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Administrivia

• Assignments• Labs• Questions??• Class questions – [email protected]

– Goes to dpd and the TA’s

• Hand in lab assignments – [email protected]– Goes to a file.

mailto:[email protected]




Adding Decimal Numbers

4 6 5 5 4 3=========

We all know that the answer is 1008 but how do we do this?


4 6 5 5 4 3=========

Start from the right5+3 = 8 and there isno carry


4 6 5 5 4 3========= 8


Next, 6+4 = 10Write the 0, carry the 1

Adding Decimal Numbers 1

4 6 5 5 4 3========= 0 8



Adding Decimal Numbers 1

4 6 5 5 4 3========= 0 8



Finally, 1+4+5 = 10Write the 0, carry the 1

Adding Decimal Numbers 1 1

4 6 5 5 4 3========= 1 0 0 8



Finally, 1+4+5 = 10Write the 0, carry the 1

Adding Decimal Numbers 1 1

4 6 5 5 4 3========= 1 0 0 8

Start from the right5+3 = 8 and there is no carryNext, 6+4 = 10Write the 0, carry the 1Finally, 1+4+5 = 10Write the 0, carry the 1

At each stage, take 2 addends (5 and 3 or 6 and 4 or 4 and 5) and a bit telling whether there is acarry and produce a sum bit and a bit telling if there is a carry.

Bad news

• It is inconvenient to add decimal numbers in a computer – Truth tables are easier than addition tables– The universal method wants to work from truth

tables

Bad news/Good News

• Bad news – It is inconvenient to add decimal numbers in a computer – Truth tables are easier than addition tables– The universal method wants to work from truth

tables

• Good news – the same thing works for binary numbers

Adding Binary Numbers

• At each stage, take 2 addends and a bit telling whether there is a carry and produce a sum bit and a bit telling if there is a carry.

Adding Binary Numbers (cont.)1 1

1 0 1

+ 1 1 1

_______________

1 1 0 0

0 + 0 =0

0 + 1 = 1

1 + 0 = 1

1 + 1 =10

Sidebar – what are binary numbers

• Base 2 rather than base 10

• Decimal numbers– We write (1234)10 to represent a decimal

number that has 4 units, 3 tens, 2 hundreds and 1 thousand.

– (1234)10 = 1 x103 + 2 x102 + 3 x101 + 4 x100

Sidebar – what are binary numbers

• Base 2 rather than base 10• Decimal numbers

– We write (1234)10 to represent a decimal number that has 4 units, 3 tens, 2 hundreds and 1 thousand.

– (1234)10 = 1 x103 + 2 x102 + 3 x101 + 4 x100

• Binary numbers– We write (1001)2 to represent a decimal number that

has 1 unit, 0 2’s, 0 4’s and 1 8.

– (1001)2 = 1 x23 + 0 x22 + 0 x21 + 1 x20


• At each stage, take 2 addends and a bit telling whether there is a carry and produce a sum bit and a bit telling if there is a carry.


At each stage, take 2 addends and a bit telling whether there is a carry and produce a sum bit and a bit telling if there is a carry.

We can build a black box to do this

addend

addend

carry bit in

Sum bit

carry bit out


• Write the numbers as– X4 X3 X2 X1 X0– Y4 Y3 Y2 Y1 Y0

• Represent the sum as– C Z4 Z3 Z2 Z1 Z0


• Write the numbers as

– X4 X3 X2 X1 X0

– Y4 Y3 Y2 Y1 Y0

• Represent the sum as

– C Z4 Z3 Z2 Z1 Z0

• Start with X0, Y0, 0 in, Z0 and C0 out

• Then X1,Y1,C0 in and Z1 and C1 out.

• And so on


At the ith stage

ith bit of X

ith bit of Y

carry bit

ith bit of Z

carry bit out

Abstraction in action -- This is a piece of a carry-ripple adder

Universal

Circuit

Carry-Ripple Adder

Z0

C0

UniversalCircuit

X0

Y0

Z1

C1

UniversalCircuit

X1

Y1

Z2

C2

UniversalCircuit

X2

Y2

X2 X1 X0 Y2 Y1 Y0============C2 Z2 Z1 Z0

0

Fixed at 0

What’s inside the box?

Zi

Cout

UniversalCircuit

Xi

Yi

Cin

What’s inside the box?

Zi

Cout

UniversalCircuit

Xi

Yi

Cin

Do the problem set and find out

Carry ripple adders

• Useful for some applications

• Too slow for other– Delay while waiting for the carry to ripple

• One solution – add larger blocks– Details on the homework assignment

Arithmetic Logical Unit (ALU)

• This is the part of the CPU that does arithmetic and comparison operations

• We’ve built a piece of the ALU• With abstraction, we could build the other parts

– Multiplication/subtraction/division

– Comparison

• Then we would write a language to let the user talk to the ALU (programming)

Pause

• Questions??

• How to build the other parts of the ALU?– Use abstraction

• Back to representing information

Representing information

• How do we represent characters?– How many characters might we want to

represent?– What characters might we want to represent?

Representing information

• How do we represent characters?– How many characters might we want to represent?

– What characters might we want to represent?• A-Z 26

• A-Z and a-z 52

• All the keys on my keyboard 104

• Maybe a power of 2? 128

• Maybe an even power of 2? 256

• Maybe an even bigger power of 2? 65536

Representing characters

• ASCII is the American Standard Code for Information Interchange. It is a 7-bit code.

• Many 8-bit codes contain ASCII as their lower half

• The ASCII standard was published by the United States of America Standards Institute (USASI) in 1968.

Unicode

• Universal Character Set (UCS) contains all characters of all other character set standards. It also guarantees round-trip compatibility, i.e., conversion tables can be built such that no information is lost when a string is converted from any other encoding to UCS and back.

• UCS contains the characters required to represent almost all known languages. This includes apart from the many languages which use extensions of the Latin script also the following scripts and

languages: Greek, Cyrillic, Hebrew, Arabic, Armenian, Gregorian, Japanese, Chinese, Hiragana, Katakana, Korean, Hangul, Devangari, Bengali, Gurmukhi, Gujarati, Oriya, Tamil, Telugu, Kannada, Malayam, Thai, Lao, Bopomofo, and a number of others. Work is going on to include further scripts like Tibetian, Khmer, Runic, Ethiopian, Hieroglyphics, various Indo-European languages, and many others.

• It’s intended to use 31 bits (32768 possible characters)

What do we do in practice

• Problems– Bits represent too little – too many are needed– Decimal numbers don’t translate well into bits

• So,– Group into blocks of 4 and 8 bits

• 8 bits = 256 characters, holds ASCII

• 8 bits make 1 byte – things are organized into bytes

• 4 bits make 1 nibble

Shorthand: Hexadecimal

0

1

2

3

4

5

6

7

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

8

9

A

B

C

D

E

F

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

Sidebar – what are hexadecimal numbers

• Base 16 rather than base 10• Decimal numbers

– We write (1234)10 to represent a decimal number that has 4 units, 3 tens, 2 hundreds and 1 thousand.

– (1234)10 = 1 x103 + 2 x102 + 3 x101 + 4 x100

• Hexadecimal numbers– We write (2AC4)16 to represent a decimal number that

has 1 units, 12 16’s, 10 256’s and 2 65,536’s

– (2AC4)16 = 1 x163 + 0 x162 + 0 x161 + 1 x160

Hexadecimal

• We can add numbers– 1+1=2, 2+2=4, 4+4=8, 4+8=C, 2+8=A, …

• We can combine 2 hexadecimal numbers to make a byte.

• It’s easier to read than 0’s and 1’s• In ASCII

– hex 41 through 5A represent A to Z– Hex 61 through 7A represent a to z

Memory

Memory

• Logic circuits we’ve seen so far have no memory. They just transform input to output.

• Can we make logic circuits that remember?

The Stubborn Guy• Matt really likes Sue, but he doesn’t like

changing his mind… so:

• Matt decides to go to the party if Sue decides to go OR if he (Matt) already feels like going.

The Stubborn Guy: Feedback• Matt decides to go to the party if Sue

decides to go OR if he (Matt) already feels like going.

ORSue

Matt

FeedbackWire



• If Sue decides to go...

ORSue

Matt1



• If Sue decides to go, Matt will go.

ORSue

Matt1

1

1



• If Sue changes her mind…

ORSue

Matt0

1

1



• If Sue changes her mind, Matt will still go!

• Once he decides to go, we can’t ever get Matt to change his mind.

ORSue

Matt0

1

1

Enter Rita• Matt doesn’t like Rita

• Matt decides to go to the party if Sue decides to go OR if he (Matt) already feels like going, UNLESS Rita decides to go.


• Matt decides to go to the party if Sue decides to go OR:

• If he (Matt) already feels like going AND Rita decides NOT to go.




OR

AND

Sue

Rita

Matt

Feedback Wire




OR

AND

Sue

Rita

Matt0

0

0

0

01




OR

AND

Sue

Rita

Matt1

0

0

0

01




OR

AND

Sue

Rita

Matt1

0

1

1

11




OR

AND

Sue

Rita

Matt0

0

1

1

11




OR

AND

Sue

Rita

Matt0 – nothing changes

0

1

1

11




OR

AND

Sue

Rita

Matt0

1

1

1

11




OR

AND

Sue

Rita

Matt0

1

1

1

10




OR

AND

Sue

Rita

Matt0

1

1

1

00




OR

AND

Sue

Rita

Matt0

1

0

1

00




OR

AND

Sue

Rita

Matt0

1

0

0

00




OR

AND

Sue

Rita

Matt0

0

0

0

00




OR

AND

Sue

Rita

Matt0

0

0

0

01

The Flip-Flop

OR

AND

Set

Reset

M

• M becomes 1 if Set is turned on

• M becomes 0 if Reset is turned on

• Otherwise (if Set and Reset are both 0), M just remembers its value

The Flip-Flop

• M becomes 1 if Set is turned on

• M becomes 0 if Reset is turned on

• Otherwise (if Set and Reset are both 0), M just remembers its value

S

R

M

AND

AND

The Data Flip-Flop

• Nothing happens unless Write = 1

S

R

MD Write

AND

AND

The Data Flip-Flop

• Nothing happens unless Write = 1

• If Write = 1, then M becomes set to D

• Once Write = 0 again, M just keeps its value.(It ignores D.)

S

R

MD Write

AND

AND

Using a Data Flip-Flop

• Initially, Write = 0. Let’s say M = 1.

S

R

MD Write 0

0

0

1

AND

AND



• First, set D to desired value, say 0.

S

R

MD Write 0

0

0

10

1

AND

AND




• Then, set Write to 1.

S

R

MD Write 1

0

1

10

1

AND

AND





• This causes M to be reset to 0.

S

R

MD Write 1

0

1

00

1

AND

AND





• This causes M to be reset to 0.

• Finally set Write back to 0. Now D irrelevant.

S

R

MD Write 0

0

0

0?

1

The Data Flip-Flop

• If Write = 0, M just keeps its value.(It ignores D.)

• If Write = 1, then M becomes set to D

D

Write

M

A subtle problem

• When Write = 1, then M = D.

• If we have some feedback between M and D,then circuit could go haywire.

D

Write

M ??

A subtle problem

• For example, suppose NOT gate connectsM and D.

• When Write = 1, M and D keep changing. We have no control.

D

Write

M NOT

1

?

?

A subtle problem

• We want to control the feedback, so thateach time we set Write to 1 and then back to 0, M stores only the last value of D

• (In this case, M should invert itself once each time we set Write to 1 and back to 0)

D

Write

M NOT

1

?

?

“Airlock” Flip-Flop

D1

W1

M1D0

W0

M0

“Outer door” “Inner door”

Write

• Two-Stage System to prevent feedback loop.

D M


D1

W1

M1D0

W0

M0


Write

• We start with Write = 0.

• Let’s say D is always NOT M.Start with D = 0, M = 1.

D M


D1

W1

M1D0

W0

M0


Write

• We start with Write = 0.

• Let’s say D is always NOT M.Start with D = 0, M = 1.

D M

0

0

1

1 1


D1

W1

M1D0

W0

M0


Write

• Want to store D in memory.

• Set Write to 1

D M

1

0

0

1 1


D1

W1

M1D0

W0

M0


Write

• Want to store D in memory.

• Set Write to 1

• “Outer” flip-flop sets M0 = D0 = 0

• “Inner” flip-flop ignores D1 since W1 = 0

D M

1

0

0

0 1


D1

W1

M1D0

W0

M0


Write

• Now, set Write back to 0

D M

0

0

1

0 1


D1

W1

M1D0

W0

M0


Write

• Now, set Write back to 0

• Now “Inner” flip-flop sets M = D1 = 0

D M

0

0

1

0 0


D1

W1

M1D0

W0

M0


Write

• Because of feedback, D might change to (NOT M) , which is 1

• But Write = 0, so “Outer” flip-flop ignores D,and so M0 stays 0.

D M

0

1

1

0 0


D1

W1

M1D0

W0

M0


Write

• So memory does not change until we “toggle” Write.

• (“toggle” means change from 0 to 1 or vice versa)

D M

0

1

1

0 0


D1

W1

M1D0

W0

M0


Write

• This is Real Memory!

D M

Memory “Register”: 4 bitsD

MW

DM

W

DM

W

DM

WWrite

Data1

Data2

Data3

Data4

Review

• We have used the Universal method to build– ALU– Memory

• Next steps– State machines to computer with memory– Building the computer– Writing a program to use the computer

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