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Tiêu đề The Digital Abstraction
Tác giả Agarwal, Lang
Trường học Massachusetts Institute of Technology
Chuyên ngành Circuits and Electronics
Thể loại Lecture
Năm xuất bản 2000
Thành phố Cambridge
Định dạng
Số trang 20
Dung lượng 156,29 KB

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6.002 CIRCUITS ANDELECTRONICS The Digital Abstraction... z Discretize matter by agreeing to observe the lumped matter discipline zAnalysis tool kit: KVL/KCL, node method, superposition,

Trang 1

6.002 CIRCUITS AND

ELECTRONICS

The Digital Abstraction

Trang 2

z Discretize matter by agreeing to

observe the lumped matter discipline

zAnalysis tool kit: KVL/KCL, node method, superposition, Thévenin, Norton

(remember superposition, Thévenin,

Norton apply only for linear circuits)

Lumped Circuit Abstraction

Trang 3

Discretize value Digital abstraction

Interestingly, we will see shortly that the tools learned in the previous three

lectures are sufficient to analyze simple digital circuits

Reading: Chapter 5 of Agarwal & Lang

Today

Trang 4

Analog signal processing

But first, why digital?

In the past …

By superposition,

The above is an “adder” circuit

2 2 1

1 1

2 1

2

R R

R V

R R

R V

+

+ +

=

If R1 = R2,

2

2 1

0

V V

V = +

1

V

1

R

2

R

+ –

2

V +–

0

V

and might represent the outputs of two

sensors, for example.

1

Trang 5

Noise Problem

noise hampers our ability to distinguish between small differences in value —

e.g between 3.1V and 3.2V

Receiver: huh?

add noise on

this wire

t

Trang 6

Value Discretization

Why is this discretization useful?

Restrict values to be one of two

HIGH 5V TRUE 1

LOW 0V FALSE 0

like two digits 0 and 1

(Remember, numbers larger than 1 can be

represented using multiple binary digits and

coding, much like using multiple decimal digits to

represent numbers greater than 9 E.g., the

binary number 101 has decimal value 5.)

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Digital System

V

R

V

noise

S

V

“0” “1” “0”

0V

2.5V

LOW

t

R

V

“0” “1” “0”

0V 2.5V

5V

t

V

V N = 0

N

V

S

V

“0” “1” “0”

With noise

V

V N = 0.2

S

V

“0” “1” “0”

0V 2.5V

5V

t

0.2V

t

Trang 8

Digital System

Better noise immunity

Lots of “noise margin”

For “1”: noise margin 5V to 2.5V = 2.5V

For “0”: noise margin 0V to 2.5V = 2.5V

Trang 9

Voltage Thresholds

and Logic Values

1 0

1

0

1

0

0V 2.5V 5V

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forbidden region

VH

V L

3V 2V

But, but, but …

What about 2.5V?

Hmmm… create “no man’s land”

or forbidden region

For example,

0V

5V

“1” V 5V

“0” 0V V

H

L

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sender receiver

But, but, but …

Where’s the noise margin?

What if the sender sent 1: ? VH

Hold the sender to tougher standards!

5V

0V

1

1

V0H

V0L

VIH VIL

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sender receiver

But, but, but …

Where’s the noise margin?

What if the sender sent 1: ?

Hold the sender to tougher standards!

5V

0V

“1” noise margin:

“0” noise margin: VIH - V0H

VIL - V0L

1

1

V0H

V0L

VIH VIL

Noise margins

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Digital systems follow static discipline : if

inputs to the digital system meet valid input

thresholds, then the system guarantees its

sender

receiver

0 1 0 1

t

5V

V0H

V0L

0V

VIH

VIL

0 1 0 1

t

5V

V0H

V0L

0V

VIH

VIL

Trang 14

Processing digital signals

Recall, we have only two values —

Map naturally to logic: T, F Can also represent numbers

1,0

Trang 15

Processing digital signals

Boolean Logic

If X is true and Y is true

Then Z is true else Z is false.

Z = X AND Y

X, Y, Z are digital signals

“0” , “1”

Z = X Y

Boolean equation

Enumerate all input combinations

Truth table representation:

Z

X Y

AND gate

Z

X Y

0 0 0

0 1 0

1 0 0

1 1 1

Trang 16

„ Adheres to static discipline

inputs alone.

Combinational gate

abstraction

Digital logic designers do not

have to care about what is

inside a gate.

Trang 17

Noise

Z

X Y

Z

Y

X

Trang 18

Z = X Y

Examples for recitation

X

t

Y

t

Z

t

Trang 19

In recitation…

Another example of a gate

If (A is true) OR (B is true)

then C is true

else C is false

C = A + B Boolean equation

OR

OR gate

C

A B

Z

X Y

NAND

Z = X Y

More gates

Inverter

Trang 20

Boolean Identities

AB + AC = A (B + C)

X 1 = X

X 0 = X

X + 1 = 1

X + 0 = X

1 = 0

0 = 1

output

B

A

Digital Circuits

Implement: output = A + B C

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