Chapter 4Hardware Basics

We assume that a computing agent will exe-

cute all the algorithms we have been discussing.

Now, we will see how this agent looks like and

how it is able to execute those algorithms.

We will first look at the very basics, such as

how to represent data, how to follow the logic

principles to design gates, and how to combine

gates to construct logic circuits that manipu-

late data. This part is called hardware design

or logic design.

1

Binary system

We are used to follow the decimal represen-

tation of numbers, e.g., 261. When we use

computers to process data, we still use this

system as the external representation, but it

will not be used as the internal representation.

Instead, the computer will use the binary num-

bering system.

2

Binary or decimal?

Question: What does 261 mean?

Answer: It means two hundreds and sixty one.

It has the following decimal representation:

261 = 2 × 100 + 6 × 10 + 1

= 2 × 102 + 6 × 101 + 1 × 100

= (261)10.

Similarly, it also could be represented in a bi-

nary system.

261 = 256 + 4 + 1 = 1 × 28 + 0 × 27

+ 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23

+ 1 × 22 + 0 × 21 + 1 × 20

= (100000101)2.

The two digits 0, 1, are called bits.

3

A conversion table

Binary Decimal Binary Decimal0 0 10000 161 1 10001 17

10 2 10010 1811 3 10011 19

100 4 10100 20101 5 10101 21110 6 10110 22111 7 10111 23

1000 8 11000 241001 9 11001 251010 10 11010 261011 11 11011 271100 12 11100 281101 13 11101 291110 14 11110 301111 15 11111 31

Generally speaking, any whole number can berepresented in this “base-2” representation.

Homework: Exercises 1–6.4

Representing fractions

Decimal numbers, such as 12.34 and -43.21,

can also be represented in the binary system,

by first converting them into the scientific sys-

tem: ±M × B±E, where M is the mantissa, B

is the base, and E is the exponent.

For example,

(5.75)10 = 4 + 1 + 0.5 + 0.25

= 1 × 22 + 0 × 21 + 1 × 20

+1 × 2−1 + 1 × 2−2

= (101.11)2.

Homework: Exercises 7.

5

A couple of notes

1. As computer is a finite device, there is a

maximal number of binary digits that can be

used to store digits, e.g., 16, 32, or 64. Once

this number is fixed, the maximal integer that

can be represented is also fixed. For example,

for a 16 bit machine, the maximum integer is

216 = 65,535. Any operation in this machine

that produces a larger integer will lead to an

overflow error.

2. Binary system can be used to represent

other things as well. For example, to represent

a signed number, we can use the leftmost bit

to represent the sign, 0 for ‘+’, and 1 for ‘-’.

Hence, -49 and 3 will be represented, respec-

tively, as (1110001) and (0000011), assuming

we use 7 digits.

Homework: Exercise 8.6

Moreover, we can also use it represent textinformation. There is no the way to do so, wejust assign a binary number, usually with 8 bits,to every text symbol we want to represent.ASCII is the most widely used representationschema. For example,

Symbol Decimal Value Binary Value! 33 00100001... ... .../ 47 001011110 48 00110000... ... ...9 57 00111001... ... ...A 65 01000001... ... ...Z 90 01011010... ... ... 126 01111110

Hence, BAD=00000010︸ ︷︷ ︸B

00000001︸ ︷︷ ︸A

00000100︸ ︷︷ ︸D

.

Homework: Exercise 9.7

UNICODE representation

Since ASCII code has only 8 bits, it can repre-

sent up to 256 different symbols, not enough

for today’s world. Thus, the new scheme of

UNICODE gains popularity since it uses 16

bits, and can represent up to 65, 536 different

symbols. It can be used to represent symbols

drawn from other alphabets, such as Russian,

Chinese, etc..

This is also the representation used in the JAVA

language.

For a full discussion of UNICODE, please have

a look at www.unicode.org.

8

Sound and images

The key to represent other forms of informa-

tion, such as sound and image, is based on thedigitization of those information.

For example, given a B/W picture, we can putit in a grid to convert it into a bunch of pixels,namely, elements of picture. We then assign

either a 1 or 0 to each pixel, depending on ifthat picture is black or white. For a color pic-

ture, besides identifying the position, we alsoneed to assign its share in three of the basic

colors: red, blue, and green, each of which istypically represented using a byte (8 bits).

Hence, a picture taken with a 3 MegaPixel digi-

tal camera using 24 bit color occupies 72 Megabits.

Homework: Complete §4.2.2 and completeExercises 10 and 11.

9

Data compression

Since the information storage needs so muchspace, the subject of data compression be-comes really important. A simple compressiontechnique is run-length encoding, which re-places a sequence of identical values v1, v2, · · · , vn

with a pair (v, n). If each v and n requires 1byte, then the total number of bytes will bereduced from n to 2, if n ≥ 2. For example,the following data

255 255 255255 0 0255 255 255255 0 0255 255 255

can be represented as the following list:

(255, 4) (0, 2) (255, 4) (0, 2) (255, 3), thusreducing the total bytes from 15 to 10, a ratioof 15/10=1.5.

10

Variable length code

In ASCII code, every character is coded in 8

bits. So, if we have a text file with 1,000 char-

acters, we have to use 8,000 bits to store it.

In reality, some characters are used more often

than the others (Think about Wheel of For-

tune). It makes sense to assign shorter codes

to those used more often, and longer codes to

those used less often. The question is how?

One approach is to assign code to symbols

based on their frequencies. For example, in

the string aaxuaxz, the frequency of a, x, u

and z are 3, 2, 1 and 1. When frequency

varies a great deal, it makes sense to assign

shortest code to the most frequently occurring

one, while assigns longest code to the least fre-

quently occurring symbol.

11

For the above example, we can assign the codesas follows: 0=a, 10=x, 110=u, 111=z). Hence,aaxuaxz will be coded as 0010110010111. Thelength is 13 bits, compared with 14, if we giveeach of them two bits. On the other hand, ifthe frequency of the four symbols are (996, 2,1, 1), then the 2 bits per code method, for a1,000 character file will lead to 2,000 bits long,while our code will lead to a file of only 1,006bits, a ratio of almost 2.

How could we decode? For 0010110010111,since we have no code ”00”, the first piece ofcode must be ”0”, which is an ’a’, the nextis also ’a’. As we do have a code ”10”, weread off an ’x’, etc.. So, we always read offthe longest possible piece from the undecodedstring. The reason that this method worksis because this is a prefix code, i.e., no codeis a prefix of another. Such a code is calledHuffman code.

Homework: Exercises 12 and 13.12

Why binary?

There is absolutely no theoretical reason why

we can’t build up a base-10 computer, or a

computer based on any notation system. The

use of binary system is wholly a reliability con-

cern.

The computer is an electronic equipment so

that the internal representation must be based

on such quantities as currents or voltages. As

a result, we must, say the least, use 10 dif-

ferent voltages to represent the 10 digits. For

example, 0v for 0, 5v for 1, 10v for 2, · · · , 45v

for 9. So, to represent 28, we uses two de-

vices, one is charged to 10v, and the other to

40v.

The problem is that voltages are not stable.

When the voltage of the second drops to 37.5,

what number does it represent?

13

The trouble is that in representing a base-10

system, we need 10 stable states. Such a sys-

tem is difficult to come by. On the other hand,

an electrical system works best in a bi-stable

environment, where there are only 2 states,

separated by “huge” energy gap. For example,

on-off, fully charged-fully discharged, charged

positively-charged negatively, etc.

In such a system, there are two stable states.

We can let one, e.g., 0v, represent the digit

0, the other, e.g., 45v, represent the digit 1.

Now, a mere 10% voltage change would not

cause any problem.

The use of the binary system for the internal

data representation significantly increases the

inherent reliability of modern computer.

14

Binary storage devices

combined with our earlier results, we know that

any information can be represented by using bi-

stable devices that 1) has two stable states; 2)

it is possible to know what state the device is

in; and 3) it is possible to switch the device

from one state to the other.

There are many such devices, such as light

switch. Thus, at least theoretically, we can

build a computer out of a bunch of light switches.

Modern computers are built on such other much

smaller, more efficient, and stable devices. We

discuss two of them: magnetic core and tran-

sistor.

15

Magnetic core

This type of device was used to construct com-

puter memories, roughly from 1955 to 1975.

It is a small magnetizable iron oxide-coated

“doughnut”, about 1/50 of an inch in its di-

ameter, with one or more wires through its

center hole. If electric current is sent through

the wire in specific way, the core will be mag-

netized in one of two ways, to represent 0 or

1.

Since magnetic fields do not change much over

time, this device is very stable.

16

Transistors

Today, transistor is the building block for com-puter memories. It is simply a tiny switch. But,it has no mechanical or moving parts. Theswitching is done electronically, rather thanmechanically. This allows it to be extremelyfast as well as small. It can be switched inabout 10-20 billions of a second, while severalmillions of transistors can fit in a space onecentimeter square.

Constructed from semi-conductor material, tran-sistors, together with some connecting wires,can be printed photographically on a layer ofsilicon to produce integrated circuits (chips),which will be used as functional parts inside acomputer.

Photographic techniques, such as VLSI andULSI, allow us to produce chip with billionsof transistor/cm2, which leads to smaller andfaster computers.

17

Boolean logic and gates

The construction of chips is based on Boolean

logic, which makes operations in base-2 possi-

ble. This logic has two variables, T(rue) and

F(alse.) Interpreting 0 as F and 1 as T, the re-

lationship between this logic and binary system

is immediately established.

There are three operations, AND(∧), OR(∨)

and NOT(¬), in Boolean logic. A Boolean

expression is essentially a combination of those

operation applied on boolean variables. For

example, (a ∧ b) ∨ ¬c is a boolean expression.

18

Boolean operations

Let a and b be two Boolean expression. Then

a∧b is true iff both a and b are true, a∨b is true

iff at least one of a and b is true, ¬a is true iff a

is false. These rules can be demonstrated with

the following truth tables, when true and false

are represented with 1, and 0, respectively.

Homework: Exercises 15 and 16.

19

Gates

A gate is an electronic device that operates ona collection of binary inputs to produce a binaryoutput. Although there are a wide range ofgates, we only consider those that implementthe three Boolean operations, as they are theonly things we need.

Below show the implementation of OR gate,one of the three basic ones, by using transis-tors.

20

Building circuits

A (combinational) circuit is a collection of (AND,

OR and NOT) gates that 1) transforms a set

of binary inputs into a set of binary outputs,

and 2) where the values of the outputs depend

only on the current values of inputs. For exam-

ple, the following diagram shows such a circuit,

where it implements the following Boolean ex-

pression:

c = a ∨ b

d = ¬((a ∨ b) ∧ (¬b))

21

What to output?

For any circuit, once the input values are de-

termined, we can use truth tables to decide

the output(s). Thus, given any circuit, we can

construct a truth table to show its behavior.

For example, below is the truth table for the

previous circuit.

a b c d0 0 0 10 1 1 11 0 1 01 1 1 1

22

Circuit construction

We study the opposite problem: how could we

construct a circuit to solve a problem?

There are a number of such circuit construc-

tion algorithms. They all consist of a few ma-

jor steps:

1. Construct a truth table for the expression.

2. Construct sub-expressions using AND and

NOT gates.

3. Combine sub-expressions using OR gates.

4. Draw diagram(s).

23

An example

1. We begin with the following truth table.

a b c O1 O20 0 0 0 10 0 1 0 00 1 0 1 10 1 1 0 11 0 0 0 01 0 1 0 01 1 0 1 11 1 1 0 0

24

Output construction

2. We fix up one output column, say O1. For

every 1 occurs in that column, we construct

a subexpression with the corresponding inputs

by using ∧, in which, if the value for an input

is 1, we use the input itself; otherwise, we use

its negation. For example, as there are only

two 1’s in the column for O1, there are two

such subexpressions, (¬a ∧ b ∧ ¬c) and (a ∧ b ∧¬c). Similarly, we have four expression for O2 :

(¬a ∧ ¬b ∧ ¬c), (¬a ∧ b ∧ ¬c), (¬a ∧ b ∧ c) and

(a ∧ b ∧ ¬c).

3. As either of these subexpressions leads to

a 1, we finish the construction by combining

these subexpression, for each output, with the

∨ operator. For example,

O1 = (¬a ∧ b ∧ ¬c) ∨ (a ∧ b ∧ ¬c).

25

Draw circuits

4. Based on the Boolean expression, we can

draw diagrams for corresponding circuits. For

example, we have the following implementa-

tion for O1.

26

Is this the simplest?

Generally speaking, there could be lots of equiv-

alent circuits for the same truth table. We

always want to have an optimal circuit, in the

sense that it uses the smallest number of gates.

For example, we actually have the following

simpler expression for O1, which leads to a

much simpler circuit.

O1 = b ∧ ¬c.

Homework: Exercises 17 and 18.

27

Another example

We want to construct a circuit to test two un-

signed binary numbers for exact equality. It will

produce a 1, if those two numbers are equal,

a 0 otherwise.

We begin with a simpler circuit, 1-CA, which

tests equality for two 1-bit numbers. We find

out that

O = (a ∧ b) ∨ (¬a ∧ ¬b),

based on the following truth table.

a b O0 0 10 1 01 0 01 1 1

28

We then combine n 1-CE’s to construct a CA

circuit to test equality for two n−bit numbers.

Homework: Exercises 19 and 20.

29

An addition circuit

We now construct a circuit to add two n− bit

binary numbers, a full adder. For example,

0 0 1 1 0 1+ 00 01 11 10 10 0

0 1 1 0 1 1

Hence, the circuit should have three inputs,

two for digits, one for a carry; and two outputs,

one for sum, the other for a carry for the next

bit.

30

Generally, we have the following truth table.

ai bi ci−1 si ci0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

Following the construction algorithm, we have

the following:

si = ((¬ai)(¬bi)ci) ∨ ((¬ai)bi(¬ci))

∨ (ai(¬bi)(¬ci)) ∨ (aibici),

ci = ((¬ai)bici) ∨ (ai(¬bi)ci)

∨ (aibi(¬ci)) ∨ (aibici).

31

Based on the expressions for si and bi, we can

draw the following diagram.

32

The n−bit adder

We combine n one-bit full adder to form ann−bit adder, as follows:

As each 1-bit adder uses 3 NOT gates, 16AND gates and 6 OR gates, a 32-bit adderuses 800 gates, or 1,504 transistors, in total.

Homework: Exercise 21.33

Control circuits

We saw how could we implement circuits for

arithmetic and logic operations, now, we have

a brief look at some of the other essential cir-

cuits: multiplexer and decoder. A multiplexer

has 2N input lines, N select lines, and 1 output

line. Each of the select line can be feed a value

of 0 or 1, the collection of which decides which

input line will be selected to send its value out.

34

Implementing a multiplexer

It is quite easy to construct such a control

device. Below is the diagram for a multiplexer

with two input lines.

When the select line sends in a value of 0, the

upper input is selected; otherwise, the lower

line is selected. The Boolean expression is

O = (I0 ∧ ¬S) ∨ (I1 ∧ S).

35

Decoder

A decoder has N input lines, numbered 0, 1,

. . . , n-1 and 2N output line, 0, 1, . . . ,2N − 1.

Again, each of the input can be either 0, or 1.

With the collection of all the inputs, it decides

which of the 2N outputs should be selected.

In application, it can be used to select which

operation unit should be used.

36

Implementing a decoder

It is also easy to design a decoder. Below is a

diagram for a decoder with 2 input lines.

Homework: Exercises 23 and 24, as well as

Challenge work 1 and 2.

37