SLIDES FOR CHAPTER 1 INTRODUCTION NUMBER SYSTEMS AND CONVERSION

©2010 Cengage Learning

SLIDES FOR

CHAPTER 1

INTRODUCTIONNUMBER SYSTEMS AND CONVERSION

Click the mouse to move to the next page.Use the ESC key to exit this chapter.

This chapter in the book includes:ObjectivesStudy Guide

1.1 Digital Systems and Switching Circuits1.2 Number Systems and Conversion1.3 Binary Arithmetic1.4 Representation of Negative Numbers1.5 Binary Codes

Problems


Figure 1-1: Switching circuit


Decimal Notation

953.7810 = 9x102 + 5x101 + 3x100 + 7x10-1 + 8x10-2

Binary

1011.112 = 1x23 + 0x22 + 1x21 + 1x20 + 1x2-1 + 1x2-2

= 8 + 0 + 2 + 1 + 1/2 + 1/4

= 11.7510

( 下標 : Base 10)

(Base 2)

Number system and Conversion 二進制、八進制、十進制、十六進制


建立一數字系統 : 以 R 為基數，用一幕級數 (power series) 來展開來獲知我們所習用的十進制 ( 詳課本 pp.9)

Octal147.38 = 1x82 + 4x81 + 7x80 + 3x8-1

= 64 + 32 + 7 + 3 x 1/8

= 103.37510

Hexadecimal Composed from 0 、 1 、 2 、 3 、 4 、 5 、 6 、 7 、 8 、 9 、 A 、B 、 C 、 D A2F16 = 10x162 + 2x161 + 15x160

= 2560+32+15

= 260710


EXAMPLE: Convert 5310 to binary.

Conversion (a)

十進制轉二進制 :整數值用連除法 ( 詳課本 p.10)


Conversion (b)

EXAMPLE: Convert .62510 to binary.

十進制轉二進制 :小數值用連乘法 ( 詳課本 p.10)

乘盡

乘小數


Conversion (c)

EXAMPLE: Convert 0.710 to binary.

乘不盡


Conversion (d)

EXAMPLE: Convert 231.34 to base 7.


Equation (1-1)

Conversion from binary to hexadecimal (and conversely) can be done by inspection because each hexadecimal digit corresponds to exactly four binary digits (bits).

Binary HexadecimalConversion

每一個 16 進位數可以轉換成等值 4- 位元二進位數每一個 8 進位數可以轉換成等值 3- 位元二進位數

(a) 4-digits 為一單位(b)不足位元以 0 補上


Addition

Add 1310 and 1110 in binary.

二進位之運算法 (Binary Arithmetic)加法


Subtraction (a)

The subtraction table for binary numbers is

0 – 0 = 00 – 1 = 1 and borrow 1 from the next column1 – 0 = 11 – 1 = 0

Borrowing 1 from a columnis equivalent to subtracting 1 from that column.

二進位之運算法 (Binary Arithmetic)減法


Subtraction (b)

EXAMPLES OF BINARY SUBTRACTION:

另一種二進位減法可以用 2’ 補數 (complement arithmetic) 來計算

©2010 Cengage LearningSubtraction (c)

A detailed analysis of the borrowing process for this example, indicating first a borrow of 1 from column 1 and then a borrow of 1 from column 2, is as follows:


Multiplication (a)

The multiplication table for binary numbers is

0 x 0 = 00 x 1 = 01 x 0 = 01 x 1 = 1

二進位之運算法 (Binary Arithmetic)乘法


Multiplication (b)

The following example illustratesmultiplication of 1310 by 1110 in binary:


Multiplication (c)

When doing binary multiplication, a common way to avoid carries greater than 1 is to add in the partial products one at a time as illustrated by the following example:

1111 multiplicand1101 multiplier1111 1st partial product

0000 2nd partial product(01111) sum of first two partial products

1111 3rd partial product (1001011) sum after adding 3rd partial product 1111 4th partial product11000011 final product (sum after adding 4th

partial product)

二進位之運算法 : 避免大於 1 之數值


Binary Division

Binary division is similar to decimal division, except it is much easier because the only two possible quotient digits are 0 and 1.

We start division by comparing the divisor with the upper bits of the dividend.

If we cannot subtract without getting a negative result, we move one place to the right and try again.

If we can subtract, we place a 1 for the quotient above the number we subtracted from and append the next dividend bit to the end of the difference and repeat this process with this modified difference until we run out of bits in the dividend.

二進位之運算法 : 除法


Binary Division

The following example illustratesdivision of 14510 by 1110 in binary:

二進位之運算法 : 不足以扣除，則往後退一位


3 Systems for representing negative numbers in binary

Sign & Magnitude: Most significant bit is the sign

Ex: – 510 = 11012

1’s Complement: = (2n – 1) - N

Ex: – 510 = (24 – 1) – 5 = 16 – 1 – 5 = 1010 = 10102

2’s Complement: N* = 2n - N

Ex: – 510 = 24 – 5 = 16 – 5 = 1110 = 10112

Section 1.4 (p. 16)

For n-bit word, the first sign is sign,remaining n-1bits represent the magnitudeSo we have 2n-1 for positive and 2n-1 for negative integers


1’s Complement: (a)For n= 4: -0 =(24-1)-0=15 =(1111) -8=(24-1)-8= 7=(0111)

(b) 正數 (N) 與負數 : 由 1 變 0 0 變 1(c) 1’s complement + 1 = 2’s complement

2’s Complement: (a)正數不變 (N)(b)負數以 2n 去扣正數 : 2n-N ( 其中 n 為數字系統之位元數 (bit))(c)所有 2’s complement 之負數，其最左邊數值為 1 。 ( 與前述之 Sign & Magnitude system 同樣具備易於辨識之優點 )

與 +7 混淆


Table 1-1: Signed Binary Integers (word length n = 4)

1’ and 2’ complement are commonly used: adv. : arithmetic unit are easy to design

由 1 變 0 0 變 1

1’ s +1


2’s Complement Addition (a)


2’s Complement Addition (b)

proof


2’s Complement Addition (c)

proof


1’s Complement Addition (b)

proof


1’s Complement Addition (c)

proof


1’s Complement Addition (d)


2’s Complement Addition (d)

Note: For 1’s and 2’s complement to get an n-bit sum is :An overflow occurs if adding two positive numbers gives a negativeanswer or if adding two negative numbers gives a positive answer: Than overflow occurs!


Binary Codes

Although most large computers work internally with binary numbers, the input-output equipment generally uses decimal numbers. Because most logic circuits only accept two-valued signals, the decimal numbers must be coded in terms of binary signals. In the simplest form of binary code, each decimal digit is replaced by its binary equivalent. For example, 937.25 is represented by:

Section 1.5 (p. 21)


Table 1–2. Binary Codes for Decimal Digits

Decimal Digit

8-4-2-1 Code (BCD)

6-3-1-1 Code

Excess-3 Code

2-out-of-5 Code

Gray Code

0 0000 0000 0011 00011 0000

1 0001 0001 0100 00101 0001

2 0010 0011 0101 00110 0011

3 0011 0100 0110 01001 0010

4 0100 0101 0111 01010 0110

5 0101 0111 1000 01100 1110

6 0110 1000 1001 10001 1010

7 0111 1001 1010 10010 1011

8 1000 1011 1011 10100 1001

9 1001 1100 1100 11000 1000

Error-checking


Table 1-3ASCII code

(incomplete)


Chapter 01-Homework

Documents

SLIDES FOR CHAPTER 1 INTRODUCTION NUMBER SYSTEMS AND CONVERSION