©2010 Cengage Learning SLIDES FOR CHAPTER 1 INTRODUCTION NUMBER 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: Objectives Study Guide 1.1Digital Systems and Switching Circuits 1.2Number Systems and Conversion 1.3Binary Arithmetic 1.4Representation of Negative Numbers 1.5Binary Codes Problems

©2010 Cengage Learning Figure 1-1: Switching circuit

©2010 Cengage Learning Decimal Notation = 9x x x x x10 -2 Binary = 1x x x x x x2 -2 = / / 4 = ( 下標 : Base 10) (Base 2) Number system and Conversion 二進制、八進制、十進制、十六進制

©2010 Cengage Learning 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) Octal = 1x x x x8 -1 = x 1 / 8 = Hexadecimal Composed from 0 、 1 、 2 、 3 、 4 、 5 、 6 、 7 、 8 、 9 、 A 、 B 、 C 、 D A2F 16 = 10x x x16 0 = =

©2010 Cengage Learning EXAMPLE: Convert to binary. Conversion (a) 十進制轉二進制 : 整數值用連除法 ( 詳課本 p.10)

©2010 Cengage Learning Conversion (b) EXAMPLE:Convert 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

©2010 Cengage Learning Conversion (c) EXAMPLE: Convert to binary. 乘不盡

©2010 Cengage Learning Conversion (d) EXAMPLE: Convert to base 7.

©2010 Cengage Learning 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  Hexadecimal Conversion 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

©2010 Cengage Learning Addition Add and in binary. 二進位之運算法 (Binary Arithmetic) 加法

©2010 Cengage Learning Subtraction (a) The subtraction table for binary numbers is 0 – 0 = 0 0 – 1 = 1and borrow 1 from the next column 1 – 0 = 1 1 – 1 = 0 Borrowing 1 from a column is equivalent to subtracting 1 from that column. 二進位之運算法 (Binary Arithmetic) 減法

©2010 Cengage Learning 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 Learning Subtraction (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:

©2010 Cengage Learning Multiplication (a) The multiplication table for binary numbers is 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1 二進位之運算法 (Binary Arithmetic) 乘法

©2010 Cengage Learning Multiplication (b) The following example illustrates multiplication of by in binary:

©2010 Cengage Learning 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: 1111multiplicand 1101multiplier 11111st partial product nd partial product (01111)sum of first two partial products rd partial product ( )sum after adding 3rd partial product th partial product final product (sum after adding 4th partial product) 二進位之運算法 : 避免大於 1 之數值

©2010 Cengage Learning 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

©2010 Cengage Learning Binary Division The following example illustrates division of by in binary: 二進位之運算法 : 不足以扣除，則往後退一位

©2010 Cengage Learning 3 Systems for representing negative numbers in binary Sign & Magnitude: Most significant bit is the sign Ex: – 5 10 = ’s Complement: = (2 n – 1) - N Ex: – 5 10 = (2 4 – 1) – 5 = 16 – 1 – 5 = = ’s Complement: N* = 2 n - N Ex: – 5 10 = 2 4 – 5 = 16 – 5 = = Section 1.4 (p. 16) For n-bit word, the first sign is sign, remaining n-1bits represent the magnitude So we have 2 n-1 for positive and 2 n-1 for negative integers

©2010 Cengage Learning 1’s Complement: (a)For n= 4: -0 =(2 4 -1)-0=15 =(1111) -8=(2 4 -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 2 n 去扣正數 : 2 n -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

©2010 Cengage Learning 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

©2010 Cengage Learning 2’s Complement Addition (a)

©2010 Cengage Learning 2’s Complement Addition (b) proof

©2010 Cengage Learning 2’s Complement Addition (c) proof

©2010 Cengage Learning 1’s Complement Addition (b) proof

©2010 Cengage Learning 1’s Complement Addition (c) proof

©2010 Cengage Learning 1’s Complement Addition (d)

©2010 Cengage Learning 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 negative answer or if adding two negative numbers gives a positive answer: Than overflow occurs!

©2010 Cengage Learning 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, is represented by: Section 1.5 (p. 21)

©2010 Cengage Learning Table 1–2. Binary Codes for Decimal Digits Error-checking

©2010 Cengage Learning Table 1-3 ASCII code (incomplete)

©2010 Cengage Learning Chapter 01-Homework