greater than

Subtrahend discard carry Minuend is less than subtrahend

10’s comp the result

and add negative sign

10’s comp

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Subtraction …Examples

2’s complement subtraction

A = 1010100, A’ = 0101100

B = 1000100, B’ = 0111100

A – B = 1010100 B – A = 1000100

+ 0111100 + 0101100

1 0010000 1110000

2’s comp - 0010000

Minuend is greater than

Subtrahend discard carry Minuend is less than subtrahend

10’s comp the result

and add negative sign

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Subtraction r-1 complement …Examples

9’s complement subtraction

A = 72532, A’ = 27467

B = 03250, B’ = 96749

A – B = 72532 B – A = 03250

+ 96749 + 27467

1 69281 30717 - 69282 + 00001

69282

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Subtraction …Examples

1’s complement subtraction

A = 1010100, A’ = 0101011

B = 1000100, B’ = 0111011

A – B = 1010100 B – A = 1000100

+ 0111011 + 0101011

1 0010000 1101111

0000001 - 0010000 0010001

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Radix Conversion The integral part of a decimal number to radix r,

repeatedly divide by r with reminders becoming ai

Convert (77)10 to binary

Integer Remainder Coefficient

77

38 1 a0

19 0 a1

9 1 a2

4 1 a3

2 0 a4

1 0 a5

0 1 a6

(77)10 = (1001101)2

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Examples Convert (173)10 to r = 7

Integer Remainder Coefficient

173

24 5 a0

3 3 a1

0 3 a2

(173)10 = (335)7

Converting from Binary to decimal

(101101)2 = 1x25 + 0x24 + 1x23 + 1x22 +0x21 +1x20 =

(45)10

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Fraction Conversion

To convert the fractional part of a number to radix r,

repeatedly multiply by r; integral parts of products

becoming ai

Convert (0.7215)10 to binary

0.7215x2 = 1.443 a-1 = 1

0.443x2 = 0.866 a-2 = 0

0.866x2 = 1.772 a-3 =1

0.772x2 = 1.544 a-4 = 1

0.544 = 1.088 a-5 = 1

(0.7215)10 = (0.10111..)2

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5 bit word (1 for sign, 4 for values)

-2+3=1

1|1110

+0|0011

0|0001

There is a carry, value is answer

7-9=-2

0|0111

+ 1|0111

0|1110

no carry value is 2scomp of answer

Answer (-) 2scomp of 1110- 0010

-5-(-3)=-2

1|1011

+0|0011

0|1110

No carry answer=(-) 2s comp of 1110=- 0010

More Examples: Binary

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323-121=202

0|323

0|879

1|202

Carry value is answer

323-532=-219

0|323

0|468

0|781

No carry value is (-)10s comp of answer

10scom - 219

111-888=-777

0|111

0|112

0|223

No carry value is (-)10s comp of answer, 10s comp -777

More Examples: Decimal

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Codes Decimal numbers are coded using binary bit patterns

Decimal BCD (8421) 2421 Excess-3 Gray

0 0000 0000 0011 0000

1 0001 0001 0100 0001

2 0010 0010 0101 0011

3 0011 0011 0110 0010

4 0100 0100 0111 0110

5 0101 0101 1000 0111

6 0110 0110 1001 0101

7 0111 0111 1010 0100

8 1000 1110 1011 1100

9 1001 1111 1100 1101

SELF-COMPLEMENTING

ONLY ONE BIT

ALLOWED TO CHANGE 9=2+4+2+1

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More on Codes

BCD

Self-complementing code: Code word of 9’s complement of N obtained

by interchanging 1’s and 0’s in the code word of N

Self-complementing Codes

N=1011=5, 9’s complement is 4=0100=1’s comp of 1011

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BCD Addition

BCD addition can be carried out as follows:

4 0100 8 1000

+8 +1000 +9 +1001

12 1100 17 10001

Add 6 0110 0110

BCD 00010010 1 0111

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Complements, Signed Arithmetic, Codes - Book Sections

Complements and Signed Arithmetic are covered in Chapter 1, Sections 1.5 through 1.7

Page 19

Binary Logic

Binary (Boolean) logic deals with binary variables and binary logic functions.

By binary, we means they can take on two discrete values. “0” means OPEN, NO, FALSE.

“1” means CLOSED, YES, TRUE.

Logic functions (expressions) produce an output as some logical function of its inputs.

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Truth Tables

We can define a logic function using a truth table.

The truth table tells us the value of the function for each possible set of input values.

Binary variable

f(x,y): binary function

f(x,y) is 1 if x and y are 00 or 11.

That is (Not(x) and Not(y)) OR (X and Y)

f(x,y)=x’.y’+x.y

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Basic Logical Operations

Three basic logical operations:

AND/OR/NOT.

NOT also called COMPLEMENT or INVERTER

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Precedence of Operations

Precedence of operations:

(), then

NOT, then

AND, then

OR

Introduce () to help clarify precedence.

f(x,y,z)=Not(x).y+z= (Not(x) and y) or z

f(x,y,z)=(x+z).y+x=(x or z) and (y or x)

Page 23

Logic Gates and Symbols

In reality, logic functions are implemented via logic gates.

We have symbols to represent the different logic functions visually (e.g., In a schematic diagram).

Logic gates are implemented via transistors.

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Logic Function and Symbol for

AND (2-input) Binary variable

f(x,y): binary function

f = x.y

f = x&y

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Logic Function and Symbol for

OR (2-input)

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Logic Function and Symbol for

NOT (1-input only/always)

How many logical functions? 7, one function for each segment.

b1,b2,b3,b4

e,g

1010

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b1 b2 b3 b4

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b1 b2 b3 b4

Logical Expressions

A=(b1’b2’b3’b4’)+

(b1’b2’b3b4’)+

(b1’b2’b3b4)+

(b1’b2b3’b4)+

(b1’b2b3b4’)+

(b1’b2b3b4)+

(b1b2’b3’b4’)+

(b1b2’b3’b4)+

(b1b2’b3b4’)

(b1b2b3’b4’)+

(b1b2b3b4’)+

(b1b2b3b4)

A’=(b1’b2’b3’b4)+(b1’b2b3’b4’)+(b1b2’b3b4)+

(b1b2b3’b4)

1

Derive expressions for B to G

Page 30

Multiple Input Logic Gates

AND/OR can have any number of inputs. Meaning is understood.

Output of n-input AND gate is 1 when every input is 1, otherwise 0.

Output of n-input OR gate is 1 when any input is 1, otherwise 0.

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Just to be clear…

Can show the truth tables for 3-input

AND/OR gates:

Simple Question: Given n-inputs, how many possible input combinations are there?

Page 32

Textbook Sections

Material introducing binary logic is in

Chapter 1, Section 1.9 of the textbook.

Page 33

Boolean Algebra

Provides a means of manipulating Boolean

logic functions.

Introduced by George Boole in 1854.

Shown to be useful/applicable to electrical

switching circuits by C. E. Shannon in 1938.

Boolean Algebra Postulate 1: Set and Operators

Define a set K containing two or more elements.

Define two binary operators:

+, also called “OR”

·, also called “AND”

Such that for any pair of elements, a and b in K, a + b and a·b also belong to K

Postulate 2: Identity Elements

There exist 0 and 1 elements in K, such that for every element a in K

a + 0 = a

a · 1 = a

Definitions:

0 is the identity element for + operation

1 is the identity element for · operation

Remember, 0 and 1 here should not be misinterpreted as 0 and 1 of ordinary algebra.

Postulate 3: Commutativity

Binary operators + and · are commutative.

That is, for any elements a and b in K:

a + b = b + a

a · b = b · a

Postulate 4: Associativity

Binary operators + and · are associative.

That is, for any elements a, b and c in K:

a + (b + c) = (a + b) + c

a · (b · c) = (a · b) · c

Postulate 5: Distributivity

Binary operator + is distributive over · and · is distributive over +.

That is, for any elements a, b and c in K:

a + (b · c) = (a + b) · (a + c)

a · (b + c) = (a · b) + (a · c)

Remember dot (·) operation is performed before + operation:

a + b · c = a + ( b · c) ≠ (a + b) · c

Postulate 6: Complement

A unary operation, complementation, exists for every element of K.

That is, for any elements a in K:

0aa

1aa

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Axioms/Postulates of Boolean

Algebra Continued Using the definitions of AND/OR and NOT

functions, we can show all the postulates

are satisfied.

Note: Postulates are facts that can be taken as true; they do not require proof.

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Theorems and Properties of

Boolean Algebra Theorems help us out in manipulating

Boolean expressions.

They must be proven from the postulates and/or

other theorems known to be proven correct.

Note: duality in relationships if we interchange + with ² and 0 with 1.

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Use of Boolean Algebra

We can use the postulates and theorems to

simplify expressions, to prove equivalence

of expressions, etc.