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Computer Organization
By
Dr. M. Khamis
Mrs. Dua’a Al Sinari
Computer Organization

The course is aimed at designing the different computer
components (circuits) and connecting these components
in a way to achieve the goals of a specific architectures.
Computer (hardware) consists of processor, memory and I/O
units.
 Processor itself consists of Arithmetic Logic Unit (ALU) and
Control unit.
All the above units are designed using primitive logic circuits.

Course Objectives
Understanding the basic Laws of Boolean algebra.
 Designing and using the basic logic devices.
 Understanding the operation of the main computer
units and their design.
 Interconnecting the various computer units to achieve
the specific architecture.
 Presenting the attributes of the different architectures.
 Programming specific architecture using its instruction
set (machine instruction).
 Explaining the Interaction between Computer hardware
and the operating system.

Course outline

The course will consist of two parts:
The first part is Logic design: in which the primitive
components, by which the different devices are designed,
are presented.
 The second part is intended for interconnecting the
components presented in first part in a way to build a logical
system (computer organization).

Part 1: Logic Design
Introduction to number systems and arithmetic
operations in binary system.
Combinational circuits:






Logic Gates (AND, OR, NOT, NOR, NAND and XOR), in this
regards we will give the truth tables and symbols for each .
Laws of Boolean algebra, deriving logical expression and
simplification.
Karnaugh maps and its use for simplification .
half and full adders and binary coded decimal adders
Part 1: Logic Design

devices include:
Decoder.

Encoder.

Multiplexers/ De-multiplexer.

Comparator.
Sequential circuits include :




Flip/Flops and counter
Design Mealy and MOORE machines.
Part 2: Computer Organization









chapter 3: computer system
chapter 7: Input/output
chapter 8: Operating System Support.
chapter 9: Computer Arithmetic
chapter 10: Instruction Sets.
chapter 11: :Instruction Sets: Addressing Modes and
Format.
chapter 12: CPU structure
chapter 16 : Control Unit
chapter 17: Micro Programmed Control Unit.
NUMBERS AND BOOLEAN ALGEBRA
Author: Abhinav Bhatele
Revised By: Dr. M. Khamis
FAll 2008

June 10th, 2008
NUMBER SYSTEMS
To get started, we’ll discuss one of the fundamental concepts
underlying digital computer design:
Number Systems and Boolean Algebra
Deep down inside, computers work with just 1s and 0s.



Computers use voltages to represent information. In modern
CPUs the voltage is usually limited to 1.6-1.8V to minimize
power consumption.
It’s convenient for us to translate these analog
1
voltages into the discrete, or digital, values 1 and 0.
But how can binary system be useful for anything?
First, we’ll see how to represent numbers with
just 1s and 0s.
 Then we’ll introduce special operations
for computing with 1s and 0s, by treating them as
the logical values “true” and “false.”

Volts
1.8
0
0
8

June 10th, 2008
TODAY’S LECTURE
Number systems
Review of binary number representation
 How to convert between binary and decimal representations
 Octal and Hex representations

Number Systems and Boolean Algebra

Basic boolean operations
AND, OR and NOT
 The idea of “Truth Table”
 Boolean functions and expressions
 Truth table for Boolean expressions

9

Numbers consist of a bunch of digits, each with a weight

2
1
.
3
1/10
7
1/100
5
1/1000
Digits
Weights
These weights are all powers of the base, which is 10.
We can rewrite this:
1
102

6
10
6
101
2
100
.
3
10-1
7
10-2
5
10-3
Digits
Weights
To find the decimal value of a number, multiply each
digit by its weight and sum the products.
Number Systems and Boolean Algebra
1
100
June 10th, 2008
DECIMAL REVIEW
(1 x 102) + (6 x 101) + (2 x 100) + (3 x 10-1) + (7 x 10-2) + (5 x 10-3) = 162.375
10

We can use the same trick to convert binary, or base 2, numbers to
decimal. This time, the weights are powers of 2.
Example: 1101.01 in binary
1
23

1
22
0
21
1
20
.
0
2-1
1
2-2
Binary digits, or bits
Weights (in base 2)
The decimal value is:
(1 x 23) + (1 x 22) + (0 x 21) + (1 x 20) + (0 x 2-1) + (1 x 2-2) =
8
+ 4
+ 0
+ 1
+
0
+ 0.25
Powers of 2:
20 = 1
24 = 16
21 = 2
25 = 32
22 = 4
26 = 64
23 = 8
27 = 128
28 = 256
29 = 512
210 = 1024
= 13.25
Useful abbreviations:
K = 210 = 1,024
M = 220 = 1,048,576
G = 230 = 1,073,741,824
Number Systems and Boolean Algebra

June 10th, 2008
CONVERTING BINARY TO DECIMAL
11


162 / 2
81 / 2
40 / 2
20 / 2
10 / 2
5/2
2/2
1/2

= 81
= 40
= 20
= 10
=5
=2
=1
=0
rem 0
rem 1
rem 0
rem 0
rem 0
rem 1
rem 0
rem 1
So, 162.37510 = 10100010.0112
0.375 x 2 = 0.750
0.750 x 2 = 1.500
0.500 x 2 = 1.000
Number Systems and Boolean Algebra

To convert a decimal integer into binary, keep dividing by 2 until the
quotient is 0. Collect the remainders in reverse order.
To convert a fraction, keep multiplying the fractional part by 2 until
it becomes 0. Collect the integer parts in forward order.
Example: 162.375:
June 10th, 2008
CONVERTING DECIMAL TO BINARY
12


162 / 10 = 16
16 / 10 = 1
1 / 10 = 0


rem 2
rem 6
rem 1
Each division strips off the rightmost digit (the
remainder). The quotient represents the remaining
digits in the number.
Similarly, to convert fractions, each multiplication
strips off the leftmost digit (the integer part). The
fraction represents the remaining digits.
0.375 x 10 = 3.750
0.750 x 10 = 7.500
0.500 x 10 = 5.000
Number Systems and Boolean Algebra
This works for converting from decimal to any base
Why? Think about converting 162.375 from decimal
to decimal.
June 10th, 2008
WHY DOES THIS WORK?
13

The hexadecimal system uses 16 digits:
0123456789ABCDEF

You can convert between base 10 and base
16 using techniques like the ones we just
showed for converting between decimal
and binary.
For our purposes, base 16 is most useful as
a “shorthand” notation for binary
numbers.
Since 16 = 24, one hexadecimal digit is
equivalent to 4 binary digits.
 It’s often easier to work with a number like B4
instead of 10110100.


Hex is frequently used to specify things like
32-bit IP addresses and 24-bit colors.
Binary
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Hex
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
Number Systems and Boolean Algebra

Decimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
June 10th, 2008
BASE 16 IS USEFUL TOO
14
Converting from hexadecimal to binary is easy: just replace each hex digit
with its equivalent 4-bit binary sequence.
261.3516 = 2
6
1
. 3
516
= 0010 0110 0001 . 0011 01012

To convert from binary to hex, make groups of 4 bits, starting from the
binary point. Add 0s to the ends of the number if needed. Then, just
convert each bit group to its corresponding hex digit.
10110100.0010112
= 1011 0100
= B
4
. 0010
.
2
11002
C16
Hex
Binary
Hex
Binary
Hex
Binary
Hex
Binary
0
0000
4
0100
8
1000
C
1100
1
0001
5
0101
9
1001
D
1101
2
0010
6
0110
A
1010
E
1110
3
0011
7
0111
B
1011
F
1111
Number Systems and Boolean Algebra

June 10th, 2008
BINARY AND HEXADECIMAL CONVERSIONS
15
2’s complement



Binary number can be represented using sign and
magnitude. If N bits are used to represent the number,
then the last bit is used to hold the sign of the number
while the other (N-1) bits are used to represent the
value.
0 is used for +ve sign and 1 is used for -ve sign.
To get the 2’s complement for any number follow the
following two steps:
Convert each bit in the value into its complement (1 to 0
and vice versa)
2. Add 1 to the result of step 1.
1.
Binary addition & subtraction
If the number is +ve keep it in sign and magnitude form,
otherwise represent the number (magnitude only) using
its 2’s complement.
 Add the binary numbers in the ordinary way as the
decimal numbers.
 The addition in decimal makes carry 1 for the next digit
for each 10 collected in the sum, and the reset which
will be less than 10 is left as result of the addition of the
corresponding bits.
 This operation continues until adding all bits with its
corresponding bits in the other number.

Binary addition & subtraction (Continued)
The addition in binary is exactly the same as decimal
with only one difference, which is, carry 1 is taken for
the next digit for each 2 collected in the sum, and the
reset which is less than 2 is left as result of the addition
of the corresponding bit.
 The addition is continued for all bits including the sign
bit, and in order to get correct answer the number must
be represented in enough number of bits.
 Any carry after the sign bit is discarded.
 The value of the negative result is represented in the 2’s
complement (i.e. the actual value is the 2’s complement
of the result once again).

Computers are binary devices.
 We’re forced to think in terms of base 2.
 Today we learned how to convert numbers between binary, decimal
and hexadecimal.

Also, we have seen:
 We use 0 and 1 as abstractions for analog voltages.
 We showed how to represent numbers using just these two signals.

Next we’ll introduce special operations for binary values and show how
those correspond to circuits.
Number Systems and Boolean Algebra

June 10th, 2008
NUMBER SYSTEMS SUMMARY
19
How logical functions can be defined for expressing
computations
 How to build circuits that implement our functions in
hardware

Number Systems and Boolean Algebra
So far, we’ve talked about how arbitrary numbers can be
represented using just the two binary values 1 and 0.
 Now we’ll interpret voltages as the logical values “true”
and “false” instead. We’ll show:

June 10th, 2008
BOOLEAN OPERATIONS
20
Earlier, we used electrical voltages to represent
two discrete values 1 and 0, from which binary numbers
can be formed.
 It’s also possible to think of voltages as representing
two logical values, true and false.
 For simplicity, we often still write digits instead:
True
June 10th, 2008
BOOLEAN VALUES

Number Systems and Boolean Algebra
1 is true
 0 is false
Volts
1.8


False
We will use this interpretation along with special
operations to design functions and hardware for doing
arbitrary computations.
0
21

Computers take inputs and produce outputs, just like
functions in math!

A function table is
unique but infinite
x
y
f(x,y)
0
…
2
…
23
…
0
…
2
…
41
…
0
…
6
…
87
…
Logical functions can be expressed in two ways:


Number Systems and Boolean Algebra
An expression is
finite but not unique
f(x,y) = 2x + y
=x+x+y
= 2(x + y/2)
= ...
June 10th, 2008
FUNCTIONS
A finite, but non-unique Boolean expression.
A truth table, which will turn out to be unique and finite.
22

June 10th, 2008
BASIC BOOLEAN OPERATIONS
There are three basic operations for logical values.
AND (product)
of two inputs
Expression:
xy, or xy
Truth table:
NOT (complement)
on one input
OR (sum) of
two inputs
x+y
x’
x
y
xy
x
y
x+y
x
x’
0
0
0
0
0
0
0
1
0
1
0
0
1
1
1
0
1
0
0
1
0
1
1
1
1
1
1
1
Number Systems and Boolean Algebra
Operation:
23

We can use these basic operations to form more complex expressions:
f(x,y,z) = (x + y’)z + x’
Some terminology and notation:
 f is the name of the function.
 (x,y,z) are the input variables, each representing 1 or 0. Listing the
inputs is optional, but sometimes helpful.
 A literal is any occurrence of an input variable or its complement. The
function above has four literals: x, y’, z, and x’.

Precedences are important, but not too difficult.
 NOT has the highest precedence, followed by AND, and then OR.
 Fully parenthesized, the function above would be kind of messy:
Number Systems and Boolean Algebra

June 10th, 2008
BOOLEAN EXPRESSIONS
f(x,y,z) = (((x +(y’))z) + x’)
24
A truth table shows all possible inputs and outputs of a function.

Remember that each input variable represents either 1 or 0.
 Because there are only a finite number of values (1 and 0), truth tables
themselves are finite.
n
 A function with n variables has 2 possible combinations of inputs.

Inputs are listed in binary order—in this example, from 000 to 111.
f(x,y,z) = (x + y’)z + x’
f(0,0,0)
f(0,0,1)
f(0,1,0)
f(0,1,1)
f(1,0,0)
f(1,0,1)
f(1,1,0)
f(1,1,1)
= (0 + 1)0 + 1
= (0 + 1)1 + 1
= (0 + 0)0 + 1
= (0 + 0)1 + 1
= (1 + 1)0 + 0
= (1 + 1)1 + 0
= (1 + 0)0 + 0
= (1 + 0)1 + 0
=1
=1
=1
=1
=0
=1
=0
=1
x
y
z
f(x,y,z)
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
1
0
1
Number Systems and Boolean Algebra

June 10th, 2008
TRUTH TABLES
25

June 10th, 2008
PRIMITIVE LOGIC GATES
Each of our basic operations can be implemented in
hardware using a primitive logic gate.
Operation:
AND (product)
of two inputs
Expression:
xy, or xy
Logic gate:
OR (sum) of
two inputs
x+y
NOT (complement)
on one input
x’
Number Systems and Boolean Algebra
Symbols for each of the logic gates are shown below.
 These gates output the product, sum or complement of their
inputs.

26


(x + y’)z + x’
Number Systems and Boolean Algebra

Any Boolean expression can be converted into a circuit by
combining basic gates in a relatively straightforward way.
The diagram below shows the inputs and outputs of each gate.
The precedences are explicit in a circuit. Clearly, we have to make
sure that the hardware does operations in the right order!
June 10th, 2008
EXPRESSIONS AND CIRCUITS
27
After finding the circuit inputs and outputs, you can come up with either an
expression or a truth table to describe what the circuit does.

You can easily convert between expressions and truth tables.
Find the circuit’s
inputs and outputs
Find a Boolean
expression
for the circuit
Number Systems and Boolean Algebra

June 10th, 2008
CIRCUIT ANALYSIS SUMMARY
Find a truth table
for the circuit
28
We can interpret high or low voltage as representing true or false.

A variable whose value can be either 1 or 0 is called a Boolean variable.

AND, OR, and NOT are the basic Boolean operations.

We can express Boolean functions with either an expression or a truth
table.

Every Boolean expression can be converted to a circuit.

Next, we’ll look at how Boolean algebra can help simplify expressions,
which in turn will lead to simpler circuits.
Number Systems and Boolean Algebra

June 10th, 2008
BOOLEAN OPERATIONS SUMMARY
29
Last time we talked about Boolean functions, Boolean expressions, and
truth tables.

Now we’ll learn how to how use Boolean algebra to simplify Booleans
expressions.

Last time, we saw this expression and converted it to a circuit:
(x + y’)z + x’
Can we make this circuit “better”?
• Cheaper: fewer gates
• Faster: fewer delays from inputs to
outputs
Number Systems and Boolean Algebra

June 10th, 2008
BOOLEAN ALGEBRA
30
Normal mathematical expressions can be simplified using the laws of
algebra

For binary systems, we can use Boolean algebra, which is superficially
similar to regular algebra

There are many differences, due to
 having only two values (0 and 1) to work with
 having a complement operation
 the OR operation is not the same as addition
Number Systems and Boolean Algebra

June 10th, 2008
EXPRESSION SIMPLIFICATION
31

1.
3.
5.
7.
9.
10.
12.
14.
16.
x+0=x
x+1=1
x+x=x
x + x’ = 1
(x’)’ = x
x+y=y+x
x + (y + z) = (x + y) + z
x(y + z) = xy + xz
(x + y)’ = x’y’
2.
4.
6.
8.
11.
13.
15.
17.
x1=x
x0=0
xx=x
x  x’ = 0
xy = yx
x(yz) = (xy)z
x + yz = (x + y)(x + z)
(xy)’ = x’ + y’
Commutative
Associative
Distributive
DeMorgan’s
Number Systems and Boolean Algebra
A Boolean algebra requires
 A set of elements B, which needs at least two elements (0 and 1)
 Two binary (two-argument) operations OR and AND
 A unary (one-argument) operation NOT
 The axioms below must always be true (textbook, p. 42)
 The magenta axioms deal with the complement operation
 Blue axioms (especially 15) are different from regular algebra
June 10th, 2008
FORMAL DEFINITION OF BOOLEAN ALGEBRA
32
The associative laws show that there is no ambiguity about a term such as
x + y + z or xyz, so we can introduce multiple-input primitive gates:

The left and right columns of axioms are duals
 exchange all ANDs with ORs, and 0s with 1s

The dual of any equation is always true
1.
3.
5.
7.
9.
10.
12.
14.
16.
x+0=x
x+1=1
x+x=x
x + x’ = 1
(x’)’ = x
x+y=y+x
x + (y + z) = (x + y) + z
x(y + z) = xy + xz
(x + y)’ = x’y’
2.
4.
6.
8.
11.
13.
15.
17.
Number Systems and Boolean Algebra

x1=x
x0=0
xx=x
x  x’ = 0
xy = yx
x(yz) = (xy)z
x + yz = (x + y)(x + z)
(xy)’ = x’ + y’
June 10th, 2008
COMMENTS ON THE AXIOMS
Commutative
Associative
Distributive
DeMorgan’s
33

We can show that these axioms are valid, given the definitions of AND, OR
and NOT

y
0
1
0
1
xy
0
0
0
1
x
0
0
1
1
y
0
1
0
1
x+y
0
1
1
1
x
0
1
x’
1
0
The first 11 axioms are easy to see from these truth tables alone. For
example, x + x’ = 1 because of the middle two lines below (where y = x’)
x
0
0
1
1
y
0
1
0
1
x+y
0
1
1
1
Number Systems and Boolean Algebra
x
0
0
1
1
June 10th, 2008
ARE THESE AXIOMS FOR REAL?
34

We can make up truth tables to prove (both parts of) DeMorgan’s law

For (x + y)’ = x’y’, we can make truth tables for (x + y)’ and for x’y’
y
0
1
0
1
x+y
0
1
1
1
(x + y)’
1
0
0
0
x
0
0
1
1
y
0
1
0
1
x’
1
1
0
0
y’
1
0
1
0
x’y’
1
0
0
0

In each table, the columns on the left (x and y) are the inputs. The columns
on the right are outputs.

In this case, we only care about the columns in blue. The other “outputs”
are just to help us find the blue columns.

Since both of the columns in blue are the same, this shows that (x + y)’ and
x’y’ are equivalent
Number Systems and Boolean Algebra
x
0
0
1
1
June 10th, 2008
PROVING THE REST OF THE AXIOMS
35

June 10th, 2008
SIMPLIFICATION WITH AXIOMS
We can now start doing some simplifications
1.
3.
5.
7.
9.
10.
12.
14.
16.
x+0=x
x+1=1
x+x=x
x + x’ = 1
(x’)’ = x
x+y=y+x
x + (y + z) = (x + y) + z
x(y + z) = xy + xz
(x + y)’ = x’y’
[ Distributive; x’y’ + x’y = x’(y’ + y) ]
[ Axiom 7; y’ + y = 1 ]
[ Axiom 2; x’1 = x’ ]
[ Distributive ]
[ Axiom 7; x’ + x = 1 ]
[ Axiom 2 ]
2.
4.
6.
8.
11.
13.
15.
17.
x1=x
x0=0
xx=x
x  x’ = 0
xy = yx
x(yz) = (xy)z
x + yz = (x + y)(x + z)
(xy)’ = x’ + y’
Commutative
Associative
Distributive
DeMorgan’s
Number Systems and Boolean Algebra
x’y’ + xyz + x’y
= x’(y’ + y) + xyz
= x’1 + xyz
= x’ + xyz
= (x’ + x)(x’ + yz)
= 1  (x’ + yz)
= x’ + yz
36
Here are two different
but equivalent circuits.

In general the one
with fewer gates is
“better”:
 It costs less to
build
 It requires less
power
 But we had to do
some work to find
the second form
Number Systems and Boolean Algebra

June 10th, 2008
LET’S COMPARE THE RESULTING CIRCUITS
37

Here are some more useful laws. Notice the duals again!
4. x(x + y) = x
5. (x + y)(x + y’) = x
6. x(x’ + y) = xy
(x + y)(x’ + z)(y + z) = (x + y)(x’ + z)
We can prove these laws by either

Making truth tables:
x
0
0
1
1

Using the axioms:
y
0
1
0
1
x’
x’y
x + x’y
x + x’y = (x + x’)(x + y)
= 1  (x + y)
=x+y
x
0
0
1
1
y
0
1
0
1
x+y
0
1
1
1
[ Distributive ]
[ x + x’ = 1 ]
[ Axiom 3 ]
Number Systems and Boolean Algebra
1. x + xy = x
2. xy + xy’ = x
3. x + x’y = x + y
xy + x’z + yz = xy + x’z

June 10th, 2008
SOME MORE LAWS
38
The complement of a function always outputs 0 where the original function
outputted 1, and 1 where the original produced 0.

In a truth table, we can just exchange 0s and 1s in the output column(s)
f(x,y,z)
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
= x(y’z’ + yz)
z
0
1
0
1
0
1
0
1
f(x,y,z)
1
1
1
1
0
0
1
0
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
f’(x,y,z)
0
0
0
0
1
1
0
1
Number Systems and Boolean Algebra

June 10th, 2008
THE COMPLEMENT OF A FUNCTION
39

f(x,y,z)
= x(y’z’ + yz)
f’(x,y,z)
= ( x(y’z’ + yz) )’
= x’ + (y’z’ + yz)’
= x’ + (y’z’)’ (yz)’
= x’ + (y + z)(y’ + z’)
[ complement both sides ]
[ because (xy)’ = x’ + y’ ]
[ because (x + y)’ = x’ y’ ]
[ because (xy)’ = x’ + y’, twice]
You can also take the dual of the function, and then complement each
literal
 If f(x,y,z) = x(y’z’ + yz)…
 … the dual of f is x + (y’ + z’)(y + z)…
 … then complementing each literal gives x’ + (y + z)(y’ + z’)…
 … so f’(x,y,z) = x’ + (y + z)(y’ + z’)
Number Systems and Boolean Algebra

You can use DeMorgan’s law to keep “pushing” the complements inwards
June 10th, 2008
COMPLEMENTING A FUNCTION ALGEBRAICALLY
40
So far:
 A bunch of Boolean algebra trickery for simplifying expressions and
circuits
 The algebra guarantees us that the simplified circuit is equivalent to the
original one

Next:
 Introducing some standard forms and terminology
 An alternative simplification method
 We’ll start using all this stuff to build and analyze bigger, more useful,
circuits
Number Systems and Boolean Algebra

June 10th, 2008
SUMMARY SO FAR
41